Euclid's Elements [free web version ed.]

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EUCLID’S ELEMENTS OF GEOMETRY The Greek text of J.L. Heiberg (1883–1885) from Euclidis Elementa, edidit et Latine interpretatus est I.L. Heiberg, in aedibus B.G. Teubneri, 1883–1885 edited, and provided with a modern English translation, by Richard Fitzpatrick

c Richard Fitzpatrick, 2007. All rights reserved.

ISBN

Contents Introduction

4

Book 1

5

Book 2

49

Book 3

69

Book 4

109

Book 5

129

Book 6

155

Book 7

193

Book 8

227

Book 9

253

Book 10

281

Book 11

423

Book 12

471

Book 13

505

Greek-English Lexicon

539

Introduction Euclid’s Elements is by far the most famous mathematical work of classical antiquity, and also has the distinction of being the world’s oldest continuously used mathematical textbook. Little is known about the author, beyond the fact that he lived in Alexandria around 300 BCE. The main subjects of the work are geometry, proportion, and number theory. Most of the theorems appearing in the Elements were not discovered by Euclid himself, but were the work of earlier Greek mathematicians such as Pythagoras (and his school), Hippocrates of Chios, Theaetetus of Athens, and Eudoxus of Cnidos. However, Euclid is generally credited with arranging these theorems in a logical manner, so as to demonstrate (admittedly, not always with the rigour demanded by modern mathematics) that they necessarily follow from five simple axioms. Euclid is also credited with devising a number of particularly ingenious proofs of previously discovered theorems: e.g., Theorem 48 in Book 1. The geometrical constructions employed in the Elements are restricted to those which can be achieved using a straight-rule and a compass. Furthermore, empirical proofs by means of measurement are strictly forbidden: i.e., any comparison of two magnitudes is restricted to saying that the magnitudes are either equal, or that one is greater than the other. The Elements consists of thirteen books. Book 1 outlines the fundamental propositions of plane geometry, including the three cases in which triangles are congruent, various theorems involving parallel lines, the theorem regarding the sum of the angles in a triangle, and the Pythagorean theorem. Book 2 is commonly said to deal with “geometric algebra”, since most of the theorems contained within it have simple algebraic interpretations. Book 3 investigates circles and their properties, and includes theorems on tangents and inscribed angles. Book 4 is concerned with regular polygons inscribed in, and circumscribed around, circles. Book 5 develops the arithmetic theory of proportion. Book 6 applies the theory of proportion to plane geometry, and contains theorems on similar figures. Book 7 deals with elementary number theory: e.g., prime numbers, greatest common denominators, etc. Book 8 is concerned with geometric series. Book 9 contains various applications of results in the previous two books, and includes theorems on the infinitude of prime numbers, as well as the sum of a geometric series. Book 10 attempts to classify incommensurable (i.e., irrational) magnitudes using the so-called “method of exhaustion”, an ancient precursor to integration. Book 11 deals with the fundamental propositions of three-dimensional geometry. Book 12 calculates the relative volumes of cones, pyramids, cylinders, and spheres using the method of exhaustion. Finally, Book 13 investigates the five so-called Platonic solids. This edition of Euclid’s Elements presents the definitive Greek text—i.e., that edited by J.L. Heiberg (1883– 1885)—accompanied by a modern English translation, as well as a Greek-English lexicon. Neither the spurious books 14 and 15, nor the extensive scholia which have been added to the Elements over the centuries, are included. The aim of the translation is to make the mathematical argument as clear and unambiguous as possible, whilst still adhering closely to the meaning of the original Greek. Text within square parenthesis (in both Greek and English) indicates material identified by Heiberg as being later interpolations to the original text (some particularly obvious or unhelpful interpolations have been omitted altogether). Text within round parenthesis (in English) indicates material which is implied, but not actually present, in the Greek text.

4

ELEMENTS BOOK 1 Fundamentals of plane geometry involving straight-lines

5

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

“Οροι.

Definitions

α΄. Σηµε‹όν ™στιν, οá µέρος οÙθέν. β΄. Γραµµ¾ δ µÁκος ¢πλατές. γ΄. ΓραµµÁς δ πέρατα σηµε‹α. δ΄. ΕÙθε‹α γραµµή ™στιν, ¼τις ™ξ ‡σου το‹ς ™φ' ˜αυτÁς σηµείοις κε‹ται. ε΄. 'Επιφάνεια δέ ™στιν, Ö µÁκος κሠπλάτος µόνον œχει. $΄. 'Επιφανείας δ πέρατα γραµµαί. ζ΄. 'Επίπεδος ™πιφάνειά ™στιν, ¼τις ™ξ ‡σου τα‹ς ™φ' ˜αυτÁς εÙθείαις κε‹ται. η΄. 'Επίπεδος δ γωνία ™στˆν ¹ ™ν ™πιπέδJ δύο γραµµîν ¡πτοµένων ¢λλήλων κሠµ¾ ™π' εÙθείας κειµένων πρÕς ¢λλήλας τîν γραµµîν κλίσις. θ΄. “Οταν δ αƒ περιέχουσαι τ¾ν γωνίαν γρᵵሠεÙθε‹αι ðσιν, εÙθύγραµµος καλε‹ται ¹ γωνία. ι΄. “Οταν δ εÙθε‹α ™π' εÙθε‹αν σταθε‹σα τ¦ς ™φεξÁς γωνίας ‡σας ¢λλήλαις ποιÍ, Ñρθ¾ ˜κατέρα τîν ‡σων γωνιîν ™στι, κሠ¹ ™φεστηκυ‹α εÙθε‹α κάθετος καλε‹ται, ™φ' ¿ν ™φέστηκεν. ια΄. 'Αµβλε‹α γωνία ™στˆν ¹ µείζων ÑρθÁς. ιβ΄. 'Οξε‹α δ ¹ ™λάσσων ÑρθÁς. ιγ΄. “Ορος ™στίν, Ó τινός ™στι πέρας. ιδ΄. ΣχÁµά ™στι τÕ Øπό τινος ½ τινων Óρων περιεχόµενον. ιε΄. Κύκλος ™στˆ σχÁµα ™πίπεδον ØπÕ µι©ς γραµµÁς περιεχόµενον [¿ καλε‹ται περιφέρεια], πρÕς ¿ν ¢φ' ˜νÕς σηµείου τîν ™ντÕς τοà σχήµατος κειµένων π©σαι αƒ προσπίπτουσαι εÙθε‹αι [πρÕς τ¾ν τοà κύκλου περιφέρειαν] ‡σαι ¢λλήλαις ε„σίν. ι$΄. Κέντρον δ τοà κύκλου τÕ σηµε‹ον καλε‹ται. ιζ΄. ∆ιάµετρος δ τοà κύκλου ™στˆν εÙθε‹ά τις δι¦ τοà κέντρου ºγµένη κሠπερατουµένη ™φ' ˜κάτερα τ¦ µέρη ØπÕ τÁς τοà κύκλου περιφερείας, ¼τις κሠδίχα τέµνει τÕν κύκλον. ιη΄. `Ηµικύκλιον δέ ™στι τÕ περιεχόµενον σχÁµα Øπό τε τÁς διαµέτρου κሠτÁς ¢πολαµβανοµένης Øπ' αÙτÁς περιφερείας. κέντρον δ τοà ¹µικυκλίου τÕ αÙτό, Ö κሠτοà κύκλου ™στίν. ιθ΄. Σχήµατα εÙθύγραµµά ™στι τ¦ ØπÕ εÙθειîν περιεχόµενα, τρίπλευρα µν τ¦ ØπÕ τριîν, τετράπλευρα δ τ¦ ØπÕ τεσσάρων, πολύπλευρα δ τ¦ ØπÕ πλειόνων À τεσσάρων εÙθειîν περιεχόµενα. κ΄. Τîν δ τριπλεύρων σχηµάτων „σόπλευρον µν τρίγωνόν ™στι τÕ τ¦ς τρε‹ς ‡σας œχον πλευράς, „σοσκελς δ τÕ τ¦ς δύο µόνας ‡σας œχον πλευράς, σκαληνÕν δ τÕ τ¦ς τρε‹ς ¢νίσους œχον πλευράς. κα΄ ”Ετι δ τîν τριπλεύρων σχηµάτων Ñρθογώνιον µν τρίγωνόν ™στι τÕ œχον Ñρθ¾ν γωνίαν, ¢µβλυγώνιον

1. A point is that of which there is no part. 2. And a line is a length without breadth. 3. And the extremities of a line are points. 4. A straight-line is whatever lies evenly with points upon itself. 5. And a surface is that which has length and breadth alone. 6. And the extremities of a surface are lines. 7. A plane surface is whatever lies evenly with straight-lines upon itself. 8. And a plane angle is the inclination of the lines, when two lines in a plane meet one another, and are not laid down straight-on with respect to one another. 9. And when the lines containing the angle are straight then the angle is called rectilinear. 10. And when a straight-line stood upon (another) straight-line makes adjacent angles (which are) equal to one another, each of the equal angles is a right-angle, and the former straight-line is called perpendicular to that upon which it stands. 11. An obtuse angle is greater than a right-angle. 12. And an acute angle is less than a right-angle. 13. A boundary is that which is the extremity of something. 14. A figure is that which is contained by some boundary or boundaries. 15. A circle is a plane figure contained by a single line [which is called a circumference], (such that) all of the straight-lines radiating towards [the circumference] from a single point lying inside the figure are equal to one another. 16. And the point is called the center of the circle. 17. And a diameter of the circle is any straight-line, being drawn through the center, which is brought to an end in each direction by the circumference of the circle. And any such (straight-line) cuts the circle in half.† 18. And a semi-circle is the figure contained by the diameter and the circumference it cuts off. And the center of the semi-circle is the same (point) as (the center of) the circle. 19. Rectilinear figures are those figures contained by straight-lines: trilateral figures being contained by three straight-lines, quadrilateral by four, and multilateral by more than four. 20. And of the trilateral figures: an equilateral triangle is that having three equal sides, an isosceles (triangle) that having only two equal sides, and a scalene (triangle) that having three unequal sides.

6

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

δ τÕ œχον ¢µβλε‹αν γωνίαν, Ñξυγώνιον δ τÕ τ¦ς τρε‹ς Ñξείας œχον γωνίας. κβ΄. Τëν δ τετραπλεύρων σχηµάτων τετράγωνον µέν ™στιν, Ö „σόπλευρόν τέ ™στι κሠÑρθογώνιον, ˜τερόµηκες δέ, Ö Ñρθογώνιον µέν, οÙκ „σόπλευρον δέ, ·όµβος δέ, Ö „σόπλευρον µέν, οÙκ Ñρθογώνιον δέ, ·οµβοειδς δ τÕ τ¦ς ¢πεναντίον πλευράς τε κሠγωνίας ‡σας ¢λλήλαις œχον, Ö οÜτε „σόπλευρόν ™στιν οÜτε Ñρθογώνιον· τ¦ δ παρ¦ ταàτα τετράπλευρα τραπέζια καλείσθω. κγ΄. Παράλληλοί ε„σιν εÙθε‹αι, α†τινες ™ν τù αÙτù ™πιπέδJ οâσαι κሠ™κβαλλόµεναι ε„ς ¥πειρον ™φ' ˜κάτερα τ¦ µέρη ™πˆ µηδέτερα συµπίπτουσιν ¢λλήλαις.

†

21. And further of the trilateral figures: a right-angled triangle is that having a right-angle, an obtuse-angled (triangle) that having an obtuse angle, and an acuteangled (triangle) that having three acute angles. 22. And of the quadrilateral figures: a square is that which is right-angled and equilateral, a rectangle that which is right-angled but not equilateral, a rhombus that which is equilateral but not right-angled, and a rhomboid that having opposite sides and angles equal to one another which is neither right-angled nor equilateral. And let quadrilateral figures besides these be called trapezia. 23. Parallel lines are straight-lines which, being in the same plane, and being produced to infinity in each direction, meet with one another in neither (of these directions).

This should really be counted as a postulate, rather than as part of a definition.

Α„τήµατα.

Postulates

α΄. 'Ηιτήσθω ¢πÕ παντÕς σηµείου ™πˆ π©ν σηµε‹ον εÙθε‹αν γραµµ¾ν ¢γαγε‹ν. β΄. Κሠπεπερασµένην εÙθε‹αν κατ¦ τÕ συνεχς ™π' εÙθείας ™κβαλε‹ν. γ΄. Κሠπαντˆ κέντρJ κሠδιαστήµατι κύκλον γράφεσθαι. δ΄. Κሠπάσας τ¦ς Ñρθ¦ς γωνίας ‡σας ¢λλήλαις εναι. ε΄. Κሠ™¦ν ε„ς δύο εÙθείας εÙθε‹α ™µπίπτουσα τ¦ς ™ντÕς κሠ™πˆ τ¦ αÙτ¦ µέρη γωνίας δύο Ñρθîν ™λάσσονας ποιÍ, ™κβαλλοµένας τ¦ς δύο εÙθείας ™π' ¥πειρον συµπίπτειν, ™φ' § µέρη ε„σˆν αƒ τîν δύο Ñρθîν ™λάσσονες.

1. Let it have been postulated to draw a straight-line from any point to any point. 2. And to produce a finite straight-line continuously in a straight-line. 3. And to draw a circle with any center and radius. 4. And that all right-angles are equal to one another. 5. And that if a straight-line falling across two (other) straight-lines makes internal angles on the same side (of itself whose sum is) less than two right-angles, then, being produced to infinity, the two (other) straight-lines meet on that side (of the original straight-line) that the (sum of the internal angles) is less than two right-angles (and do not meet on the other side).†

†

This postulate effectively specifies that we are dealing with the geometry of flat, rather than curved, space.

Κοινሠœννοιαι.

Common Notions

α΄. Τ¦ τù αÙτù ‡σα κሠ¢λλήλοις ™στˆν ‡σα. β΄. Κሠ™¦ν ‡σοις ‡σα προστεθÍ, τ¦ Óλα ™στˆν ‡σα. γ΄. Κሠ™¦ν ¢πÕ ‡σων Šσα ¢φαιρεθÍ, τ¦ καταλειπόµενά ™στιν ‡σα. δ΄. Κሠτ¦ ™φαρµόζοντα ™π' ¢λλήλα ‡σα ¢λλήλοις ™στίν. ε΄. ΚሠτÕ Óλον τοà µέρους µε‹ζόν [™στιν].

1. Things equal to the same thing are also equal to one another. 2. And if equal things are added to equal things then the wholes are equal. 3. And if equal things are subtracted from equal things then the remainders are equal.† 4. And things coinciding with one another are equal to one another. 5. And the whole [is] greater than the part.

†

As an obvious extension of C.N.s 2 & 3—if equal things are added or subtracted from the two sides of an inequality then the inequality remains

an inequality of the same type.

7

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 α΄.

Proposition 1

'Επˆ τÁς δοθείσης εÙθείας πεπερασµένης τρίγωνον „σόπλευρον συστήσασθαι.

To construct an equilateral triangle on a given finite straight-line.

Γ

∆

Α

C

Β

Ε

D

”Εστω ¹ δοθε‹σα εÙθε‹α πεπερασµένη ¹ ΑΒ. ∆ε‹ δ¾ ™πˆ τÁς ΑΒ εÙθείας τρίγωνον „σόπλευρον συστήσασθαι. ΚέντρJ µν τù Α διαστήµατι δ τù ΑΒ κύκλος γεγράφθω Ð ΒΓ∆, κሠπάλιν κέντρJ µν τù Β διαστήµατι δ τù ΒΑ κύκλος γεγράφθω Ð ΑΓΕ, κሠ¢πÕ τοà Γ σηµείου, καθ' Ö τέµνουσιν ¢λλήλους οƒ κύκλοι, ™πί τ¦ Α, Β σηµε‹α ™πεζεύχθωσαν εÙθε‹αι αƒ ΓΑ, ΓΒ. Κሠ™πεˆ τÕ Α σηµε‹ον κέντρον ™στˆ τοà Γ∆Β κύκλου, ‡ση ™στˆν ¹ ΑΓ τÍ ΑΒ· πάλιν, ™πεˆ τÕ Β σηµε‹ον κέντρον ™στˆ τοà ΓΑΕ κύκλου, ‡ση ™στˆν ¹ ΒΓ τÍ ΒΑ. ™δείχθη δ κሠ¹ ΓΑ τÍ ΑΒ ‡ση· ˜κατέρα ¥ρα τîν ΓΑ, ΓΒ τÍ ΑΒ ™στιν ‡ση. τ¦ δ τù αÙτù ‡σα κሠ¢λλήλοις ™στˆν ‡σα· κሠ¹ ΓΑ ¥ρα τÍ ΓΒ ™στιν ‡ση· αƒ τρε‹ς ¥ρα αƒ ΓΑ, ΑΒ, ΒΓ ‡σαι ¢λλήλαις ε„σίν. 'Ισόπλευρον ¤ρα ™στˆ τÕ ΑΒΓ τρίγωνον. κሠσυνέσταται ™πˆ τÁς δοθείσης εÙθείας πεπερασµένης τÁς ΑΒ· Óπερ œδει ποιÁσαι.

†

A

B

E

Let AB be the given finite straight-line. So it is required to construct an equilateral triangle on the straight-line AB. Let the circle BCD with center A and radius AB have been drawn [Post. 3], and again let the circle ACE with center B and radius BA have been drawn [Post. 3]. And let the straight-lines CA and CB have been joined from the point C, where the circles cut one another,† to the points A and B (respectively) [Post. 1]. And since the point A is the center of the circle CDB, AC is equal to AB [Def. 1.15]. Again, since the point B is the center of the circle CAE, BC is equal to BA [Def. 1.15]. But CA was also shown (to be) equal to AB. Thus, CA and CB are each equal to AB. But things equal to the same thing are also equal to one another [C.N. 1]. Thus, CA is also equal to CB. Thus, the three (straightlines) CA, AB, and BC are equal to one another. Thus, the triangle ABC is equilateral, and has been constructed on the given finite straight-line AB. (Which is) the very thing it was required to do.

The assumption that the circles do indeed cut one another should be counted as an additional postulate. There is also an implicit assumption

that two straight-lines cannot share a common segment.

β΄.

Proposition 2†

ΠρÕς τù δοθέντι σηµείJ τÍ δοθείσV εÙθείv ‡σην εÙθε‹αν θέσθαι. ”Εστω τÕ µν δοθν σηµε‹ον τÕ Α, ¹ δ δοθε‹σα εÙθε‹α ¹ ΒΓ· δε‹ δ¾ πρÕς τù Α σηµείJ τÍ δοθείσV εÙθείv τÍ ΒΓ ‡σην εÙθε‹αν θέσθαι. 'Επεζεύχθω γ¦ρ ¢πÕ τοà Α σηµείου ™πί τÕ Β σηµε‹ον εÙθε‹α ¹ ΑΒ, κሠσυνεστάτω ™π' αÙτÁς τρίγωνον „σόπλευρον τÕ ∆ΑΒ, κሠ™κβεβλήσθωσαν ™π' εÙθείας τα‹ς ∆Α, ∆Β εÙθε‹αι αƒ ΑΕ, ΒΖ, κሠκέντρJ µν τù Β διαστήµατι δ τù ΒΓ κύκλος γεγράφθω Ð ΓΗΘ, κሠπάλιν κέντρJ τù ∆ κሠδιαστήµατι τù ∆Η κύκλος

To place a straight-line equal to a given straight-line at a given point. Let A be the given point, and BC the given straightline. So it is required to place a straight-line at point A equal to the given straight-line BC. For let the straight-line AB have been joined from point A to point B [Post. 1], and let the equilateral triangle DAB have been been constructed upon it [Prop. 1.1]. And let the straight-lines AE and BF have been produced in a straight-line with DA and DB (respectively) [Post. 2]. And let the circle CGH with center B and ra8

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

γεγράφθω Ð ΗΚΛ.

dius BC have been drawn [Post. 3], and again let the circle GKL with center D and radius DG have been drawn [Post. 3].

Γ

C

Θ

H

Κ

K

∆

D

Β

B

Α

A

Η

G

Ζ

F

Λ

L

Ε

E

'Επεˆ οâν τÕ Β σηµε‹ον κέντρον ™στˆ τοà ΓΗΘ, ‡ση ™στˆν ¹ ΒΓ τÍ ΒΗ. πάλιν, ™πεˆ τÕ ∆ σηµε‹ον κέντρον ™στˆ τοà ΗΚΛ κύκλου, ‡ση ™στˆν ¹ ∆Λ τÍ ∆Η, ïν ¹ ∆Α τÍ ∆Β ‡ση ™στίν. λοιπ¾ ¥ρα ¹ ΑΛ λοιπÍ τÍ ΒΗ ™στιν ‡ση. ™δείχθη δ κሠ¹ ΒΓ τÍ ΒΗ ‡ση· ˜κατέρα ¥ρα τîν ΑΛ, ΒΓ τÍ ΒΗ ™στιν ‡ση. τ¦ δ τù αÙτù ‡σα κሠ¢λλήλοις ™στˆν ‡σα· κሠ¹ ΑΛ ¥ρα τÍ ΒΓ ™στιν ‡ση. ΠρÕς ¥ρα τù δοθέντι σηµείJ τù Α τÍ δοθείσV εÙθείv τÍ ΒΓ ‡ση εÙθε‹α κε‹ται ¹ ΑΛ· Óπερ œδει ποιÁσαι.

†

Therefore, since the point B is the center of (the circle) CGH, BC is equal to BG [Def. 1.15]. Again, since the point D is the center of the circle GKL, DL is equal to DG [Def. 1.15]. And within these, DA is equal to DB. Thus, the remainder AL is equal to the remainder BG [C.N. 3]. But BC was also shown (to be) equal to BG. Thus, AL and BC are each equal to BG. But things equal to the same thing are also equal to one another [C.N. 1]. Thus, AL is also equal to BC. Thus, the straight-line AL, equal to the given straightline BC, has been placed at the given point A. (Which is) the very thing it was required to do.

This proposition admits of a number of different cases, depending on the relative positions of the point A and the line BC. In such situations,

Euclid invariably only considers one particular case—usually, the most difficult—and leaves the remaining cases as exercises for the reader.

γ΄.

Proposition 3

∆ύο δοθεισîν εÙθειîν ¢νίσων ¢πÕ τÁς µείζονος τÍ ™λάσσονι ‡σην εÙθε‹αν ¢φελε‹ν. ”Εστωσαν αƒ δοθε‹σαι δύο εÙθε‹αι ¥νισοι αƒ ΑΒ, Γ, ïν µείζων œστω ¹ ΑΒ· δε‹ δ¾ ¢πÕ τÁς µείζονος τÁς ΑΒ τÍ ™λάσσονι τÍ Γ ‡σην εÙθε‹αν ¢φελε‹ν. Κείσθω πρÕς τù Α σηµείJ τÍ Γ εÙθείv ‡ση ¹ Α∆· κሠκέντρJ µν τù Α διαστήµατι δ τù Α∆ κύκλος γεγράφθω Ð ∆ΕΖ. Κሠ™πεˆ τÕ Α σηµε‹ον κέντρον ™στˆ τοà ∆ΕΖ κύκλου, ‡ση ™στˆν ¹ ΑΕ τÍ Α∆· ¢λλ¦ κሠ¹ Γ τÍ Α∆ ™στιν ‡ση. ˜κατέρα ¥ρα τîν ΑΕ, Γ τÍ Α∆ ™στιν ‡ση· éστε κሠ¹ ΑΕ τÍ Γ ™στιν ‡ση.

For two given unequal straight-lines, to cut off from the greater a straight-line equal to the lesser. Let AB and C be the two given unequal straight-lines, of which let the greater be AB. So it is required to cut off a straight-line equal to the lesser C from the greater AB. Let the line AD, equal to the straight-line C, have been placed at point A [Prop. 1.2]. And let the circle DEF have been drawn with center A and radius AD [Post. 3]. And since point A is the center of circle DEF , AE is equal to AD [Def. 1.15]. But, C is also equal to AD. Thus, AE and C are each equal to AD. So AE is also

9

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 equal to C [C.N. 1].

Γ

C

∆

D

Ε

Α

E

Β

B

A

Ζ

F

∆ύο ¥ρα δοθεισîν εÙθειîν ¢νίσων τîν ΑΒ, Γ ¢πÕ Thus, for two given unequal straight-lines, AB and C, τÁς µείζονος τÁς ΑΒ τÍ ™λάσσονι τÍ Γ ‡ση ¢φÇρηται ¹ the (straight-line) AE, equal to the lesser C, has been cut ΑΕ· Óπερ œδει ποιÁσαι. off from the greater AB. (Which is) the very thing it was required to do.

δ΄.

Proposition 4

'Ε¦ν δύο τρίγωνα τ¦ς δύο πλευρ¦ς [τα‹ς] δυσˆ πλευρα‹ς ‡σας œχV ˜κατέραν ˜κατέρv κሠτ¾ν γωνίαν τÍ γωνίv ‡σην œχV τ¾ν ØπÕ τîν ‡σων εÙθειîν περιεχοµένην, κሠτ¾ν βάσιν τÊ βάσει ‡σην ›ξει, κሠτÕ τρίγωνον τù τριγώνJ ‡σον œσται, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται ˜κατέρα ˜κατέρv, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν.

If two triangles have two corresponding sides equal, and have the angles enclosed by the equal sides equal, then they will also have equal bases, and the two triangles will be equal, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles.

Α

Β

∆

Γ

Ε

D

A

Ζ

B

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ τ¦ς δύο πλευρ¦ς τ¦ς ΑΒ, ΑΓ τα‹ς δυσˆ πλευρα‹ς τα‹ς ∆Ε, ∆Ζ ‡σας œχοντα ˜κατέραν ˜κατέρv τ¾ν µν ΑΒ τÍ ∆Ε τ¾ν δ ΑΓ τÍ ∆Ζ κሠγωνίαν τ¾ν ØπÕ ΒΑΓ γωνίv τÍ ØπÕ Ε∆Ζ ‡σην. λέγω, Óτι κሠβάσις ¹ ΒΓ βάσει τÍ ΕΖ ‡ση ™στίν, κሠτÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ ‡σον œσται, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται ˜κατέρα ˜κατέρv, Øφ' §ς αƒ Šσαι πλευρሠØποτείνουσιν, ¹ µν ØπÕ ΑΒΓ τÍ ØπÕ ∆ΕΖ, ¹ δ ØπÕ ΑΓΒ τÍ ØπÕ ∆ΖΕ. 'Εφαρµοζοµένου γ¦ρ τοà ΑΒΓ τριγώνου ™πˆ τÕ ∆ΕΖ τρίγωνον κሠτιθεµένου τοà µν Α σηµείου ™πˆ τÕ ∆ σηµε‹ον τÁς δ ΑΒ εÙθείας ™πˆ τ¾ν ∆Ε, ™φαρµόσει κሠτÕ Β σηµε‹ον ™πˆ τÕ Ε δι¦ τÕ ‡σην εναι τ¾ν ΑΒ τÍ ∆Ε· ™φαρµοσάσης δ¾ τÁς ΑΒ ™πˆ τ¾ν ∆Ε ™φαρµόσει

C

E

F

Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF , respectively. (That is) AB to DE, and AC to DF . And (let) the angle BAC (be) equal to the angle EDF . I say that the base BC is also equal to the base EF , and triangle ABC will be equal to triangle DEF , and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles. (That is) ABC to DEF , and ACB to DF E. Let the triangle ABC be applied to the triangle DEF ,† the point A being placed on the point D, and the straight-line AB on DE. The point B will also coincide with E, on account of AB being equal to DE. So (because of) AB coinciding with DE, the straight-line

10

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

κሠ¹ ΑΓ εÙθε‹α ™πˆ τ¾ν ∆Ζ δι¦ τÕ ‡σην εναι τ¾ν ØπÕ ΒΑΓ γωνίαν τÍ ØπÕ Ε∆Ζ· éστε κሠτÕ Γ σηµε‹ον ™πˆ τÕ Ζ σηµε‹ον ™φαρµόσει δι¦ τÕ ‡σην πάλιν εναι τ¾ν ΑΓ τÍ ∆Ζ. ¢λλ¦ µ¾ν κሠτÕ Β ™πˆ τÕ Ε ™φηρµόκει· éστε βάσις ¹ ΒΓ ™πˆ βάσιν τ¾ν ΕΖ ™φαρµόσει. ε„ γ¦ρ τοà µν Β ™πˆ τÕ Ε ™φαρµόσαντος τοà δ Γ ™πˆ τÕ Ζ ¹ ΒΓ βάσις ™πˆ τ¾ν ΕΖ οÙκ ™φαρµόσει, δύο εÙθε‹αι χωρίον περιέξουσιν· Óπερ ™στˆν ¢δύνατον. ™φαρµόσει ¥ρα ¹ ΒΓ βάσις ™πˆ τ¾ν ΕΖ κሠ‡ση αÙτÍ œσται· éστε κሠÓλον τÕ ΑΒΓ τρίγωνον ™πˆ Óλον τÕ ∆ΕΖ τρίγωνον ™φαρµόσει κሠ‡σον αÙτù œσται, καˆ αƒ λοιπሠγωνίαι ™πˆ τ¦ς λοιπ¦ς γωνίας ™φαρµόσουσι κሠ‡σαι αÙτα‹ς œσονται, ¹ µν ØπÕ ΑΒΓ τÍ ØπÕ ∆ΕΖ ¹ δ ØπÕ ΑΓΒ τÍ ØπÕ ∆ΖΕ. 'Ε¦ν ¥ρα δύο τρίγωνα τ¦ς δύο πλευρ¦ς [τα‹ς] δύο πλευρα‹ς ‡σας œχV ˜κατέραν ˜κατέρv κሠτ¾ν γωνίαν τÍ γωνίv ‡σην œχV τ¾ν ØπÕ τîν ‡σων εÙθειîν περιεχοµένην, κሠτ¾ν βάσιν τÊ βάσει ‡σην ›ξει, κሠτÕ τρίγωνον τù τριγώνJ ‡σον œσται, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται ˜κατέρα ˜κατέρv, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν· Óπερ œδει δε‹ξαι.

AC will also coincide with DF , on account of the angle BAC being equal to EDF . So the point C will also coincide with the point F , again on account of AC being equal to DF . But, point B certainly also coincided with point E, so that the base BC will coincide with the base EF . For if B coincides with E, and C with F , and the base BC does not coincide with EF , then two straightlines will encompass an area. The very thing is impossible [Post. 1].‡ Thus, the base BC will coincide with EF , and will be equal to it [C.N. 4]. So the whole triangle ABC will coincide with the whole triangle DEF , and will be equal to it [C.N. 4]. And the remaining angles will coincide with the remaining angles, and will be equal to them [C.N. 4]. (That is) ABC to DEF , and ACB to DF E [C.N. 4]. Thus, if two triangles have two corresponding sides equal, and have the angles enclosed by the equal sides equal, then they will also have equal bases, and the two triangles will be equal, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles. (Which is) the very thing it was required to show.

†

The application of one figure to another should be counted as an additional postulate.

‡

Since Post. 1 implicitly assumes that the straight-line joining two given points is unique.

ε΄.

Proposition 5

Τîν „σοσκελîν τριγώνων αƒ τρÕς τÍ βάσει γωνίαι ‡σαι For isosceles triangles, the angles at the base are equal ¢λλήλαις ε„σίν, κሠπροσεκβληθεισîν τîν ‡σων εÙθειîν to one another, and if the equal sides are produced then αƒ ØπÕ τ¾ν βάσιν γωνίαι ‡σαι ¢λλήλαις œσονται. the angles under the base will be equal to one another. Α A

Β Ζ ∆

Γ

B

Η

F

Ε

”Εστω τρίγωνον „σοσκελς τÕ ΑΒΓ ‡σην œχον τ¾ν ΑΒ πλευρ¦ν τÍ ΑΓ πλευρ´, κሠπροσεκβεβλήσθωσαν ™π' εÙθείας τα‹ς ΑΒ, ΑΓ εÙθε‹αι αƒ Β∆, ΓΕ· λέγω, Óτι ¹ µν ØπÕ ΑΒΓ γωνία τÍ ØπÕ ΑΓΒ ‡ση ™στίν, ¹ δ ØπÕ ΓΒ∆ τÍ ØπÕ ΒΓΕ. Ε„λήφθω γ¦ρ ™πˆ τÁς Β∆ τυχÕν σηµε‹ον τÕ Ζ, κሠ¢φVρήσθω ¢πÕ τÁς µείζονος τÁς ΑΕ τÍ ™λάσσονι τÍ ΑΖ ‡ση ¹ ΑΗ, κሠ™πεζεύχθωσαν αƒ ΖΓ, ΗΒ εÙθε‹αι. 'Επεˆ οâν ‡ση ™στˆν ¹ µν ΑΖ τÍ ΑΗ ¹ δ ΑΒ τÍ

C G

E D Let ABC be an isosceles triangle having the side AB equal to the side AC, and let the straight-lines BD and CE have been produced in a straight-line with AB and AC (respectively) [Post. 2]. I say that the angle ABC is equal to ACB, and (angle) CBD to BCE. For let the point F have been taken somewhere on BD, and let AG have been cut off from the greater AE, equal to the lesser AF [Prop. 1.3]. Also, let the straightlines F C and GB have been joined [Post. 1].

11

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

ΑΓ, δύο δ¾ αƒ ΖΑ, ΑΓ δυσˆ τα‹ς ΗΑ, ΑΒ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνίαν κοιν¾ν περιέχουσι τ¾ν ØπÕ ΖΑΗ· βάσις ¥ρα ¹ ΖΓ βάσει τÍ ΗΒ ‡ση ™στίν, κሠτÕ ΑΖΓ τρίγωνον τù ΑΗΒ τριγώνJ ‡σον œσται, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται ˜κατέρα ˜κατέρv, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν, ¹ µν ØπÕ ΑΓΖ τÍ ØπÕ ΑΒΗ, ¹ δ ØπÕ ΑΖΓ τÍ ØπÕ ΑΗΒ. κሠ™πεˆ Óλη ¹ ΑΖ ÓλV τÍ ΑΗ ™στιν ‡ση, ïν ¹ ΑΒ τÍ ΑΓ ™στιν ‡ση, λοιπ¾ ¥ρα ¹ ΒΖ λοιπÍ τÍ ΓΗ ™στιν ‡ση. ™δείχθη δ κሠ¹ ΖΓ τÍ ΗΒ ‡ση· δύο δ¾ αƒ ΒΖ, ΖΓ δυσˆ τα‹ς ΓΗ, ΗΒ „σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΒΖΓ γωνίv τV ØπÕ ΓΗΒ ‡ση, κሠβάσις αÙτîν κοιν¾ ¹ ΒΓ· κሠτÕ ΒΖΓ ¥ρα τρίγωνον τù ΓΗΒ τριγώνJ ‡σον œσται, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται ˜κατέρα ˜κατέρv, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν· ‡ση ¥ρα ™στˆν ¹ µν ØπÕ ΖΒΓ τÍ ØπÕ ΗΓΒ ¹ δ ØπÕ ΒΓΖ τÍ ØπÕ ΓΒΗ. ™πεˆ οâν Óλη ¹ ØπÕ ΑΒΗ γωνία ÓλV τÍ ØπÕ ΑΓΖ γωνίv ™δείχθη ‡ση, ïν ¹ ØπÕ ΓΒΗ τÍ ØπÕ ΒΓΖ ‡ση, λοιπ¾ ¥ρα ¹ ØπÕ ΑΒΓ λοιπÍ τÍ ØπÕ ΑΓΒ ™στιν ‡ση· καί ε„σι πρÕς τÍ βάσει τοà ΑΒΓ τριγώνου. ™δείχθη δ κሠ¹ ØπÕ ΖΒΓ τÍ ØπÕ ΗΓΒ ‡ση· καί ε„σιν ØπÕ τ¾ν βάσιν. Τîν ¥ρα „σοσκελîν τριγώνων αƒ τρÕς τÍ βάσει γωνίαι ‡σαι ¢λλήλαις ε„σίν, κሠπροσεκβληθεισîν τîν ‡σων εÙθειîν αƒ ØπÕ τ¾ν βάσιν γωνίαι ‡σαι ¢λλήλαις œσονται· Óπερ œδει δε‹ξαι.

In fact, since AF is equal to AG, and AB to AC, the two (straight-lines) F A, AC are equal to the two (straight-lines) GA, AB, respectively. They also encompass a common angle F AG. Thus, the base F C is equal to the base GB, and the triangle AF C will be equal to the triangle AGB, and the remaining angles subtendend by the equal sides will be equal to the corresponding remaining angles [Prop. 1.4]. (That is) ACF to ABG, and AF C to AGB. And since the whole of AF is equal to the whole of AG, within which AB is equal to AC, the remainder BF is thus equal to the remainder CG [C.N. 3]. But F C was also shown (to be) equal to GB. So the two (straightlines) BF , F C are equal to the two (straight-lines) CG, GB, respectively, and the angle BF C (is) equal to the angle CGB, and the base BC is common to them. Thus, the triangle BF C will be equal to the triangle CGB, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles [Prop. 1.4]. Thus, F BC is equal to GCB, and BCF to CBG. Therefore, since the whole angle ABG was shown (to be) equal to the whole angle ACF , within which CBG is equal to BCF , the remainder ABC is thus equal to the remainder ACB [C.N. 3]. And they are at the base of triangle ABC. And F BC was also shown (to be) equal to GCB. And they are under the base. Thus, for isosceles triangles, the angles at the base are equal to one another, and if the equal sides are produced then the angles under the base will be equal to one another. (Which is) the very thing it was required to show.

$΄.

Proposition 6

'Ε¦ν τριγώνου αƒ δØο γωνίαι ‡σαι ¢λλήλαις ðσιν, If a triangle has two angles equal to one another then καˆ αƒ ØπÕ τ¦ς ‡σας γωνίας Øποτείνουσαι πλευρሠ‡σαι the sides subtending the equal angles will also be equal ¢λλήλαις œσονται. to one another.

Α

A

∆

Β

D

Γ

B

C

”Εστω τρίγωνον τÕ ΑΒΓ ‡σην œχον τ¾ν ØπÕ ΑΒΓ Let ABC be a triangle having the angle ABC equal γωνίαν τÍ ØπÕ ΑΓΒ γωνίv· λέγω, Óτι κሠπλευρ¦ ¹ ΑΒ to the angle ACB. I say that side AB is also equal to side πλευρ´ τÍ ΑΓ ™στιν ‡ση. AC. Ε„ γ¦ρ ¥νισός ™στιν ¹ ΑΒ τÍ ΑΓ, ¹ ˜τέρα αÙτîν For if AB is unequal to AC then one of them is µείζων ™στίν. œστω µείζων ¹ ΑΒ, κሠ¢φVρήσθω ¢πÕ greater. Let AB be greater. And let DB, equal to

12

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

τÁς µείζονος τÁς ΑΒ τÍ ™λάττονι τÍ ΑΓ ‡ση ¹ ∆Β, κሠ™πεζεύχθω ¹ ∆Γ. 'Επεˆ οâν ‡ση ™στˆν ¹ ∆Β τÍ ΑΓ κοιν¾ δ ¹ ΒΓ, δύο δ¾ αƒ ∆Β, ΒΓ δύο τα‹ς ΑΓ, ΓΒ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv, κሠγωνία ¹ ØπÕ ∆ΒΓ γωνιv τÍ ØπÕ ΑΓΒ ™στιν ‡ση· βάσις ¥ρα ¹ ∆Γ βάσει τÍ ΑΒ ‡ση ™στίν, κሠτÕ ∆ΒΓ τρίγωνον τù ΑΓΒ τριγώνJ ‡σον œσται, τÕ œλασσον τù µείζονι· Óπερ ¥τοπον· οÙκ ¥ρα ¥νισός ™στιν ¹ ΑΒ τÍ ΑΓ· ‡ση ¥ρα. 'Ε¦ν ¥ρα τριγώνου αƒ δØο γωνίαι ‡σαι ¢λλήλαις ðσιν, καˆ αƒ ØπÕ τ¦ς ‡σας γωνίας Øποτείνουσαι πλευρሠ‡σαι ¢λλήλαις œσονται· Óπερ œδει δε‹ξαι.

†

the lesser AC, have been cut off from the greater AB [Prop. 1.3]. And let DC have been joined [Post. 1]. Therefore, since DB is equal to AC, and BC (is) common, the two sides DB, BC are equal to the two sides AC, CB, respectively, and the angle DBC is equal to the angle ACB. Thus, the base DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB [Prop. 1.4], the lesser to the greater. The very notion (is) absurd [C.N. 5]. Thus, AB is not unequal to AC. Thus, (it is) equal.† Thus, if a triangle has two angles equal to one another then the sides subtending the equal angles will also be equal to one another. (Which is) the very thing it was required to show.

Here, use is made of the previously unmentioned common notion that if two quantities are not unequal then they must be equal. Later on, use

is made of the closely related common notion that if two quantities are not greater than or less than one another, respectively, then they must be equal to one another.

ζ΄.

Proposition 7

'Επˆ τÁς αÙτÁς εÙθείας δύο τα‹ς αÙτα‹ς εÙθείαις ¥λλαι On the same straight-line, two other straight-lines δύο εÙθε‹αι ‡σαι ˜κατέρα ˜κατέρv οÙ συσταθήσονται equal, respectively, to two (given) straight-lines (which πρÕς ¥λλJ κሠ¥λλJ σηµείJ ™πˆ τ¦ αÙτ¦ µέρη τ¦ αÙτ¦ meet) cannot be constructed (meeting) at a different πέρατα œχουσαι τα‹ς ™ξ ¢ρχÁς εÙθείαις. point on the same side (of the straight-line), but having the same ends as the given straight-lines.

Γ

C

∆

Α

D

Β

A

Ε„ γ¦ρ δυνατόν, ™πˆ τÁς αÙτÁς εÙθείας τÁς ΑΒ δύο τα‹ς αÙτα‹ς εÙθείαις τα‹ς ΑΓ, ΓΒ ¥λλαι δύο εÙθε‹αι αƒ Α∆, ∆Β ‡σαι ˜κατέρα ˜κατερv συνεστάτωσαν πρÕς ¥λλJ κሠ¥λλJ σηµείJ τù τε Γ κሠ∆ ™πˆ τ¦ αÙτ¦ µέρη τ¦ αÙτ¦ πέρατα œχουσαι, éστε ‡σην εναι τÁν µν ΓΑ τÍ ∆Α τÕ αÙτÕ πέρας œχουσαν αÙτÍ τÕ Α, τ¾ν δ ΓΒ τÍ ∆Β τÕ αÙτÕ πέρας œχουσαν αÙτÍ τÕ Β, κሠ™πεζεύχθω ¹ Γ∆. 'Επεˆ οâν ‡ση ™στˆν ¹ ΑΓ τÍ Α∆, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ΑΓ∆ τÍ ØπÕ Α∆Γ· µείζων ¥ρα ¹ ØπÕ Α∆Γ τÁς ØπÕ ∆ΓΒ· πολλù ¥ρα ¹ ØπÕ Γ∆Β µείζων ™στί τÁς ØπÕ ∆ΓΒ. πάλιν ™πεˆ ‡ση ™στˆν ¹ ΓΒ τÍ ∆Β, ‡ση ™στˆ κሠγωνία ¹ ØπÕ Γ∆Β γωνίv τÍ ØπÕ ∆ΓΒ. ™δείχθη δ αÙτÁς κሠπολλù µείζων· Óπερ ™στˆν ¢δύατον. ΟÙκ ¥ρα ™πˆ τÁς αÙτÁς εÙθείας δύο τα‹ς αÙτα‹ς εÙθείαις ¥λλαι δύο εÙθε‹αι ‡σαι ˜κατέρα ˜κατέρv συ-

B

For, if possible, let the two straight-lines AD, DB, equal to two (given) straight-lines AC, CB, respectively, have been constructed on the same straight-line AB, meeting at different points, C and D, on the same side (of AB), and having the same ends (on AB). So CA and DA are equal, having the same ends at A, and CB and DB are equal, having the same ends at B. And let CD have been joined [Post. 1]. Therefore, since AC is equal to AD, the angle ACD is also equal to angle ADC [Prop. 1.5]. Thus, ADC (is) greater than DCB [C.N. 5]. Thus, CDB is much greater than DCB [C.N. 5]. Again, since CB is equal to DB, the angle CDB is also equal to angle DCB [Prop. 1.5]. But it was shown that the former (angle) is also much greater (than the latter). The very thing is impossible. Thus, on the same straight-line, two other straight-

13

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

σταθήσονται πρÕς ¥λλJ κሠ¥λλJ σηµείJ ™πˆ τ¦ αÙτ¦ µέρη τ¦ αÙτ¦ πέρατα œχουσαι τα‹ς ™ξ ¢ρχÁς εÙθείαις· Óπερ œδει δε‹ξαι.

lines equal, respectively, to two (given) straight-lines (which meet) cannot be constructed (meeting) at a different point on the same side (of the straight-line), but having the same ends as the given straight-lines. (Which is) the very thing it was required to show.

η΄.

Proposition 8

'Ε¦ν δύο τρίγωνα τ¦ς δύο πλευρ¦ς [τα‹ς] δύο If two triangles have two corresponding sides equal, πλευρα‹ς ‡σας œχV ˜κατέραν ˜κατέρv, œχV δ κሠτ¾ν and also have equal bases, then the angles encompassed βάσιν τÍ βάσει ‡σην, κሠτ¾ν γωνίαν τÍ γωνίv ‡σην ›ξει by the equal straight-lines will also be equal. τ¾ν ØπÕ τîν ‡σων εØθειîν περιεχοµένην.

∆

Α

D

A

Ζ

Γ Β

Η

F

C

Ε

B

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ τ¦ς δύο πλευρ¦ς τ¦ς ΑΒ, ΑΓ τα‹ς δύο πλευρα‹ς τα‹ς ∆Ε, ∆Ζ ‡σας œχοντα ˜κατέραν ˜κατέρv, τ¾ν µν ΑΒ τÍ ∆Ε τ¾ν δ ΑΓ τÍ ∆Ζ· ™χέτω δ κሠβάσιν τ¾ν ΒΓ βάσει τÍ ΕΖ ‡σην· λέγω, Óτι κሠγωνία ¹ ØπÕ ΒΑΓ γωνίv τÍ ØπÕ Ε∆Ζ ™στιν ‡ση. 'Εφαρµοζοµένου γ¦ρ τοà ΑΒΓ τριγώνου ™πˆ τÕ ∆ΕΖ τρίγωνον κሠτιθεµένου τοà µν Β σηµείου ™πˆ τÕ Ε σηµε‹ον τÁς δ ΒΓ εÙθείας ™πˆ τ¾ν ΕΖ ™φαρµόσει κሠτÕ Γ σηµε‹ον ™πˆ τÕ Ζ δι¦ τÕ ‡σην εναι τ¾ν ΒΓ τÍ ΕΖ· ™φαρµοσάσης δ¾ τÁς ΒΓ ™πˆ τ¾ν ΕΖ ™φαρµόσουσι καˆ αƒ ΒΑ, ΓΑ ™πˆ τ¦ς Ε∆, ∆Ζ. ε„ γ¦ρ βάσις µν ¹ ΒΓ ™πˆ βάσιν τ¾ν ΕΖ ™φαρµόσει, αƒ δ ΒΑ, ΑΓ πλευρሠ™πˆ τ¦ς Ε∆, ∆Ζ οÙκ ™φαρµόσουσιν ¢λλ¦ παραλλάξουσιν æς αƒ ΕΗ, ΗΖ, συσταθήσονται ™πˆ τÁς αÙτÁς εÙθείας δύο τα‹ς αÙτα‹ς εÙθείαις ¥λλαι δύο εÙθε‹αι ‡σαι ˜κατέρα ˜κατέρv πρÕς ¥λλJ κሠ¥λλJ σηµείJ ™πˆ τ¦ αÙτ¦ µέρη τ¦ αÙτ¦ πέρατα œχουσαι. οÙ συνίστανται δέ· οÙκ ¥ρα ™φαρµοζοµένης τÁς ΒΓ βάσεως ™πˆ τ¾ν ΕΖ βάσιν οÙκ ™φαρµόσουσι καˆ αƒ ΒΑ, ΑΓ πλευρሠ™πˆ τ¦ς Ε∆, ∆Ζ. ™φαρµόσουσιν ¥ρα· éστε κሠγωνία ¹ ØπÕ ΒΑΓ ™πˆ γωνίαν τ¾ν ØπÕ Ε∆Ζ ™φαρµόσει κሠ‡ση αÙτÍ œσται. 'Ε¦ν ¥ρα δύο τρίγωνα τ¦ς δύο πλευρ¦ς [τα‹ς] δύο πλευρα‹ς ‡σας œχV ˜κατέραν ˜κατέρv κሠτ¾ν βάσιν τÍ βάσει ‡σην œχV, κሠτ¾ν γωνίαν τÍ γωνίv ‡σην ›ξει τ¾ν ØπÕ τîν ‡σων εØθειîν περιεχοµένην· Óπερ œδει δε‹ξαι.

G

E

Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF , respectively. (That is) AB to DE, and AC to DF . Let them also have the base BC equal to the base EF . I say that the angle BAC is also equal to the angle EDF . For if triangle ABC is applied to triangle DEF , the point B being placed on point E, and the straight-line BC on EF , point C will also coincide with F , on account of BC being equal to EF . So (because of) BC coinciding with EF , (the sides) BA and CA will also coincide with ED and DF (respectively). For if base BC coincides with base EF , but the sides AB and AC do not coincide with ED and DF (respectively), but miss like EG and GF (in the above figure), then we will have constructed upon the same straight-line, two other straight-lines equal, respectively, to two (given) straight-lines, and (meeting) at a different point on the same side (of the straight-line), but having the same ends. But (such straight-lines) cannot be constructed [Prop. 1.7]. Thus, the base BC being applied to the base EF , the sides BA and AC cannot not coincide with ED and DF (respectively). Thus, they will coincide. So the angle BAC will also coincide with angle EDF , and they will be equal [C.N. 4]. Thus, if two triangles have two corresponding sides equal, and have equal bases, then the angles encompassed by the equal straight-lines will also be equal. (Which is) the very thing it was required to show.

14

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 θ΄.

Proposition 9

Τ¾ν δοθε‹σαν γωνίαν εÙθύγραµµον δίχα τεµε‹ν.

To cut a given rectilinear angle in half.

Α

∆

Β

A

Ε

Ζ

D

Γ

B

E

F

C

”Εστω ¹ δοθε‹σα γωνία εÙθύγραµµος ¹ ØπÕ ΒΑΓ. δε‹ δ¾ αÙτ¾ν δίχα τεµε‹ν. Ε„λήφθω ™πˆ τÁς ΑΒ τυχÕν σηµε‹ον τÕ ∆, κሠ¢φVρήσθω ¢πÕ τÁς ΑΓ τÍ Α∆ ‡ση ¹ ΑΕ, κሠ™πεζεύχθω ¹ ∆Ε, κሠσυνεστάτω ™πˆ τÁς ∆Ε τρίγωνον „σόπλευρον τÕ ∆ΕΖ, κሠ™πεζεύχθω ¹ ΑΖ· λέγω, Óτι ¹ ØπÕ ΒΑΓ γωνία δίχα τέτµηται ØπÕ τÁς ΑΖ εØθείας. 'Επεˆ γ¦ρ ‡ση ™στˆν ¹ Α∆ τÍ ΑΕ, κοιν¾ δ ¹ ΑΖ, δύο δ¾ αƒ ∆Α, ΑΖ δυσˆ τα‹ς ΕΑ, ΑΖ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv. κሠβάσις ¹ ∆Ζ βάσει τÍ ΕΖ ‡ση ™στίν· γωνία ¥ρα ¹ ØπÕ ∆ΑΖ γωνίv τÍ ØπÕ ΕΑΖ ‡ση ™στίν. `Η ¥ρα δοθε‹σα γωνία εÙθύγραµµος ¹ ØπÕ ΒΑΓ δίχα τέτµηται ØπÕ τÁς ΑΖ εÙθείας· Óπερ œδει ποιÁσαι.

Let BAC be the given rectilinear angle. So it is required to cut it in half. Let the point D have been taken somewhere on AB, and let AE, equal to AD, have been cut off from AC [Prop. 1.3], and let DE have been joined. And let the equilateral triangle DEF have been constructed upon DE [Prop. 1.1], and let AF have been joined. I say that the angle BAC has been cut in half by the straight-line AF . For since AD is equal to AE, and AF is common, the two (straight-lines) DA, AF are equal to the two (straight-lines) EA, AF , respectively. And the base DF is equal to the base EF . Thus, angle DAF is equal to angle EAF [Prop. 1.8]. Thus, the given rectilinear angle BAC has been cut in half by the straight-line AF . (Which is) the very thing it was required to do.

ι΄.

Proposition 10

Τ¾ν δοθε‹σαν εÙθε‹αν πεπερασµένην δίχα τεµε‹ν. ”Εστω ¹ δοθε‹σα εÙθε‹α πεπερασµένη ¹ ΑΒ· δε‹ δ¾ τ¾ν ΑΒ εÙθε‹αν πεπερασµένην δίχα τεµε‹ν. Συνεστάτω ™π' αÙτÁς τρίγωνον „σόπλευρον τÕ ΑΒΓ, κሠτετµήσθω ¹ ØπÕ ΑΓΒ γωνία δίχα τÍ Γ∆ εÙθείv· λέγω, Óτι ¹ ΑΒ εÙθε‹α δίχα τέτµηται κατ¦ τÕ ∆ σηµε‹ον. 'Επεˆ γ¦ρ ‡ση ™στˆν ¹ ΑΓ τÍ ΓΒ, κοιν¾ δ ¹ Γ∆, δύο δ¾ αƒ ΑΓ, Γ∆ δύο τα‹ς ΒΓ, Γ∆ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΑΓ∆ γωνίv τÍ ØπÕ ΒΓ∆ ‡ση ™στίν· βάσις ¥ρα ¹ Α∆ βάσει τÍ Β∆ ‡ση ™στίν.

To cut a given finite straight-line in half. Let AB be the given finite straight-line. So it is required to cut the finite straight-line AB in half. Let the equilateral triangle ABC have been constructed upon (AB) [Prop. 1.1], and let the angle ACB have been cut in half by the straight-line CD [Prop. 1.9]. I say that the straight-line AB has been cut in half at point D. For since AC is equal to CB, and CD (is) common, the two (straight-lines) AC, CD are equal to the two (straight-lines) BC, CD, respectively. And the angle ACD is equal to the angle BCD. Thus, the base AD is equal to the base BD [Prop. 1.4].

15

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

Γ

Α

C

Β

∆

A

B

D

`Η ¥ρα δοθε‹σα εÙθε‹α πεπερασµένη ¹ ΑΒ δίχα τέτµηται κατ¦ τÕ ∆· Óπερ œδει ποιÁσαι.

Thus, the given finite straight-line AB has been cut in half at (point) D. (Which is) the very thing it was required to do.

ια΄.

Proposition 11

ΠΤÍ δοθείσV εÙθείv ¢πÕ τοà πρÕς αÙτV δοθέντος σηµείου πρÕς Ñρθ¦ς γωνίας εÙθε‹αν γραµµ¾ν ¢γαγε‹ν.

To draw a straight-line at right-angles to a given straight-line from a given point on it.

Ζ

F

Α

Β ∆

Γ

B

A

Ε

D

”Εστω ¹ µν δοθε‹σα εÙθε‹α ¹ ΑΒ τÕ δ δοθν σηµε‹ον ™π' αÙτÁς τÕ Γ· δε‹ δ¾ ¢πÕ τοà Γ σηµείου τÍ ΑΒ εÙθείv πρÕς Ñρθ¦ς γωνίας εÙθε‹αν γραµµ¾ν ¢γαγε‹ν. Ε„λήφθω ™πˆ τÁς ΑΓ τυχÕν σηµε‹ον τÕ ∆, κሠκείσθω τÍ Γ∆ ‡ση ¹ ΓΕ, κሠσυνεστάτω ™πˆ τÁς ∆Ε τρίγωνον „σόπλευρον τÕ Ζ∆Ε, κሠ™πεζεύχθω ¹ ΖΓ· λέγω, Óτι τÍ δοθείσV εÙθείv τÍ ΑΒ ¢πÕ τοà πρÕς αÙτÍ δοθέντος σηµείου τοà Γ πρÕς Ñρθ¦ς γωνίας εÙθε‹α γραµµ¾ Ãκται ¹ ΖΓ. 'Επεˆ γ¦ρ ‡ση ™στˆν ¹ ∆Γ τÍ ΓΕ, κοιν¾ δ ¹ ΓΖ, δύο δ¾ αƒ ∆Γ, ΓΖ δυσˆ τα‹ς ΕΓ, ΓΖ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠβάσις ¹ ∆Ζ βάσει τÍ ΖΕ ‡ση ™στίν· γωνία ¥ρα ¹ ØπÕ ∆ΓΖ γωνίv τÍ ØπÕ ΕΓΖ ‡ση ™στίν· καί ε„σιν ™φεξÁς. Óταν δ εÙθε‹α ™π' εÙθε‹αν σταθε‹σα τ¦ς ™φεξÁς γωνίας ‡σας ¢λλήλαις ποιÍ, Ñρθ¾ ˜κατέρα τîν ‡σων γωνιîν ™στιν· Ñρθ¾ ¥ρα ™στˆν ˜κατέρα τîν ØπÕ ∆ΓΖ, ΖΓΕ. ΤÍ ¥ρα δοθείσV εÙθείv τÍ ΑΒ ¢πÕ τοà πρÕς αÙτÍ δοθέντος σηµείου τοà Γ πρÕς Ñρθ¦ς γωνίας εÙθε‹α γραµµ¾ Ãκται ¹ ΓΖ· Óπερ œδει ποιÁσαι.

C

E

Let AB be the given straight-line, and C the given point on it. So it is required to draw a straight-line from the point C at right-angles to the straight-line AB. Let the point D be have been taken somewhere on AC, and let CE be made equal to CD [Prop. 1.3], and let the equilateral triangle F DE have been constructed on DE [Prop. 1.1], and let F C have been joined. I say that the straight-line F C has been drawn at right-angles to the given straight-line AB from the given point C on it. For since DC is equal to CE, and CF is common, the two (straight-lines) DC, CF are equal to the two (straight-lines), EC, CF , respectively. And the base DF is equal to the base F E. Thus, the angle DCF is equal to the angle ECF [Prop. 1.8], and they are adjacent. But when a straight-line stood on a(nother) straight-line makes the adjacent angles equal to one another, each of the equal angles is a right-angle [Def. 1.10]. Thus, each of the (angles) DCF and F CE is a right-angle. Thus, the straight-line CF has been drawn at right-

16

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 angles to the given straight-line AB from the given point C on it. (Which is) the very thing it was required to do.

ιβ΄.

Proposition 12

'Επˆ τ¾ν δοθε‹σαν εÙθε‹αν ¥πειρον ¢πÕ τοà δοθέντος σηµείου, Ö µή ™στιν ™π' αÙτÁς, κάθετον εÙθε‹αν γραµµ¾ν ¢γαγε‹ν.

To draw a straight-line perpendicular to a given infinite straight-line from a given point which is not on it.

Ζ

F

Γ

C

Α

Η

Θ ∆

Ε

Β

A

B G

H

E D

”Εστω ¹ µν δοθε‹σα εÙθε‹α ¥πειρος ¹ ΑΒ τÕ δ δοθν σηµε‹ον, Ö µή ™στιν ™π' αÙτÁς, τÕ Γ· δε‹ δ¾ ™πˆ τ¾ν δοθε‹σαν εÙθε‹αν ¥πειρον τ¾ν ΑΒ ¢πÕ τοà δοθέντος σηµείου τοà Γ, Ö µή ™στιν ™π' αÙτÁς, κάθετον εÙθε‹αν γραµµ¾ν ¢γαγε‹ν. Ε„λήφθω γ¦ρ ™πˆ τ¦ ›τερα µέρη τÁς ΑΒ εÙθείας τυχÕν σηµε‹ον τÕ ∆, κሠκέντρJ µν τù Γ διαστήµατι δ τù Γ∆ κύκλος γεγράφθω Ð ΕΖΗ, κሠτετµήσθω ¹ ΕΗ εÙθε‹α δίχα κατ¦ τÕ Θ, κሠ™πεζεύχθωσαν αƒ ΓΗ, ΓΘ, ΓΕ εØθε‹αι· λέγω, Óτι ™πˆ τ¾ν δοθε‹σαν εÙθε‹αν ¥πειρον τ¾ν ΑΒ ¢πÕ τοà δοθέντος σηµείου τοà Γ, Ö µή ™στιν ™π' αÙτÁς, κάθετος Ãκται ¹ ΓΘ. 'Επεˆ γ¦ρ ‡ση ™στˆν ¹ ΗΘ τÍ ΘΕ, κοιν¾ δ ¹ ΘΓ, δύο δ¾ αƒ ΗΘ, ΘΓ δύο τα‹ς ΕΘ, ΘΓ ‡σαι εƒσˆν ˜κατέρα ˜κατέρv· κሠβάσις ¹ ΓΗ βάσει τÍ ΓΕ ™στιν ‡ση· γωνία ¥ρα ¹ ØπÕ ΓΘΗ γωνίv τÍ ØπÕ ΕΘΓ ™στιν ‡ση. καί ε„σιν ™φεξÁς. Óταν δ εÙθε‹α ™π' εÙθε‹αν σταθε‹σα τ¦ς ™φεξÁς γωνίας ‡σας ¢λλήλαις ποιÍ, Ñρθ¾ ˜κατέρα τîν ‡σων γωνιîν ™στιν, κሠ¹ ™φεστηκυ‹α εÙθε‹α κάθετος καλε‹ται ™φ' ¿ν ™φέστηκεν. 'Επˆ τ¾ν δοθε‹σαν ¥ρα εÙθε‹αν ¥πειρον τ¾ν ΑΒ ¢πÕ τοà δοθέντος σηµείου τοà Γ, Ö µή ™στιν ™π' αÙτÁς, κάθετος Ãκται ¹ ΓΘ· Óπερ œδει ποιÁσαι.

Let AB be the given infinite straight-line and C the given point, which is not on (AB). So it is required to draw a straight-line perpendicular to the given infinite straight-line AB from the given point C, which is not on (AB). For let point D have been taken somewhere on the other side (to C) of the straight-line AB, and let the circle EF G have been drawn with center C and radius CD [Post. 3], and let the straight-line EG have been cut in half at (point) H [Prop. 1.10], and let the straight-lines CG, CH, and CE have been joined. I say that a (straightline) CH has been drawn perpendicular to the given infinite straight-line AB from the given point C, which is not on (AB). For since GH is equal to HE, and HC (is) common, the two (straight-lines) GH, HC are equal to the two straight-lines EH, HC, respectively, and the base CG is equal to the base CE. Thus, the angle CHG is equal to the angle EHC [Prop. 1.8], and they are adjacent. But when a straight-line stood on a(nother) straight-line makes the adjacent angles equal to one another, each of the equal angles is a right-angle, and the former straightline is called perpendicular to that upon which it stands [Def. 1.10]. Thus, the (straight-line) CH has been drawn perpendicular to the given infinite straight-line AB from the given point C, which is not on (AB). (Which is) the very thing it was required to do.

17

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 ιγ΄.

Proposition 13

'Ε¦ν εÙθε‹α ™π' εÙθε‹αν σταθε‹σα γωνίας ποιÍ, ½τοι δύο Ñρθ¦ς À δυσˆν Ñρθα‹ς ‡σας ποιήσει.

If a straight-line stood on a(nother) straight-line makes angles, it will certainly either make two rightangles, or (angles whose sum is) equal to two rightangles.

Ε

∆

E

Α

Γ

Β

D

A

B

C

ΕÙθε‹α γάρ τις ¹ ΑΒ ™π' εÙθε‹αν τ¾ν Γ∆ σταθε‹σα γωνίας ποιείτω τ¦ς ØπÕ ΓΒΑ, ΑΒ∆· λγω, Óτι αƒ ØπÕ ΓΒΑ, ΑΒ∆ γωνίαι ½τοι δύο Ñρθαί ε„σιν À δυσˆν Ñρθα‹ς ‡σαι. Ε„ µν οâν ‡ση ™στˆν ¹ ØπÕ ΓΒΑ τÍ ØπÕ ΑΒ∆, δύο Ñρθαί ε„σιν. ε„ δ οÜ, ½χθω ¢πÕ τοà Β σηµείου τÍ Γ∆ [εÙθείv] πρÕς Ñρθ¦ς ¹ ΒΕ· αƒ ¥ρα ØπÕ ΓΒΕ, ΕΒ∆ δύο Ñρθαί ε„σιν· κሠ™πεˆ ¹ ØπÕ ΓΒΕ δυσˆ τα‹ς ØπÕ ΓΒΑ, ΑΒΕ ‡ση ™στίν, κοιν¾ προσκείσθω ¹ ØπÕ ΕΒ∆· αƒ ¥ρα ØπÕ ΓΒΕ, ΕΒ∆ τρισˆ τα‹ς ØπÕ ΓΒΑ, ΑΒΕ, ΕΒ∆ ‡σαι ε„σίν. πάλιν, ™πεˆ ¹ ØπÕ ∆ΒΑ δυσˆ τα‹ς ØπÕ ∆ΒΕ, ΕΒΑ ‡ση ™στίν, κοιν¾ προσκείσθω ¹ ØπÕ ΑΒΓ· αƒ ¥ρα Øπό ∆ΒΑ, ΑΒΓ τρισˆ τα‹ς ØπÕ ∆ΒΕ, ΕΒΑ, ΑΒΓ ‡σαι ε„σίν. ™δείχθησαν δ καˆ αƒ ØπÕ ΓΒΕ, ΕΒ∆ τρισˆ τα‹ς αÙτα‹ς ‡σαι· τ¦ δ τù αÙτù ‡σα κሠ¢λλήλοις ™στˆν ‡σα· καˆ αƒ ØπÕ ΓΒΕ, ΕΒ∆ ¥ρα τα‹ς ØπÕ ∆ΒΑ, ΑΒΓ ‡σαι ε„σίν· ¢λλ¦ αƒ ØπÕ ΓΒΕ, ΕΒ∆ δύο Ñρθαί ε„σιν· καˆ αƒ ØπÕ ∆ΒΑ, ΑΒΓ ¥ρα δυσˆν Ñρθα‹ς ‡σαι ε„σίν. 'Ε¦ν ¥ρα εÙθε‹α ™π' εÙθε‹αν σταθε‹σα γωνίας ποιÍ, ½τοι δύο Ñρθ¦ς À δυσˆν Ñρθα‹ς ‡σας ποιήσει· Óπερ œδει δε‹ξαι.

For let some straight-line AB stood on the straightline CD make the angles CBA and ABD. I say that the angles CBA and ABD are certainly either two rightangles, or (have a sum) equal to two right-angles. In fact, if CBA is equal to ABD then they are two right-angles [Def. 1.10]. But, if not, let BE have been drawn from the point B at right-angles to [the straightline] CD [Prop. 1.11]. Thus, CBE and EBD are two right-angles. And since CBE is equal to the two (angles) CBA and ABE, let EBD have been added to both. Thus, the (sum of the angles) CBE and EBD is equal to the (sum of the) three (angles) CBA, ABE, and EBD [C.N. 2]. Again, since DBA is equal to the two (angles) DBE and EBA, let ABC have been added to both. Thus, the (sum of the angles) DBA and ABC is equal to the (sum of the) three (angles) DBE, EBA, and ABC [C.N. 2]. But (the sum of) CBE and EBD was also shown (to be) equal to the (sum of the) same three (angles). And things equal to the same thing are also equal to one another [C.N. 1]. Therefore, (the sum of) CBE and EBD is also equal to (the sum of) DBA and ABC. But, (the sum of) CBE and EBD is two right-angles. Thus, (the sum of) ABD and ABC is also equal to two right-angles. Thus, if a straight-line stood on a(nother) straightline makes angles, it will certainly either make two rightangles, or (angles whose sum is) equal to two rightangles. (Which is) the very thing it was required to show.

ιδ΄.

Proposition 14

'Ε¦ν πρός τινι εÙθείv κሠτù πρÕς αÙτÍ σηµείJ δύο εÙθε‹αι µ¾ ™πˆ τ¦ αÙτ¦ µέρη κείµεναι τ¦ς ™φεξÁς γωνίας

If two straight-lines, not lying on the same side, make adjacent angles (whose sum is) equal to two right-angles

18

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

δυσˆν Ñρθα‹ς ‡σας ποιîσιν, ™π' εÙθείας œσονται ¢λλήλαις αƒ εÙθε‹αι.

Α

Γ

at the same point on some straight-line, then the two straight-lines will be straight-on (with respect) to one another.

Ε

Β

A

∆

C

E

B

D

ΠρÕς γάρ τινι εÙθείv τÍ ΑΒ κሠτù πρÕς αÙτÍ σηµείJ τù Β δύο εÙθε‹αι αƒ ΒΓ, Β∆ µ¾ ™πˆ τ¦ αÙτ¦ µέρη κείµεναι τ¦ς ™φεξÁς γωνίας τ¦ς ØπÕ ΑΒΓ, ΑΒ∆ δύο Ñρθα‹ς ‡σας ποιείτωσαν· λέγω, Óτι ™π' εÙθείας ™στˆ τÍ ΓΒ ¹ Β∆. Ε„ γ¦ρ µή ™στι τÍ ΒΓ ™π' εÙθείας ¹ Β∆, œστω τÍ ΓΒ ™π' εÙθείας ¹ ΒΕ. 'Επεˆ οâν εÙθε‹α ¹ ΑΒ ™π' εÙθε‹αν τ¾ν ΓΒΕ ™φέστηκεν, αƒ ¥ρα ØπÕ ΑΒΓ, ΑΒΕ γωνίαι δύο Ñρθα‹ς ‡σαι ε„σίν· ε„σˆ δ καˆ αƒ ØπÕ ΑΒΓ, ΑΒ∆ δύο Ñρθα‹ς ‡σαι· αƒ ¥ρα ØπÕ ΓΒΑ, ΑΒΕ τα‹ς ØπÕ ΓΒΑ, ΑΒ∆ ‡σαι ε„σίν. κοιν¾ ¢φVρήσθω ¹ ØπÕ ΓΒΑ· λοιπ¾ ¥ρα ¹ ØπÕ ΑΒΕ λοιπÍ τÍ ØπÕ ΑΒ∆ ™στιν ‡ση, ¹ ™λάσσων τÍ µείζονι· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ™π' εÙθείας ™στˆν ¹ ΒΕ τÍ ΓΒ. еοίως δ¾ δείξοµεν, Óτι οÙδ ¥λλη τις πλ¾ν τÁς Β∆· ™π' εÙθείας ¥ρα ™στˆν ¹ ΓΒ τÍ Β∆. 'Ε¦ν ¥ρα πρός τινι εÙθείv κሠτù πρÕς αÙτÍ σηµείJ δύο εÙθε‹αι µ¾ ™πˆ αÙτ¦ µέρη κείµεναι τ¦ς ™φεξÁς γωνίας δυσˆν Ñρθα‹ς ‡σας ποιîσιν, ™π' εÙθείας œσονται ¢λλήλαις αƒ εÙθε‹αι· Óπερ œδει δε‹ξαι.

For let two straight-lines BC and BD, not lying on the same side, make adjacent angles ABC and ABD (whose sum is) equal to two right-angles at the same point B on some straight-line AB. I say that BD is straight-on with respect to CB. For if BD is not straight-on to BC then let BE be straight-on to CB. Therefore, since the straight-line AB stands on the straight-line CBE, the (sum of the) angles ABC and ABE is thus equal to two right-angles [Prop. 1.13]. But (the sum of) ABC and ABD is also equal to two rightangles. Thus, (the sum of angles) CBA and ABE is equal to (the sum of angles) CBA and ABD [C.N. 1]. Let (angle) CBA have been subtracted from both. Thus, the remainder ABE is equal to the remainder ABD [C.N. 3], the lesser to the greater. The very thing is impossible. Thus, BE is not straight-on with respect to CB. Similarly, we can show that neither (is) any other (straightline) than BD. Thus, CB is straight-on with respect to BD. Thus, if two straight-lines, not lying on the same side, make adjacent angles (whose sum is) equal to two rightangles at the same point on some straight-line, then the two straight-lines will be straight-on (with respect) to one another. (Which is) the very thing it was required to show.

ιε΄.

Proposition 15

'Ε¦ν δύο εÙθε‹αι τέµνωσιν ¢λλήλας, τ¦ς κατ¦ κορυφ¾ν γωνίας ‡σας ¢λλήλαις ποιοàσιν. ∆ύο γ¦ρ εÙθε‹αι αƒ ΑΒ, Γ∆ τεµνέτωσαν ¢λλήλας κατ¦ τÕ Ε σηµε‹ον· λέγω, Óτι ‡ση ™στˆν ¹ µν ØπÕ ΑΕΓ γωνία τÍ ØπÕ ∆ΕΒ, ¹ δ ØπÕ ΓΕΒ τÍ ØπÕ ΑΕ∆. 'Επεˆ γ¦ρ εÙθε‹α ¹ ΑΕ ™π' εÙθε‹αν τ¾ν Γ∆ ™φέστηκε γωνίας ποιοàσα τ¦ς ØπÕ ΓΕΑ, ΑΕ∆, αƒ ¥ρα ØπÕ ΓΕΑ, ΑΕ∆ γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν. πάλιν, ™πεˆ εÙθε‹α

If two straight-lines cut one another then they make the vertically opposite angles equal to one another. For let the two straight-lines AB and CD cut one another at the point E. I say that angle AEC is equal to (angle) DEB, and (angle) CEB to (angle) AED. For since the straight-line AE stands on the straightline CD, making the angles CEA and AED, the (sum of the) angles CEA and AED is thus equal to two right-

19

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

¹ ∆Ε ™π' εÙθε‹αν τ¾ν ΑΒ ™φέστηκε γωνίας ποιοàσα τ¦ς ØπÕ ΑΕ∆, ∆ΕΒ, αƒ ¥ρα ØπÕ ΑΕ∆, ∆ΕΒ γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν. ™δείχθησαν δ καˆ αƒ ØπÕ ΓΕΑ, ΑΕ∆ δυσˆν Ñρθα‹ς ‡σαι· ¡ι ¥ρα ØπÕ ΓΕΑ, ΑΕ∆ τα‹ς ØπÕ ΑΕ∆, ∆ΕΒ ‡σαι ε„σίν. κοιν¾ ¢φVρήσθω ¹ ØπÕ ΑΕ∆· λοιπ¾ ¥ρα ¹ ØπÕ ΓΕΑ λοιπÍ τÍ ØπÕ ΒΕ∆ ‡ση ™στίν· еοίως δ¾ δειχθήσεται, Óτι καˆ αƒ ØπÕ ΓΕΒ, ∆ΕΑ ‡σαι ε„σίν.

angles [Prop. 1.13]. Again, since the straight-line DE stands on the straight-line AB, making the angles AED and DEB, the (sum of the) angles AED and DEB is thus equal to two right-angles [Prop. 1.13]. But (the sum of) CEA and AED was also shown (to be) equal to two right-angles. Thus, (the sum of) CEA and AED is equal to (the sum of) AED and DEB [C.N. 1]. Let AED have been subtracted from both. Thus, the remainder CEA is equal to the remainder BED [C.N. 3]. Similarly, it can be shown that CEB and DEA are also equal.

Α

A

Ε ∆

E Γ

D

C

Β

B

'Ε¦ν ¥ρα δύο εÙθε‹αι τέµνωσιν ¢λλήλας, τ¦ς κατ¦ Thus, if two straight-lines cut one another then they κορυφ¾ν γωνίας ‡σας ¢λλήλαις ποιοàσιν· Óπερ œδει make the vertically opposite angles equal to one another. δε‹ξαι. (Which is) the very thing it was required to show.

ι$΄.

Proposition 16

ΠαντÕς τριγώνου µι©ς τîν πλευρîν προσεκβληθείσης ¹ ™κτÕς γωνία ˜κατέρας τîν ™ντÕς κሠ¢πεναντίον γωνιîν µείζων ™στίν. ”Εστω τρίγωνον τÕ ΑΒΓ, κሠπροσεκβεβλήσθω αÙτοà µία πλευρ¦ ¹ ΒΓ ™πˆ τÕ ∆· λγω, Óτι ¹ ™κτÕς γωνία ¹ ØπÕ ΑΓ∆ µείζων ™στˆν ˜κατέρας τîν ™ντÕς κሠ¢πεναντίον τîν ØπÕ ΓΒΑ, ΒΑΓ γωνιîν. Τετµήσθω ¹ ΑΓ δίχα κατ¦ τÕ Ε, κሠ™πιζευχθε‹σα ¹ ΒΕ ™κβεβλήσθω ™π' εÙθείας ™πˆ τÕ Ζ, κሠκείσθω τÍ ΒΕ ‡ση ¹ ΕΖ, κሠ™πεζεύχθω ¹ ΖΓ, κሠδιήχθω ¹ ΑΓ ™πˆ τÕ Η. 'Επεˆ οâν ‡ση ™στˆν ¹ µν ΑΕ τÍ ΕΓ, ¹ δ ΒΕ τÍ ΕΖ, δύο δ¾ αƒ ΑΕ, ΕΒ δυσˆ τα‹ς ΓΕ, ΕΖ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΑΕΒ γωνίv τÍ ØπÕ ΖΕΓ ‡ση ™στίν· κατ¦ κορυφ¾ν γάρ· βάσις ¥ρα ¹ ΑΒ βάσει τÍ ΖΓ ‡ση ™στίν, κሠτÕ ΑΒΕ τρίγωνον τù ΖΕΓ τριγώνJ ™στˆν ‡σον, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι ε„σˆν ˜κατέρα ˜κατέρv, Øφ' §ς αƒ ‡σας πλευρሠØποτείνουσιν· ‡ση ¥ρα ™στˆν ¹ ØπÕ ΒΑΕ τÍ ØπÕ ΕΓΖ. µείζων δέ ™στιν ¹ ØπÕ ΕΓ∆ τÁς ØπÕ ΕΓΖ· µείζων ¥ρα ¹ ØπÕ ΑΓ∆ τÁς ØπÕ ΒΑΕ. `Οµοίως δ¾ τÁς ΒΓ τετµηµένης δίχα δειχθήσεται

For any triangle, when one of the sides is produced, the external angle is greater than each of the internal and opposite angles. Let ABC be a triangle, and let one of its sides BC have been produced to D. I say that the external angle ACD is greater than each of the internal and opposite angles, CBA and BAC. Let the (straight-line) AC have been cut in half at (point) E [Prop. 1.10]. And BE being joined, let it have been produced in a straight-line to (point) F .† And let EF be made equal to BE [Prop. 1.3], and let F C have been joined, and let AC have been drawn through to (point) G. Therefore, since AE is equal to EC, and BE to EF , the two (straight-lines) AE, EB are equal to the two (straight-lines) CE, EF , respectively. Also, angle AEB is equal to angle F EC, for (they are) vertically opposite [Prop. 1.15]. Thus, the base AB is equal to the base F C, and the triangle ABE is equal to the triangle F EC, and the remaining angles subtended by the equal sides are equal to the corresponding remaining angles [Prop. 1.4].

20

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

κሠ¹ ØπÕ ΒΓΗ, τουτέστιν ¹ ØπÕ ΑΓ∆, µείζων κሠτÁς ØπÕ ΑΒΓ.

Α

Thus, BAE is equal to ECF . But ECD is greater than ECF . Thus, ACD is greater than BAE. Similarly, by having cut BC in half, it can be shown (that) BCG—that is to say, ACD—(is) also greater than ABC.

Ζ

A

Ε

Β

E

∆

Γ

B

D C

Η

G

ΠαντÕς ¥ρα τριγώνου µι©ς τîν πλευρîν προσεκβληθείσης ¹ ™κτÕς γωνία ˜κατέρας τîν ™ντÕς κሠ¢πεναντίον γωνιîν µείζων ™στίν· Óπερ œδει δε‹ξαι. †

F

Thus, for any triangle, when one of the sides is produced, the external angle is greater than each of the internal and opposite angles. (Which is) the very thing it was required to show.

The implicit assumption that the point F lies in the interior of the angle ABC should be counted as an additional postulate.

ιζ΄.

Proposition 17

ΠαντÕς τριγώνου αƒ δύο γωνίαι δύο Ñρθîν ™λάσσονές For any triangle, (the sum of any) two angles is less ε„σι πάντÍ µεταλαµβανόµεναι. than two right-angles, (the angles) being taken up in any (possible way).

Α

A

Β

Γ

∆

B

”Εστω τρίγωνον τÕ ΑΒΓ· λέγω, Óτι τοà ΑΒΓ τριγώνου αƒ δύο γωνίαι δύο Ñρθîν ™λάττονές ε„σι πάντV µεταλαµβανόµεναι. 'Εκβεβλήσθω γ¦ρ ¹ ΒΓ ™πˆ τÕ ∆. Κሠ™πεˆ τριγώνου τοà ΑΒΓ ™κτός ™στι γωνία ¹ ØπÕ ΑΓ∆, µείζων ™στˆ τÁς ™ντÕς κሠ¢πεναντίον τÁς ØπÕ ΑΒΓ. κοιν¾ προσκείσθω ¹ ØπÕ ΑΓΒ· αƒ ¥ρα ØπÕ ΑΓ∆, ΑΓΒ τîν ØπÕ ΑΒΓ, ΒΓΑ µείζονές ε„σιν. ¢λλ' αƒ ØπÕ ΑΓ∆,

C

D

Let ABC be a triangle. I say that (the sum of any) two angles of triangle ABC is less than two right-angles, (the angles) being taken up in any (possible way). For let BC have been produced to D. And since the angle ACD is external to triangle ABC, it is greater than the internal and opposite angle ABC [Prop. 1.16]. Let ACB have been added to both. Thus, the (sum of the angles) ACD and ACB is greater than

21

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

ΑΓΒ δύο Ñρθα‹ς ‡σαι ε„σίν· αƒ ¥ρα ØπÕ ΑΒΓ, ΒΓΑ δύο Ñρθîν ™λάσσονές ε„σιν. еοίως δ¾ δείξοµεν, Óτι καˆ αƒ ØπÕ ΒΑΓ, ΑΓΒ δύο Ñρθîν ™λάσσονές ε„σι κሠœτι αƒ ØπÕ ΓΑΒ, ΑΒΓ. ΠαντÕς ¥ρα τριγώνου αƒ δύο γωνίαι δύο Ñρθîν ™λάσσονές ε„σι πάντÍ µεταλαµβανόµεναι· Óπερ œδει δε‹ξαι.

the (sum of the angles) ABC and BCA. But, (the sum of) ACD and ACB is equal to two right-angles [Prop. 1.13]. Thus, (the sum of) ABC and BCA is less than two rightangles. Similarly, we can show that (the sum of) BAC and ACB is also less than two right-angles, and again (that the sum of) CAB and ABC (is less than two rightangles). Thus, for any triangle, (the sum of any) two angles is less than two right-angles, (the angles) being taken up in any (possible way). (Which is) the very thing it was required to show.

ιη΄.

Proposition 18

ΠαντÕς τριγώνου ¹ µείζων πλευρ¦ τ¾ν µείζονα γωνίαν Øποτείνει.

For any triangle, the greater side subtends the greater angle.

Α

A

∆

D

Γ

Β

B

C

”Εστω γ¦ρ τρίγωνον τÕ ΑΒΓ µείζονα œχον τ¾ν ΑΓ πλευρ¦ν τÁς ΑΒ· λέγω, Óτι κሠγωνία ¹ ØπÕ ΑΒΓ µείζων ™στˆ τÁς ØπÕ ΒΓΑ· 'Επεˆ γ¦ρ µείζων ™στˆν ¹ ΑΓ τÁς ΑΒ, κείσθω τÍ ΑΒ ‡ση ¹ Α∆, κሠ™πεζεύχθω ¹ Β∆. Κሠ™πεˆ τριγώνου τοà ΒΓ∆ ™κτός ™στι γωνία ¹ ØπÕ Α∆Β, µείζων ™στˆ τÁς ™ντÕς κሠ¢πεναντίον τÁς ØπÕ ∆ΓΒ· ‡ση δ ¹ ØπÕ Α∆Β τÍ ØπÕ ΑΒ∆, ™πεˆ κሠπλευρ¦ ¹ ΑΒ τÍ Α∆ ™στιν ‡ση· µείζων ¥ρα κሠ¹ ØπÕ ΑΒ∆ τÁς ØπÕ ΑΓΒ· πολλù ¥ρα ¹ ØπÕ ΑΒΓ µείζων ™στˆ τÁς ØπÕ ΑΓΒ. ΠαντÕς ¥ρα τριγώνου ¹ µείζων πλευρ¦ τ¾ν µείζονα γωνίαν Øποτείνει· Óπερ œδει δε‹ξαι.

For let ABC be a triangle having side AC greater than AB. I say that angle ABC is also greater than BCA. For since AC is greater than AB, let AD be made equal to AB [Prop. 1.3], and let BD have been joined. And since angle ADB is external to triangle BCD, it is greater than the internal and opposite (angle) DCB [Prop. 1.16]. But ADB (is) equal to ABD, since side AB is also equal to side AD [Prop. 1.5]. Thus, ABD is also greater than ACB. Thus, ABC is much greater than ACB. Thus, for any triangle, the greater side subtends the greater angle. (Which is) the very thing it was required to show.

ιθ΄.

Proposition 19

ΠαντÕς τριγώνου ØπÕ τ¾ν µείζονα γωνίαν ¹ µείζων πλευρ¦ Øποτείνει. ”Εστω τρίγωνον τÕ ΑΒΓ µείζονα œχον τ¾ν ØπÕ ΑΒΓ γωνίαν τÁς ØπÕ ΒΓΑ· λέγω, Óτι κሠπλευρ¦ ¹ ΑΓ πλευρ©ς τÁς ΑΒ µείζων ™στίν. Ε„ γ¦ρ µή, ½τοι ‡ση ™στˆν ¹ ΑΓ τÍ ΑΒ À ™λάσσων· ‡ση µν οâν οÙκ œστιν ¹ ΑΓ τÍ ΑΒ· ‡ση γ¦ρ ¨ν Ãν κሠγωνία ¹ ØπÕ ΑΒΓ τÍ ØπÕ ΑΓΒ· οÙκ œστι δέ· οÙκ ¥ρα ‡ση ™στˆν ¹ ΑΓ τÍ ΑΒ. οÙδ µ¾ν ™λάσσων ™στˆν ¹ ΑΓ

For any triangle, the greater angle is subtended by the greater side. Let ABC be a triangle having the angle ABC greater than BCA. I say that side AC is also greater than side AB. For if not, AC is certainly either equal to, or less than, AB. In fact, AC is not equal to AB. For then angle ABC would also have been equal to ACB [Prop. 1.5]. But it is not. Thus, AC is not equal to AB. Neither, indeed, is AC

22

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

τÁς ΑΒ· ™λάσσων γ¦ρ ¨ν Ãν κሠγωνία ¹ ØπÕ ΑΒΓ τÁς ØπÕ ΑΓΒ· οÙκ œστι δέ· οÙκ ¥ρα ™λάσσων ™στˆν ¹ ΑΓ τÁς ΑΒ. ™δείχθη δέ, Óτι οÙδ ‡ση ™στίν. µείζων ¥ρα ™στˆν ¹ ΑΓ τÁς ΑΒ.

less than AB. For then angle ABC would also have been less than ACB [Prop. 1.18]. But it is not. Thus, AC is not less than AB. But it was shown that (AC) is also not equal (to AB). Thus, AC is greater than AB.

Α

A

Β

B

Γ

C

ΠαντÕς ¥ρα τριγώνου ØπÕ τ¾ν µείζονα γωνίαν ¹ Thus, for any triangle, the greater angle is subtended µείζων πλευρ¦ Øποτείνει· Óπερ œδει δε‹ξαι. by the greater side. (Which is) the very thing it was required to show.

κ΄.

Proposition 20

ΠαντÕς τριγώνου αƒ δύο πλευρሠτÁς λοιπÁς µείζονές ε„σι πάντV µεταλαµβανόµεναι.

For any triangle, (the sum of any) two sides is greater than the remaining (side), (the sides) being taken up in any (possible way).

∆

D

Α

Β

A

Γ

B

”Εστω γ¦ρ τρίγωνον τÕ ΑΒΓ· λέγω, Óτι τοà ΑΒΓ τριγώνου αƒ δύο πλευρሠτÁς λοιπÁς µείζονές ε„σι παντV µεταλαµβανόµεναι, αƒ µν ΒΑ, ΑΓ τÁς ΒΓ, αƒ δ ΑΒ, ΒΓ τÁς ΑΓ, αƒ δ ΒΓ, ΓΑ τÁς ΑΒ. ∆ιήχθω γ¦ρ ¹ ΒΑ ™πˆ τÕ ∆ σηµε‹ον, κሠκείσθω τÍ ΓΑ ‡ση ¹ Α∆, κሠ™πεζεύχθω ¹ ∆Γ. 'Επεˆ οâν ‡ση ™στˆν ¹ ∆Α τÍ ΑΓ, ‡ση ™στˆ κሠγωνία ¹ ØπÕ Α∆Γ τÍ ØπÕ ΑΓ∆· µείζων ¥ρα ¹ ØπÕ ΒΓ∆ τÁς

C

For let ABC be a triangle. I say that for triangle ABC (the sum of any) two sides is greater than the remaining (side), (the sides) being taken up in any (possible way). (So), (the sum of) BA and AC (is greater) than BC, (the sum of) AB and BC than AC, and (the sum of) BC and CA than AB. For let BA have been drawn through to point D, and let AD be made equal to CA [Prop. 1.3], and let DC

23

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

ØπÕ Α∆Γ· κሠ™πεˆ τρίγωνόν ™στι τÕ ∆ΓΒ µείζονα œχον τ¾ν ØπÕ ΒΓ∆ γωνίαν τÁς ØπÕ Β∆Γ, ØπÕ δ τ¾ν µείζονα γωνίαν ¹ µείζων πλευρ¦ Øποτείνει, ¹ ∆Β ¥ρα τÁς ΒΓ ™στι µείζων. ‡ση δ ¹ ∆Α τÍ ΑΓ· µείζονες ¥ρα αƒ ΒΑ, ΑΓ τÁς ΒΓ· еοίως δ¾ δείξοµεν, Óτι καˆ αƒ µν ΑΒ, ΒΓ τÁς ΓΑ µείζονές ε„σιν, αƒ δ ΒΓ, ΓΑ τÁς ΑΒ. ΠαντÕς ¥ρα τριγώνου αƒ δύο πλευρሠτÁς λοιπÁς µείζονές ε„σι πάντV µεταλαµβανόµεναι· Óπερ œδει δε‹ξαι.

have been joined. Therefore, since DA is equal to AC, the angle ADC is also equal to ACD [Prop. 1.5]. Thus, BCD is greater than ADC. And since triangle DCB has the angle BCD greater than BDC, and the greater angle subtends the greater side [Prop. 1.19], DB is thus greater than BC. But DA is equal to AC. Thus, (the sum of) BA and AC is greater than BC. Similarly, we can show that (the sum of) AB and BC is also greater than CA, and (the sum of) BC and CA than AB. Thus, for any triangle, (the sum of any) two sides is greater than the remaining (side), (the sides) being taken up in any (possible way). (Which is) the very thing it was required to show.

κα΄.

Proposition 21

'Ε¦ν τριγώνου ™πˆ µι©ς τîν πλευρîν ¢πÕ τîν If two internal straight-lines are constructed on one περάτων δύο εÙθε‹αι ™ντÕς συσταθîσιν, αƒ συσταθε‹σαι of the sides of a triangle, from its ends, the constructed τîν λοιπîν τοà τριγώνου δύο πλευρîν ™λάττονες µν (straight-lines) will be less than the two remaining sides œσονται, µείζονα δ γωνίαν περιέξουσιν. of the triangle, but will encompass a greater angle.

Α

A Ε

E

∆

Β

D

Γ

B

Τριγώνου γ¦ρ τοà ΑΒΓ ™πˆ µι©ς τîν πλευρîν τÁς ΒΓ ¢πÕ τîν περάτων τîν Β, Γ δύο εÙθε‹αι ™ντÕς συνεστάτωσαν αƒ Β∆, ∆Γ· λέγω, Óτι αƒ Β∆, ∆Γ τîν λοιπîν τοà τριγώνου δύο πλευρîν τîν ΒΑ, ΑΓ ™λάσσονες µέν ε„σιν, µείζονα δ γωνίαν περιέχουσι τ¾ν ØπÕ Β∆Γ τÁς ØπÕ ΒΑΓ. ∆ιήχθω γ¦ρ ¹ Β∆ ™πˆ τÕ Ε. κሠ™πεˆ παντÕς τριγώνου αƒ δύο πλευρሠτÁς λοιπÁς µείζονές ε„σιν, τοà ΑΒΕ ¥ρα τριγώνου αƒ δύο πλευραˆ αƒ ΑΒ, ΑΕ τÁς ΒΕ µείζονές ε„σιν· κοιν¾ προσκείσθω ¹ ΕΓ· αƒ ¥ρα ΒΑ, ΑΓ τîν ΒΕ, ΕΓ µείζονές ε„σιν. πάλιν, ™πεˆ τοà ΓΕ∆ τριγώνου αƒ δύο πλευραˆ αƒ ΓΕ, Ε∆ τÁς Γ∆ µείζονές ε„σιν, κοιν¾ προσκείσθω ¹ ∆Β· αƒ ΓΕ, ΕΒ ¥ρα τîν Γ∆, ∆Β µείζονές ε„σιν. ¢λλ¦ τîν ΒΕ, ΕΓ µείζονες ™δείχθησαν αƒ ΒΑ, ΑΓ· πολλù ¥ρα αƒ ΒΑ, ΑΓ τîν Β∆, ∆Γ µείζονές ε„σιν. Πάλιν, ™πεˆ παντÕς τριγώνου ¹ ™κτÕς γωνία τÁς ™ντÕς κሠ¢πεναντίον µείζων ™στίν, τοà Γ∆Ε ¥ρα τριγώνου ¹ ™κτÕς γωνία ¹ ØπÕ Β∆Γ µείζων ™στˆ τÁς ØπÕ ΓΕ∆. δι¦ ταÙτ¦ τοίνυν κሠτοà ΑΒΕ τριγώνου ¹ ™κτÕς γωνία ¹ ØπÕ ΓΕΒ µείζων ™στˆ τÁς ØπÕ ΒΑΓ. ¢λλ¦ τÁς ØπÕ ΓΕΒ µείζων ™δείχθη ¹ ØπÕ Β∆Γ· πολλù ¥ρα ¹ ØπÕ Β∆Γ

C

For let the two internal straight-lines BD and DC have been constructed on one of the sides BC of the triangle ABC, from its ends B and C (respectively). I say that BD and DC are less than the (sum of the) two remaining sides of the triangle BA and AC, but encompass an angle BDC greater than BAC. For let BD have been drawn through to E. And since for every triangle (the sum of any) two sides is greater than the remaining (side) [Prop. 1.20], for triangle ABE the (sum of the) two sides AB and AE is thus greater than BE. Let EC have been added to both. Thus, (the sum of) BA and AC is greater than (the sum of) BE and EC. Again, since in triangle CED the (sum of the) two sides CE and ED is greater than CD, let DB have been added to both. Thus, (the sum of) CE and EB is greater than (the sum of) CD and DB. But, (the sum of) BA and AC was shown (to be) greater than (the sum of) BE and EC. Thus, (the sum of) BA and AC is much greater than (the sum of) BD and DC. Again, since for every triangle the external angle is greater than the internal and opposite (angles) [Prop.

24

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

µείζων ™στˆ τÁς ØπÕ ΒΑΓ. 'Ε¦ν ¥ρα τριγώνου ™πˆ µι©ς τîν πλευρîν ¢πÕ τîν περάτων δύο εÙθε‹αι ™ντÕς συσταθîσιν, αƒ συσταθε‹σαι τîν λοιπîν τοà τριγώνου δύο πλευρîν ™λάττονες µέν ε„σιν, µείζονα δ γωνίαν περιέχουσιν· Óπερ œδει δε‹ξαι.

1.16], for triangle CDE the external angle BDC is thus greater than CED. Accordingly, for the same (reason), the external angle CEB of the triangle ABE is also greater than BAC. But, BDC was shown (to be) greater than CEB. Thus, BDC is much greater than BAC. Thus, if two internal straight-lines are constructed on one of the sides of a triangle, from its ends, the constructed (straight-lines) are less than the two remaining sides of the triangle, but encompass a greater angle. (Which is) the very thing it was required to show.

κβ΄.

Proposition 22

'Εκ τριîν εÙθειîν, α† ε„σιν ‡σαι τρισˆ τα‹ς δοθείσαις [εÙθείαις], τρίγωνον συστήσασθαι· δε‹ δ τ¦ς δύο τÁς λοιπÁς µείζονας εναι πάντV µεταλαµβανοµένας [δι¦ τÕ κሠπαντÕς τριγώνου τ¦ς δύο πλευρ¦ς τÁς λοιπÁς µείζονας εναι πάντV µεταλαµβανοµένας].

To construct a triangle from three straight-lines which are equal to three given [straight-lines]. It is necessary for (the sum of) two (of the straight-lines) to be greater than the remaining (one), (the straight-lines) being taken up in any (possible way) [on account of the (fact that) for every triangle (the sum of any) two sides is greater than the remaining (one), (the sides) being taken up in any (possible way) [Prop. 1.20] ].

Α Β Γ

A B C Κ

∆

Ζ

Η

K

Θ

Ε

D

Λ

F

G

H

E

L

”Εστωσαν αƒ δοθε‹σαι τρε‹ς εÙθε‹αι αƒ Α, Β, Γ, ïν αƒ δύο τÁς λοιπÁς µείζονες œστωσαν πάντV µεταλαµβανόµεναι, αƒ µν Α, Β τÁς Γ, αƒ δ Α, Γ τÁς Β, κሠœτι αƒ Β, Γ τÁς Α· δε‹ δ¾ ™κ τîν ‡σων τα‹ς Α, Β, Γ τρίγωνον συστήσασθαι. 'Εκκείσθω τις εÙθε‹α ¹ ∆Ε πεπερασµένη µν κατ¦ τÕ ∆ ¥πειρος δ κατ¦ τÕ Ε, κሠκείσθω τÍ µν Α ‡ση ¹ ∆Ζ, τÍ δ Β ‡ση ¹ ΖΗ, τÍ δ Γ ‡ση ¹ ΗΘ· κሠκέντρJ µν τù Ζ, διαστήµατι δ τù Ζ∆ κύκλος γεγράφθω Ð ∆ΚΛ· πάλιν κέντρJ µν τù Η, διαστήµατι δ τù ΗΘ κύκλος γεγράφθω Ð ΚΛΘ, κሠ™πεζεύχθωσαν αƒ ΚΖ, ΚΗ· λέγω, Óτι ™κ τριîν εÙθειîν τîν ‡σων τα‹ς Α, Β, Γ τρίγωνον συνέσταται τÕ ΚΖΗ. 'Επεˆ γ¦ρ τÕ Ζ σηµε‹ον κέντρον ™στˆ τοà ∆ΚΛ κύκλου, ‡ση ™στˆν ¹ Ζ∆ τÍ ΖΚ· ¢λλ¦ ¹ Ζ∆ τÍ Α ™στιν ‡ση. κሠ¹ ΚΖ ¥ρα τÍ Α ™στιν ‡ση. πάλιν, ™πεˆ τÕ Η

Let A, B, and C be the three given straight-lines, of which let (the sum of any) two be greater than the remaining (one), (the straight-lines) being taken up in (any possible way). (Thus), (the sum of) A and B (is greater) than C, (the sum of) A and C than B, and also (the sum of) B and C than A. So it is required to construct a triangle from (straight-lines) equal to A, B, and C. Let some straight-line DE be set out, terminated at D, and infinite in the direction of E. And let DF made equal to A [Prop. 1.3], and F G equal to B [Prop. 1.3], and GH equal to C [Prop. 1.3]. And let the circle DKL have been drawn with center F and radius F D. Again, let the circle KLH have been drawn with center G and radius GH. And let KF and KG have been joined. I say that the triangle KF G has been constructed from three straight-lines equal to A, B, and C.

25

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

σηµε‹ον κέντρον ™στˆ τοà ΛΚΘ κύκλου, ‡ση ™στˆν ¹ ΗΘ τÍ ΗΚ· ¢λλ¦ ¹ ΗΘ τÍ Γ ™στιν ‡ση· κሠ¹ ΚΗ ¥ρα τÍ Γ ™στιν ‡ση. ™στˆ δ κሠ¹ ΖΗ τÍ Β ‡ση· αƒ τρε‹ς ¥ρα εÙθε‹αι αƒ ΚΖ, ΖΗ, ΗΚ τρισˆ τα‹ς Α, Β, Γ ‡σαι ε„σίν. 'Εκ τριîν ¥ρα εÙθειîν τîν ΚΖ, ΖΗ, ΗΚ, α† ε„σιν ‡σαι τρισˆ τα‹ς δοθείσαις εÙθείαις τα‹ς Α, Β, Γ, τρίγωνον συνέσταται τÕ ΚΖΗ· Óπερ œδει ποιÁσαι.

For since point F is the center of the circle DKL, F D is equal to F K. But, F D is equal to A. Thus, KF is also equal to A. Again, since point G is the center of the circle LKH, GH is equal to GK. But, GH is equal to C. Thus, KG is also equal to C. And F G is equal to B. Thus, the three straight-lines KF , F G, and GK are equal to A, B, and C (respectively). Thus, the triangle KF G has been constructed from the three straight-lines KF , F G, and GK, which are equal to the three given straight-lines A, B, and C (respectively). (Which is) the very thing it was required to do.

κγ΄.

Proposition 23

ΠρÕς τÍ δοθείσV εÙθείv κሠτù πρÕς αÙτV σηµείJ τÍ To construct a rectilinear angle equal to a given rectiδοθείσV γωνίv εÙθυγράµµJ ‡σην γωνίαν εÙθύγραµµον linear angle at a (given) point on a given straight-line. συστήσασθαι.

∆

D

Γ

C Ε

E

Ζ

Α

F

Η

Β

A

”Εστω ¹ µν δοθε‹σα εÙθε‹α ¹ ΑΒ, τÕ δ πρÕς αÙτÍ σηµε‹ον τÕ Α, ¹ δ δοθε‹σα γωνία εÙθύγραµµος ¹ ØπÕ ∆ΓΕ· δε‹ δ¾ πρÕς τÍ δοθε‹σV εÙθείv τÍ ΑΒ κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ δοθείσV γωνίv εÙθυγράµµJ τÍ ØπÕ ∆ΓΕ ‡σην γωνίαν εÙθύγραµµον συστήσασθαι. Ε„λήφθω ™φ' ˜κατέρας τîν Γ∆, ΓΕ τυχόντα σηµε‹α τ¦ ∆, Ε, κሠ™πεζεύχθω ¹ ∆Ε· κሠ™κ τριîν εÙθειîν, α† ε„σιν ‡σαι τρισˆ τα‹ς Γ∆, ∆Ε, ΓΕ, τρίγωνον συνεστάτω τÕ ΑΖΗ, éστε ‡σην εναι τ¾ν µν Γ∆ τÍ ΑΖ, τ¾ν δ ΓΕ τÍ ΑΗ, κሠœτι τ¾ν ∆Ε τÍ ΖΗ. 'Επεˆ οâν δύο αƒ ∆Γ, ΓΕ δύο τα‹ς ΖΑ, ΑΗ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv, κሠβάσις ¹ ∆Ε βάσει τÍ ΖΗ ‡ση, γωνία ¥ρα ¹ ØπÕ ∆ΓΕ γωνίv τÍ ØπÕ ΖΑΗ ™στιν ‡ση. ΠρÕς ¥ρα τÍ δοθείσV εÙθείv τÍ ΑΒ κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ δοθείσV γωνίv εÙθυγράµµJ τÍ ØπÕ ∆ΓΕ ‡ση γωνία εÙθύγραµµος συνέσταται ¹ ØπÕ ΖΑΗ· Óπερ œδει ποιÁσαι.

G

B

Let AB be the given straight-line, A the (given) point on it, and DCE the given rectilinear angle. So it is required to construct a rectilinear angle equal to the given rectilinear angle DCE at the (given) point A on the given straight-line AB. Let the points D and E have been taken somewhere on each of the (straight-lines) CD and CE (respectively), and let DE have been joined. And let the triangle AF G have been constructed from three straight-lines which are equal to CD, DE, and CE, such that CD is equal to AF , CE to AG, and also DE to F G [Prop. 1.22]. Therefore, since the two (straight-lines) DC, CE are equal to the two straight-lines F A, AG, respectively, and the base DE is equal to the base F G, the angle DCE is thus equal to the angle F AG [Prop. 1.8]. Thus, the rectilinear angle F AG, equal to the given rectilinear angle DCE, has been constructed at the (given) point A on the given straight-line AB. (Which is) the very thing it was required to do.

26

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 κδ΄.

Proposition 24

'Ε¦ν δύο τρίγωνα τ¦ς δύο πλευρ¦ς [τα‹ς] δύο If two triangles have two sides equal to two sides, reπλευρα‹ς ‡σας œχV ˜κατέραν ˜κατέρv, τ¾ν δ γωνίαν spectively, but (one) has the angle encompassed by the τÁς γωνίας µείζονα œχV τ¾ν ØπÕ τîν ‡σων εÙθειîν πε- equal straight-lines greater than the (corresponding) anριεχοµένην, κሠτ¾ν βάσιν τÁς βάσεως µείζονα ›ξει. gle (in the other), then (the former triangle) will also have a base greater than the base (of the latter).

Α

∆

A

D

Ε

E

Β Γ

B Η

Ζ

C

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ τ¦ς δύο πλευρ¦ς τ¦ς ΑΒ, ΑΓ τα‹ς δύο πλευρα‹ς τα‹ς ∆Ε, ∆Ζ ‡σας œχοντα ˜κατέραν ˜κατέρv, τ¾ν µν ΑΒ τÍ ∆Ε τ¾ν δ ΑΓ τÍ ∆Ζ, ¹ δ πρÕς τù Α γωνία τÁς πρÕς τù ∆ γωνίας µείζων œστω· λέγω, Óτι κሠβάσις ¹ ΒΓ βάσεως τÁς ΕΖ µείζων ™στίν. 'Επεˆ γ¦ρ µείζων ¹ ØπÕ ΒΑΓ γωνία τÁς ØπÕ Ε∆Ζ γωνίας, συνεστάτω πρÕς τÍ ∆Ε εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù ∆ τÍ ØπÕ ΒΑΓ γωνίv ‡ση ¹ ØπÕ Ε∆Η, κሠκείσθω Ðποτέρv τîν ΑΓ, ∆Ζ ‡ση ¹ ∆Η, κሠ™πεζεύχθωσαν αƒ ΕΗ, ΖΗ. 'Επεˆ οâν ‡ση ™στˆν ¹ µν ΑΒ τÍ ∆Ε, ¹ δ ΑΓ τÍ ∆Η, δύο δ¾ αƒ ΒΑ, ΑΓ δυσˆ τα‹ς Ε∆, ∆Η ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΒΑΓ γωνίv τÍ ØπÕ Ε∆Η ‡ση· βάσις ¥ρα ¹ ΒΓ βάσει τÍ ΕΗ ™στιν ‡ση. πάλιν, ™πεˆ ‡ση ™στˆν ¹ ∆Ζ τÍ ∆Η, ‡ση ™στˆ κሠ¹ ØπÕ ∆ΗΖ γωνία τÍ ØπÕ ∆ΖΗ· µείζων ¥ρα ¹ ØπÕ ∆ΖΗ τÁς ØπÕ ΕΗΖ· πολλù ¥ρα µείζων ™στˆν ¹ ØπÕ ΕΖΗ τÁς ØπÕ ΕΗΖ. κሠ™πεˆ τρίγωνόν ™στι τÕ ΕΖΗ µείζονα œχον τ¾ν ØπÕ ΕΖΗ γωνίαν τÁς ØπÕ ΕΗΖ, ØπÕ δ τ¾ν µείζονα γωνίαν ¹ µείζων πλευρ¦ Øποτείνει, µείζων ¥ρα κሠπλευρ¦ ¹ ΕΗ τÁς ΕΖ. ‡ση δ ¹ ΕΗ τÍ ΒΓ· µείζων ¥ρα κሠ¹ ΒΓ τÁς ΕΖ. 'Ε¦ν ¥ρα δύο τρίγωνα τ¦ς δύο πλευρ¦ς δυσˆ πλευρα‹ς ‡σας œχV ˜κατέραν ˜κατέρv, τ¾ν δ γωνίαν τÁς γωνίας µείζονα œχV τ¾ν ØπÕ τîν ‡σων εÙθειîν περιεχοµένην, κሠτ¾ν βάσιν τÁς βάσεως µείζονα ›ξει· Óπερ œδει δε‹ξαι.

G

F

Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF , respectively. (That is), AB to DE, and AC to DF . Let them also have the angle at A greater than the angle at D. I say that the base BC is greater than the base EF . For since angle BAC is greater than angle EDF , let (angle) EDG, equal to angle BAC, have been constructed at point D on the straight-line DE [Prop. 1.23]. And let DG be made equal to either of AC or DF [Prop. 1.3], and let EG and F G have been joined. Therefore, since AB is equal to DE and AC to DG, the two (straight-lines) BA, AC are equal to the two (straight-lines) ED, DG, respectively. Also the angle BAC is equal to the angle EDG. Thus, the base BC is equal to the base EG [Prop. 1.4]. Again, since DF is equal to DG, angle DGF is also equal to angle DF G [Prop. 1.5]. Thus, DF G (is) greater than EGF . Thus, EF G is much greater than EGF . And since triangle EF G has angle EF G greater than EGF , and the greater angle subtends the greater side [Prop. 1.19], side EG (is) thus also greater than EF . But EG (is) equal to BC. Thus, BC (is) also greater than EF . Thus, if two triangles have two sides equal to two sides, respectively, but (one) has the angle encompassed by the equal straight-lines greater than the (corresponding) angle (in the other), then (the former triangle) will also have a base greater than the base (of the latter). (Which is) the very thing it was required to show.

27

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 κε΄.

Proposition 25

'Ε¦ν δύο τρίγωνα τ¦ς δύο πλευρ¦ς δυσˆ πλευρα‹ς If two triangles have two sides equal to two sides, ‡σας œχV ˜κατέραν ˜κατέρv, τ¾ν δ βασίν τÁς βάσεως respectively, but (one) has a base greater than the base µείζονα œχV, κሠτ¾ν γωνίαν τÁς γωνίας µείζονα ›ξει τ¾ν (of the other), then (the former triangle) will also have ØπÕ τîν ‡σων εÙθειîν περιεχοµένην. the angle encompassed by the equal straight-lines greater than the (corresponding) angle (in the latter).

Α

A Γ

C

∆

D

Β

B

Ε

Ζ

E

F

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ τ¦ς δύο πλευρ¦ς τ¦ς ΑΒ, ΑΓ τα‹ς δύο πλευρα‹ς τα‹ς ∆Ε, ∆Ζ ‡σας œχοντα ˜κατέραν ˜κατέρv, τ¾ν µν ΑΒ τÍ ∆Ε, τ¾ν δ ΑΓ τÍ ∆Ζ· βάσις δ ¹ ΒΓ βάσεως τÁς ΕΖ µείζων œστω· λέγω, Óτι κሠγωνία ¹ ØπÕ ΒΑΓ γωνίας τÁς ØπÕ Ε∆Ζ µείζων ™στίν. Ε„ γ¦ρ µή, ½τοι ‡ση ™στˆν αÙτÍ À ™λάσσων· ‡ση µν οâν οÙκ œστιν ¹ ØπÕ ΒΑΓ τÍ ØπÕ Ε∆Ζ· ‡ση γ¦ρ ¨ν Ãν κሠβάσις ¹ ΒΓ βάσει τÍ ΕΖ· οÙκ œστι δέ. οÙκ ¥ρα ‡ση ™στˆ γωνία ¹ ØπÕ ΒΑΓ τÍ ØπÕ Ε∆Ζ· οÙδ µ¾ν ™λάσσων ™στˆν ¹ ØπÕ ΒΑΓ τÁς ØπÕ Ε∆Ζ· ™λάσσων γ¦ρ ¨ν Ãν κሠβάσις ¹ ΒΓ βάσεως τÁς ΕΖ· οÙκ œστι δέ· οÙκ ¥ρα ™λάσσων ™στˆν ¹ ØπÕ ΒΑΓ γωνία τÁς ØπÕ Ε∆Ζ. ™δείχθη δέ, Óτι οÙδ ‡ση· µείζων ¥ρα ™στˆν ¹ ØπÕ ΒΑΓ τÁς ØπÕ Ε∆Ζ. 'Ε¦ν ¥ρα δύο τρίγωνα τ¦ς δύο πλευρ¦ς δυσˆ πλευρα‹ς ‡σας œχV ˜κατέραν ˜κάτερv, τ¾ν δ βασίν τÁς βάσεως µείζονα œχV, κሠτ¾ν γωνίαν τÁς γωνίας µείζονα ›ξει τ¾ν ØπÕ τîν ‡σων εÙθειîν περιεχοµένην· Óπερ œδει δε‹ξαι.

Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF , respectively (That is), AB to DE, and AC to DF . And let the base BC be greater than the base EF . I say that angle BAC is also greater than EDF . For if not, (BAC) is certainly either equal to, or less than, (EDF ). In fact, BAC is not equal to EDF . For then the base BC would also have been equal to EF [Prop. 1.4]. But it is not. Thus, angle BAC is not equal to EDF . Neither, indeed, is BAC less than EDF . For then the base BC would also have been less than EF [Prop. 1.24]. But it is not. Thus, angle BAC is not less than EDF . But it was shown that (BAC is) also not equal (to EDF ). Thus, BAC is greater than EDF . Thus, if two triangles have two sides equal to two sides, respectively, but (one) has a base greater than the base (of the other), then (the former triangle) will also have the angle encompassed by the equal straight-lines greater than the (corresponding) angle (in the latter). (Which is) the very thing it was required to show.

κ$΄.

Proposition 26

'Ε¦ν δύο τρίγωνα τ¦ς δύο γωνίας δυσˆ γωνίαις ‡σας œχV ˜καρέραν ˜καρέρv κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην ½τοι τ¾ν πρÕς τα‹ς ‡σαις γωνίαις À τ¾ν Øποτείνουσαν ØπÕ µίαν τîν ‡σων γωνιîν, κሠτ¦ς λοιπ¦ς πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξει [˜κατέραν ˜κατέρv] κሠτ¾ν λοιπ¾ν γωνίαν τÍ λοιπÍ γωνίv. ”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ τ¦ς δύο γωνίας τ¦ς ØπÕ ΑΒΓ, ΒΓΑ δυσˆ τα‹ς ØπÕ ∆ΕΖ, ΕΖ∆ ‡σας œχοντα ˜κατέραν ˜κατέρv, τ¾ν µν ØπÕ ΑΒΓ τÍ ØπÕ

If two triangles have two angles equal to two angles, respectively, and one side equal to one side—in fact, either that by the equal angles, or that subtending one of the equal angles—then (the triangles) will also have the remaining sides equal to the [corresponding] remaining sides, and the remaining angle (equal) to the remaining angle. Let ABC and DEF be two triangles having the two angles ABC and BCA equal to the two (angles) DEF

28

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

∆ΕΖ, τ¾ν δ ØπÕ ΒΓΑ τÍ ØπÕ ΕΖ∆· ™χέτω δ κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην, πρότερον τ¾ν πρÕς τα‹ς ‡σαις γωνίαις τ¾ν ΒΓ τÍ ΕΖ· λέγω, Óτι κሠτ¦ς λοιπ¦ς πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξει ˜κατέραν ˜κατέρv, τ¾ν µν ΑΒ τÍ ∆Ε τ¾ν δ ΑΓ τÍ ∆Ζ, κሠτ¾ν λοιπ¾ν γωνίαν τÍ λοιπÍ γωνίv, τ¾ν ØπÕ ΒΑΓ τÍ ØπÕ Ε∆Ζ.

and EF D, respectively. (That is) ABC to DEF , and BCA to EF D. And let them also have one side equal to one side. First of all, the (side) by the equal angles. (That is) BC (equal) to EF . I say that the remaining sides will be equal to the corresponding remaining sides. (That is) AB to DE, and AC to DF . And the remaining angle (will be equal) to the remaining angle. (That is) BAC to EDF .

∆

D

Α Η Β

A Ε

G

Ζ

E

Θ Γ

B

Ε„ γ¦ρ ¥νισός ™στιν ¹ ΑΒ τÍ ∆Ε, µία αÙτîν µείζων ™στίν. œστω µείζων ¹ ΑΒ, κሠκείσθω τÍ ∆Ε ‡ση ¹ ΒΗ, κሠ™πεζεύχθω ¹ ΗΓ. 'Επεˆ οâν ‡ση ™στˆν ¹ µν ΒΗ τÍ ∆Ε, ¹ δ ΒΓ τÍ ΕΖ, δύο δ¾ αƒ ΒΗ, ΒΓ δυσˆ τα‹ς ∆Ε, ΕΖ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΗΒΓ γωνίv τÍ ØπÕ ∆ΕΖ ‡ση ™στίν· βάσις ¥ρα ¹ ΗΓ βάσει τÍ ∆Ζ ‡ση ™στίν, κሠτÕ ΗΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ ‡σον ™στίν, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται, Øφ' §ς αƒ ‡σας πλευρሠØποτείνουσιν· ‡ση ¥ρα ¹ ØπÕ ΗΓΒ γωνία τÍ ØπÕ ∆ΖΕ. ¢λλ¦ ¹ ØπÕ ∆ΖΕ τÍ ØπÕ ΒΓΑ Øπόκειται ‡ση· κሠ¹ ØπÕ ΒΓΗ ¥ρα τÍ ØπÕ ΒΓΑ ‡ση ™στίν, ¹ ™λάσσων τÍ µείζονι· Óπερ ¢δύνατον. οÙκ ¥ρα ¥νισός ™στιν ¹ ΑΒ τÍ ∆Ε. ‡ση ¥ρα. œστι δ κሠ¹ ΒΓ τÍ ΕΖ ‡ση· δύο δ¾ αƒ ΑΒ, ΒΓ δυσˆ τα‹ς ∆Ε, ΕΖ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΑΒΓ γωνίv τÍ ØπÕ ∆ΕΖ ™στιν ‡ση· βάσις ¥ρα ¹ ΑΓ βάσει τÍ ∆Ζ ‡ση ™στίν, κሠλοιπ¾ γωνία ¹ ØπÕ ΒΑΓ τÍ λοιπÍ γωνίv τÍ ØπÕ Ε∆Ζ ‡ση ™στίν. 'Αλλ¦ δ¾ πάλιν œστωσαν αƒ ØπÕ τ¦ς ‡σας γωνίας πλευρሠØποτείνουσαι ‡σαι, æς ¹ ΑΒ τÍ ∆Ε· λέγω πάλιν, Óτι καˆ αƒ λοιπሠπλευρሠτα‹ς λοιπα‹ς πλευρα‹ς ‡σας œσονται, ¹ µν ΑΓ τÍ ∆Ζ, ¹ δ ΒΓ τÍ ΕΖ κሠœτι ¹ λοιπ¾ γωνία ¹ ØπÕ ΒΑΓ τÍ λοιπÍ γωνίv τÍ ØπÕ Ε∆Ζ ‡ση ™στίν. Ε„ γ¦ρ ¥νισός ™στιν ¹ ΒΓ τÍ ΕΖ, µία αÙτîν µείζων ™στίν. œστω µείζων, ε„ δυνατόν, ¹ ΒΓ, κሠκείσθω τÍ ΕΖ ‡ση ¹ ΒΘ, κሠ™πεζεύχθω ¹ ΑΘ. κሠ™πι ‡ση ™στˆν ¹ µν ΒΘ τÍ ΕΖ ¹ δ ΑΒ τÍ ∆Ε, δύο δ¾ αƒ ΑΒ, ΒΘ δυσˆ τα‹ς ∆Ε, ΕΖ ‡σαι ε„σˆν ˜κατέρα ˜καρέρv· κሠγωνίας ‡σας περιέχουσιν· βάσις ¥ρα ¹ ΑΘ βάσει τÍ ∆Ζ ‡ση ™στίν, κሠτÕ ΑΒΘ τρίγωνον τù ∆ΕΖ τριγώνJ ‡σον ™στίν, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται, Øφ' §ς αƒ ‡σας πλευρሠØποτείνουσιν· ‡ση ¥ρα ™στˆν ¹ ØπÕ ΒΘΑ

H

F

C

For if AB is unequal to DE then one of them is greater. Let AB be greater, and let BG be made equal to DE [Prop. 1.3], and let GC have been joined. Therefore, since BG is equal to DE, and BC to EF , the two (straight-lines) GB, BC † are equal to the two (straight-lines) DE, EF , respectively. And angle GBC is equal to angle DEF . Thus, the base GC is equal to the base DF , and triangle GBC is equal to triangle DEF , and the remaining angles subtended by the equal sides will be equal to the (corresponding) remaining angles [Prop. 1.4]. Thus, GCB (is equal) to DF E. But, DF E was assumed (to be) equal to BCA. Thus, BCG is also equal to BCA, the lesser to the greater. The very thing (is) impossible. Thus, AB is not unequal to DE. Thus, (it is) equal. And BC is also equal to EF . So the two (straight-lines) AB, BC are equal to the two (straightlines) DE, EF , respectively. And angle ABC is equal to angle DEF . Thus, the base AC is equal to the base DF , and the remaining angle BAC is equal to the remaining angle EDF [Prop. 1.4]. But, again, let the sides subtending the equal angles be equal: for instance, (let) AB (be equal) to DE. Again, I say that the remaining sides will be equal to the remaining sides. (That is) AC to DF , and BC to EF . Furthermore, the remaining angle BAC is equal to the remaining angle EDF . For if BC is unequal to EF then one of them is greater. If possible, let BC be greater. And let BH be made equal to EF [Prop. 1.3], and let AH have been joined. And since BH is equal to EF , and AB to DE, the two (straight-lines) AB, BH are equal to the two (straight-lines) DE, EF , respectively. And the angles they encompass (are also equal). Thus, the base AH is

29

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

γωνία τÍ ØπÕ ΕΖ∆. ¢λλ¦ ¹ ØπÕ ΕΖ∆ τÍ ØπÕ ΒΓΑ ™στιν ‡ση· τριγώνου δ¾ τοà ΑΘΓ ¹ ™κτÕς γωνία ¹ ØπÕ ΒΘΑ ‡ση ™στˆ τÍ ™ντÕς κሠ¢πεναντίον τÍ ØπÕ ΒΓΑ· Óπερ ¢δύνατον. οÙκ ¥ρα ¥νισός ™στιν ¹ ΒΓ τÍ ΕΖ· ‡ση ¥ρα. ™στˆ δ κሠ¹ ΑΒ τÍ ∆Ε ‡ση. δύο δ¾ αƒ ΑΒ, ΒΓ δύο τα‹ς ∆Ε, ΕΖ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνίας ‡σας περιέχουσι· βάσις ¥ρα ¹ ΑΓ βάσει τÍ ∆Ζ ‡ση ™στίν, κሠτÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ ‡σον κሠλοιπ¾ γωνία ¹ ØπÕ ΒΑΓ τÍ λοιπÊ γωνίv τÍ ØπÕ Ε∆Ζ ‡ση. 'Ε¦ν ¥ρα δύο τρίγωνα τ¦ς δύο γωνίας δυσˆ γωνίαις ‡σας œχV ˜καρέραν ˜καρέρv κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην ½τοι τ¾ν πρÕς τα‹ς ‡σαις γωνίαις, À τ¾ν Øποτείνουσαν ØπÕ µίαν τîν ‡σων γωνιîν, κሠτ¦ς λοιπ¦ς πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξει κሠτ¾ν λοιπ¾ν γωνίαν τÍ λοιπÍ γωνίv· Óπερ œδει δε‹ξαι.

†

equal to the base DF , and the triangle ABH is equal to the triangle DEF , and the remaining angles subtended by the equal sides will be equal to the (corresponding) remaining angles [Prop. 1.4]. Thus, angle BHA is equal to EF D. But, EF D is equal to BCA. So, for triangle AHC, the external angle BHA is equal to the internal and opposite angle BCA. The very thing (is) impossible [Prop. 1.16]. Thus, BC is not unequal to EF . Thus, (it is) equal. And AB is also equal to DE. So the two (straight-lines) AB, BC are equal to the two (straightlines) DE, EF , respectively. And they encompass equal angles. Thus, the base AC is equal to the base DF , and triangle ABC (is) equal to triangle DEF , and the remaining angle BAC (is) equal to the remaining angle EDF [Prop. 1.4]. Thus, if two triangles have two angles equal to two angles, respectively, and one side equal to one side—in fact, either that by the equal angles, or that subtending one of the equal angles—then (the triangles) will also have the remaining sides equal to the (corresponding) remaining sides, and the remaining angle (equal) to the remaining angle. (Which is) the very thing it was required to show.

The Greek text has “BG, BC”, which is obviously a mistake.

κζ΄.

Proposition 27

'Ε¦ν ε„ς δύο εÙθείας εÙθε‹α ™µπίπτουσα τ¦ς ™ναλλ¦ξ γωνίας ‡σας ¢λλήλαις ποιÍ, παράλληλοι œσονται ¢λλήλαις αƒ εÙθε‹αι.

If a straight-line falling across two straight-lines makes the alternate angles equal to one another then the (two) straight-lines will be parallel to one another.

Α

Ε

Β

A

E

B

Η Γ

Ζ

G

∆

C

Ε„ς γ¦ρ δύο εÙθείας τ¦ς ΑΒ, Γ∆ εÙθε‹α ™µπίπτουσα ¹ ΕΖ τ¦ς ™ναλλ¦ξ γωνίας τ¦ς ØπÕ ΑΕΖ, ΕΖ∆ ‡σας ¢λλήλαις ποιείτω· λέγω, Óτι παράλληλός ™στιν ¹ ΑΒ τÍ Γ∆. Ε„ γ¦ρ µή, ™κβαλλόµεναι αƒ ΑΒ, Γ∆ συµπεσοàνται ½τοι ™πˆ τ¦ Β, ∆ µέρη À ™πˆ τ¦ Α, Γ. ™κβεβλήσθωσαν κሠσυµπιπτέτωσαν ™πˆ τ¦ Β, ∆ µέρη κατ¦ τÕ Η. τριγώνου δ¾ τοà ΗΕΖ ¹ ™κτÕς γωνία ¹ ØπÕ ΑΕΖ ‡ση ™στˆ τÍ ™ντÕς κሠ¢πεναντίον τÍ ØπÕ ΕΖΗ· Óπερ ™στˆν ¢δύνατον· οÙκ ¥ρα αƒ ΑΒ, ∆Γ ™κβαλλόµεναι συµπεσοàνται ™πˆ τ¦ Β, ∆ µέρη. еοίως δ¾ δειχθήσεται, Óτι οÙδ ™πˆ τ¦ Α,

F

D

For let the straight-line EF , falling across the two straight-lines AB and CD, make the alternate angles AEF and EF D equal to one another. I say that AB and CD are parallel. For if not, being produced, AB and CD will certainly meet together: either in the direction of B and D, or (in the direction) of A and C [Def. 1.23]. Let them have been produced, and let them meet together in the direction of B and D at (point) G. So, for the triangle GEF , the external angle AEF is equal to the interior and opposite (angle) EF G. The very thing is impossible

30

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

Γ· αƒ δ ™πˆ µηδέτερα τ¦ µέρη συµπίπτουσαι παράλληλοί ε„σιν· παράλληλος ¥ρα ™στˆν ¹ ΑΒ τÍ Γ∆. 'Ε¦ν ¥ρα ε„ς δύο εÙθείας εÙθε‹α ™µπίπτουσα τ¦ς ™ναλλ¦ξ γωνίας ‡σας ¢λλήλαις ποιÍ, παράλληλοι œσονται αƒ εÙθε‹αι· Óπερ œδει δε‹ξαι.

[Prop. 1.16]. Thus, being produced, AB and DC will not meet together in the direction of B and D. Similarly, it can be shown that neither (will they meet together) in (the direction of) A and C. But (straight-lines) meeting in neither direction are parallel [Def. 1.23]. Thus, AB and CD are parallel. Thus, if a straight-line falling across two straight-lines makes the alternate angles equal to one another then the (two) straight-lines will be parallel (to one another). (Which is) the very thing it was required to show.

κη΄.

Proposition 28

'Ε¦ν ε„ς δύο εÙθείας εÙθε‹α ™µπίπτουσα τ¾ν ™κτÕς If a straight-line falling across two straight-lines γωνίαν τÍ ™ντÕς κሠ¢πεναντίον κሠ™πˆ τ¦ αÙτ¦ µέρη makes the external angle equal to the internal and oppo‡σην ποιÍ À τ¦ς ™ντÕς κሠ™πˆ τ¦ αÙτ¦ µέρη δυσˆν Ñρθα‹ς site angle on the same side, or (makes) the (sum of the) ‡σας, παράλληλοι œσονται ¢λλήλαις αƒ εÙθε‹αι. internal (angles) on the same side equal to two rightangles, then the (two) straight-lines will be parallel to one another.

Ε Α

Γ

E Η

Θ

Β

A

∆

C

Ζ

G

B

H

D

F

Ε„ς γ¦ρ δύο εØθείας τ¦ς ΑΒ, Γ∆ εÙθε‹α ™µπίπτουσα ¹ ΕΖ τ¾ν ™κτÕς γωνίαν τ¾ν ØπÕ ΕΗΒ τÍ ™ντÕς κሠ¢πεναντίον γωνίv τÍ ØπÕ ΗΘ∆ ‡σην ποιείτω À τ¦ς ™ντÕς κሠ™πˆ τ¦ αÙτ¦ µέρη τ¦ς ØπÕ ΒΗΘ, ΗΘ∆ δυσˆν Ñρθα‹ς ‡σας· λέγω, Óτι παράλληλός ™στιν ¹ ΑΒ τÍ Γ∆. 'Επεˆ γ¦ρ ‡ση ™στˆν ¹ ØπÕ ΕΗΒ τÍ ØπÕ ΗΘ∆, ¢λλ¦ ¹ ØπÕ ΕΗΒ τÍ ØπÕ ΑΗΘ ™στιν ‡ση, κሠ¹ ØπÕ ΑΗΘ ¥ρα τÍ ØπÕ ΗΘ∆ ™στιν ‡ση· καί ε„σιν ™ναλλάξ· παράλληλος ¥ρα ™στˆν ¹ ΑΒ τÍ Γ∆. Πάλιν, ™πεˆ αƒ ØπÕ ΒΗΘ, ΗΘ∆ δύο Ñρθα‹ς ‡σαι ε„σίν, ε„σˆ δ καˆ αƒ ØπÕ ΑΗΘ, ΒΗΘ δυσˆν Ñρθα‹ς ‡σαι, αƒ ¥ρα ØπÕ ΑΗΘ, ΒΗΘ τα‹ς ØπÕ ΒΗΘ, ΗΘ∆ ‡σαι ε„σίν· κοιν¾ ¢φVρήσθω ¹ ØπÕ ΒΗΘ· λοιπ¾ ¥ρα ¹ ØπÕ ΑΗΘ λοιπÍ τÍ ØπÕ ΗΘ∆ ™στιν ‡ση· καί ε„σιν ™ναλλάξ· παράλληλος ¥ρα ™στˆν ¹ ΑΒ τÍ Γ∆. 'Ε¦ν ¥ρα ε„ς δύο εÙθείας εÙθε‹α ™µπίπτουσα τ¾ν ™κτÕς γωνίαν τÍ ™ντÕς κሠ¢πεναντίον κሠ™πˆ τ¦ αÙτ¦ µέρη ‡σην ποιÍ À τ¦ς ™ντÕς κሠ™πˆ τ¦ αÙτ¦ µέρη δυσˆν Ñρθα‹ς ‡σας, παράλληλοι œσονται αƒ εÙθε‹αι· Óπερ œδει

For let EF , falling across the two straight-lines AB and CD, make the external angle EGB equal to the internal and opposite angle GHD, or the (sum of the) internal (angles) on the same side, BGH and GHD, equal to two right-angles. I say that AB is parallel to CD. For since (in the first case) EGB is equal to GHD, but EGB is equal to AGH [Prop. 1.15], AGH is thus also equal to GHD. And they are alternate (angles). Thus, AB is parallel to CD [Prop. 1.27]. Again, since (in the second case, the sum of) BGH and GHD is equal to two right-angles, and (the sum of) AGH and BGH is also equal to two right-angles [Prop. 1.13], (the sum of) AGH and BGH is thus equal to (the sum of) BGH and GHD. Let BGH have been subtracted from both. Thus, the remainder AGH is equal to the remainder GHD. And they are alternate (angles). Thus, AB is parallel to CD [Prop. 1.27]. Thus, if a straight-line falling across two straight-lines makes the external angle equal to the internal and oppo-

31

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

δε‹ξαι.

site angle on the same side, or (makes) the (sum of the) internal (angles) on the same side equal to two rightangles, then the (two) straight-lines will be parallel (to one another). (Which is) the very thing it was required to show.

κθ΄.

Proposition 29

`Η ε„ς τ¦ς παραλλήλους εÙθείας εÙθε‹α ™µπίπτουσα A straight-line falling across parallel straight-lines τάς τε ™ναλλ¦ξ γωνίας ‡σας ¢λλήλαις ποιε‹ κሠτ¾ν ™κτÕς makes the alternate angles equal to one another, the exτÍ ™ντÕς κሠ¢πεναντίον ‡σην κሠτ¦ς ™ντÕς κሠ™πˆ τ¦ ternal (angle) equal to the internal and opposite (angle), αÙτ¦ µέρη δυσˆν Ñρθα‹ς ‡σας. and the (sum of the) internal (angles) on the same side equal to two right-angles.

Ε Α

Γ

E Η

Θ

Β

A

∆

C

Ζ

G

B

H

D

F

Ε„ς γ¦ρ παραλλήλους εÙθείας τ¦ς ΑΒ, Γ∆ εÙθε‹α For let the straight-line EF fall across the parallel ™µπιπτέτω ¹ ΕΖ· λέγω, Óτι τ¦ς ™ναλλ¦ξ γωνίας τ¦ς ØπÕ straight-lines AB and CD. I say that it makes the alterΑΗΘ, ΗΘ∆ ‡σας ποιε‹ κሠτ¾ν ™κτÕς γωνίαν τ¾ν ØπÕ nate angles, AGH and GHD, equal, the external angle ΕΗΒ τÍ ™ντÕς κሠ¢πεναντίον τÍ ØπÕ ΗΘ∆ ‡σην κሠτ¦ς EGB equal to the internal and opposite (angle) GHD, ™ντÕς κሠ™πˆ τ¦ αÙτ¦ µέρη τ¦ς ØπÕ ΒΗΘ, ΗΘ∆ δυσˆν and the (sum of the) internal (angles) on the same side, Ñρθα‹ς ‡σας. BGH and GHD, equal to two right-angles. Ε„ γ¦ρ ¥νισός ™στιν ¹ ØπÕ ΑΗΘ τÍ ØπÕ ΗΘ∆, µία For if AGH is unequal to GHD then one of them is αÙτîν µείζων ™στίν. œστω µείζων ¹ ØπÕ ΑΗΘ· κοιν¾ greater. Let AGH be greater. Let BGH have been added προσκείσθω ¹ ØπÕ ΒΗΘ· αƒ ¥ρα ØπÕ ΑΗΘ, ΒΗΘ τîν to both. Thus, (the sum of) AGH and BGH is greater ØπÕ ΒΗΘ, ΗΘ∆ µείζονές ε„σιν. ¢λλ¦ αƒ ØπÕ ΑΗΘ, than (the sum of) BGH and GHD. But, (the sum of) ΒΗΘ δυσˆν Ñρθα‹ς ‡σαι ε„σίν. [καˆ] αƒ ¥ρα ØπÕ ΒΗΘ, AGH and BGH is equal to two right-angles [Prop 1.13]. ΗΘ∆ δύο Ñρθîν ™λάσσονές ε„σιν. αƒ δ ¢π' ™λασσόνων Thus, (the sum of) BGH and GHD is [also] less than À δύο Ñρθîν ™κβαλλόµεναι ε„ς ¥πειρον συµπίπουσιν· αƒ two right-angles. But (straight-lines) being produced to ¥ρα ΑΒ, Γ∆ ™κβαλλόµεναι ε„ς ¥πειρον συµπεσοàνται· οÙ infinity from (internal angles whose sum is) less than two συµπίπτουσι δ δι¦ τÕ παραλλήλους αØτ¦ς Øποκε‹σθαι· right-angles meet together [Post. 5]. Thus, AB and CD, οÙκ ¥ρα ¥νισός ™στιν ¹ ØπÕ ΑΗΘ τÍ ØπÕ ΗΘ∆· ‡ση being produced to infinity, will meet together. But they do ¥ρα. ¢λλ¦ ¹ ØπÕ ΑΗΘ τÍ ØπÕ ΕΗΒ ™στιν ‡ση· κሠ¹ ØπÕ not meet, on account of them (initially) being assumed ΕΗΒ ¥ρα τÍ ØπÕ ΗΘ∆ ™στιν ‡ση· κοιν¾ προσκείσθω ¹ parallel (to one another) [Def. 1.23]. Thus, AGH is not ØπÕ ΒΗΘ· αƒ ¥ρα ØπÕ ΕΗΒ, ΒΗΘ τα‹ς ØπÕ ΒΗΘ, unequal to GHD. Thus, (it is) equal. But, AGH is equal ΗΘ∆ ‡σαι ε„σίν. ¢λλ¦ αƒ ØπÕ ΕΗΒ, ΒΗΘ δύο Ñρθα‹ς to EGB [Prop. 1.15]. And EGB is thus also equal to ‡σαι ε„σίν· καˆ αƒ ØπÕ ΒΗΘ, ΗΘ∆ ¥ρα δύο Ñρθα‹ς ‡σαι GHD. Let BGH be added to both. Thus, (the sum of) ε„σίν. EGB and BGH is equal to (the sum of) BGH and GHD. `Η ¥ρα ε„ς τ¦ς παραλλήλους εÙθείας εÙθε‹α ™µπίπτουσα But, (the sum of) EGB and BGH is equal to two rightτάς τε ™ναλλ¦ξ γωνίας ‡σας ¢λλήλαις ποιε‹ κሠτ¾ν ™κτÕς angles [Prop. 1.13]. Thus, (the sum of) BGH and GHD

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ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

τÍ ™ντÕς κሠ¢πεναντίον ‡σην κሠτ¦ς ™ντÕς κሠ™πˆ τ¦ is also equal to two right-angles. αÙτ¦ µέρη δυσˆν Ñρθα‹ς ‡σας· Óπερ œδει δε‹ξαι. Thus, a straight-line falling across parallel straightlines makes the alternate angles equal to one another, the external (angle) equal to the internal and opposite (angle), and the (sum of the) internal (angles) on the same side equal to two right-angles. (Which is) the very thing it was required to show.

λ΄.

Proposition 30

Αƒ τÍ αÙτÍ εÙθείv παράλληλοι κሠ¢λλήλαις ε„σˆ (Straight-lines) parallel to the same straight-line are παράλληλοι. also parallel to one another.

Η

Α Θ

Ε Γ

Κ

Β

A

Ζ

E

∆

C

G H K

B F D

”Εστω ˜κατέρα τîν ΑΒ, Γ∆ τÍ ΕΖ παράλληλος· λέγω, Óτι κሠ¹ ΑΒ τÍ Γ∆ ™στι παράλληλος. 'Εµπιπτέτω γ¦ρ ε„ς αÙτ¦ς εÙθε‹α ¹ ΗΚ. Κሠ™πεˆ ε„ς παραλλήλους εÙθείας τ¦ς ΑΒ, ΕΖ εÙθε‹α ™µπέπτωκεν ¹ ΗΚ, ‡ση ¥ρα ¹ ØπÕ ΑΗΚ τÍ ØπÕ ΗΘΖ. πάλιν, ™πεˆ ε„ς παραλλήλους εÙθείας τ¦ς ΕΖ, Γ∆ εÙθε‹α ™µπέπτωκεν ¹ ΗΚ, ‡ση ™στˆν ¹ ØπÕ ΗΘΖ τÍ ØπÕ ΗΚ∆. ™δείχθη δ κሠ¹ ØπÕ ΑΗΚ τÍ ØπÕ ΗΘΖ ‡ση. κሠ¹ ØπÕ ΑΗΚ ¥ρα τÍ ØπÕ ΗΚ∆ ™στιν ‡ση· καί εƒσιν ™ναλλάξ. παράλληλος ¥ρα ™στˆν ¹ ΑΒ τÍ Γ∆. [Αƒ ¥ρα τÍ αÙτÍ εÙθείv παράλληλοι κሠ¢λλήλαις ε„σˆ παράλληλοι·] Óπερ œδει δε‹ξαι.

Let each of the (straight-lines) AB and CD be parallel to EF . I say that AB is also parallel to CD. For let the straight-line GK fall across (AB, CD, and EF ). And since GK has fallen across the parallel straightlines AB and EF , (angle) AGK (is) thus equal to GHF [Prop. 1.29]. Again, since GK has fallen across the parallel straight-lines EF and CD, (angle) GHF is equal to GKD [Prop. 1.29]. But AGK was also shown (to be) equal to GHF . Thus, AGK is also equal to GKD. And they are alternate (angles). Thus, AB is parallel to CD [Prop. 1.27]. [Thus, (straight-lines) parallel to the same straightline are also parallel to one another.] (Which is) the very thing it was required to show.

λα΄.

Proposition 31

∆ι¦ τοà δοθέντος σηµείου τÍ δοθείσV εÙθείv παράλληλον εÙθε‹αν γραµµ¾ν ¢γαγε‹ν. ”Εστω τÕ µν δοθν σηµε‹ον τÕ Α, ¹ δ δοθε‹σα εÙθε‹α ¹ ΒΓ· δε‹ δ¾ δι¦ τοà Α σηµείου τÍ ΒΓ εÙθείv παράλληλον εÙθε‹αν γραµµ¾ν ¢γαγε‹ν. Ε„λήφθω ™πˆ τÁς ΒΓ τυχÕν σηµε‹ον τÕ ∆, κሠ™πεζεύχθω ¹ Α∆· κሠσυνεστάτω πρÕς τÍ ∆Α εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ ØπÕ Α∆Γ γωνίv ‡ση ¹ ØπÕ ∆ΑΕ· κሠ™κβεβλήσθω ™π' εÙθείας τÍ ΕΑ εÙθε‹α

To draw a straight-line parallel to a given straight-line, through a given point. Let A be the given point, and BC the given straightline. So it is required to draw a straight-line parallel to the straight-line BC, through the point A. Let the point D have been taken somewhere on BC, and let AD have been joined. And let (angle) DAE, equal to angle ADC, have been constructed at the point A on the straight-line DA [Prop. 1.23]. And let the

33

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

¹ ΑΖ.

straight-line AF have been produced in a straight-line with EA.

Α

Ε

Β

∆

Ζ

E

Γ

B

A

F

C D

Κሠ™πεˆ ε„ς δύο εÙθείας τ¦ς ΒΓ, ΕΖ εÙθε‹α ™µπίπτουσα ¹ Α∆ τ¦ς ™ναλλ¦ξ γωνίας τ¦ς ØπÕ ΕΑ∆, Α∆Γ ‡σας ¢λλήλαις πεποίηκεν, παράλληλος ¥ρα ™στˆν ¹ ΕΑΖ τÍ ΒΓ. ∆ι¦ τοà δοθέντος ¥ρα σηµείου τοà Α τÍ δοθείσV εÙθείv τÍ ΒΓ παράλληλος εÙθε‹α γραµµ¾ Ãκται ¹ ΕΑΖ· Óπερ œδει ποιÁσαι.

And since the straight-line AD, (in) falling across the two straight-lines BC and EF , has made the alternate angles EAD and ADC equal to one another, EAF is thus parallel to BC [Prop. 1.27]. Thus, the straight-line EAF has been drawn parallel to the given straight-line BC, through the given point A. (Which is) the very thing it was required to do.

λβ΄.

Proposition 32

ΠαντÕς τριγώνου µι©ς τîν πλευρîν προσεκβληθείσης For any triangle, (if) one of the sides (is) produced ¹ ™κτÕς γωνία δυσˆ τα‹ς ™ντÕς κሠ¢πεναντίον ‡ση ™στίν, (then) the external angle is equal to the (sum of the) two καˆ αƒ ™ντÕς τοà τριγώνου τρε‹ς γωνίαι δυσˆν Ñρθα‹ς ‡σαι internal and opposite (angles), and the (sum of the) three ε„σίν. internal angles of the triangle is equal to two right-angles.

Α

Β

Ε

Γ

A

∆

B

”Εστω τρίγωνον τÕ ΑΒΓ, κሠπροσεκβεβλήσθω αÙτοà µία πλευρ¦ ¹ ΒΓ ™πˆ τÕ ∆· λέγω, Óτι ¹ ™κτÕς γωνία ¹ ØπÕ ΑΓ∆ ‡ση ™στˆ δυσˆ τα‹ς ™ντÕς κሠ¢πεναντίον τα‹ς ØπÕ ΓΑΒ, ΑΒΓ, καˆ αƒ ™ντÕς τοà τριγώνου τρε‹ς γωνίαι αƒ ØπÕ ΑΒΓ, ΒΓΑ, ΓΑΒ δυσˆν Ñρθα‹ς ‡σαι ε„σίν. ”Ηχθω γ¦ρ δι¦ τοà Γ σηµείου τÍ ΑΒ εÙθείv παράλληλος ¹ ΓΕ. Κሠ™πεˆ παράλληλός ™στιν ¹ ΑΒ τÍ ΓΕ, κሠε„ς αÙτ¦ς ™µπέπτωκεν ¹ ΑΓ, αƒ ™ναλλ¦ξ γωνίαι αƒ ØπÕ ΒΑΓ, ΑΓΕ ‡σαι ¢λλήλαις ε„σίν. πάλιν, ™πεˆ παράλληλός ™στιν ¹ ΑΒ τÍ ΓΕ, κሠε„ς αÙτ¦ς ™µπέπτωκεν εÙθε‹α ¹ Β∆, ¹ ™κτÕς γωνία ¹ ØπÕ ΕΓ∆ ‡ση ™στˆ τÍ ™ντÕς κሠ¢πεναντίον τÍ ØπÕ ΑΒΓ. ™δείχθη δ κሠ¹ ØπÕ ΑΓΕ τÍ ØπÕ ΒΑΓ ‡ση· Óλη ¥ρα ¹ ØπÕ ΑΓ∆ γωνία ‡ση ™στˆ δυσˆ τα‹ς ™ντÕς

E

C

D

Let ABC be a triangle, and let one of its sides BC have been produced to D. I say that the external angle ACD is equal to the (sum of the) two internal and opposite angles CAB and ABC, and the (sum of the) three internal angles of the triangle—ABC, BCA, and CAB— is equal to two right-angles. For let CE have been drawn through point C parallel to the straight-line AB [Prop. 1.31]. And since AB is parallel to CE, and AC has fallen across them, the alternate angles BAC and ACE are equal to one another [Prop. 1.29]. Again, since AB is parallel to CE, and the straight-line BD has fallen across them, the external angle ECD is equal to the internal and opposite (angle) ABC [Prop. 1.29]. But ACE was also shown (to be) equal to BAC. Thus, the whole an-

34

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

κሠ¢πεναντίον τα‹ς ØπÕ ΒΑΓ, ΑΒΓ. Κοιν¾ προσκείσθω ¹ ØπÕ ΑΓΒ· αƒ ¥ρα ØπÕ ΑΓ∆, ΑΓΒ τρισˆ τα‹ς ØπÕ ΑΒΓ, ΒΓΑ, ΓΑΒ ‡σαι ε„σίν. ¢λλ' αƒ ØπÕ ΑΓ∆, ΑΓΒ δυσˆν Ñρθα‹ς ‡σαι ε„σίν· καˆ αƒ ØπÕ ΑΓΒ, ΓΒΑ, ΓΑΒ ¥ρα δυσˆν Ñρθα‹ς ‡σαι ε„σίν. ΠαντÕς ¥ρα τριγώνου µι©ς τîν πλευρîν προσεκβληθείσης ¹ ™κτÕς γωνία δυσˆ τα‹ς ™ντÕς κሠ¢πεναντίον ‡ση ™στίν, καˆ αƒ ™ντÕς τοà τριγώνου τρε‹ς γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν· Óπερ œδει δε‹ξαι.

gle ACD is equal to the (sum of the) two internal and opposite (angles) BAC and ABC. Let ACB have been added to both. Thus, (the sum of) ACD and ACB is equal to the (sum of the) three (angles) ABC, BCA, and CAB. But, (the sum of) ACD and ACB is equal to two right-angles [Prop. 1.13]. Thus, (the sum of) ACB, CBA, and CAB is also equal to two right-angles. Thus, for any triangle, (if) one of the sides (is) produced (then) the external angle is equal to the (sum of the) two internal and opposite (angles), and the (sum of the) three internal angles of the triangle is equal to two right-angles. (Which is) the very thing it was required to show.

λγ΄.

Proposition 33

Αƒ τ¦ς ‡σας τε κሠπαραλλήλους ™πˆ τ¦ αÙτ¦ µέρη Straight-lines joining equal and parallel (straight™πιζευγνύουσαι εÙθε‹αι κሠαÙτሠ‡σας τε κሠπαράλληλοί lines) on the same sides are themselves also equal and ε„σιν. parallel.

Β

∆

Α

B

Γ

D

”Εστωσαν ‡σαι τε κሠπαράλληλοι αƒ ΑΒ, Γ∆, κሠ™πιζευγνύτωσαν αÙτ¦ς ™πˆ τ¦ αÙτ¦ µέρη εÙθε‹αι αƒ ΑΓ, Β∆· λέγω, Óτι καˆ αƒ ΑΓ, Β∆ ‡σαι τε κሠπαράλληλοί ε„σιν. 'Επεζεύχθω ¹ ΒΓ. κሠ™πεˆ παράλληλός ™στιν ¹ ΑΒ τÍ Γ∆, κሠε„ς αÙτ¦ς ™µπέπτωκεν ¹ ΒΓ, αƒ ™ναλλ¦ξ γωνίαι αƒ ØπÕ ΑΒΓ, ΒΓ∆ ‡σαι ¢λλήλαις ε„σίν. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΒ τÍ Γ∆ κοιν¾ δ ¹ ΒΓ, δύο δ¾ αƒ ΑΒ, ΒΓ δύο τα‹ς ΒΓ, Γ∆ ‡σαι ε„σίν· κሠγωνία ¹ ØπÕ ΑΒΓ γωνίv τÍ ØπÕ ΒΓ∆ ‡ση· βάσις ¥ρα ¹ ΑΓ βάσει τÍ Β∆ ™στιν ‡ση, κሠτÕ ΑΒΓ τρίγωνον τù ΒΓ∆ τριγώνJ ‡σον ™στίν, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται ˜κατέρα ˜κατέρv, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν· ‡ση ¥ρα ¹ ØπÕ ΑΓΒ γωνία τÍ ØπÕ ΓΒ∆. κሠ™πεˆ ε„ς δύο εÙθείας τ¦ς ΑΓ, Β∆ εÙθε‹α ™µπίπτουσα ¹ ΒΓ τ¦ς ™ναλλ¦ξ γωνίας ‡σας ¢λλήλαις πεποίηκεν, παράλληλος ¥ρα ™στˆν ¹ ΑΓ τÍ Β∆. ™δείχθη δ αÙτÍ κሠ‡ση. Αƒ ¥ρα τ¦ς ‡σας τε κሠπαραλλήλους ™πˆ τ¦ αÙτ¦ µέρη ™πιζευγνύουσαι εÙθε‹αι κሠαÙτሠ‡σαι τε κሠπαράλληλοί ε„σιν· Óπερ œδει δε‹ξαι.

A

C

Let AB and CD be equal and parallel (straight-lines), and let the straight-lines AC and BD join them on the same sides. I say that AC and BD are also equal and parallel. Let BC have been joined. And since AB is parallel to CD, and BC has fallen across them, the alternate angles ABC and BCD are equal to one another [Prop. 1.29]. And since AB and CD are equal, and BC is common, the two (straight-lines) AB, BC are equal to the two (straight-lines) DC, CB.† And the angle ABC is equal to the angle BCD. Thus, the base AC is equal to the base BD, and triangle ABC is equal to triangle ACD, and the remaining angles will be equal to the corresponding remaining angles subtended by the equal sides [Prop. 1.4]. Thus, angle ACB is equal to CBD. Also, since the straight-line BC, (in) falling across the two straight-lines AC and BD, has made the alternate angles (ACB and CBD) equal to one another, AC is thus parallel to BD [Prop. 1.27]. And (AC) was also shown (to be) equal to (BD). Thus, straight-lines joining equal and parallel (straight-

35

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 lines) on the same sides are themselves also equal and parallel. (Which is) the very thing it was required to show.

†

The Greek text has “BC, CD”, which is obviously a mistake.

λδ΄.

Proposition 34

Τîν παραλληλογράµµων χωρίων αƒ ¢πεναντίον For parallelogrammic figures, the opposite sides and anπλευραί τε κሠγωνίαι ‡σαι ¢λλήλαις ε„σίν, κሠ¹ gles are equal to one another, and a diagonal cuts them διάµετρος αÙτ¦ δίχα τέµνει. in half.

Α

Γ

Β

A

∆

C

”Εστω παραλληλόγραµµον χωρίον τÕ ΑΓ∆Β, διάµετρος δ αÙτοà ¹ ΒΓ· λέγω, Óτι τοà ΑΓ∆Β παραλληλογράµµου αƒ ¢πεναντίον πλευραί τε κሠγωνίαι ‡σαι ¢λλήλαις ε„σίν, κሠ¹ ΒΓ διάµετρος αÙτÕ δίχα τέµνει. 'Επεˆ γ¦ρ παράλληλός ™στιν ¹ ΑΒ τÍ Γ∆, κሠε„ς αÙτ¦ς ™µπέπτωκεν εÙθε‹α ¹ ΒΓ, αƒ ™ναλλ¦ξ γωνιάι αƒ ØπÕ ΑΒΓ, ΒΓ∆ ‡σαι ¢λλήλαις ε„σίν. πάλιν ™πεˆ παράλληλός ™στιν ¹ ΑΓ τÍ Β∆, κሠε„ς αÙτ¦ς ™µπέπτωκεν ¹ ΒΓ, αƒ ™ναλλ¦ξ γωνίαι αƒ ØπÕ ΑΓΒ, ΓΒ∆ ‡σας ¢λλήλαις ε„σίν. δύο δ¾ τρίγωνά ™στι τ¦ ΑΒΓ, ΒΓ∆ τ¦ς δύο γωνίας τ¦ς ØπÕ ΑΒΓ, ΒΓΑ δυσˆ τα‹ς ØπÕ ΒΓ∆, ΓΒ∆ ‡σας œχοντα ˜κατέραν ˜κατέρv κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην τ¾ν πρÕς τα‹ς ‡σαις γωνίαις κοιν¾ν αÙτîν τ¾ν ΒΓ· κሠτ¦ς λοιπ¦ς ¥ρα πλευρ¦ς τα‹ς λοιπα‹ς ‡σας ›ξει ˜κατέραν ˜κατέρv κሠτ¾ν λοιπ¾ν γωνίαν τÍ λοιπÍ γωνίv· ‡ση ¥ρα ¹ µν ΑΒ πλευρ¦ τÍ Γ∆, ¹ δ ΑΓ τÍ Β∆, κሠœτι ‡ση ™στˆν ¹ ØπÕ ΒΑΓ γωνία τÍ ØπÕ Γ∆Β. κሠ™πεˆ ‡ση ™στˆν ¹ µν ØπÕ ΑΒΓ γωνία τÍ ØπÕ ΒΓ∆, ¹ δ ØπÕ ΓΒ∆ τÍ ØπÕ ΑΓΒ, Óλη ¥ρα ¹ ØπÕ ΑΒ∆ ÓλV τÍ ØπÕ ΑΓ∆ ™στιν ‡ση. ™δείχθη δ κሠ¹ ØπÕ ΒΑΓ τÍ ØπÕ Γ∆Β ‡ση. Τîν ¥ρα παραλληλογράµµων χωρίων αƒ ¢πεναντίον πλευραί τε κሠγωνίαι ‡σαι ¢λλήλαις ε„σίν. Λέγω δή, Óτι κሠ¹ διάµετρος αÙτ¦ δίχα τέµνει. ™πεˆ γ¦ρ ‡ση ™στˆν ¹ ΑΒ τÍ Γ∆, κοιν¾ δ ¹ ΒΓ, δύο δ¾ αƒ ΑΒ, ΒΓ δυσˆ τα‹ς Γ∆, ΒΓ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΑΒΓ γωνίv τÍ ØπÕ ΒΓ∆ ‡ση. κሠβάσις ¥ρα ¹ ΑΓ τÍ ∆Β ‡ση. κሠτÕ ΑΒΓ [¥ρα] τρίγωνον τù ΒΓ∆ τριγώνJ ‡σον ™στίν. `Η ¥ρα ΒΓ διάµετρος δίχα τέµνει τÕ ΑΒΓ∆ παραλληλόγραµµον· Óπερ œδει δε‹ξαι.

B

D

Let ACDB be a parallelogrammic figure, and BC its diagonal. I say that for parallelogram ACDB, the opposite sides and angles are equal to one another, and the diagonal BC cuts it in half. For since AB is parallel to CD, and the straight-line BC has fallen across them, the alternate angles ABC and BCD are equal to one another [Prop. 1.29]. Again, since AC is parallel to BD, and BC has fallen across them, the alternate angles ACB and CBD are equal to one another [Prop. 1.29]. So ABC and BCD are two triangles having the two angles ABC and BCA equal to the two (angles) BCD and CBD, respectively, and one side equal to one side—the (one) common to the equal angles, (namely) BC. Thus, they will also have the remaining sides equal to the corresponding remaining (sides), and the remaining angle (equal) to the remaining angle [Prop. 1.26]. Thus, side AB is equal to CD, and AC to BD. Furthermore, angle BAC is equal to CDB. And since angle ABC is equal to BCD, and CBD to ACB, the whole (angle) ABD is thus equal to the whole (angle) ACD. And BAC was also shown (to be) equal to CDB. Thus, for parallelogrammic figures, the opposite sides and angles are equal to one another. And, I also say that a diagonal cuts them in half. For since AB is equal to CD, and BC (is) common, the two (straight-lines) AB, BC are equal to the two (straightlines) DC, CB † , respectively. And angle ABC is equal to angle BCD. Thus, the base AC (is) also equal to DB [Prop. 1.4]. Also, triangle ABC is equal to triangle BCD [Prop. 1.4].

36

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 Thus, the diagonal BC cuts the parallelogram ACDB ‡ in half. (Which is) the very thing it was required to show.

† ‡

The Greek text has “CD, BC”, which is obviously a mistake. The Greek text has “ABCD”, which is obviously a mistake.

λε΄.

Proposition 35

Τ¦ παραλληλόγραµµα τ¦ ™πˆ τÁς αÙτÁς βάσεως Ôντα κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ‡σα ¢λλήλοις ™στίν.

Parallelograms which are on the same base and between the same parallels are equal† to one another.

Α

Ε

∆

Ζ

A

D

Η

Β

F

G

Γ

B

”Εστω παραλληλόγραµµα τ¦ ΑΒΓ∆, ΕΒΓΖ ™πˆ τÁς αÙτÁς βάσεως τÁς ΒΓ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ΑΖ, ΒΓ· λέγω, Óτι ‡σον ™στˆ τÕ ΑΒΓ∆ τù ΕΒΓΖ παραλληλογράµµJ. 'Επεˆ γ¦ρ παραλληλόγραµµόν ™στι τÕ ΑΒΓ∆, ‡ση ™στˆν ¹ Α∆ τÍ ΒΓ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΕΖ τÍ ΒΓ ™στιν ‡ση· éστε κሠ¹ Α∆ τÍ ΕΖ ™στιν ‡ση· κሠκοιν¾ ¹ ∆Ε· Óλη ¥ρα ¹ ΑΕ ÓλV τÍ ∆Ζ ™στιν ‡ση. œστι δ κሠ¹ ΑΒ τÍ ∆Γ ‡ση· δύο δ¾ αƒ ΕΑ, ΑΒ δύο τα‹ς Ζ∆, ∆Γ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ Ζ∆Γ γωνίv τÍ ØπÕ ΕΑΒ ™στιν ‡ση ¹ ™κτÕς τÍ ™ντός· βάσις ¥ρα ¹ ΕΒ βάσει τÍ ΖΓ ‡ση ™στίν, κሠτÕ ΕΑΒ τρίγωνον τù ∆ΖΓ τριγώνJ ‡σον œσται· κοινÕν ¢φVρήσθω τÕ ∆ΗΕ· λοιπÕν ¥ρα τÕ ΑΒΗ∆ τραπέζιον λοιπù τù ΕΗΓΖ τραπεζίJ ™στˆν ‡σον· κοινÕν προσκείσθω τÕ ΗΒΓ τρίγωνον· Óλον ¥ρα τÕ ΑΒΓ∆ παραλληλόγραµµον ÓλJ τù ΕΒΓΖ παραλληλογράµµJ ‡σον ™στίν. Τ¦ ¥ρα παραλληλόγραµµα τ¦ ™πˆ τÁς αÙτÁς βάσεως Ôντα κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ‡σα ¢λλήλοις ™στίν· Óπερ œδει δε‹ξαι. †

E

C

Let ABCD and EBCF be parallelograms on the same base BC, and between the same parallels AF and BC. I say that ABCD is equal to parallelogram EBCF . For since ABCD is a parallelogram, AD is equal to BC [Prop. 1.34]. So, for the same (reasons), EF is also equal to BC. So AD is also equal to EF . And DE is common. Thus, the whole (straight-line) AE is equal to the whole (straight-line) DF . And AB is also equal to DC. So the two (straight-lines) EA, AB are equal to the two (straight-lines) F D, DC, respectively. And angle F DC is equal to angle EAB, the external to the internal [Prop. 1.29]. Thus, the base EB is equal to the base F C, and triangle EAB will be equal to triangle DF C [Prop. 1.4]. Let DGE have been taken away from both. Thus, the remaining trapezium ABGD is equal to the remaining trapezium EGCF . Let triangle GBC have been added to both. Thus, the whole parallelogram ABCD is equal to the whole parallelogram EBCF . Thus, parallelograms which are on the same base and between the same parallels are equal to one another. (Which is) the very thing it was required to show.

Here, for the first time, “equal” means “equal in area”, rather than “congruent”.

λ$΄.

Proposition 36

Τ¦ παραλληλόγραµµα τ¦ ™πˆ ‡σων βάσεων Ôντα κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ‡σα ¢λλήλοις ™στίν. ”Εστω παραλληλόγραµµα τ¦ ΑΒΓ∆, ΕΖΗΘ ™πˆ ‡σων βάσεων Ôντα τîν ΒΓ, ΖΗ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ΑΘ, ΒΗ· λέγω, Óτι ‡σον ™στˆ τÕ ΑΒΓ∆ παραλληλόγραµµον τù ΕΖΗΘ.

Parallelograms which are on equal bases and between the same parallels are equal to one another. Let ABCD and EF GH be parallelograms which are on the equal bases BC and F G, and (are) between the same parallels AH and BG. I say that the parallelogram ABCD is equal to EF GH.

37

ΣΤΟΙΧΕΙΩΝ α΄. Α

Β

ELEMENTS BOOK 1

∆

Ε

Θ

Γ

Ζ

A

Η

B

D

E

H

C

F

G

'Επεζεύχθωσαν γ¦ρ αƒ ΒΕ, ΓΘ. κሠ™πεˆ ‡ση ™στˆν ¹ ΒΓ τÍ ΖΗ, ¢λλ¦ ¹ ΖΗ τÍ ΕΘ ™στιν ‡ση, κሠ¹ ΒΓ ¥ρα τÍ ΕΘ ™στιν ‡ση. ε„σˆ δ κሠπαράλληλοι. κሠ™πιζευγνύουσιν αÙτ¦ς αƒ ΕΒ, ΘΓ· αƒ δ τ¦ς ‡σας τε κሠπαραλλήλους ™πˆ τ¦ αÙτ¦ µέρη ™πιζευγνύουσαι ‡σαι τε κሠπαράλληλοί ε„σι [καˆ αƒ ΕΒ, ΘΓ ¥ρα ‡σας τέ ε„σι κሠπαράλληλοι]. παραλληλόγραµµον ¥ρα ™στˆ τÕ ΕΒΓΘ. καί ™στιν ‡σον τù ΑΒΓ∆· βάσιν τε γ¦ρ αÙτù τ¾ν αÙτ¾ν œχει τ¾ν ΒΓ, κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ™στˆν αÙτù τα‹ς ΒΓ, ΑΘ. δˆα τ¦ αÙτ¦ δ¾ κሠτÕ ΕΖΗΘ τù αÙτù τù ΕΒΓΘ ™στιν ‡σον· éστε κሠτÕ ΑΒΓ∆ παραλληλόγραµµον τù ΕΖΗΘ ™στιν ‡σον. Τ¦ ¥ρα παραλληλόγραµµα τ¦ ™πˆ ‡σων βάσεων Ôντα κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ‡σα ¢λλήλοις ™στίν· Óπερ œδει δε‹ξαι.

For let BE and CH have been joined. And since BC and F G are equal, but F G and EH are equal [Prop. 1.34], BC and EH are thus also equal. And they are also parallel, and EB and HC join them. But (straight-lines) joining equal and parallel (straight-lines) on the same sides are (themselves) equal and parallel [Prop. 1.33] [thus, EB and HC are also equal and parallel]. Thus, EBCH is a parallelogram [Prop. 1.34], and is equal to ABCD. For it has the same base, BC, as (ABCD), and is between the same parallels, BC and AH, as (ABCD) [Prop. 1.35]. So, for the same (reasons), EF GH is also equal to the same (parallelogram) EBCH [Prop. 1.34]. So that the parallelogram ABCD is also equal to EF GH. Thus, parallelograms which are on equal bases and between the same parallels are equal to one another. (Which is) the very thing it was required to show.

λζ΄.

Proposition 37

Τ¦ τρίγωνα τ¦ ™πˆ τÁς αÙτÁς βάσεως Ôντα κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ‡σα ¢λλήλοις ™στίν.

Triangles which are on the same base and between the same parallels are equal to one another.

Α

Ε

Β

∆

Ζ

A

E

Γ

F

B

”Εστω τρίγωνα τ¦ ΑΒΓ, ∆ΒΓ ™πˆ τÁς αÙτÁς βάσεως τÁς ΒΓ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς Α∆, ΒΓ· λέγω, Óτι ‡σον ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΒΓ τριγώνJ. 'Εκβεβλήσθω ¹ Α∆ ™φ' ˜κάτερα τ¦ µέρη ™πˆ τ¦ Ε, Ζ, κሠδι¦ µν τοà Β τÍ ΓΑ παράλληλος ½χθω ¹ ΒΕ, δˆα δ τοà Γ τÍ Β∆ παράλληλος ½χθω ¹ ΓΖ. παραλληλόγραµµον ¥ρα ™στˆν ˜κάτερον τîν ΕΒΓΑ, ∆ΒΓΖ· καί ε„σιν ‡σα· ™πί τε γ¦ρ τÁς αÙτÁς βάσεώς ε„σι τÁς ΒΓ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ΒΓ, ΕΖ· καί ™στι τοà µν ΕΒΓΑ παραλληλογράµµου ¼µισυ τÕ ΑΒΓ τρίγωνον· ¹ γ¦ρ ΑΒ διάµετρος αÙτÕ δίχα τέµνει· τοà δ ∆ΒΓΖ παραλληλογράµµου ¼µισυ τÕ ∆ΒΓ τρίγωνον· ¹ γ¦ρ ∆Γ

D

C

Let ABC and DBC be triangles on the same base BC, and between the same parallels AD and BC. I say that triangle ABC is equal to triangle DBC. Let AD have been produced in each direction to E and F , and let the (straight-line) BE have been drawn through B parallel to CA [Prop. 1.31], and let the (straight-line) CF have been drawn through C parallel to BD [Prop. 1.31]. Thus, EBCA and DBCF are both parallelograms, and are equal. For they are on the same base BC, and between the same parallels BC and EF [Prop. 1.35]. And the triangle ABC is half of the parallelogram EBCA. For the diagonal AB cuts the latter in

38

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

διάµετρος αÙτÕ δίχα τέµνει. [τ¦ δ τîν ‡σων ¹µίση ‡σα ¢λλήλοις ™στίν]. ‡σον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΒΓ τριγώνJ. Τ¦ ¥ρα τρίγωνα τ¦ ™πˆ τÁς αÙτÁς βάσεως Ôντα κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ‡σα ¢λλήλοις ™στίν· Óπερ œδει δε‹ξαι.

†

half [Prop. 1.34]. And the triangle DBC (is) half of the parallelogram DBCF . For the diagonal DC cuts the latter in half [Prop. 1.34]. [And the halves of equal things are equal to one another.]† Thus, triangle ABC is equal to triangle DBC. Thus, triangles which are on the same base and between the same parallels are equal to one another. (Which is) the very thing it was required to show.

This is an additional common notion.

λη΄.

Proposition 38

Τ¦ τρίγωνα τ¦ ™πˆ ‡σων βάσεων Ôντα κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ‡σα ¢λλήλοις ™στίν.

Triangles which are on equal bases and between the same parallels are equal to one another.

Η

Β

Α

∆

Γ

Θ

Ε

G

Ζ

B

A

D

C

H

E

F

”Εστω τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ ™πˆ ‡σων βάσεων τîν ΒΓ, ΕΖ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ΒΖ, Α∆· λέγω, Óτι ‡σον ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ. 'Εκβεβλήσθω γ¦ρ ¹ Α∆ ™φ' ˜κάτερα τ¦ µέρη ™πˆ τ¦ Η, Θ, κሠδι¦ µν τοà Β τÍ ΓΑ παράλληλος ½χθω ¹ ΒΗ, δˆα δ τοà Ζ τÍ ∆Ε παράλληλος ½χθω ¹ ΖΘ. παραλληλόγραµµον ¥ρα ™στˆν ˜κάτερον τîν ΗΒΓΑ, ∆ΕΖΘ· κሠ‡σον τÕ ΗΒΓΑ τù ∆ΕΖΘ· ™πί τε γ¦ρ ‡σων βάσεών ε„σι τîν ΒΓ, ΕΖ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ΒΖ, ΗΘ· και΄ ™στι τοà µν ΗΒΓΑ παραλληλογράµµου ¼µισυ τÕ ΑΒΓ τρίγωνον. ¹ γ¦ρ ΑΒ διάµετρος αÙτÕ δίχα τέµνει· τοà δ ∆ΕΖΘ παραλληλογράµµου ¼µισυ τÕ ΖΕ∆ τρίγωνον· ¹ γ¦ρ ∆Ζ δίαµετρος αÙτÕ δίχα τέµνει [τ¦ δ τîν ‡σων ¹µίση ‡σα ¢λλήλοις ™στίν]. ‡σον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ. Τ¦ ¥ρα τρίγωνα τ¦ ™πˆ ‡σων βάσεων Ôντα κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ‡σα ¢λλήλοις ™στίν· Óπερ œδει δε‹ξαι.

Let ABC and DEF be triangles on the equal bases BC and EF , and between the same parallels BF and AD. I say that triangle ABC is equal to triangle DEF . For let AD have been produced in each direction to G and H, and let the (straight-line) BG have been drawn through B parallel to CA [Prop. 1.31], and let the (straight-line) F H have been drawn through F parallel to DE [Prop. 1.31]. Thus, GBCA and DEF H are each parallelograms. And GBCA is equal to DEF H. For they are on the equal bases BC and EF , and between the same parallels BF and GH [Prop. 1.36]. And triangle ABC is half of the parallelogram GBCA. For the diagonal AB cuts the latter in half [Prop. 1.34]. And triangle F ED (is) half of parallelogram DEF H. For the diagonal DF cuts the latter in half. [And the halves of equal things are equal to one another]. Thus, triangle ABC is equal to triangle DEF . Thus, triangles which are on equal bases and between the same parallels are equal to one another. (Which is) the very thing it was required to show.

λθ΄.

Proposition 39

Τ¦ ‡σα τρίγωνα τ¦ ™πˆ τÁς αÙτÁς βάσεως Ôντα κሠEqual triangles which are on the same base, and on ™πˆ τ¦ αÙτ¦ µέρη κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ™στίν. the same side, are also between the same parallels. ”Εστω ‡σα τρίγωνα τ¦ ΑΒΓ, ∆ΒΓ ™πˆ τÁς αÙτÁς Let ABC and DBC be equal triangles which are on βάσεως Ôντα κሠ™πˆ τ¦ αÙτ¦ µέρη τÁς ΒΓ· λέγω, Óτι the same base BC, and on the same side. I say that they

39

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ™στίν.

are also between the same parallels.

∆

Α

D

A

Ε

E

Β

Γ

B

C

'Επεζεύχθω γ¦ρ ¹ Α∆· λέγω, Óτι παράλληλός ™στιν ¹ Α∆ τÍ ΒΓ. Ε„ γ¦ρ µή, ½χθω δι¦ τοà Α σηµείου τÍ ΒΓ εÙθείv παράλληλος ¹ ΑΕ, κሠ™πεζεύχθω ¹ ΕΓ. ‡σον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ΕΒΓ τριγώνJ· ™πί τε γ¦ρ τÁς αÙτÁς βάσεώς ™στιν αÙτù τÁς ΒΓ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις. ¢λλ¦ τÕ ΑΒΓ τù ∆ΒΓ ™στιν ‡σον· κሠτÕ ∆ΒΓ ¥ρα τù ΕΒΓ ‡σον ™στˆ τÕ µε‹ζον τù ™λάσσονι· Óπερ ™στˆν ¢δύνατον· οÙκ ¥ρα παράλληλός ™στιν ¹ ΑΕ τÍ ΒΓ. еοίως δ¾ δείξοµεν, Óτι οÙδ' ¥λλη τις πλ¾ν τÁς Α∆· ¹ Α∆ ¥ρα τÍ ΒΓ ™στι παράλληλος. Τ¦ ¥ρα ‡σα τρίγωνα τ¦ ™πˆ τÁς αÙτÁς βάσεως Ôντα κሠ™πˆ τ¦ αÙτ¦ µέρη κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ™στίν· Óπερ œδει δε‹ξαι.

For let AD have been joined. I say that AD and AC are parallel. For, if not, let AE have been drawn through point A parallel to the straight-line BC [Prop. 1.31], and let EC have been joined. Thus, triangle ABC is equal to triangle EBC. For it is on the same base as it, BC, and between the same parallels [Prop. 1.37]. But ABC is equal to DBC. Thus, DBC is also equal to EBC, the greater to the lesser. The very thing is impossible. Thus, AE is not parallel to BC. Similarly, we can show that neither (is) any other (straight-line) than AD. Thus, AD is parallel to BC. Thus, equal triangles which are on the same base, and on the same side, are also between the same parallels. (Which is) the very thing it was required to show.

µ΄.

Proposition 40†

Τ¦ ‡σα τρίγωνα τ¦ ™πˆ ‡σων βάσεων Ôντα κሠ™πˆ τ¦ Equal triangles which are on equal bases, and on the αÙτ¦ µέρη κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ™στίν. same side, are also between the same parallels.

Α

∆

A

D

Ζ

Β

Γ

F

Ε

B

”Εστω ‡σα τρίγωνα τ¦ ΑΒΓ, Γ∆Ε ™πˆ ‡σων βάσεων τîν ΒΓ, ΓΕ κሠ™πˆ τ¦ αÙτ¦ µέρη. λέγω, Óτι κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ™στίν. 'Επεζεύχθω γ¦ρ ¹ Α∆· λέγω, Óτι παράλληλός ™στιν ¹ Α∆ τÍ ΒΕ. Ε„ γ¦ρ µή, ½χθω δι¦ τοà Α τÍ ΒΕ παράλληλος ¹ ΑΖ, κሠ™πεζεύχθω ¹ ΖΕ. ‡σον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ΖΓΕ τριγώνJ· ™πί τε γ¦ρ ‡σων βάσεών ε„σι τîν ΒΓ, ΓΕ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ΒΕ, ΑΖ. ¢λλ¦ τÕ ΑΒΓ τρίγωνον ‡σον ™στˆ τù ∆ΓΕ [τρίγωνJ]· κሠτÕ ∆ΓΕ ¥ρα [τρίγωνον] ‡σον ™στˆ τù ΖΓΕ

C

E

Let ABC and CDE be equal triangles on the equal bases BC and CE (respectively), and on the same side. I say that they are also between the same parallels. For let AD have been joined. I say that AD is parallel to BE. For if not, let AF have been drawn through A parallel to BE [Prop. 1.31], and let F E have been joined. Thus, triangle ABC is equal to triangle F CE. For they are on equal bases, BC and CE, and between the same parallels, BE and AF [Prop. 1.38]. But, triangle ABC is equal to [triangle] DCE. Thus, [triangle] DCE is also equal to

40

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

τριγώνJ τÕ µε‹ζον τù ™λάσσονι· Óπερ ™στˆν ¢δύνατον· οÙκ ¥ρα παράλληλος ¹ ΑΖ τÍ ΒΕ. еοίως δ¾ δείξοµεν, Óτι οÙδ' ¥λλη τις πλ¾ν τÁς Α∆· ¹ Α∆ ¥ρα τÍ ΒΕ ™στι παράλληλος. Τ¦ ¥ρα ‡σα τρίγωνα τ¦ ™πˆ ‡σων βάσεων Ôντα κሠ™πˆ τ¦ αÙτ¦ µέρη κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ™στίν· Óπερ œδει δε‹ξαι. †

triangle F CE, the greater to the lesser. The very thing is impossible. Thus, AF is not parallel to BE. Similarly, we can show that neither (is) any other (straight-line) than AD. Thus, AD is parallel to BE. Thus, equal triangles which are on equal bases, and on the same side, are also between the same parallels. (Which is) the very thing it was required to show.

This whole proposition is regarded by Heiberg as a relatively early interpolation to the original text.

µα΄.

Proposition 41

'Ε¦ν παραλληλόγραµµον τριγώνJ βάσιν τε œχV τ¾ν If a parallelogram has the same base as a triangle, and αÙτ¾ν κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις Ï, διπλάσιόν ™στί is between the same parallels, then the parallelogram is τÕ παραλληλόγραµµον τοà τριγώνου. double (the area) of the triangle.

Α

Β

∆

Ε

A

Γ

B

D

E

C

Παραλληλόγραµµον γ¦ρ τÕ ΑΒΓ∆ τριγώνJ τù ΕΒΓ βάσιν τε ™χέτω τ¾ν αÙτ¾ν τ¾ν ΒΓ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις œστω τα‹ς ΒΓ, ΑΕ· λέγω, Óτι διπλάσιόν ™στι τÕ ΑΒΓ∆ παραλληλόγραµµον τοà ΒΕΓ τριγώνου. 'Επεζεύχθω γ¦ρ ¹ ΑΓ. ‡σον δή ™στι τÕ ΑΒΓ τρίγωνον τù ΕΒΓ τριγώνJ· ™πί τε γ¦ρ τÁς αÙτÁς βάσεώς ™στιν αÙτù τÁς ΒΓ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ΒΓ, ΑΕ. ¢λλ¦ τÕ ΑΒΓ∆ παραλληλόγραµµον διπλάσιόν œστι τοà ΑΒΓ τριγώνου· ¹ γ¦ρ ΑΓ διάµετρος αÙτÕ δίχα τέµνει· éστε τÕ ΑΒΓ∆ παραλληλόγραµµον κሠτοà ΕΒΓ τριγώνου ™στˆ διπλάσιον. 'Ε¦ν ¥ρα παραλληλόγραµµον τριγώνJ βάσιν τε œχV τ¾ν αÙτ¾ν κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις Ï, διπλάσιόν ™στί τÕ παραλληλόγραµµον τοà τριγώνου· Óπερ œδει δε‹ξαι.

For let parallelogram ABCD have the same base BC as triangle EBC, and let it be between the same parallels, BC and AE. I say that parallelogram ABCD is double (the area) of triangle BEC. For let AC have been joined. So triangle ABC is equal to triangle EBC. For it is on the same base, BC, as (EBC), and between the same parallels, BC and AE [Prop. 1.37]. But, parallelogram ABCD is double (the area) of triangle ABC. For the diagonal AC cuts the former in half [Prop. 1.34]. So parallelogram ABCD is also double (the area) of triangle EBC. Thus, if a parallelogram has the same base as a triangle, and is between the same parallels, then the parallelogram is double (the area) of the triangle. (Which is) the very thing it was required to show.

µβ΄.

Proposition 42

Τù δοθέντι τριγώνJ ‡σον παραλληλόγραµµον συστήTo construct a parallelogram equal to a given triangle σασθαι ™ν τÍ δοθείσV γωνίv εÙθυγράµµJ. in a given rectilinear angle. ”Εστω τÕ µν δοθν τρίγωνον τÕ ΑΒΓ, ¹ δ δοθε‹σα Let ABC be the given triangle, and D the given rectiγωνία εÙθύγραµµος ¹ ∆· δε‹ δ¾ τù ΑΒΓ τριγώνJ linear angle. So it is required to construct a parallelogram ‡σον παραλληλόγραµµον συστήσασθαι ™ν τÍ ∆ γωνίv equal to triangle ABC in the rectilinear angle D. εÙθυγράµµJ.

41

ΣΤΟΙΧΕΙΩΝ α΄. ∆

ELEMENTS BOOK 1

D

Α

Β

Ζ

Ε

Η

A

Γ

B

G

F

E

C

Τετµήσθω ¹ ΒΓ δίχα κατ¦ τÕ Ε, κሠ™πεζεύχθω ¹ ΑΕ, κሠσυνεστάτω πρÕς τÍ ΕΓ εÙθείv κሠτù πρÕς αÙτV σηµείJ τù Ε τÍ ∆ γωνίv ‡ση ¹ ØπÕ ΓΕΖ, κሠδι¦ µν τοà Α τÍ ΕΓ παράλληλος ½χθω ¹ ΑΗ, δι¦ δ τοà Γ τÍ ΕΖ παράλληλος ½χθω ¹ ΓΗ· παραλληλόγραµµον ¥ρα ™στˆ τÕ ΖΕΓΗ. κሠ™πεˆ ‡ση ™στˆν ¹ ΒΕ τÍ ΕΓ, ‡σον ™στˆ κሠτÕ ΑΒΕ τρίγωνον τù ΑΕΓ τριγώνJ· ™πί τε γ¦ρ ‡σων βάσεών ε„σι τîν ΒΕ, ΕΓ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ΒΓ, ΑΗ· διπλάσιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τοà ΑΕΓ τριγώνου. œστι δ κሠτÕ ΖΕΓΗ παραλληλόγραµµον διπλάσιον τοà ΑΕΓ τριγώνου· βάσιν τε γ¦ρ αÙτù τ¾ν αÙτ¾ν œχει κሠ™ν τα‹ς αÙτα‹ς ™στιν αÙτJ παραλλήλοις· ‡σον ¥ρα ™στˆ τÕ ΖΕΓΗ παραλληλόγραµµον τù ΑΒΓ τριγώνJ. κሠœχει τ¾ν ØπÕ ΓΕΖ γωνίαν ‡σην τÍ δοθείσV τÍ ∆. Τù ¥ρα δοθέντι τριγώνJ τù ΑΒΓ ‡σον παραλληλόγραµµον συνέσταται τÕ ΖΕΓΗ ™ν γωνίv τÍ ØπÕ ΓΕΖ, ¼τις ™στˆν ‡ση τÍ ∆· Óπερ œδει ποιÁσαι.

Let BC have been cut in half at E [Prop. 1.10], and let AE have been joined. And let (angle) CEF , equal to angle D, have been constructed at the point E on the straight-line EC [Prop. 1.23]. And let AG have been drawn through A parallel to EC [Prop. 1.31], and let CG have been drawn through C parallel to EF [Prop. 1.31]. Thus, F ECG is a parallelogram. And since BE is equal to EC, triangle ABE is also equal to triangle AEC. For they are on the equal bases, BE and EC, and between the same parallels, BC and AG [Prop. 1.38]. Thus, triangle ABC is double (the area) of triangle AEC. And parallelogram F ECG is also double (the area) of triangle AEC. For it has the same base as (AEC), and is between the same parallels as (AEC) [Prop. 1.41]. Thus, parallelogram F ECG is equal to triangle ABC. (F ECG) also has the angle CEF equal to the given (angle) D. Thus, parallelogram F ECG, equal to the given triangle ABC, has been constructed in the angle CEF , which is equal to D. (Which is) the very thing it was required to do.

µγ΄.

Proposition 43

ΠαντÕς παραλληλογράµµου τîν περˆ τ¾ν διάµετρον παραλληλογράµµων τ¦ παραπληρώµατα ‡σα ¢λλήλοις ™στίν. ”Εστω παραλληλόγραµµον τÕ ΑΒΓ∆, διάµετρος δ αÙτοà ¹ ΑΓ, περˆ δ τ¾ν ΑΓ παραλληλόγραµµα µν œστω τ¦ ΕΘ, ΖΗ, τ¦ δ λεγόµενα παραπληρώµατα τ¦ ΒΚ, Κ∆· λέγω, Óτι ‡σον ™στˆ τÕ ΒΚ παραπλήρωµα τù Κ∆ παραπληρώµατι. 'Επεˆ γ¦ρ παραλληλόγραµµόν ™στι τÕ ΑΒΓ∆, διάµετρος δ αÙτοà ¹ ΑΓ, ‡σον ™στˆ τÕ ΑΒΓ τρίγωνον τù ΑΓ∆ τριγώνJ. πάλιν, ™πεˆ παραλληλόγραµµόν ™στι τÕ ΕΘ, διάµετρος δ αÙτοà ™στιν ¹ ΑΚ, ‡σον ™στˆ τÕ ΑΕΚ τρίγωνον τù ΑΘΚ τριγώνJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ΚΖΓ τρίγωνον τù ΚΗΓ ™στιν ‡σον. ™πεˆ οâν τÕ µν ΑΕΚ τρίγωνον τù ΑΘΚ τριγώνJ ™στˆν ‡σον, τÕ δ ΚΖΓ τù ΚΗΓ, τÕ ΑΕΚ τρίγωνον µετ¦ τοà ΚΗΓ ‡σον ™στˆ τù ΑΘΚ τριγώνJ µετ¦ τοà ΚΖΓ· œστι δ κሠÓλον

For any parallelogram, the complements of the parallelograms about the diagonal are equal to one another. Let ABCD be a parallelogram, and AC its diagonal. And let EH and F G be the parallelograms about AC, and BK and KD the so-called complements (about AC). I say that the complement BK is equal to the complement KD. For since ABCD is a parallelogram, and AC its diagonal, triangle ABC is equal to triangle ACD [Prop. 1.34]. Again, since EH is a parallelogram, and AK is its diagonal, triangle AEK is equal to triangle AHK [Prop. 1.34]. So, for the same (reasons), triangle KF C is also equal to (triangle) KGC. Therefore, since triangle AEK is equal to triangle AHK, and KF C to KGC, triangle AEK plus KGC is equal to triangle AHK plus KF C. And the whole triangle ABC is also equal to the whole (triangle) ADC. Thus, the remaining complement BK is equal to

42

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

τÕ ΑΒΓ τρίγωνον ÓλJ τù Α∆Γ ‡σον· λοιπÕν ¥ρα τÕ ΒΚ the remaining complement KD. παραπλήρωµα λοιπù τù Κ∆ παραπληρώµατί ™στιν ‡σον.

Α

Θ Κ

Ε

Β

∆

A

Ζ

B

D

K

E

Γ

Η

H

F

G

C

ΠαντÕς ¥ρα παραλληλογράµµου χωρίου τîν περˆ Thus, for any parallelogramic figure, the compleτ¾ν διάµετρον παραλληλογράµµων τ¦ παραπληρώµατα ments of the parallelograms about the diagonal are equal ‡σα ¢λλήλοις ™στίν· Óπερ œδει δε‹ξαι. to one another. (Which is) the very thing it was required to show.

µδ΄.

Proposition 44

Παρ¦ τ¾ν δοθε‹σαν εÙθε‹αν τù δοθέντι τριγώνJ ‡σον παραλληλόγραµµον παραβαλε‹ν ™ν τÍ δοθείσV γωνίv εÙθυγράµµJ.

To apply a parallelogram equal to a given triangle to a given straight-line in a given rectilinear angle.

∆

D

Γ Ζ

Κ

Ε

Η Θ

C

Μ

Β Α

F

E

G

Λ

H

”Εστω ¹ µν δοθε‹σα εÙθε‹α ¹ ΑΒ, τÕ δ δοθν τρίγωνον τÕ Γ, ¹ δ δοθε‹σα γωνία εÙθύγραµµος ¹ ∆· δε‹ δ¾ παρ¦ τ¾ν δοθε‹σαν εÙθε‹αν τ¾ν ΑΒ τù δοθέντι τριγώνJ τù Γ ‡σον παραλληλόγραµµον παραβαλείν ™ν ‡σV τÍ ∆ γωνίv. Συνεστάτω τù Γ τριγώνJ ‡σον παραλληλόγραµµον τÕ ΒΕΖΗ ™ν γωνίv τÍ ØπÕ ΕΒΗ, ¼ ™στιν ‡ση τÍ ∆· κሠκείσθω éστε ™π' εÙθείας εναι τ¾ν ΒΕ τÍ ΑΒ, κሠδιήχθω ¹ ΖΗ ™πˆ τÕ Θ, κሠδι¦ τοà Α Ðποτέρv τîν ΒΗ, ΕΖ παράλληλος ½χθω ¹ ΑΘ, κሠ™πεζεύχθω ¹ ΘΒ. κሠ™πεˆ ε„ς παραλλήλους τ¦ς ΑΘ, ΕΖ εÙθε‹α ™νέπεσεν ¹ ΘΖ, αƒ ¥ρα ØπÕ ΑΘΖ, ΘΖΕ γωνίαι δυσˆν

K

M

B A

L

Let AB be the given straight-line, C the given triangle, and D the given rectilinear angle. So it is required to apply a parallelogram equal to the given triangle C to the given straight-line AB in an angle equal to D. Let the parallelogram BEF G, equal to the triangle C, have been constructed in the angle EBG, which is equal to D [Prop. 1.42]. And let it have been placed so that BE is straight-on to AB.† And let F G have been drawn through to H, and let AH have been drawn through A parallel to either of BG or EF [Prop. 1.31], and let HB have been joined. And since the straight-line HF falls across the parallel-lines AH and EF , the (sum of the)

43

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

Ñρθα‹ς ε„σιν ‡σαι. αƒ ¥ρα ØπÕ ΒΘΗ, ΗΖΕ δύο Ñρθîν ™λάσσονές ε„σιν· αƒ δ ¢πÕ ™λασσόνων À δύο Ñρθîν ε„ς ¥πειρον ™κβαλλόµεναι συµπίπτουσιν· αƒ ΘΒ, ΖΕ ¥ρα ™κβαλλόµεναι συµπεσοàνται. ™κβεβλήσθωσαν κሠσυµπιπτέτωσαν κατ¦ τÕ Κ, κሠδι¦ τοà Κ σηµείου Ðποτέρv τîν ΕΑ, ΖΘ παράλληλος ½χθω ¹ ΚΛ, κሠ™κβεβλήσθωσαν αƒ ΘΑ, ΗΒ ™πˆ τ¦ Λ, Μ σηµε‹α. παραλληλόγραµµον ¥ρα ™στˆ τÕ ΘΛΚΖ, διάµετρος δ αÙτοà ¹ ΘΚ, περˆ δ τ¾ν ΘΚ παραλληλόγραµµα µν τ¦ ΑΗ, ΜΕ, τ¦ δ λεγόµενα παραπληρώµατα τ¦ ΛΒ, ΒΖ· ‡σον ¥ρα ™στˆ τÕ ΛΒ τù ΒΖ. ¢λλ¦ τÕ ΒΖ τù Γ τριγώνJ ™στˆν ‡σον· κሠτÕ ΛΒ ¥ρα τù Γ ™στιν ‡σον. κሠ™πεˆ ‡ση ™στˆν ¹ ØπÕ ΗΒΕ γωνία τÍ ØπÕ ΑΒΜ, ¢λλ¦ ¹ ØπÕ ΗΒΕ τÍ ∆ ™στιν ‡ση, κሠ¹ ØπÕ ΑΒΜ ¥ρα τÍ ∆ γωνίv ™στˆν ‡ση. Παρ¦ τ¾ν δοθε‹σαν ¥ρα εÙθε‹αν τ¾ν ΑΒ τù δοθέντι τριγώνJ τù Γ ‡σον παραλληλόγραµµον παραβέβληται τÕ ΛΒ ™ν γωνίv τÍ ØπÕ ΑΒΜ, ¼ ™στιν ‡ση τÍ ∆· Óπερ œδει ποιÁσαι.

†

angles AHF and HF E is thus equal to two right-angles [Prop. 1.29]. Thus, (the sum of) BHG and GF E is less than two right-angles. And (straight-lines) produced to infinity from (internal angles whose sum is) less than two right-angles meet together [Post. 5]. Thus, being produced, HB and F E will meet together. Let them have been produced, and let them meet together at K. And let KL have been drawn through point K parallel to either of EA or F H [Prop. 1.31]. And let HA and GB have been produced to points L and M (respectively). Thus, HLKF is a parallelogram, and HK its diagonal. And AG and M E (are) parallelograms, and LB and BF the so-called complements, about HK. Thus, LB is equal to BF [Prop. 1.43]. But, BF is equal to triangle C. Thus, LB is also equal to C. Also, since angle GBE is equal to ABM [Prop. 1.15], but GBE is equal to D, ABM is thus also equal to angle D. Thus, the parallelogram LB, equal to the given triangle C, has been applied to the given straight-line AB in the angle ABM , which is equal to D. (Which is) the very thing it was required to do.

This can be achieved using Props. 1.3, 1.23, and 1.31.

µε΄.

Proposition 45

Τö δοθέντι εÙθυγράµµJ ‡σον παραλληλόγραµµον συστήσασθαι ™ν τÍ δοθείσV γωνίv εÙθυγράµµJ.

To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle.

∆

D Γ

C

Α

A

Ζ

Κ

Η

Θ

Β

Ε

E B

Λ

F

Μ

K

”Εστω τÕ µν δοθν εÙθύγραµµον τÕ ΑΒΓ∆, ¹ δ δοθε‹σα γωνία εÙθύγραµµος ¹ Ε· δε‹ δ¾ τù ΑΒΓ∆ εÙθυγράµµJ ‡σον παραλληλόγραµµον συστήσασθαι ™ν τÍ δοθείσV γωνίv τÍ Ε. 'Επεζεύχθω ¹ ∆Β, κሠσυνεστάτω τù ΑΒ∆ τριγώνJ ‡σον παραλληλόγραµµον τÕ ΖΘ ™ν τÍ ØπÕ ΘΚΖ γωνίv, ¼ ™στιν ‡ση τÍ Ε· κሠπαραβεβλήσθω παρ¦ τ¾ν ΗΘ

G

H

L

M

Let ABCD be the given rectilinear figure,† and E the given rectilinear angle. So it is required to construct a parallelogram equal to the rectilinear figure ABCD in the given angle E. Let DB have been joined, and let the parallelogram F H, equal to the triangle ABD, have been constructed in the angle HKF , which is equal to E [Prop. 1.42]. And let

44

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

εÙθε‹αν τù ∆ΒΓ τριγώνJ ‡σον παραλληλόγραµµον τÕ ΗΜ ™ν τÍ ØπÕ ΗΘΜ γωνίv, ¼ ™στιν ‡ση τÍ Ε. κሠ™πεˆ ¹ Ε γωνία ˜κατέρv τîν ØπÕ ΘΚΖ, ΗΘΜ ™στιν ‡ση, κሠ¹ ØπÕ ΘΚΖ ¥ρα τÍ ØπÕ ΗΘΜ ™στιν ‡ση. κοιν¾ προσκείσθω ¹ ØπÕ ΚΘΗ· αƒ ¥ρα ØπÕ ΖΚΘ, ΚΘΗ τα‹ς ØπÕ ΚΘΗ, ΗΘΜ ‡σαι ε„σίν. ¢λλ' αƒ ØπÕ ΖΚΘ, ΚΘΗ δυσˆν Ñρθα‹ς ‡σαι ε„σίν· καˆ αƒ ØπÕ ΚΘΗ, ΗΘΜ ¥ρα δύο Ñρθα‹ς ‡σας ε„σίν. πρÕς δή τινι εÙθε‹v τÍ ΗΘ κሠτù πρÕς αÙτÍ σηµείJ τù Θ δύο εÙθε‹αι αƒ ΚΘ, ΘΜ µ¾ ™πˆ τ¦ αÙτ¦ µέρη κείµεναι τ¦ς ™φεξÁς γωνίας δύο Ñρθα‹ς ‡σας ποιοàσιν· ™π' εÙθείας ¥ρα ™στˆν ¹ ΚΘ τÍ ΘΜ· κሠ™πεˆ ε„ς παραλλήλους τ¦ς ΚΜ, ΖΗ εÙθε‹α ™νέπεσεν ¹ ΘΗ, αƒ ™ναλλ¦ξ γωνίαι αƒ ØπÕ ΜΘΗ, ΘΗΖ ‡σαι ¢λλήλαις ε„σίν. κοιν¾ προσκείσθω ¹ ØπÕ ΘΗΛ· αƒ ¥ρα ØπÕ ΜΘΗ, ΘΗΛ τα‹ς ØπÕ ΘΗΖ, ΘΗΛ ‡σαι ε„σιν. ¢λλ' αƒ ØπÕ ΜΘΗ, ΘΗΛ δύο Ñρθα‹ς ‡σαι ε„σίν· καˆ αƒ ØπÕ ΘΗΖ, ΘΗΛ ¥ρα δύο Ñρθα‹ς ‡σαι ε„σίν· ™π' εÙθείας ¥ρα ™στˆν ¹ ΖΗ τÍ ΗΛ. κሠ™πεˆ ¹ ΖΚ τÍ ΘΗ ‡ση τε κሠπαράλληλός ™στιν, ¢λλ¦ κሠ¹ ΘΗ τÍ ΜΛ, κሠ¹ ΚΖ ¥ρα τÍ ΜΛ ‡ση τε κሠπαράλληλός ™στιν· κሠ™πιζευγνύουσιν αÙτ¦ς εÙθε‹αι αƒ ΚΜ, ΖΛ· καˆ αƒ ΚΜ, ΖΛ ¥ρα ‡σαι τε κሠπαράλληλοί ε„σιν· παραλληλόγραµµον ¥ρα ™στˆ τÕ ΚΖΛΜ. κሠ™πεˆ ‡σον ™στˆ τÕ µν ΑΒ∆ τρίγωνον τù ΖΘ παραλληλογράµµJ, τÕ δ ∆ΒΓ τù ΗΜ, Óλον ¥ρα τÕ ΑΒΓ∆ εÙθύγραµµον ÓλJ τù ΚΖΛΜ παραλληλογράµµJ ™στˆν ‡σον. Τù ¥ρα δοθέντι εÙθυγράµµJ τù ΑΒΓ∆ ‡σον παραλληλόγραµµον συνέσταται τÕ ΚΖΛΜ ™ν γωνίv τÍ ØπÕ ΖΚΜ, ¼ ™στιν ‡ση τÍ δοθείσV τÍ Ε· Óπερ œδει ποιÁσαι.

†

the parallelogram GM , equal to the triangle DBC, have been applied to the straight-line GH in the angle GHM , which is equal to E [Prop. 1.44]. And since angle E is equal to each of (angles) HKF and GHM , (angle) HKF is thus also equal to GHM . Let KHG have been added to both. Thus, (the sum of) F KH and KHG is equal to (the sum of) KHG and GHM . But, (the sum of) F KH and KHG is equal to two right-angles [Prop. 1.29]. Thus, (the sum of) KHG and GHM is also equal to two rightangles. So two straight-lines, KH and HM , not lying on the same side, make the (sum of the) adjacent angles equal to two right-angles at the point H on some straightline GH. Thus, KH is straight-on to HM [Prop. 1.14]. And since the straight-line HG falls across the parallellines KM and F G, the alternate angles M HG and HGF are equal to one another [Prop. 1.29]. Let HGL have been added to both. Thus, (the sum of) M HG and HGL is equal to (the sum of) HGF and HGL. But, (the sum of) M HG and HGL is equal to two right-angles [Prop. 1.29]. Thus, (the sum of) HGF and HGL is also equal to two right-angles. Thus, F G is straight-on to GL [Prop. 1.14]. And since F K is equal and parallel to HG [Prop. 1.34], but also HG to M L [Prop. 1.34], KF is thus also equal and parallel to M L [Prop. 1.30]. And the straight-lines KM and F L join them. Thus, KM and F L are equal and parallel as well [Prop. 1.33]. Thus, KF LM is a parallelogram. And since triangle ABD is equal to parallelogram F H, and DBC to GM , the whole rectilinear figure ABCD is thus equal to the whole parallelogram KF LM . Thus, the parallelogram KF LM , equal to the given rectilinear figure ABCD, has been constructed in the angle F KM , which is equal to the given (angle) E. (Which is) the very thing it was required to do.

The proof is only given for a four-sided figure. However, the extension to many-sided figures is trivial.

µ$΄.

Proposition 46

'ΑπÑ τÁς δοθείσης εÙθείας τετράγωνον ¢ναγράψαι. ”Εστω ¹ δοθε‹σα εÙθε‹α ¹ ΑΒ· δε‹ δ¾ ¢πÕ τÁς ΑΒ εÙθείας τετράγωνον ¢ναγράψαι. ”Ηχθω τÍ ΑΒ εÙθείv ¢πÕ τοà πρÕς αÙτÍ σηµείου τοà Α πρÕς Ñρθ¦ς ¹ ΑΓ, κሠκείσθω τÍ ΑΒ ‡ση ¹ Α∆· κሠδι¦ µν τοà ∆ σηµείου τÍ ΑΒ παράλληλος ½χθω ¹ ∆Ε, δι¦ δ τοà Β σηµείου τÍ Α∆ παράλληλος ½χθω ¹ ΒΕ. παραλληλόγραµµον ¥ρα ™στˆ τÕ Α∆ΕΒ· ‡ση ¥ρα ™στˆν ¹ µν ΑΒ τÍ ∆Ε, ¹ δ Α∆ τÍ ΒΕ. ¢λλ¦ ¹ ΑΒ τÍ Α∆ ™στιν ‡ση· αƒ τέσσαρες ¥ρα αƒ ΒΑ, Α∆, ∆Ε, ΕΒ ‡σαι ¢λλήλαις ε„σίν· „σόπλευρον ¥ρα ™στˆ τÕ Α∆ΕΒ παραλληλόγραµµον. λέγω δή, Óτι κሠÑρθογώνιον. ™πεˆ γ¦ρ ε„ς παραλλήλους τ¦ς ΑΒ, ∆Ε εÙθε‹α ™νέπεσεν ¹ Α∆,

To describe a square on a given straight-line. Let AB be the given straight-line. So it is required to describe a square on the straight-line AB. Let AC have been drawn at right-angles to the straight-line AB from the point A on it [Prop. 1.11], and let AD have been made equal to AB [Prop. 1.3]. And let DE have been drawn through point D parallel to AB [Prop. 1.31], and let BE have been drawn through point B parallel to AD [Prop. 1.31]. Thus, ADEB is a parallelogram. Thus, AB is equal to DE, and AD to BE [Prop. 1.34]. But, AB is equal to AD. Thus, the four (sides) BA, AD, DE, and EB are equal to one another. Thus, the parallelogram ADEB is equilateral. So

45

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

αƒ ¥ρα ØπÕ ΒΑ∆, Α∆Ε γωνίαι δύο Ñρθα‹ς ‡σαι ε„σίν. Ñρθ¾ δ ¹ ØπÕ ΒΑ∆· Ñρθ¾ ¥ρα κሠ¹ ØπÕ Α∆Ε. τîν δ παραλληλογράµµων χωρίων αƒ ¢πεναντίον πλευραί τε κሠγωνίαι ‡σαι ¢λλήλαις ε„σίν· Ñρθ¾ ¥ρα κሠ˜κατέρα τîν ¢πεναντίον τîν ØπÕ ΑΒΕ, ΒΕ∆ γωνιîν· Ñρθογώνιον ¥ρα ™στˆ τÕ Α∆ΕΒ. ™δείχθη δ κሠ„σόπλευρον.

I say that (it is) also right-angled. For since the straightline AD falls across the parallel-lines AB and DE, the (sum of the) angles BAD and ADE is equal to two rightangles [Prop. 1.29]. But BAD (is a) right-angle. Thus, ADE (is) also a right-angle. And for parallelogrammic figures, the opposite sides and angles are equal to one another [Prop. 1.34]. Thus, each of the opposite angles ABE and BED (are) also right-angles. Thus, ADEB is right-angled. And it was also shown (to be) equilateral.

Γ

C

∆

Ε

Α

D

Β

A

E

B

Τετράγωνον ¥ρα ™στίν· καί ™στιν ¢πÕ τÁς ΑΒ εÙθείας ¢ναγεγραµµένον· Óπερ œδει ποιÁσαι.

Thus, (ADEB) is a square [Def. 1.22]. And it is described on the straight-line AB. (Which is) the very thing it was required to do.

µζ΄.

Proposition 47

'Εν το‹ς Ñρθογωνίοις τριγώνοις τÕ ¢πÕ τÁς τ¾ν Ñρθ¾ν γωνίαν Øποτεινούσης πλευρ©ς τετράγωνον ‡σον ™στˆ το‹ς ¢πÕ τîν τ¾ν Ñρθ¾ν γωνίαν περιεχουσîν πλευρîν τετραγώνοις. ”Εστω τρίγωνον Ñρθογώνιον τÕ ΑΒΓ Ñρθ¾ν œχον τ¾ν ØπÕ ΒΑΓ γωνίαν· λέγω, Óτι τÕ ¢πÕ τÁς ΒΓ τετράγωνον ‡σον ™στˆ το‹ς ¢πÕ τîν ΒΑ, ΑΓ τετραγώνοις. 'Αναγεγράφθω γ¦ρ ¢πÕ µν τÁς ΒΓ τετράγωνον τÕ Β∆ΕΓ, ¢πÕ δ τîν ΒΑ, ΑΓ τ¦ ΗΒ, ΘΓ, κሠδι¦ τοà Α Ðποτέρv τîν Β∆, ΓΕ παράλληλος ½χθω ¹ ΑΛ· κሠ™πεζεύχθωσαν αƒ Α∆, ΖΓ. κሠ™πεˆ Ñρθή ™στιν ˜κατέρα τîν ØπÕ ΒΑΓ, ΒΑΗ γωνιîν, πρÕς δή τινι εÙθείv τÍ ΒΑ κሠτù πρÕς αÙτÍ σηµείJ τù Α δύο εÙθε‹αι αƒ ΑΓ, ΑΗ µ¾ ™πˆ τ¦ αÙτ¦ µέρη κείµεναι τ¦ς ™φεξÁς γωνίας δυσˆν Ñρθα‹ς ‡σας ποιοàσιν· ™π' εÙθείας ¥ρα ™στˆν ¹ ΓΑ τÍ ΑΗ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΒΑ τÍ ΑΘ ™στιν ™π' εÙθείας. κሠ™πεˆ ‡ση ™στˆν ¹ ØπÕ ∆ΒΓ γωνία τÍ ØπÕ ΖΒΑ· Ñρθ¾ γ¦ρ ˜κατέρα· κοιν¾ προσκείσθω ¹ ØπÕ ΑΒΓ· Óλη ¥ρα ¹ ØπÕ ∆ΒΑ ÓλV τÍ ØπÕ ΖΒΓ ™στιν ‡ση. κሠ™πεˆ ‡ση ™στˆν ¹ µν ∆Β τÍ ΒΓ, ¹ δ ΖΒ τÍ ΒΑ, δύο δ¾ αƒ ∆Β, ΒΑ δύο τα‹ς ΖΒ, ΒΓ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία

In a right-angled triangle, the square on the side subtending the right-angle is equal to the (sum of the) squares on the sides surrounding the right-angle. Let ABC be a right-angled triangle having the rightangle BAC. I say that the square on BC is equal to the (sum of the) squares on BA and AC. For let the square BDEC have been described on BC, and (the squares) GB and HC on AB and AC (respectively) [Prop. 1.46]. And let AL have been drawn through point A parallel to either of BD or CE [Prop. 1.31]. And let AD and F C have been joined. And since angles BAC and BAG are each right-angles, then two straight-lines AC and AG, not lying on the same side, make the (sum of the) adjacent angles equal to two right-angles at the same point A on some straight-line BA . Thus, CA is straight-on to AG [Prop. 1.14]. So, for the same (reasons), BA is also straight-on to AH. And since angle DBC is equal to F BA, for (they are) both right-angles, let ABC have been added to both. Thus, the whole (angle) DBA is equal to the whole (angle) F BC. And since DB is equal to BC, and F B to BA,

46

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1

¹ ØπÕ ∆ΒΑ γωνίv τÍ ØπÕ ΖΒΓ ‡ση· βάσις ¥ρα ¹ Α∆ βάσει τÍ ΖΓ [™στιν] ‡ση, κሠτÕ ΑΒ∆ τρίγωνον τù ΖΒΓ τριγώνJ ™στˆν ‡σον· καί [™στι] τοà µν ΑΒ∆ τριγώνου διπλάσιον τÕ ΒΛ παραλληλόγραµµον· βάσιν τε γ¦ρ τ¾ν αÙτ¾ν œχουσι τ¾ν Β∆ κሠ™ν τα‹ς αÙτα‹ς ε„σι παραλλήλοις τα‹ς Β∆, ΑΛ· τοà δ ΖΒΓ τριγώνου διπλάσιον τÕ ΗΒ τετράγωνον· βάσιν τε γ¦ρ πάλιν τ¾ν αÙτ¾ν œχουσι τ¾ν ΖΒ κሠ™ν τα‹ς αÙτα‹ς ε„σι παραλλήλοις τα‹ς ΖΒ, ΗΓ. [τ¦ δ τîν ‡σων διπλάσια ‡σα ¢λλήλοις ™στίν·] ‡σον ¥ρα ™στˆ κሠτÕ ΒΛ παραλληλόγραµµον τù ΗΒ τετραγώνJ. еοίως δ¾ ™πιζευγνυµένων τîν ΑΕ, ΒΚ δειχθήσεται κሠτÕ ΓΛ παραλληλόγραµµον ‡σον τù ΘΓ τετραγώνJ· Óλον ¥ρα τÕ Β∆ΕΓ τετράγωνον δυσˆ το‹ς ΗΒ, ΘΓ τετραγώνοις ‡σον ™στίν. καί ™στι τÕ µν Β∆ΕΓ τετράγωνον ¢πÕ τÁς ΒΓ ¢ναγραφέν, τ¦ δ ΗΒ, ΘΓ ¢πÕ τîν ΒΑ, ΑΓ. τÕ ¥ρα ¢πÕ τÁς ΒΓ πλευρ©ς τετράγωνον ‡σον ™στˆ το‹ς ¢πÕ τîν ΒΑ, ΑΓ πλευρîν τετραγώνοις.

the two (straight-lines) DB, BA are equal to the two (straight-lines) CB, BF ,† respectively. And angle DBA (is) equal to angle F BC. Thus, the base AD [is] equal to the base F C, and the triangle ABD is equal to the triangle F BC [Prop. 1.4]. And parallelogram BL [is] double (the area) of triangle ABD. For they have the same base, BD, and are between the same parallels, BD and AL [Prop. 1.41]. And parallelogram GB is double (the area) of triangle F BC. For again they have the same base, F B, and are between the same parallels, F B and GC [Prop. 1.41]. [And the doubles of equal things are equal to one another.]‡ Thus, the parallelogram BL is also equal to the square GB. So, similarly, AE and BK being joined, the parallelogram CL can be shown (to be) equal to the square HC. Thus, the whole square BDEC is equal to the (sum of the) two squares GB and HC. And the square BDEC is described on BC, and the (squares) GB and HC on BA and AC (respectively). Thus, the square on the side BC is equal to the (sum of the) squares on the sides BA and AC.

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D L E ∆ Λ Ε 'Εν ¥ρα το‹ς Ñρθογωνίοις τριγώνοις τÕ ¢πÕ τÁς Thus, in a right-angled triangle, the square on the τ¾ν Ñρθ¾ν γωνίαν Øποτεινούσης πλευρ©ς τετράγωνον side subtending the right-angle is equal to the (sum of ‡σον ™στˆ το‹ς ¢πÕ τîν τ¾ν Ñρθ¾ν [γωνίαν] περιεχουσîν the) squares on the sides surrounding the right-[angle]. πλευρîν τετραγώνοις· Óπερ œδει δε‹ξαι. (Which is) the very thing it was required to show. †

The Greek text has “F B, BC”, which is obviously a mistake.

‡

This is an additional common notion.

47

ΣΤΟΙΧΕΙΩΝ α΄.

ELEMENTS BOOK 1 µη΄.

Proposition 48

'Ε¦ν τριγώνου τÕ ¢πÕ µι©ς τîν πλευρîν τετράγωνον If the square on one of the sides of a triangle is equal ‡σον Ï το‹ς ¢πÕ τîν λοιπîν τοà τριγώνου δύο πλευρîν to the (sum of the) squares on the remaining sides of the τετραγώνοις, ¹ περιεχοµένη γωνία ØπÕ τîν λοιπîν τοà triangle then the angle contained by the remaining sides τριγώνου δύο πλευρîν Ñρθή ™στιν. of the triangle is a right-angle.

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Τριγώνου γ¦ρ τοà ΑΒΓ τÕ ¢πÕ µι©ς τÁς ΒΓ πλευρ©ς τετράγωνον ‡σον œστω το‹ς ¢πÕ τîν ΒΑ, ΑΓ πλευρîν τετραγώνοις· λέγω, Óτι Ñρθή ™στιν ¹ ØπÕ ΒΑΓ γωνία. ”Ηχθω γ¦ρ ¢πÕ τοà Α σηµείου τÍ ΑΓ εÙθείv πρÕς Ñρθ¦ς ¹ Α∆ κሠκείσθω τÍ ΒΑ ‡ση ¹ Α∆, κሠ™πεζεύχθω ¹ ∆Γ. ™πεˆ ‡ση ™στˆν ¹ ∆Α τÍ ΑΒ, ‡σον ™στˆ κሠτÕ ¢πÕ τÁς ∆Α τετράγωνον τù ¢πÕ τÁς ΑΒ τετραγώνJ. κοινÕν προσκείσθω τÕ ¢πÕ τÁς ΑΓ τετράγωνον· τ¦ ¥ρα ¢πÕ τîν ∆Α, ΑΓ τετράγωνα ‡σα ™στˆ το‹ς ¢πÕ τîν ΒΑ, ΑΓ τετραγώνοις. ¢λλ¦ το‹ς µν ¢πÕ τîν ∆Α, ΑΓ ‡σον ™στˆ τÕ ¢πÕ τÁς ∆Γ· Ñρθ¾ γάρ ™στιν ¹ ØπÕ ∆ΑΓ γωνία· το‹ς δ ¢πÕ τîν ΒΑ, ΑΓ ‡σον ™στˆ τÕ ¢πÕ τÁς ΒΓ· Øπόκειται γάρ· τÕ ¥ρα ¢πÕ τÁς ∆Γ τετράγωνον ‡σον ™στˆ τù ¢πÕ τÁς ΒΓ τετραγώνJ· éστε κሠπλευρ¦ ¹ ∆Γ τÍ ΒΓ ™στιν ‡ση· κሠ™πεˆ ‡ση ™στˆν ¹ ∆Α τÍ ΑΒ, κοιν¾ δ ¹ ΑΓ, δύο δ¾ αƒ ∆Α, ΑΓ δύο τα‹ς ΒΑ, ΑΓ ‡σαι ε„σίν· κሠβάσις ¹ ∆Γ βάσει τÍ ΒΓ ‡ση· γωνία ¥ρα ¹ ØπÕ ∆ΑΓ γωνίv τÍ ØπÕ ΒΑΓ [™στιν] ‡ση. Ñρθ¾ δ ¹ ØπÕ ∆ΑΓ· Ñρθ¾ ¥ρα κሠ¹ ØπÕ ΒΑΓ. 'Ε¦ν ¢ρ¦ τριγώνου τÕ ¢πÕ µι©ς τîν πλευρîν τετράγωνον ‡σον Ï το‹ς ¢πÕ τîν λοιπîν τοà τριγώνου δύο πλευρîν τετραγώνοις, ¹ περιεχοµένη γωνία ØπÕ τîν λοιπîν τοà τριγώνου δύο πλευρîν Ñρθή ™στιν· Óπερ œδει δε‹ξαι.

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For let the square on one of the sides, BC, of triangle ABC be equal to the (sum of the) squares on the sides BA and AC. I say that angle BAC is a right-angle. For let AD have been drawn from point A at rightangles to the straight-line AC [Prop. 1.11], and let AD have been made equal to BA [Prop. 1.3], and let DC have been joined. Since DA is equal to AB, the square on DA is thus also equal to the square on AB.† Let the square on AC have been added to both. Thus, the (sum of the) squares on DA and AC is equal to the (sum of the) squares on BA and AC. But, the (sum of the squares) on DA and AC is equal to the (square) on DC. For angle DAC is a right-angle [Prop. 1.47]. But, the (sum of the squares) on BA and AC is equal to the (square) on BC. For (that) was assumed. Thus, the square on DC is equal to the square on BC. So DC is also equal to BC. And since DA is equal to AB, and AC (is) common, the two (straight-lines) DA, AC are equal to the two (straight-lines) BA, AC. And the base DC is equal to the base BC. Thus, angle DAC [is] equal to angle BAC [Prop. 1.8]. But DAC is a right-angle. Thus, BAC is also a right-angle. Thus, if the square on one of the sides of a triangle is equal to the (sum of the) squares on the remaining sides of the triangle then the angle contained by the remaining sides of the triangle is a right-angle. (Which is) the very thing it was required to show.

Here, use is made of the additional common notion that the squares of equal things are themselves equal. Later on, the inverse notion is used.

48

ELEMENTS BOOK 2 Fundamentals of geometric algebra

49

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

“Οροι.

Definitions

α΄. Π©ν παραλληλόγραµµον Ñρθογώνιον περιέχεσθαι λέγεται ØπÕ δύο τîν τ¾ν Ñρθ¾ν γωνίαν περιεχουσîν εÙθειîν. β΄. ΠαντÕς δ παραλληλογράµµου χωρίου τîν περˆ τ¾ν διάµετρον αÙτοà παραλληλογράµµων žν Ðποιονοàν σÝν το‹ς δυσˆ παραπληρώµασι γνώµων καλείσθω.

1. Any right-angled parallelogram is said to be contained by the two straight-lines containing a(ny) rightangle. 2. And for any parallelogrammic figure, let any one whatsoever of the parallelograms about its diagonal, (taken) with its two complements, be called a gnomon.

α΄.

Proposition 1†

'Ε¦ν ðσι δύο εÙθε‹αι, τµηθÍ δ ¹ ˜τέρα αÙτîν ε„ς Ðσαδηποτοàν τµήµατα, τÕ περιεχόµενον Ñρθογώνιον ØπÕ τîν δύο εÙθειîν ‡σον ™στˆ το‹ς Øπό τε τÁς ¢τµήτου κሠ˜κάστου τîν τµηµάτων περιεχοµένοις Ñρθογωνίοις.

If there are two straight-lines, and one of them is cut into any number of pieces whatsoever, then the rectangle contained by the two straight-lines is equal to the (sum of the) rectangles contained by the uncut (straight-line), and every one of the pieces (of the cut straight-line).

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”Εστωσαν δύο εÙθε‹αι αƒ Α, ΒΓ, κሠτετµήσθω ¹ ΒΓ, æς œτυχεν, κατ¦ τ¦ ∆, Ε σηµε‹α· λέγω, Óτι τÕ ØπÕ τîν Α, ΒΓ περιεχοµένον Ñρθογώνιον ‡σον ™στˆ τù τε ØπÕ τîν Α, Β∆ περιεχοµένJ ÑρθογωνίJ κሠτù ØπÕ τîν Α, ∆Ε κሠœτι τù ØπÕ τîν Α, ΕΓ. ”Ηχθω γ¦ρ ¢πÕ τοà Β τÍ ΒΓ πρÕς Ñρθ¦ς ¹ ΒΖ, κሠκείσθω τÍ Α ‡ση ¹ ΒΗ, κሠδι¦ µν τοà Η τÍ ΒΓ παράλληλος ½χθω ¹ ΗΘ, δι¦ δ τîν ∆, Ε, Γ τÍ ΒΗ παράλληλοι ½χθωσαν αƒ ∆Κ, ΕΛ, ΓΘ. ”Ισον δή ™στι τÕ ΒΘ το‹ς ΒΚ, ∆Λ, ΕΘ. καί ™στι τÕ µν ΒΘ τÕ ØπÕ τîν Α, ΒΓ· περιέχεται µν γ¦ρ ØπÕ τîν ΗΒ, ΒΓ, ‡ση δ ¹ ΒΗ τÍ Α· τÕ δ ΒΚ τÕ ØπÕ τîν Α, Β∆· περιέχεται µν γ¦ρ ØπÕ τîν ΗΒ, Β∆, ‡ση δ ¹ ΒΗ τÍ Α. τÕ δ ∆Λ τÕ ØπÕ τîν Α, ∆Ε· ‡ση γ¦ρ ¹ ∆Κ, τουτέστιν ¹ ΒΗ, τÍ Α. κሠœτι еοίως τÕ ΕΘ τÕ ØπÕ τîν Α, ΕΓ· τÕ ¥ρα ØπÕ τîν Α, ΒΓ ‡σον ™στˆ τù τε ØπÕ Α, Β∆ κሠτù ØπÕ Α, ∆Ε κሠœτι τù ØπÕ Α, ΕΓ. 'Ε¦ν ¥ρα ðσι δύο εÙθε‹αι, τµηθÍ δ ¹ ˜τέρα αÙτîν ε„ς Ðσαδηποτοàν τµήµατα, τÕ περιεχόµενον Ñρθογώνιον ØπÕ τîν δύο εÙθειîν ‡σον ™στˆ το‹ς Øπό τε τÁς ¢τµήτου κሠ˜κάστου τîν τµηµάτων περιεχοµένοις Ñρθογωνίοις·

Let A and BC be the two straight-lines, and let BC be cut, at random, at points D and E. I say that the rectangle contained by A and BC is equal to the rectangle(s) contained by A and BD, by A and DE, and, finally, by A and EC. For let BF have been drawn from point B, at rightangles to BC [Prop. 1.11], and let BG be made equal to A [Prop. 1.3], and let GH have been drawn through (point) G, parallel to BC [Prop. 1.31], and let DK, EL, and CH have been drawn through (points) D, E, and C (respectively), parallel to BG [Prop. 1.31]. So the (rectangle) BH is equal to the (rectangles) BK, DL, and EH. And BH is the (rectangle contained) by A and BC. For it is contained by GB and BC, and BG (is) equal to A. And BK (is) the (rectangle contained) by A and BD. For it is contained by GB and BD, and BG (is) equal to A. And DL (is) the (rectangle contained) by A and DE. For DK, that is to say BG [Prop. 1.34], (is) equal to A. Similarly, EH (is) the (rectangle contained) by A and EC. Thus, the (rectangle contained) by A and BC is equal to the (rectangles contained) by A and BD,

50

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

Óπερ œδει δε‹ξαι.

†

by A and DE, and, finally, by A and EC. Thus, if there are two straight-lines, and one of them is cut into any number of pieces whatsoever, then the rectangle contained by the two straight-lines is equal to the (sum of the) rectangles contained by the uncut (straight-line), and every one of the pieces (of the cut straight-line). (Which is) the very thing it was required to show.

This proposition is a geometric version of the algebraic identity: a (b + c + d + · · · ) = a b + a c + a d + · · · .

Proposition 2†

β΄.

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ ØπÕ If a straight-line is cut at random, then the (sum of τÁς Óλης κሠ˜κατέρου τîν τµηµάτων περιεχόµενον the) rectangle(s) contained by the whole (straight-line), Ñρθογώνιον ‡σον ™στˆ τù ¢πÕ τÁς Óλης τετραγώνJ. and each of the pieces (of the straight-line), is equal to the square on the whole.

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ΕÙθε‹α γ¦ρ ¹ ΑΒ τετµήσθω, æς œτυχεν, κατ¦ τÕ Γ σηµε‹ον· λέγω, Óτι τÕ ØπÕ τîν ΑΒ, ΒΓ περιεχόµενον Ñρθογώνιον µετ¦ τοà ØπÕ ΒΑ, ΑΓ περιεχοµένου Ñρθογωνίου ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ τετραγώνJ. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ Α∆ΕΒ, κሠ½χθω δˆα τοà Γ Ðποτέρv τîν Α∆, ΒΕ παράλληλος ¹ ΓΖ. ”Ισον δή ™στˆ τÕ ΑΕ το‹ς ΑΖ, ΓΕ. καί ™στι τÕ µν ΑΕ τÕ ¢πÕ τÁς ΑΒ τετράγωνον, τÕ δ ΑΖ τÕ ØπÕ τîν ΒΑ, ΑΓ περιεχόµενον Ñρθογώνιον· περιέχεται µν γ¦ρ ØπÕ τîν ∆Α, ΑΓ, ‡ση δ ¹ Α∆ τÍ ΑΒ· τÕ δ ΓΕ τÕ ØπÕ τîν ΑΒ, ΒΓ· ‡ση γ¦ρ ¹ ΒΕ τÍ ΑΒ. τÕ ¥ρα ØπÕ τîν ΒΑ, ΑΓ µετ¦ τοà ØπÕ τîν ΑΒ, ΒΓ ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ τετραγώνJ. 'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ ØπÕ τÁς Óλης κሠ˜κατέρου τîν τµηµάτων περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ¢πÕ τÁς Óλης τετραγώνJ· Óπερ œδει δε‹ξαι.

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For let the straight-line AB have been cut, at random, at point C. I say that the rectangle contained by AB and BC, plus the rectangle contained by BA and AC, is equal to the square on AB. For let the square ADEB have been described on AB [Prop. 1.46], and let CF have been drawn through C, parallel to either of AD or BE [Prop. 1.31]. So the (square) AE is equal to the (rectangles) AF and CE. And AE is the square on AB. And AF (is) the rectangle contained by the (straight-lines) BA and AC. For it is contained by DA and AC, and AD (is) equal to AB. And CE (is) the (rectangle contained) by AB and BC. For BE (is) equal to AB. Thus, the (rectangle contained) by BA and AC, plus the (rectangle contained) by AB and BC, is equal to the square on AB. Thus, if a straight-line is cut at random, then the (sum of the) rectangle(s) contained by the whole (straightline), and each of the pieces (of the straight-line), is equal to the square on the whole. (Which is) the very thing it was required to show.

51

ΣΤΟΙΧΕΙΩΝ β΄. †

ELEMENTS BOOK 2

This proposition is a geometric version of the algebraic identity: a b + a c = a2 if a = b + c.

γ΄.

Proposition 3†

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ ØπÕ τÁς Óλης κሠ˜νÕς τîν τµηµάτων περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù τε ØπÕ τîν τµηµάτων περιεχοµένJ ÑρθογωνίJ κሠτù ¢πÕ τοà προειρηµένου τµήµατος τετραγώνJ.

If a straight-line is cut at random, then the rectangle contained by the whole (straight-line), and one of the pieces (of the straight-line), is equal to the rectangle contained by (both of) the pieces, and the square on the aforementioned piece.

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ΕÙθε‹α γ¦ρ ¹ ΑΒ τετµήσθω, æς œτυχεν, κατ¦ τÕ Γ· λέγω, Óτι τÕ ØπÕ τîν ΑΒ, ΒΓ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù τε ØπÕ τîν ΑΓ, ΓΒ περιεχοµένJ ÑρθογωνίJ µετ¦ τοà ¢πÕ τÁς ΒΓ τετραγώνου. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΓΒ τετράγωνον τÕ Γ∆ΕΒ, κሠδιήχθω ¹ Ε∆ ™πˆ τÕ Ζ, κሠδι¦ τοà Α Ðποτέρv τîν Γ∆, ΒΕ παράλληλος ½χθω ¹ ΑΖ. ‡σον δή ™στι τÕ ΑΕ το‹ς Α∆, ΓΕ· καί ™στι τÕ µν ΑΕ τÕ ØπÕ τîν ΑΒ, ΒΓ περιεχόµενον Ñρθογώνιον· περιέχεται µν γ¦ρ ØπÕ τîν ΑΒ, ΒΕ, ‡ση δ ¹ ΒΕ τÍ ΒΓ· τÕ δ Α∆ τÕ ØπÕ τîν ΑΓ, ΓΒ· ‡ση γ¦ρ ¹ ∆Γ τÍ ΓΒ· τÕ δ ∆Β τÕ ¢πÕ τÁς ΓΒ τετράγωνον· τÕ ¥ρα ØπÕ τîν ΑΒ, ΒΓ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ØπÕ τîν ΑΓ, ΓΒ περιεχοµένJ ÑρθογωνίJ µετ¦ τοà ¢πÕ τÁς ΒΓ τετραγώνου. 'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ ØπÕ τÁς Óλης κሠ˜νÕς τîν τµηµάτων περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù τε ØπÕ τîν τµηµάτων περιεχοµένJ ÑρθογωνίJ κሠτù ¢πÕ τοà προειρηµένου τµήµατος τετραγώνJ· Óπερ œδει δε‹ξαι.

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For let the straight-line AB have been cut, at random, at (point) C. I say that the rectangle contained by AB and BC is equal to the rectangle contained by AC and CB, plus the square on BC. For let the square CDEB have been described on CB [Prop. 1.46], and let ED have been drawn through to F , and let AF have been drawn through A, parallel to either of CD or BE [Prop. 1.31]. So the (rectangle) AE is equal to the (rectangle) AD and the (square) CE. And AE is the rectangle contained by AB and BC. For it is contained by AB and BE, and BE (is) equal to BC. And AD (is) the (rectangle contained) by AC and CB. For DC (is) equal to CB. And DB (is) the square on CB. Thus, the rectangle contained by AB and BC is equal to the rectangle contained by AC and CB, plus the square on BC. Thus, if a straight-line is cut at random, then the rectangle contained by the whole (straight-line), and one of the pieces (of the straight-line), is equal to the rectangle contained by (both of) the pieces, and the square on the aforementioned piece. (Which is) the very thing it was required to show.

This proposition is a geometric version of the algebraic identity: (a + b) a = a b + a2 .

δ΄.

Proposition 4†

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ ¢πÕ τÁς Óλης τετράγωνον ‡σον ™στˆ το‹ς τε ¢πÕ τîν τµηµάτων τετραγώνοις κሠτù δˆς ØπÕ τîν τµηµάτων περιεχοµένJ

If a straight-line is cut at random, then the square on the whole (straight-line) is equal to the (sum of the) squares on the pieces (of the straight-line), and twice the

52

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

ÑρθογωνίJ.

rectangle contained by the pieces.

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ΕÙθε‹α γ¦ρ γραµµ¾ ¹ ΑΒ τετµήσθω, æς œτυχεν, κατ¦ τÕ Γ. λέγω, Óτι τÕ ¢πÕ τÁς ΑΒ τετράγωνον ‡σον ™στˆ το‹ς τε ¢πÕ τîν ΑΓ, ΓΒ τετραγώνοις κሠτù δˆς ØπÕ τîν ΑΓ, ΓΒ περιεχοµένJ ÑρθογωνίJ. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ Α∆ΕΒ, κሠ™πεζεύχθω ¹ Β∆, κሠδι¦ µν τοà Γ Ðπορέρv τîν Α∆, ΕΒ παράλληλος ½χθω ¹ ΓΖ, δι¦ δ τοà Η Ðποτέρv τîν ΑΒ, ∆Ε παράλληλος ½χθω ¹ ΘΚ. κሠ™πεˆ παράλληλός ™στιν ¹ ΓΖ τÍ Α∆, κሠε„ς αÙτ¦ς ™µπέπτωκεν ¹ Β∆, ¹ ™κτÕς γωνία ¹ ØπÕ ΓΗΒ ‡ση ™στˆ τÍ ™ντÕς κሠ¢πεναντίον τÍ ØπÕ Α∆Β. ¢λλ' ¹ ØπÕ Α∆Β τÍ ØπÕ ΑΒ∆ ™στιν ‡ση, ™πεˆ κሠπλευρ¦ ¹ ΒΑ τÍ Α∆ ™στιν ‡ση· κሠ¹ ØπÕ ΓΗΒ ¥ρα γωνιά τÍ ØπÕ ΗΒΓ ™στιν ‡ση· éστε κሠπλευρ¦ ¹ ΒΓ πλευρ´ τÍ ΓΗ ™στιν ‡ση· ¢λλ' ¹ µν ΓΒ τÍ ΗΚ ™στιν ‡ση. ¹ δ ΓΗ τÍ ΚΒ· κሠ¹ ΗΚ ¥ρα τÍ ΚΒ ™στιν ‡ση· „σόπλευρον ¥ρα ™στˆ τÕ ΓΗΚΒ. λέγω δή, Óτι κሠÑρθογώνιον. ™πεˆ γ¦ρ παράλληλός ™στιν ¹ ΓΗ τÍ ΒΚ [κሠε„ς αÙτ¦ς ™µπέπτωκεν εÙθε‹α ¹ ΓΒ], αƒ ¥ρα ØπÕ ΚΒΓ, ΗΓΒ γωνίαι δύο Ñρθα‹ς ε„σιν ‡σαι. Ñρθ¾ δ ¹ ØπÕ ΚΒΓ· Ñρθ¾ ¥ρα κሠ¹ ØπÕ ΒΓΗ· éστε καˆ αƒ ¢πεναντίον αƒ ØπÕ ΓΗΚ, ΗΚΒ Ñρθαί ε„σιν. Ñρθογώνιον ¥ρα ™στˆ τÕ ΓΗΚΒ· ™δείχθη δ κሠ„σόπλευρον· τετράγωνον ¥ρα ™στίν· καί ™στιν ¢πÕ τÁς ΓΒ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ΘΖ τετράγωνόν ™στιν· καί ™στιν ¢πÕ τÁς ΘΗ, τουτέστιν [¢πÕ] τÁς ΑΓ· τ¦ ¥ρα ΘΖ, ΚΓ τετράγωνα ¢πÕ τîν ΑΓ, ΓΒ ε„σιν. κሠ™πεˆ ‡σον ™στˆ τÕ ΑΗ τù ΗΕ, καί ™στι τÕ ΑΗ τÕ ØπÕ τîν ΑΓ, ΓΒ· ‡ση γ¦ρ ¹ ΗΓ τÍ ΓΒ· κሠτÕ ΗΕ ¥ρα ‡σον ™στˆ τù ØπÕ ΑΓ, ΓΒ· τ¦ ¥ρα ΑΗ, ΗΕ ‡σα ™στˆ τù δˆς ØπÕ τîν ΑΓ, ΓΒ. œστι δ κሠτ¦ ΘΖ, ΓΚ τετράγωνα ¢πÕ τîν ΑΓ, ΓΒ· τ¦ ¥ρα τέσσαρα τ¦ ΘΖ, ΓΚ, ΑΗ, ΗΕ ‡σα ™στˆ το‹ς τε ¢πÕ τîν ΑΓ, ΓΒ τετραγώνοις κሠτù δˆς ØπÕ τîν ΑΓ, ΓΒ περιεχοµένJ ÑρθογωνίJ. ¢λλ¦ τ¦ ΘΖ, ΓΚ, ΑΗ, ΗΕ Óλον ™στˆ τÕ Α∆ΕΒ, Ó ™στιν ¢πÕ τÁς ΑΒ τετράγωνον· τÕ ¥ρα ¢πÕ τÁς ΑΒ τετράγωνον ‡σον ™στˆ το‹ς τε ¢πÕ τîν ΑΓ, ΓΒ τετραγώνοις κሠτù δˆς ØπÕ τîν ΑΓ, ΓΒ περιεχοµένJ

C G

F

B K

E

For let the straight-line AB have been cut, at random, at (point) C. I say that the square on AB is equal to the (sum of the) squares on AC and CB, and twice the rectangle contained by AC and CB. For let the square ADEB have been described on AB [Prop. 1.46], and let BD have been joined, and let CF have been drawn through C, parallel to either of AD or EB [Prop. 1.31], and let HK have been drawn through G, parallel to either of AB or DE [Prop. 1.31]. And since CF is parallel to AD, and BD has fallen across them, the external angle CGB is equal to the internal and opposite (angle) ADB [Prop. 1.29]. But, ADB is equal to ABD, since the side BA is also equal to AD [Prop. 1.5]. Thus, angle CGB is also equal to GBC. So the side BC is equal to the side CG [Prop. 1.6]. But, CB is equal to GK, and CG to KB [Prop. 1.34]. Thus, GK is also equal to KB. Thus, CGKB is equilateral. So I say that (it is) also right-angled. For since CG is parallel to BK [and the straight-line CB has fallen across them], the angles KBC and GCB are thus equal to two right-angles [Prop. 1.29]. But KBC (is) a right-angle. Thus, BCG (is) also a rightangle. So the opposite (angles) CGK and GKB are also right-angles [Prop. 1.34]. Thus, CGKB is right-angled. And it was also shown (to be) equilateral. Thus, it is a square. And it is on CB. So, for the same (reasons), HF is also a square. And it is on HG, that is to say [on] AC [Prop. 1.34]. Thus, the squares HF and KC are on AC and CB (respectively). And the (rectangle) AG is equal to the (rectangle) GE [Prop. 1.43]. And AG is the (rectangle contained) by AC and CB. For CG (is) equal to CB. Thus, GE is also equal to the (rectangle contained) by AC and CB. Thus, the (rectangles) AG and GE are equal to twice the (rectangle contained) by AC and CB. And HF and CK are the squares on AC and CB (respectively). Thus, the four (figures) HF , CK, AG, and GE are equal to the (sum of the) squares on

53

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

ÑρθογωνίJ. 'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ ¢πÕ τÁς Óλης τετράγωνον ‡σον ™στˆ το‹ς τε ¢πÕ τîν τµηµάτων τετραγώνοις κሠτù δˆς ØπÕ τîν τµηµάτων περιεχοµένJ ÑρθογωνίJ· Óπερ œδει δε‹ξαι.

†

AC and BC, and twice the rectangle contained by AC and CB. But, the (figures) HF , CK, AG, and GE are (equivalent to) the whole of ADEB, which is the square on AB. Thus, the square on AB is equal to the (sum of the) squares on AC and CB, and twice the rectangle contained by AC and CB. Thus, if a straight-line is cut at random, then the square on the whole (straight-line) is equal to the (sum of the) squares on the pieces (of the straight-line), and twice the rectangle contained by the pieces. (Which is) the very thing it was required to show.

This proposition is a geometric version of the algebraic identity: (a + b)2 = a2 + b2 + 2 a b.

Proposition 5‡

ε΄.

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ ε„ς ‡σα κሠ¥νισα, τÕ If a straight-line is cut into equal and unequal (pieces), ØπÕ τîν ¢νίσων τÁς Óλης τµηµάτων περιεχόµενον then the rectangle contained by the unequal pieces of the Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς µεταξÝ τîν τοµîν τε- whole (straight-line), plus the square on the difference τραγώνου ‡σον ™στˆ τù ¢πÕ τÁς ¹µισείας τετραγώνJ. between the (equal and unequal) pieces, is equal to the square on half (of the straight-line).

Α

Γ ∆

Β

A

C D

O

Ν

Κ

Μ

Θ

Λ Ξ Ε Η

B

H

Μ

K

Ζ

ΕÙθε‹α γάρ τις ¹ ΑΒ τετµήσθω ε„ς µν ‡σα κατ¦ τÕ Γ, ε„ς δ ¥νισα κατ¦ τÕ ∆· λέγω, Óτι τÕ ØπÕ τîν Α∆, ∆Β περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς Γ∆ τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς ΓΒ τετραγώνJ. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΓΒ τετράγωνον τÕ ΓΕΖΒ, κሠ™πεζεύχθω ¹ ΒΕ, κሠδι¦ µν τοà ∆ Ðποτέρv τîν ΓΕ, ΒΖ παράλληλος ½χθω ¹ ∆Η, δι¦ δ τοà Θ Ðποτέρv τîν ΑΒ, ΕΖ παράλληλος πάλιν ½χθω ¹ ΚΜ, κሠπάλιν δι¦ τοà Α Ðποτέρv τîν ΓΛ, ΒΜ παράλληλος ½χθω ¹ ΑΚ. κሠ™πεˆ ‡σον ™στˆ τÕ ΓΘ παραπλήρωµα τù ΘΖ παραπληρώµατι, κοινÕν προσκείσθω τÕ ∆Μ· Óλον ¥ρα τÕ ΓΜ ÓλJ τù ∆Ζ ‡σον ™στίν. ¢λλ¦ τÕ ΓΜ τù ΑΛ ‡σον ™στίν, ™πεˆ κሠ¹ ΑΓ τÍ ΓΒ ™στιν ‡ση· κሠτÕ ΑΛ ¥ρα τù ∆Ζ ‡σον ™στίν. κοινÕν προσκείσθω τÕ ΓΘ· Óλον ¥ρα τÕ ΑΘ τù ΜΝΞ† γνώµονι ‡σον ™στίν. ¢λλ¦ τÕ ΑΘ τÕ ØπÕ τîν Α∆, ∆Β ™στιν· ‡ση γ¦ρ ¹ ∆Θ τÍ ∆Β· καˆ Ð ΜΝΞ ¥ρα γνώµων ‡σος ™στˆ τù ØπÕ Α∆, ∆Β. κοινÕν προσκείσθω τÕ ΛΗ, Ó ™στιν ‡σον τù ¢πÕ τÁς Γ∆· Ð ¥ρα ΜΝΞ γνώµων κሠτÕ ΛΗ ‡σα ™στˆ τù ØπÕ τîν Α∆, ∆Β

N

L P E G

M F

For let any straight-line AB have been cut—equally at C, and unequally at D. I say that the rectangle contained by AD and DB, plus the square on CD, is equal to the square on CB. For let the square CEF B have been described on CB [Prop. 1.46], and let BE have been joined, and let DG have been drawn through D, parallel to either of CE or BF [Prop. 1.31], and again let KM have been drawn through H, parallel to either of AB or EF [Prop. 1.31], and again let AK have been drawn through A, parallel to either of CL or BM [Prop. 1.31]. And since the complement CH is equal to the complement HF [Prop. 1.43], let the (square) DM have been added to both. Thus, the whole (rectangle) CM is equal to the whole (rectangle) DF . But, (rectangle) CM is equal to (rectangle) AL, since AC is also equal to CB [Prop. 1.36]. Thus, (rectangle) AL is also equal to (rectangle) DF . Let (rectangle) CH have been added to both. Thus, the whole (rectangle) AH is equal to the gnomon N OP . But, AH

54

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

περιεχοµένJ ÑρθογωνίJ κሠτù ¢πÕ τÁς Γ∆ τετραγώνJ. ¢λλ¦ Ð ΜΝΞ γνώµων κሠτÕ ΛΗ Óλον ™στˆ τÕ ΓΕΖΒ τετράγωνον, Ó ™στιν ¢πÕ τÁς ΓΒ· τÕ ¥ρα ØπÕ τîν Α∆, ∆Β περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς Γ∆ τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς ΓΒ τετραγώνJ. 'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµηθÍ ε„ς ‡σα κሠ¥νισα, τÕ ØπÕ τîν ¢νίσων τÁς Óλης τµηµάτων περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς µεταξÝ τîν τοµîν τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς ¹µισείας τετραγώνJ· Óπερ œδει δε‹ξαι.

is the (rectangle contained) by AD and DB. For DH (is) equal to DB. Thus, the gnomon N OP is also equal to the (rectangle contained) by AD and DB. Let LG, which is equal to the (square) on CD, have been added to both. Thus, the gnomon N OP and the (square) LG are equal to the rectangle contained by AD and DB, and the square on CD. But, the gnomon N OP and the (square) LG is (equivalent to) the whole square CEF B, which is on CB. Thus, the rectangle contained by AD and DB, plus the square on CD, is equal to the square on CB. Thus, if a straight-line is cut into equal and unequal (pieces), then the rectangle contained by the unequal pieces of the whole (straight-line), plus the square on the difference between the (equal and unequal) pieces, is equal to the square on half (of the straight-line). (Which is) the very thing it was required to show.

†

Note the (presumably mistaken) double use of the label M in the Greek text.

‡

This proposition is a geometric version of the algebraic identity: a b + [(a + b)/2 − b]2 = [(a + b)/2]2 .

$΄.

Proposition 6†

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ δίχα, προστεθÍ δέ τις αÙτÍ εÙθε‹α ™π' εÙθείας, τÕ ØπÕ τÁς Óλης σÝν τÍ προσκειµένV κሠτÁς προσκειµένης περιεχόµενον Ñρθόγώνιον µετ¦ τοà ¢πÕ τÁς ¹µισείας τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς συγκειµένης œκ τε τÁς ¹µισείας κሠτÁς προσκειµένης τετραγώνJ.

If a straight-line is cut in half, and any straight-line added to it straight-on, then the rectangle contained by the whole (straight-line) with the (straight-line) having being added, and the (straight-line) having being added, plus the square on half (of the original straight-line), is equal to the square on the sum of half (of the original straight-line) and the (straight-line) having been added.

Α

Γ

Β

∆

A

C

B

Ξ

Κ

Λ

Ε

Ν

Θ

H

Μ

K

N

L

Ο

Η

D O

M

P

Ζ

E

ΕÙθε‹α γάρ τις ¹ ΑΒ τετµήσθω δίχα κατ¦ τÕ Γ σηµε‹ον, προσκείσθω δέ τις αÙτÍ εÙθε‹α ™π' εÙθείας ¹ Β∆· λέγω, Óτι τÕ ØπÕ τîν Α∆, ∆Β περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς ΓΒ τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς Γ∆ τετραγώνJ. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς Γ∆ τετράγωνον τÕ ΓΕΖ∆, κሠ™πεζεύχθω ¹ ∆Ε, κሠδι¦ µν τοà Β σηµείου Ðποτέρv τîν ΕΓ, ∆Ζ παράλληλος ½χθω ¹ ΒΗ, δι¦ δ τοà Θ σηµείου Ðποτέρv τîν ΑΒ, ΕΖ παράλληλος ½χθω ¹ ΚΜ, κሠœτι δι¦ τοà Α Ðποτέρv τîν ΓΛ, ∆Μ παράλληλος ½χθω ¹ ΑΚ.

G

F

For let any straight-line AB have been cut in half at point C, and let any straight-line BD have been added to it straight-on. I say that the rectangle contained by AD and DB, plus the square on CB, is equal to the square on CD. For let the square CEF D have been described on CD [Prop. 1.46], and let DE have been joined, and let BG have been drawn through point B, parallel to either of EC or DF [Prop. 1.31], and let KM have been drawn through point H, parallel to either of AB or EF [Prop. 1.31], and finally let AK have been drawn

55

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

'Επεˆ οâν ‡ση ™στˆν ¹ ΑΓ τÍ ΓΒ, ‡σον ™στˆ κሠτÕ ΑΛ τù ΓΘ. ¢λλ¦ τÕ ΓΘ τù ΘΖ ‡σον ™στίν. κሠτÕ ΑΛ ¥ρα τù ΘΖ ™στιν ‡σον. κοινÕν προσκείσθω τÕ ΓΜ· Óλον ¥ρα τÕ ΑΜ τù ΝΞΟ γνώµονί ™στιν ‡σον. ¢λλ¦ τÕ ΑΜ ™στι τÕ ØπÕ τîν Α∆, ∆Β· ‡ση γάρ ™στιν ¹ ∆Μ τÍ ∆Β· καˆ Ð ΝΞΟ ¥ρα γνώµων ‡σος ™στˆ τù ØπÕ τîν Α∆, ∆Β [περιεχοµένJ ÑρθογωνίJ]. κοινÕν προσκείσθω τÕ ΛΗ, Ó ™στιν ‡σον τù ¢πÕ τÁς ΒΓ τετραγώνJ· τÕ ¥ρα ØπÕ τîν Α∆, ∆Β περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς ΓΒ τετραγώνου ‡σον ™στˆ τù ΝΞΟ γνώµονι κሠτù ΛΗ. ¢λλ¦ Ð ΝΞΟ γνώµων κሠτÕ ΛΗ Óλον ™στˆ τÕ ΓΕΖ∆ τετράγωνον, Ó ™στιν ¢πÕ τÁς Γ∆· τÕ ¥ρα ØπÕ τîν Α∆, ∆Β περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς ΓΒ τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς Γ∆ τετραγώνJ. 'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµηθÍ δίχα, προστεθÍ δέ τις αÙτÍ εÙθε‹α ™π' εÙθείας, τÕ ØπÕ τÁς Óλης σÝν τÍ προσκειµένV κሠτÁς προσκειµένης περιεχόµενον Ñρθόγώνιον µετ¦ τοà ¢πÕ τÁς ¹µισείας τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς συγκειµένης œκ τε τÁς ¹µισείας κሠτÁς προσκειµένης τετραγώνJ· Óπερ œδει δε‹ξαι.

†

through A, parallel to either of CL or DM [Prop. 1.31]. Therefore, since AC is equal to CB, (rectangle) AL is also equal to (rectangle) CH [Prop. 1.36]. But, (rectangle) CH is equal to (rectangle) HF [Prop. 1.43]. Thus, (rectangle) AL is also equal to (rectangle) HF . Let (rectangle) CM have been added to both. Thus, the whole (rectangle) AM is equal to the gnomon N OP . But, AM is the (rectangle contained) by AD and DB. For DM is equal to DB. Thus, gnomon N OP is also equal to the [rectangle contained] by AD and DB. Let LG, which is equal to the square on BC, have been added to both. Thus, the rectangle contained by AD and DB, plus the square on CB, is equal to the gnomon N OP , and the (square) LG. But the gnomon N OP and the (square) LG is (equivalent to) the whole square CEF D, which is on CD. Thus, the rectangle contained by AD and DB, plus the square on CB, is equal to the square on CD. Thus, if a straight-line is cut in half, and any straightline added to it straight-on, then the rectangle contained by the whole (straight-line) with the (straight-line) having being added, and the (straight-line) having being added, plus the square on half (of the original straightline), is equal to the square on the sum of half (of the original straight-line) and the (straight-line) having been added. (Which is) the very thing it was required to show.

This proposition is a geometric version of the algebraic identity: (2 a + b) b + a2 = (a + b)2 .

ζ΄.

Proposition 7†

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ ¢πÕ τÁς Óλης κሠτÕ ¢φ' ˜νÕς τîν τµηµάτων τ¦ συναµφότερα τετράγωνα ‡σα ™στˆ τù τε δˆς ØπÕ τÁς Óλης κሠτοà ε„ρηµένου τµήµατος περιεχοµένJ ÑρθογωνίJ κሠτù ¢πÕ τοà λοιποà τµήµατος τετραγώνJ.

If a straight-line is cut at random, then the sum of the squares on the whole (straight-line), and one of the pieces (of the straight-line), is equal to twice the rectangle contained by the whole, and the said piece, and the square on the remaining piece.

Α

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Β

A

C

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Η

Κ

L

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H

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∆

Ν

B

K

G

F

M

Ε

D

N

E

ΕÙθε‹α γάρ τις ¹ ΑΒ τετµήσθω, æς œτυχεν, κατ¦ τÕ For let any straight-line AB have been cut, at random, Γ σηµε‹ον· λέγω, Óτι τ¦ ¢πÕ τîν ΑΒ, ΒΓ τετράγωνα ‡σα at point C. I say that the (sum of the) squares on AB and ™στˆ τù τε δˆς ØπÕ τîν ΑΒ, ΒΓ περιεχοµένJ ÑρθογωνίJ BC is equal to twice the rectangle contained by AB and 56

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

κሠτù ¢πÕ τÁς ΓΑ τετραγώνJ. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ Α∆ΕΒ· κሠκαταγεγράφθω τÕ σχÁµα. 'Επεˆ οâν ‡σον ™στˆ τÕ ΑΗ τù ΗΕ, κοινÕν προσκείσθω τÕ ΓΖ· Óλον ¥ρα τÕ ΑΖ ÓλJ τù ΓΕ ‡σον ™στίν· τ¦ ¥ρα ΑΖ, ΓΕ διπλάσιά ™στι τοà ΑΖ. ¢λλ¦ τ¦ ΑΖ, ΓΕ Ð ΚΛΜ ™στι γνώµων κሠτÕ ΓΖ τετράγωνον· Ð ΚΛΜ ¥ρα γνώµων κሠτÕ ΓΖ διπλάσιά ™στι τοà ΑΖ. œστι δ τοà ΑΖ διπλάσιον κሠτÕ δˆς ØπÕ τîν ΑΒ, ΒΓ· ‡ση γ¦ρ ¹ ΒΖ τÍ ΒΓ· Ð ¥ρα ΚΛΜ γνώµων κሠτÕ ΓΖ τετράγωνον ‡σον ™στˆ τù δˆς ØπÕ τîν ΑΒ, ΒΓ. κοινÕν προσκείσθω τÕ ∆Η, Ó ™στιν ¢πÕ τÁς ΑΓ τετράγωνον· Ð ¥ρα ΚΛΜ γνώµων κሠτ¦ ΒΗ, Η∆ τετράγωνα ‡σα ™στˆ τù τε δˆς ØπÕ τîν ΑΒ, ΒΓ περιεχοµένJ ÑρθογωνίJ κሠτù ¢πÕ τÁς ΑΓ τετραγώνJ. ¢λλ¦ Ð ΚΛΜ γνώµων κሠτ¦ ΒΗ, Η∆ τετράγωνα Óλον ™στˆ τÕ Α∆ΕΒ κሠτÕ ΓΖ, ¤ ™στιν ¢πÕ τîν ΑΒ, ΒΓ τετράγωνα· τ¦ ¥ρα ¢πÕ τîν ΑΒ, ΒΓ τετράγωνα ‡σα ™στˆ τù [τε] δˆς ØπÕ τîν ΑΒ, ΒΓ περιεχοµένJ ÑρθογωνίJ µετ¦ τοà ¢πÕ τÁς ΑΓ τετραγώνου. 'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ ¢πÕ τÁς Óλης κሠτÕ ¢φ' ˜νÕς τîν τµηµάτων τ¦ συναµφότερα τετράγωνα ‡σα ™στˆ τù τε δˆς ØπÕ τÁς Óλης κሠτοà ε„ρηµένου τµήµατος περιεχοµένJ ÑρθογωνίJ κሠτù ¢πÕ τοà λοιποà τµήµατος τετραγώνJ· Óπερ œδει δε‹ξαι.

†

BC, and the square on CA. For let the square ADEB have been described on AB [Prop. 1.46], and let the (rest of) the figure have been drawn. Therefore, since (rectangle) AG is equal to (rectangle) GE [Prop. 1.43], let the (square) CF have been added to both. Thus, the whole (rectangle) AF is equal to the whole (rectangle) CE. Thus, (rectangle) AF plus (rectangle) CE is double (rectangle) AF . But, (rectangle) AF plus (rectangle) CE is the gnomon KLM , and the square CF . Thus, the gnomon KLM , and the square CF , is double the (rectangle) AF . But double the (rectangle) AF is also twice the (rectangle contained) by AB and BC. For BF (is) equal to BC. Thus, the gnomon KLM , and the square CF , are equal to twice the (rectangle contained) by AB and BC. Let DG, which is the square on AC, have been added to both. Thus, the gnomon KLM , and the squares BG and GD, are equal to twice the rectangle contained by AB and BC, and the square on AC. But, the gnomon KLM and the squares BG and GD is (equivalent to) the whole of ADEB and CF , which are the squares on AB and BC (respectively). Thus, the (sum of the) squares on AB and BC is equal to twice the rectangle contained by AB and BC, and the square on AC. Thus, if a straight-line is cut at random, then the sum of the squares on the whole (straight-line), and one of the pieces (of the straight-line), is equal to twice the rectangle contained by the whole, and the said piece, and the square on the remaining piece. (Which is) the very thing it was required to show.

This proposition is a geometric version of the algebraic identity: (a + b)2 + a2 = 2 (a + b) a + b2 .

η΄.

Proposition 8†

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ τετράκις ØπÕ τÁς Óλης κሠ˜νÕς τîν τµηµάτων περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τοà λοιποà τµήµατος τετραγώνου ‡σον ™στˆ τù ¢πό τε τÁς Óλης κሠτοà ε„ρηµένου τµήµατος æς ¢πÕ µι©ς ¢ναγραφέντι τετραγώνJ. ΕÙθε‹α γάρ τις ¹ ΑΒ τετµήσθω, æς œτυχεν, κατ¦ τÕ Γ σηµε‹ον· λέγω, Óτι τÕ τετράκις ØπÕ τîν ΑΒ, ΒΓ περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς ΑΓ τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ, ΒΓ æς ¢πÕ µι©ς ¢ναγραφέντι τετραγώνJ. 'Εκβεβλήσθω γ¦ρ ™π' εÙθείας [τÍ ΑΒ εÙθε‹α] ¹ Β∆, κሠκείσθω τÍ ΓΒ ‡ση ¹ Β∆, κሠ¢ναγεγράφθω ¢πÕ τÁς Α∆ τετράγωνον τÕ ΑΕΖ∆, κሠκαταγεγράφθω διπλοàν τÕ σχÁµα.

If a straight-line is cut at random, then four times the rectangle contained by the whole (straight-line), and one of the pieces (of the straight-line), plus the square on the remaining piece, is equal to the square described on the whole and the former piece, as on one (complete straightline). For let any straight-line AB have been cut, at random, at point C. I say that four times the rectangle contained by AB and BC, plus the square on AC, is equal to the square described on AB and BC, as on one (complete straight-line). For let BD have been produced in a straight-line [with the straight-line AB], and let BD be made equal to BC [Prop. 1.3], and let the square AEF D have been described on AD [Prop. 1.46], and let the (rest of the) figure have been drawn double.

57

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

Α

Γ

Β

∆

A

C

B

Τ

Η

Μ Ξ

Σ Υ

Ε

Π

Θ

Κ Ρ

Λ

D T

Ν

M

Ο

O

G K S

Q

R

N P

U

Ζ

E

'Επεˆ οâν ‡ση ™στˆν ¹ ΓΒ τÍ Β∆, ¢λλ¦ ¹ µν ΓΒ τÍ ΗΚ ™στιν ‡ση, ¹ δ Β∆ τÍ ΚΝ, κሠ¹ ΗΚ ¥ρα τÍ ΚΝ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΠΡ τÍ ΡΟ ™στιν ‡ση. κሠ™πεˆ ‡ση ™στˆν ¹ ΒΓ τÊ Β∆, ¹ δ ΗΚ τÍ ΚΝ, ‡σον ¥ρα ™στˆ κሠτÕ µν ΓΚ τù Κ∆, τÕ δ ΗΡ τù ΡΝ. ¢λλ¦ τÕ ΓΚ τù ΡΝ ™στιν ‡σον· παραπληρώµατα γ¦ρ τοà ΓΟ παραλληλογράµµου· κሠτÕ Κ∆ ¥ρα τù ΗΡ ‡σον ™στίν· τ¦ τέσσαρα ¥ρα τ¦ ∆Κ, ΓΚ, ΗΡ, ΡΝ ‡σα ¢λλήλοις ™στίν. τ¦ τέσσαρα ¥ρα τετραπλάσιά ™στι τοà ΓΚ. πάλιν ™πεˆ ‡ση ™στˆν ¹ ΓΒ τÍ Β∆, ¢λλ¦ ¹ µν Β∆ τÍ ΒΚ, τουτέστι τÍ ΓΗ ‡ση, ¹ δ ΓΒ τÍ ΗΚ, τουτέστι τÍ ΗΠ, ™στιν ‡ση, κሠ¹ ΓΗ ¥ρα τÍ ΗΠ ‡ση ™στίν. κሠ™πεˆ ‡ση ™στˆν ¹ µν ΓΗ τÍ ΗΠ, ¹ δ ΠΡ τÍ ΡΟ, ‡σον ™στˆ κሠτÕ µν ΑΗ τù ΜΠ, τÕ δ ΠΛ τù ΡΖ. ¢λλ¦ τÕ ΜΠ τù ΠΛ ™στιν ‡σον· παραπληρώµατα γ¦ρ τοà ΜΛ παραλληλογράµµου· κሠτÕ ΑΗ ¥ρα τù ΡΖ ‡σον ™στίν· τ¦ τέσσαρα ¥ρα τ¦ ΑΗ, ΜΠ, ΠΛ, ΡΖ ‡σα ¢λλήλοις ™στίν· τ¦ τέσσαρα ¥ρα τοà ΑΗ ™στι τετραπλάσια. ™δείχθη δ κሠτ¦ τέσσαρα τ¦ ΓΚ, Κ∆, ΗΡ, ΡΝ τοà ΓΚ τετραπλάσια· τ¦ ¥ρα Ñκτώ, § περιέχει τÕν ΣΤΥ γνώµονα, τετραπλάσιά ™στι τοà ΑΚ. κሠ™πεˆ τÕ ΑΚ τÕ ØπÕ τîν ΑΒ, Β∆ ™στιν· ‡ση γ¦ρ ¹ ΒΚ τÍ Β∆· τÕ ¥ρα τετράκις ØπÕ τîν ΑΒ, Β∆ τετραπλάσιόν ™στι τοà ΑΚ. ™δείχθη δ τοà ΑΚ τετραπλάσιος καˆ Ð ΣΤΥ γνώµων· τÕ ¥ρα τετράκις ØπÕ τîν ΑΒ, Β∆ ‡σον ™στˆ τù ΣΤΥ γνώµονι. κοινÕν προσκείσθω τÕ ΞΘ, Ó ™στιν ‡σον τù ¢πÕ τÁς ΑΓ τετραγώνJ· τÕ ¥ρα τετράκις ØπÕ τîν ΑΒ, Β∆ περιεχόµενων Ñρθογώνιον µετ¦ τοà ¢πÕ ΑΓ τετραγώνου ‡σον ™στˆ τù ΣΤΥ γνώµονι κሠτù ΞΘ. ¢λλ¦ Ð ΣΤΥ γνώµων κሠτÕ ΞΘ Óλον ™στˆ τÕ ΑΕΖ∆ τετραγώνον, Ó ™στιν ¢πÕ τÁς Α∆· τÕ ¥ρα τετράκις ØπÕ τîν ΑΒ, Β∆ µετ¦ τοà ¢πÕ ΑΓ ‡σον ™στˆ τù ¢πÕ Α∆ τετραγώνJ· ‡ση δ ¹ Β∆ τÍ ΒΓ. τÕ ¥ρα τετράκις ØπÕ τîν ΑΒ, ΒΓ περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ ΑΓ τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς Α∆, τουτέστι τù ¢πÕ τÁς ΑΒ κሠΒΓ æς ¢πÕ µι©ς ¢ναγραφέντι τετραγώνJ.

H

L

F

Therefore, since CB is equal to BD, but CB is equal to GK [Prop. 1.34], and BD to KN [Prop. 1.34], GK is thus also equal to KN . So, for the same (reasons), QR is equal to RP . And since BC is equal to BD, and GK to KN , (square) CK is thus also equal to (square) KD, and (square) GR to (square) RN [Prop. 1.36]. But, (square) CK is equal to (square) RN . For (they are) complements in the parallelogram CP [Prop. 1.43]. Thus, (square) KD is also equal to (square) GR. Thus, the four (squares) DK, CK, GR, and RN are equal to one another. Thus, the four (taken together) are quadruple (square) CK. Again, since CB is equal to BD, but BD (is) equal to BK—that is to say, CG—and CB is equal to GK—that is to say, GQ—CG is thus also equal to GQ. And since CG is equal to GQ, and QR to RP , (rectangle) AG is also equal to (rectangle) M Q, and (rectangle) QL to (rectangle) RF [Prop. 1.36]. But, (rectangle) M Q is equal to (rectangle) QL. For (they are) complements in the parallelogram M L [Prop. 1.43]. Thus, (rectangle) AG is also equal to (rectangle) RF . Thus, the four (rectangles) AG, M Q, QL, and RF are equal to one another. Thus, the four (taken together) are quadruple (rectangle) AG. And it was also shown that the four (squares) DK, CK, GR, and RN (taken together are) quadruple (square) CK. Thus, the eight (figures taken together), which comprise the gnomon ST U , are quadruple (rectangle) AK. And since AK is the (rectangle contained) by AB and BD, for BK (is) equal to BD, four times the (rectangle contained) by AB and BD is quadruple (rectangle) AK. But quadruple (rectangle) AK was also shown (to be equal to) the gnomon ST U . Thus, four times the (rectangle contained) by AB and BD is equal to the gnomon ST U . Let OH, which is equal to the square on AC, have been added to both. Thus, four times the rectangle contained by AB and BD, plus the square on AC, is equal to the gnomon ST U , and the

58

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµηθÍ, æς œτυχεν, τÕ τετράκις ØπÕ τÁς Óλης κሠ˜νÕς τîν τµηµάτων περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τοà λοιποà τµήµατος τετραγώνου ‡σου ™στˆ τù ¢πό τε τÁς Óλης κሠτοà ε„ρηµένου τµήµατος æς ¢πÕ µι©ς ¢ναγραφέντι τετραγώνJ· Óπερ œδει δε‹ξαι.

†

(square) OH. But, the gnomon ST U and the (square) OH is (equivalent to) the whole square AEF D, which is on AD. Thus, four times the (rectangle contained) by AB and BD, plus the (square) on AC, is equal to the square on AD. And BD (is) equal to BC. Thus, four times the rectangle contained by AB and BD, plus the square on AC, is equal to the (square) on AD, that is to say the square described on AB and BC, as on one (complete straight-line). Thus, if a straight-line is cut at random, then four times the rectangle contained by the whole (straightline), and one of the pieces (of the straight-line), plus the square on the remaining piece, is equal to the square described on the whole and the former piece, as on one (complete straight-line). (Which is) the very thing it was required to show.

This proposition is a geometric version of the algebraic identity: 4 (a + b) a + b2 = [(a + b) + a]2 .

Proposition 9†

θ΄.

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ ε„ς ‡σα κሠ¥νισα, τ¦ ¢πÕ If a straight-line is cut into equal and unequal (pieces), τîν ¢νίσων τÁς Óλης τµηµάτων τετράγωνα διπλάσιά ™στι then the (sum of the) squares on the unequal pieces of the τοà τε ¢πÕ τÁς ¹µισείας κሠτοà ¢πÕ τÁς µεταξÝ τîν whole (straight-line) is double the (sum of the) square on τοµîν τετραγώνου. half (the straight-line), and (the square) on the difference between the (equal and unequal) pieces.

Ε Η

Α

Γ

E

Ζ

∆

G

Β

A

ΕÙθε‹α γάρ τις ¹ ΑΒ τετµήσθω ε„ς µν ‡σα κατ¦ τÕ Γ, εƒς δ ¥νισα κατ¦ τÕ ∆· λέγω, Óτι τ¦ ¢πÕ τîν Α∆, ∆Β τετράγωνα διπλάσιά ™στι τîν ¢πÕ τîν ΑΓ, Γ∆ τετραγώνων. ”Ηχθω γ¦ρ ¢πÕ τοà Γ τÍ ΑΒ πρÕς Ñρθ¦ς ¹ ΓΕ, κሠκείσθω ‡ση ˜κατέρv τîν ΑΓ, ΓΒ, κሠ™πεζεύχθωσαν αƒ ΕΑ, ΕΒ, κሠδι¦ µν τοà ∆ τÍ ΕΓ παράλληλος ½χθω ¹ ∆Ζ, δι¦ δ τοà Ζ τÍ ΑΒ ¹ ΖΗ, κሠ™πεζεύχθω ¹ ΑΖ. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΓ τÍ ΓΕ, ‡ση ™στˆ κሠ¹ ØπÕ ΕΑΓ γωνία τÍ ØπÕ ΑΕΓ. κሠ™πεˆ Ñρθή ™στιν ¹ πρÕς τù Γ, λοιπሠ¥ρα αƒ ØπÕ ΕΑΓ, ΑΕΓ µι´ ÑρθÍ ‡σαι ε„σίν· καί ε„σιν ‡σαι· ¹µίσεια ¥ρα ÑρθÁς ™στιν ˜κατέρα τîν ØπÕ ΓΕΑ, ΓΑΕ. δˆα τ¦ αÙτ¦ δ¾ κሠ˜κατέρα τîν ØπÕ ΓΕΒ, ΕΒΓ ¹µίσειά ™στιν ÑρθÁς· Óλη ¥ρα ¹ ØπÕ ΑΕΒ Ñρθή

C

F

D

B

For let any straight-line AB have been cut—equally at C, and unequally at D. I say that the (sum of the) squares on AD and DB is double the (sum of the squares) on AC and CD. For let CE have been drawn from (point) C, at rightangles to AB [Prop. 1.11], and let it be made equal to each of AC and CB [Prop. 1.3], and let EA and EB have been joined. And let DF have been drawn through (point) D, parallel to EC [Prop. 1.31], and (let) F G (have been drawn) through (point) F , (parallel) to AB [Prop. 1.31]. And let AF have been joined. And since AC is equal to CE, the angle EAC is also equal to the (angle) AEC [Prop. 1.5]. And since the (angle) at C is a right-angle, the (sum of the) remaining angles (of tri-

59

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

™στιν. κሠ™πεˆ ¹ ØπÕ ΗΕΖ ¹µίσειά ™στιν ÑρθÁς, Ñρθ¾ δ ¹ ØπÕ ΕΗΖ· ‡ση γάρ ™στι τÍ ™ντÕς κሠ¢πεναντίον τÍ ØπÕ ΕΓΒ· λοιπ¾ ¥ρα ¹ ØπÕ ΕΖΗ ¹µίσειά ™στιν ÑρθÁς· ‡ση ¥ρα [™στˆν] ¹ ØπÕ ΗΕΖ γωνία τÍ ØπÕ ΕΖΗ· éστε κሠπλευρ¦ ¹ ΕΗ τÍ ΗΖ ™στιν ‡ση. πάλιν ™πεˆ ¹ πρÕς τù Β γωνία ¹µίσειά ™στιν ÑρθÁς, Ñρθ¾ δ ¹ ØπÕ Ζ∆Β· ‡ση γ¦ρ πάλιν ™στˆ τÍ ™ντÕς κሠ¢πεναντίον τÍ ØπÕ ΕΓΒ· λοιπ¾ ¥ρα ¹ ØπÕ ΒΖ∆ ¹µίσειά ™στιν ÑρθÁς· ‡ση ¥ρα ¹ πρÕς τù Β γωνία τÍ ØπÕ ∆ΖΒ· éστε κሠπλευρ¦ ¹ Ζ∆ πλευρ´ τÍ ∆Β ™στιν ‡ση. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΓ τÍ ΓΕ, ‡σον ™στˆ κሠτÕ ¢πÕ ΑΓ τù ¢πÕ ΓΕ· τ¦ ¥ρα ¢πÕ τîν ΑΓ, ΓΕ τετράγωνα διπλάσιά ™στι τοà ¢πÕ ΑΓ. το‹ς δ ¢πÕ τîν ΑΓ, ΓΕ ‡σον ™στˆ τÕ ¢πÕ τÁς ΕΑ τετράγωνον· Ñρθ¾ γ¦ρ ¹ ØπÕ ΑΓΕ γωνία· τÕ ¥ρα ¢πÕ τÁς ΕΑ διπλάσιόν ™στι τοà ¢πÕ τÁς ΑΓ. πάλιν, ™πεˆ ‡ση ™στˆν ¹ ΕΗ τÍ ΗΖ, ‡σον κሠτÕ ¢πÕ τÁς ΕΗ τù ¢πÕ τÁς ΗΖ· τ¦ ¥ρα ¢πÕ τîν ΕΗ, ΗΖ τετράγωνα διπλάσιά ™στι τοà ¢πÕ τÁς ΗΖ τετραγώνου. το‹ς δ ¢πÕ τîν ΕΗ, ΗΖ τετραγώνοις ‡σον ™στˆ τÕ ¢πÕ τÁς ΕΖ τετράγωνον· τÕ ¥ρα ¢πÕ τÁς ΕΖ τετράγωνον διπλάσιόν ™στι τοà ¢πÕ τÁς ΗΖ. ‡ση δ ¹ ΗΖ τÍ Γ∆· τÕ ¥ρα ¢πÕ τÁς ΕΖ διπλάσιόν ™στι τοà ¢πÕ τÁς Γ∆. œστι δ κሠτÕ ¢πÕ τÁς ΕΑ διπλάσιον τοà ¢πÕ τÁς ΑΓ· τ¦ ¥ρα ¢πÕ τîν ΑΕ, ΕΖ τετράγωνα διπλάσιά ™στι τîν ¢πÕ τîν ΑΓ, Γ∆ τετραγώνων. το‹ς δ ¢πÕ τîν ΑΕ, ΕΖ ‡σον ™στˆ τÕ ¢πÕ τÁς ΑΖ τετράγωνον· Ñρθ¾ γάρ ™στιν ¹ ØπÕ ΑΕΖ γωνία· τÕ ¥ρα ¢πÕ τÁς ΑΖ τετράγωνον διπλάσιόν ™στι τîν ¢πÕ τîν ΑΓ, Γ∆. τù δ ¢πÕ τÁς ΑΖ ‡σα τ¦ ¢πÕ τîν Α∆, ∆Ζ· Ñρθ¾ γ¦ρ ¹ πρÕς τù ∆ γωνία· τ¦ ¥ρα ¢πÕ τîν Α∆, ∆Ζ διπλάσιά ™στι τîν ¢πÕ τîν ΑΓ, Γ∆ τετραγώνων. ‡ση δ ¹ ∆Ζ τÍ ∆Β· τ¦ ¥ρα ¢πÕ τîν Α∆, ∆Β τετράγωνα διπλάσιά ™στι τîν ¢πÕ τîν ΑΓ, Γ∆ τετράγώνων. 'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµηθÍ ε„ς ‡σα κሠ¥νισα, τ¦ ¢πÕ τîν ¢νίσων τÁς Óλης τµηµάτων τετράγωνα διπλάσιά ™στι τοà τε ¢πÕ τÁς ¹µισείας κሠτοà ¢πÕ τÁς µεταξÝ τîν τοµîν τετραγώνου· Óπερ œδει δε‹ξαι.

angle AEC), EAC and AEC, is thus equal to one rightangle [Prop. 1.32]. And they are equal. Thus, (angles) CEA and CAE are each half a right-angle. So, for the same (reasons), (angles) CEB and EBC are also each half a right-angle. Thus, the whole (angle) AEB is a right-angle. And since GEF is half a right-angle, and EGF (is) a right-angle—for it is equal to the internal and opposite (angle) ECB [Prop. 1.29]—the remaining (angle) EF G is thus half a right-angle [Prop. 1.32]. Thus, angle GEF [is] equal to EF G. So the side EG is also equal to the (side) GF [Prop. 1.6]. Again, since the angle at B is half a right-angle, and (angle) F DB (is) a right-angle—for again it is equal to the internal and opposite (angle) ECB [Prop. 1.29]—the remaining (angle) BF D is half a right-angle [Prop. 1.32]. Thus, the angle at B (is) equal to DF B. So the side F D is also equal to the side DB [Prop. 1.6]. And since AC is equal to CE, the (square) on AC (is) also equal to the (square) on CE. Thus, the (sum of the) squares on AC and CE is double the (square) on AC. And the square on EA is equal to the (sum of the) squares on AC and CE. For angle ACE (is) a right-angle [Prop. 1.47]. Thus, the (square) on EA is double the (square) on AC. Again, since EG is equal to GF , the (square) on EG (is) also equal to the (square) on GF . Thus, the (sum of the squares) on EG and GF is double the square on GF . And the square on EF is equal to the (sum of the) squares on EG and GF [Prop. 1.47]. Thus, the square on EF is double the (square) on GF . And GF (is) equal to CD [Prop. 1.34]. Thus, the (square) on EF is double the (square) on CD. And the (square) on EA is also double the (square) on AC. Thus, the (sum of the) squares on AE and EF is double the (sum of the) squares on AC and CD. And the square on AF is equal to the (sum of the squares) on AE and EF . For the angle AEF is a right-angle [Prop. 1.47]. Thus, the square on AF is double the (sum of the squares) on AC and CD. And the (sum of the squares) on AD and DF (is) equal to the (square) on AF . For the angle at D is a right-angle [Prop. 1.47]. Thus, the (sum of the squares) on AD and DF is double the (sum of the) squares on AC and CD. And DF (is) equal to DB. Thus, the (sum of the) squares on AD and DB is double the (sum of the) squares on AC and CD. Thus, if a straight-line is cut into equal and unequal (pieces), then the (sum of the) squares on the unequal pieces of the whole (straight-line) is double the (sum of the) square on half (the straight-line), and (the square) on the difference between the (equal and unequal) pieces. (Which is) the very thing it was required to show.

60

ΣΤΟΙΧΕΙΩΝ β΄. †

ELEMENTS BOOK 2

This proposition is a geometric version of the algebraic identity: a2 + b2 = 2[([a + b]/2)2 + ([a + b]/2 − b)2 ].

ι΄.

Proposition 10†

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ δίχα, προστεθÍ δέ τις αÙτÍ εÙθε‹α ™π' εÙθείας, τÕ ¢πÕ τÁς Óλης σÝν τÍ προσκειµένV κሠτÕ ¢πÕ τÁς προσκειµένης τ¦ συναµφότερα τετράγωνα διπλάσιά ™στι τοà τε ¢πÕ τÁς ¹µισείας κሠτοà ¢πÕ τÁς συγκειµένης œκ τε τÁς ¹µισείας κሠτÁς προσκειµένης æς ¢πÕ µι©ς ¢ναγραφέντος τετραγώνου.

If a straight-line is cut in half, and any straight-line added to it straight-on, then the sum of the square on the whole (straight-line) with the (straight-line) having been added, and the (square) on the (straight-line) having been added, is double the (sum of the square) on half (the straight-line), and the square described on the sum of half (the straight-line) and (straight-line) having been added, as on one (complete straight-line).

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ΕÙθε‹α γάρ τις ¹ ΑΒ τετµήσθω δίχα κατ¦ τÕ Γ, προσκείσθω δέ τις αÙτÍ εÙθε‹α ™π' εÙθείας ¹ Β∆· λέγω, Óτι τ¦ ¢πÕ τîν Α∆, ∆Β τετράγωνα διπλάσιά ™στι τîν ¢πÕ τîν ΑΓ, Γ∆ τετραγώνων. ”Ηχθω γ¦ρ ¢πÕ τοà Γ σηµείου τÍ ΑΒ πρÕς Ñρθ¦ς ¹ ΓΕ, κሠκείσθω ‡ση ˜κατέρv τîν ΑΓ, ΓΒ, κሠ™πεζεύχθωσαν αƒ ΕΑ, ΕΒ· κሠδι¦ µν τοà Ε τÍ Α∆ παράλληλος ½χθω ¹ ΕΖ, δι¦ δ τοà ∆ τÍ ΓΕ παράλληλος ½χθω ¹ Ζ∆. κሠ™πεˆ ε„ς παραλλήλους εÙθείας τ¦ς ΕΓ, Ζ∆ εÙθε‹ά τις ™νέπεσεν ¹ ΕΖ, αƒ ØπÕ ΓΕΖ, ΕΖ∆ ¥ρα δυσˆν Ñρθα‹ς ‡σαι ε„σίν· αƒ ¥ρα ØπÕ ΖΕΒ, ΕΖ∆ δύο Ñρθîν ™λάσσονές ε„σιν· αƒ δ ¢π' ™λασσόνων À δύο Ñρθîν ™κβαλλόµεναι συµπίπτουσιν· αƒ ¥ρα ΕΒ, Ζ∆ ™κβαλλόµεναι ™πˆ τ¦ Β, ∆ µέρη συµπεσοàνται. ™κβεβλήσθωσαν κሠσυµπιπτέτωσαν κατ¦ τÕ Η, κሠ™πεζεύχθω ¹ ΑΗ. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΓ τÍ ΓΕ, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ΕΑΓ τÍ ØπÕ ΑΕΓ· κሠÑρθ¾ ¹ πρÕς τù Γ· ¹µίσεια ¥ρα ÑρθÁς [™στιν] ˜κατέρα τîν ØπÕ ΕΑΓ, ΑΕΓ. δι¦ τ¦ αÙτ¦ δ¾ κሠ˜κατέρα τîν ØπÕ ΓΕΒ, ΕΒΓ ¹µίσειά ™στιν ÑρθÁς· Ñρθ¾ ¥ρα ™στˆν ¹ ØπÕ ΑΕΒ. κሠ™πεˆ ¹µίσεια ÑρθÁς ™στιν ¹ ØπÕ ΕΒΓ, ¹µίσεια ¥ρα ÑρθÁς κሠ¹ ØπÕ ∆ΒΗ. œστι δ κሠ¹ ØπÕ Β∆Η Ñρθή· ‡ση γάρ ™στι τÍ ØπÕ ∆ΓΕ· ™ναλλ¦ξ γάρ· λοιπ¾ ¤ρα ¹ ØπÕ ∆ΗΒ ¹µίσειά ™στιν ÑρθÁς· ¹ ¥ρα ØπÕ ∆ΗΒ τÍ ØπÕ ∆ΒΗ ™στιν ‡ση· éστε κሠπλευρ¦ ¹ Β∆ πλευρ´ τÍ Η∆ ™στιν ‡ση. πάλιν, ™πεˆ ¹ ØπÕ ΕΗΖ ¹µίσειά ™στιν ÑρθÁς, Ñρθ¾ δ ¹ πρÕς τù Ζ· ‡ση γάρ ™στι τÍ ¢πεναντίον τÍ πρÕς

For let any straight-line AB have been cut in half at (point) C, and let any straight-line BD have been added to it straight-on. I say that the (sum of the) squares on AD and DB is double the (sum of the) squares on AC and CD. For let CE have been drawn from point C, at rightangles to AB [Prop. 1.11], and let it be made equal to each of AC and CB [Prop. 1.3], and let EA and EB have been joined. And let EF have been drawn through E, parallel to AD [Prop. 1.31], and let F D have been drawn through D, parallel to CE [Prop. 1.31]. And since the straight-lines EC and F D (are) parallel, and some straight-line EF falls across (them), the (internal angles) CEF and EF D are thus equal to two right-angles [Prop. 1.29]. Thus, F EB and EF D are less than two right-angles. And (straight-lines) produced from (internal angles whose sum is) less than two right-angles meet together [Post. 5]. Thus, being produced in the direction of B and D, the (straight-lines) EB and F D will meet. Let them have been produced, and let them meet together at G, and let AG have been joined. And since AC is equal to CE, angle EAC is also equal to (angle) AEC [Prop. 1.5]. And the (angle) at C (is) a rightangle. Thus, EAC and AEC [are] each half a right-angle [Prop. 1.32]. So, for the same (reasons), CEB and EBC are also each half a right-angle. Thus, (angle) AEB is a right-angle. And since EBC is half a right-angle, DBG

61

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

τù Γ· λοιπ¾ ¥ρα ¹ ØπÕ ΖΕΗ ¹µίσειά ™στιν ÑρθÁς· ‡ση ¥ρα ¹ ØπÕ ΕΗΖ γωνία τÍ ØπÕ ΖΕΗ· éστε κሠπλευρ¦ ¹ ΗΖ πλευρ´ τÍ ΕΖ ™στιν ‡ση. κሠ™πεˆ [‡ση ™στˆν ¹ ΕΓ τÍ ΓΑ], ‡σον ™στˆ [καˆ] τÕ ¢πÕ τÁς ΕΓ τετράγωνον τù ¢πÕ τÁς ΓΑ τετραγώνJ· τ¦ ¥ρα ¢πÕ τîν ΕΓ, ΓΑ τετράγωνα διπλάσιά ™στι τοà ¢πÕ τÁς ΓΑ τετραγώνου. το‹ς δ ¢πÕ τîν ΕΓ, ΓΑ ‡σον ™στˆ τÕ ¢πÕ τÁς ΕΑ· τÕ ¥ρα ¢πÕ τÁς ΕΑ τετράγωνον διπλάσιόν ™στι τοà ¢πÕ τÁς ΑΓ τετραγώνου. πάλιν, ™πεˆ ‡ση ™στˆν ¹ ΖΗ τÍ ΕΖ, ‡σον ™στˆ κሠτÕ ¢πÕ τÁς ΖΗ τù ¢πÕ τÁς ΖΕ· τ¦ ¥ρα ¢πÕ τîν ΗΖ, ΖΕ διπλάσιά ™στι τοà ¢πÕ τÁς ΕΖ. το‹ς δ ¢πÕ τîν ΗΖ, ΖΕ ‡σον ™στˆ τÕ ¢πÕ τÁς ΕΗ· τÕ ¥ρα ¢πÕ τÁς ΕΗ διπλάσιόν ™στι τοà ¢πÕ τÁς ΕΖ. ‡ση δ ¹ ΕΖ τÍ Γ∆· τÕ ¥ρα ¢πÕ τÁς ΕΗ τετράγωνον διπλάσιόν ™στι τοà ¢πÕ τÁς Γ∆. ™δείχθη δ κሠτÕ ¢πÕ τÁς ΕΑ διπλάσιον τοà ¢πÕ τÁς ΑΓ· τ¦ ¥ρα ¢πÕ τîν ΑΕ, ΕΗ τετράγωνα διπλάσιά ™στι τîν ¢πÕ τîν ΑΓ, Γ∆ τετραγώνων. το‹ς δ ¢πÕ τîν ΑΕ, ΕΗ τετραγώνοις ‡σον ™στˆ τÕ ¢πÕ τÁς ΑΗ τετράγωνον· τÕ ¥ρα ¢πÕ τÁς ΑΗ διπλάσιόν ™στι τîν ¢πÕ τîν ΑΓ, Γ∆. τù δ ¢πÕ τÁς ΑΗ ‡σα ™στˆ τ¦ ¢πÕ τîν Α∆, ∆Η· τ¦ ¥ρα ¢πÕ τîν Α∆, ∆Η [τετράγωνα] διπλάσιά ™στι τîν ¢πÕ τîν ΑΓ, Γ∆ [τετραγώνων]. ‡ση δ ¹ ∆Η τÍ ∆Β· τ¦ ¥ρα ¢πÕ τîν Α∆, ∆Β [τετράγωνα] διπλάσιά ™στι τîν ¢πÕ τîν ΑΓ, Γ∆ τετραγώνων. 'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµηθÍ δίχα, προστεθÍ δέ τις αÙτÍ εÙθε‹α ™π' εÙθείας, τÕ ¢πÕ τÁς Óλης σÝν τÍ προσκειµένV κሠτÕ ¢πÕ τÁς προσκειµένης τ¦ συναµφότερα τετράγωνα διπλάσιά ™στι τοà τε ¢πÕ τÁς ¹µισείας κሠτοà ¢πÕ τÁς συγκειµένης œκ τε τÁς ¹µισείας κሠτÁς προσκειµένης æς ¢πÕ µι©ς ¢ναγραφέντος τετραγώνου· Óπερ œδει δε‹ξαι.

†

(is) thus also half a right-angle [Prop. 1.15]. And BDG is also a right-angle. For it is equal to DCE. For (they are) alternate (angles) [Prop. 1.29]. Thus, the remaining (angle) DGB is half a right-angle. Thus, DGB is equal to DBG. So side BD is also equal to side GD [Prop. 1.6]. Again, since EGF is half a right-angle, and the (angle) at F (is) a right-angle, for it is equal to the opposite (angle) at C [Prop. 1.34], the remaining (angle) F EG is thus half a right-angle. Thus, angle EGF (is) equal to F EG. So the side GF is also equal to the side EF [Prop. 1.6]. And since [EC is equal to CA] the square on EC is [also] equal to the square on CA. Thus, the (sum of the) squares on EC and CA is double the square on CA. And the (square) on EA is equal to the (sum of the squares) on EC and CA [Prop. 1.47]. Thus, the square on EA is double the square on AC. Again, since F G is equal to EF , the (square) on F G is also equal to the (square) on F E. Thus, the (sum of the squares) on GF and F E is double the (square) on EF . And the (square) on EG is equal to the (sum of the squares) on GF and F E [Prop. 1.47]. Thus, the (square) on EG is double the (square) on EF . And EF (is) equal to CD [Prop. 1.34]. Thus, the square on EG is double the (square) on CD. But it was also shown that the (square) on EA (is) double the (square) on AC. Thus, the (sum of the) squares on AE and EG is double the (sum of the) squares on AC and CD. And the square on AG is equal to the (sum of the) squares on AE and EG [Prop. 1.47]. Thus, the (square) on AG is double the (sum of the squares) on AC and CD. And the (square) on AG is equal to the (sum of the squares) on AD and DG [Prop. 1.47]. Thus, the (sum of the) [squares] on AD and DG is double the (sum of the) [squares] on AC and CD. And DG (is) equal to DB. Thus, the (sum of the) [squares] on AD and DB is double the (sum of the) squares on AC and CD. Thus, if a straight-line is cut in half, and any straightline added to it straight-on, then the sum of the square on the whole (straight-line) with the (straight-line) having been added, and the (square) on the (straight-line) having been added, is double the (sum of the square) on half (the straight-line), and the square described on the sum of half (the straight-line) and (straight-line) having been added, as on one (complete straight-line). (Which is) the very thing it was required to show.

This proposition is a geometric version of the algebraic identity: (2 a + b)2 + b2 = 2 [a2 + (a + b)2 ].

ια΄.

Proposition 11†

Τ¾ν δοθε‹σαν εÙθε‹αν τεµε‹ν éστε τÕ ØπÕ τÁς Óλης κሠτοà ˜τέρου τîν τµηµάτων περιεχόµενον Ñρθογώνιον

To cut a given straight-line, so that the rectangle contained by the whole (straight-line), and one of the pieces

62

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

‡σον εναι τù ¢πÕ τοà λοιποà τµήµατος τετραγώνJ.

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(of the straight-line), is equal to the square on the remaining piece.

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∆

C

”Εστω ¹ δοθε‹σα εÙθε‹α ¹ ΑΒ· δε‹ δ¾ τ¾ν ΑΒ τεµε‹ν éστε τÕ ØπÕ τÁς Óλης κሠτοà ˜τέρου τîν τµηµάτων περιεχόµενον Ñρθογώνιον ‡σον εναι τù ¢πÕ τοà λοιποà τµήµατος τετραγώνJ. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ ΑΒ∆Γ, κሠτετµήσθω ¹ ΑΓ δίχα κατ¦ τÕ Ε σηµε‹ον, κሠ™πεζεύχθω ¹ ΒΕ, κሠδιήχθω ¹ ΓΑ ™πˆ τÕ Ζ, κሠκείσθω τÍ ΒΕ ‡ση ¹ ΕΖ, κሠ¢ναγεγράφθω ¢πÕ τÁς ΑΖ τετράγωνον τÕ ΖΘ, κሠδιήχθω ¹ ΗΘ ™πˆ τÕ Κ· λέγω, Óτι ¹ ΑΒ τέτµηται κατ¦ τÕ Θ, éστε τÕ ØπÕ τîν ΑΒ, ΒΘ περιεχόµενον Ñρθογώνιον ‡σον ποιε‹ν τù ¢πÕ τÁς ΑΘ τετραγώνJ. 'Επεˆ γ¦ρ εÙθε‹α ¹ ΑΓ τέτµηται δίχα κατ¦ τÕ Ε, πρόσκειται δ αÙτÍ ¹ ΖΑ, τÕ ¥ρα ØπÕ τîν ΓΖ, ΖΑ περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς ΑΕ τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς ΕΖ τετραγώνJ. ‡ση δ ¹ ΕΖ τÍ ΕΒ· τÕ ¥ρα ØπÕ τîν ΓΖ, ΖΑ µετ¦ τοà ¢πÕ τÁς ΑΕ ‡σον ™στˆ τù ¢πÕ ΕΒ. ¢λλ¦ τù ¢πÕ ΕΒ ‡σα ™στˆ τ¦ ¢πÕ τîν ΒΑ, ΑΕ· Ñρθ¾ γ¦ρ ¹ πρÕς τù Α γωνία· τÕ ¥ρα ØπÕ τîν ΓΖ, ΖΑ µετ¦ τοà ¢πÕ τÁς ΑΕ ‡σον ™στˆ το‹ς ¢πÕ τîν ΒΑ, ΑΕ. κοινÕν ¢φVρήσθω τÕ ¢πÕ τÁς ΑΕ· λοιπÕν ¥ρα τÕ ØπÕ τîν ΓΖ, ΖΑ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ τετραγώνJ. καί ™στι τÕ µν ØπÕ τîν ΓΖ, ΖΑ τÕ ΖΚ· ‡ση γ¦ρ ¹ ΑΖ τÍ ΖΗ· τÕ δ ¢πÕ τÁς ΑΒ τÕ Α∆· τÕ ¥ρα ΖΚ ‡σον ™στˆ τù Α∆. κοινÕν ¢ρVρήσθω τÕ ΑΚ· λοιπÕν ¥ρα τÕ ΖΘ τù Θ∆ ‡σον ™στίν. καί ™στι τÕ µν Θ∆ τÕ ØπÕ τîν ΑΒ, ΒΘ· ‡ση γ¦ρ ¹ ΑΒ τÍ Β∆· τÕ δ ΖΘ τÕ ¢πÕ τÁς ΑΘ· τÕ ¥ρα ØπÕ τîν ΑΒ, ΒΘ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ¢πÕ ΘΑ τετραγώνJ. `Η ¥ρα δοθε‹σα εÙθε‹α ¹ ΑΒ τέτµηται κατ¦ τÕ Θ éστε τÕ ØπÕ τîν ΑΒ, ΒΘ περιεχόµενον Ñρθογώνιον

K

D

Let AB be the given straight-line. So it is required to cut AB, such that the rectangle contained by the whole (straight-line), and one of the pieces (of the straightline), is equal to the square on the remaining piece. For let the square ABDC have been described on AB [Prop. 1.46], and let AC have been cut in half at point E [Prop. 1.10], and let BE have been joined. And let CA have been drawn through to (point) F , and let EF be made equal to BE [Prop. 1.3]. And let the square F H have been described on AF [Prop. 1.46], and let GH have been drawn through to (point) K. I say that AB has been cut at H, so as to make the rectangle contained by AB and BH equal to the square on AH. For since the straight-line AC has been cut in half at E, and F A has been added to it, the rectangle contained by CF and F A, plus the square on AE, is thus equal to the square on EF [Prop. 2.6]. And EF (is) equal to EB. Thus, the (rectangle contained) by CF and F A, plus the (square) on AE, is equal to the (square) on EB. But, the (sum of the squares) on BA and AE is equal to the (square) on EB. For the angle at A (is) a right-angle [Prop. 1.47]. Thus, the (rectangle contained) by CF and F A, plus the (square) on AE, is equal to the (sum of the squares) on BA and AE. Let the square on AE have been subtracted from both. Thus, the remaining rectangle contained by CF and F A is equal to the square on AB. And F K is the (rectangle contained) by CF and F A. For AF (is) equal to F G. And AD (is) the (square) on AB. Thus, the (rectangle) F K is equal to the (square) AD. Let (rectangle) AK have been subtracted from both. Thus, the remaining (square) F H is equal to the (rectangle) HD. And HD is the (rectangle contained) by AB

63

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

‡σον ποιε‹ν τù ¢πÕ τÁς ΘΑ τετραγώνJ· Óπερ œδει and BH. For AB (is) equal to BD. And F H (is) the ποιÁσαι. (square) on AH. Thus, the rectangle contained by AB and BH is equal to the square on HA. Thus, the given straight-line AB has been cut at (point) H, so as to make the rectangle contained by AB and BH equal to the square on HA. (Which is) the very thing it was required to do. †

This manner of cutting a straight-line—so that the ratio of the whole to the larger piece is equal to the ratio of the larger to the smaller piece—is sometimes called the “Golden Section”.

ιβ΄.

Proposition 12†

'Εν το‹ς ¢µβλυγωνίοις τριγώνοις τÕ ¢πÕ τÁς τ¾ν ¢µβλε‹αν γωνίαν Øποτεινούσης πλευρ©ς τετράγωνον µε‹ζόν ™στι τîν ¢πÕ τîν τ¾ν ¢µβλε‹αν γωνίαν περιεχουσîν πλευρîν τετραγώνων τù περιεχοµένJ δˆς ØπÕ τε µι©ς τîν περˆ τ¾ν ¢µβλε‹αν γωνίαν, ™φ' ¿ν ¹ κάθετος πίπτει, κሠτÁς ¢πολαµβανοµένης ™κτÕς ØπÕ τÁς καθέτου πρÕς τÍ ¢µβλείv γωνίv.

In obtuse-angled triangles, the square on the side subtending the obtuse angle is greater than the (sum of the) squares on the sides containing the obtuse angle by twice the (rectangle) contained by one of the sides around the obtuse angle, to which a perpendicular (straight-line) falls, and the (straight-line) cut off outside (the triangle) by the perpendicular (straight-line) towards the obtuse angle.

Β

∆

B

Α

Γ

D

”Εστω ¢µβλυγώνιον τρίγωνον τÕ ΑΒΓ ¢µβλε‹αν œχον τ¾ν ØπÕ ΒΑΓ, κሠ½χθω ¢πÕ τοà Β σηµείου ™πˆ τ¾ν ΓΑ ™κβληθε‹σαν κάθετος ¹ Β∆. λέγω, Óτι τÕ ¢πÕ τÁς ΒΓ τετράγωνον µε‹ζόν ™στι τîν ¢πÕ τîν ΒΑ, ΑΓ τετραγώνων τù δˆς ØπÕ τîν ΓΑ, Α∆ περιεχοµένJ ÑρθογωνίJ. 'Επεˆ γ¦ρ εÙθε‹α ¹ Γ∆ τέτµηται, æς œτυχεν, κατ¦ τÕ Α σηµε‹ον, τÕ ¥ρα ¢πÕ τÁς ∆Γ ‡σον ™στˆ το‹ς ¢πÕ τîν ΓΑ, Α∆ τετραγώνοις κሠτù δˆς ØπÕ τîν ΓΑ, Α∆ περιεχοµένJ ÑρθογωνίJ. κοινÕν προσκείσθω τÕ ¢πÕ τÁς ∆Β· τ¦ ¥ρα ¢πÕ τîν Γ∆, ∆Β ‡ση ™στˆ το‹ς τε ¢πÕ τîν ΓΑ, Α∆, ∆Β τετραγώνοις κሠτù δˆς ØπÕ τîν ΓΑ, Α∆ [περιεχοµένJ ÑρθογωνίJ]. ¢λλ¦ το‹ς µν ¢πÕ τîν Γ∆, ∆Β ‡σον ™στˆ τÕ ¢πÕ τÁς ΓΒ· Ñρθ¾ γ¦ρ ¹ προς τù ∆ γωνία· το‹ς δ ¢πÕ τîν Α∆, ∆Β ‡σον τÕ ¢πÕ τÁς ΑΒ· τÕ ¥ρα ¢πÕ τÁς ΓΒ τετράγωνον ‡σον ™στˆ το‹ς τε ¢πÕ τîν ΓΑ, ΑΒ τετραγώνοις κሠτù δˆς ØπÕ τîν ΓΑ, Α∆ περιεχοµένJ ÑρθογωνίJ· éστε τÕ ¢πÕ τÁς ΓΒ τετράγωνον τîν ¢πÕ τîν ΓΑ, ΑΒ τετραγώνων µε‹ζόν

A

C

Let ABC be an obtuse-angled triangle, having the obtuse angle BAC. And let BD be drawn from point B, perpendicular to CA produced [Prop. 1.12]. I say that the square on BC is greater than the (sum of the) squares on BA and AC, by twice the rectangle contained by CA and AD. For since the straight-line CD has been cut, at random, at point A, the (square) on DC is thus equal to the (sum of the) squares on CA and AD, and twice the rectangle contained by CA and AD [Prop. 2.4]. Let the (square) on DB have been added to both. Thus, the (sum of the squares) on CD and DB is equal to the (sum of the) squares on CA, AD, and DB, and twice the [rectangle contained] by CA and AD. But, the (sum of the squares) on CD and DB is equal to the (square) on CB. For the angle at D (is) a right-angle [Prop. 1.47]. And the (sum of the squares) on AD and DB (is) equal to the (square) on AB [Prop. 1.47]. Thus, the square on CB is equal to the (sum of the) squares on CA and AB,

64

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

™στι τù δˆς ØπÕ τîν ΓΑ, Α∆ περιεχοµένJ ÑρθογωνίJ. 'Εν ¥ρα το‹ς ¢µβλυγωνίοις τριγώνοις τÕ ¢πÕ τÁς τ¾ν ¢µβλε‹αν γωνίαν Øποτεινούσης πλευρ©ς τετράγωνον µε‹ζόν ™στι τîν ¢πÕ τîν τ¾ν ¢µβλε‹αν γωνίαν περιεχουσîν πλευρîν τετραγώνων τù περιχοµένJ δˆς Øπό τε µι©ς τîν περˆ τ¾ν ¢µβλε‹αν γωνίαν, ™φ' ¿ν ¹ κάθετος πίπτει, κሠτÁς ¢πολαµβανοµένης ™κτÕς ØπÕ τÁς καθέτου πρÕς τÍ ¢µβλείv γωνίv· Óπερ œδει δε‹ξαι.

†

and twice the rectangle contained by CA and AD. So the square on CB is greater than the (sum of the) squares on CA and AB by twice the rectangle contained by CA and AD. Thus, in obtuse-angled triangles, the square on the side subtending the obtuse angle is greater than the (sum of the) squares on the sides containing the obtuse angle by twice the (rectangle) contained by one of the sides around the obtuse angle, to which a perpendicular (straight-line) falls, and the (straight-line) cut off outside (the triangle) by the perpendicular (straight-line) towards the obtuse angle. (Which is) the very thing it was required to show.

This proposition is equivalent to the well-known cosine formula: BC 2 = AB 2 + AC 2 − 2 AB AC cos BAC, since cos BAC = −AD/AB.

ιγ΄.

Proposition 13†

'Εν το‹ς Ñξυγωνίοις τριγώνοις τÕ ¢πÕ τÁς τ¾ν Ñξε‹αν γωνίαν Øποτεινούσης πλευρ©ς τετράγωνον œλαττόν ™στι τîν ¢πÕ τîν τ¾ν Ñξε‹αν γωνίαν περιεχουσîν πλευρîν τετραγώνων τù περιεχοµένJ δˆς ØπÕ τε µι©ς τîν περˆ τ¾ν Ñξε‹αν γωνίαν, ™φ' ¿ν ¹ κάθετος πίπτει, κሠτÁς ¢πολαµβανοµένης ™ντÕς ØπÕ τÁς καθέτου πρÕς τÍ Ñξείv γωνίv.

In acute-angled triangles, the square on the side subtending the acute angle is less than the (sum of the) squares on the sides containing the acute angle by twice the (rectangle) contained by one of the sides around the acute angle, to which a perpendicular (straight-line) falls, and the (straight-line) cut off inside (the triangle) by the perpendicular (straight-line) towards the acute angle.

Α

Β

∆

A

Γ

B

”Εστω Ñξυγώνιον τρίγωνον τÕ ΑΒΓ Ñξε‹αν œχον τ¾ν πρÕς τù Β γωνίαν, κሠ½χθω ¢πÕ τοà Α σηµείου ™πˆ τ¾ν ΒΓ κάθετος ¹ Α∆· λέγω, Óτι τÕ ¢πÕ τÁς ΑΓ τετράγωνον œλαττόν ™στι τîν ¢πÕ τîν ΓΒ, ΒΑ τετραγώνων τù δˆς ØπÕ τîν ΓΒ, Β∆ περιεχοµένJ ÑρθογωνίJ. 'Επεˆ γ¦ρ εÙθε‹α ¹ ΓΒ τέτµηται, æς œτυχεν, κατ¦ τÕ ∆, τ¦ ¥ρα ¢πÕ τîν ΓΒ, Β∆ τετράγωνα ‡σα ™στˆ τù τε δˆς ØπÕ τîν ΓΒ, Β∆ περιεχοµένJ ÑρθογωνίJ κሠτù ¢πÕ τÁς ∆Γ τετραγώνJ. κοινÕν προσκείσθω τÕ ¢πÕ τÁς ∆Α τετράγωνον· τ¦ ¥ρα ¢πÕ τîν ΓΒ, Β∆, ∆Α τετράγωνα ‡σα ™στˆ τù τε δˆς ØπÕ τîν ΓΒ, Β∆ περιεχοµένJ ÑρθογωνίJ κሠτο‹ς ¢πÕ τîν Α∆, ∆Γ τετραγώνιος. ¢λλ¦ το‹ς µν ¢πÕ τîν Β∆, ∆Α ‡σον τÕ ¢πÕ τÁς ΑΒ· Ñρθ¾ γ¦ρ ¹ πρÕς τù ∆ γωνίv· το‹ς δ ¢πÕ τîν Α∆, ∆Γ ‡σον τÕ ¢πÕ τÁς

D

C

Let ABC be an acute-angled triangle, having an acute angle at (point) B. And let AD have been drawn from point A, perpendicular to BC [Prop. 1.12]. I say that the square on AC is less than the (sum of the) squares on CB and AB, by twice the rectangle contained by CB and BD. For since the straight-line CB has been cut, at random, at (point) D, the (sum of the) squares on CB and BD is thus equal to twice the rectangle contained by CB and BD, and the square on DC [Prop. 2.7]. Let the square on DA have been added to both. Thus, the (sum of the) squares on CB, BD, and DA is equal to twice the rectangle contained by CB and BD, and the (sum of the) squares on AD and DC. But, the (square) on AB

65

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

ΑΓ· τ¦ ¥ρα ¢πÕ τîν ΓΒ, ΒΑ ‡σα ™στˆ τù τε ¢πÕ τÁς ΑΓ κሠτù δˆς ØπÕ τîν ΓΒ, Β∆· éστε µόνον τÕ ¢πÕ τÁς ΑΓ œλαττόν ™στι τîν ¢πÕ τîν ΓΒ, ΒΑ τετραγώνων τù δˆς ØπÕ τîν ΓΒ, Β∆ περιεχοµένJ ÑρθογωνίJ. 'Εν ¥ρα το‹ς Ñξυγωνίοις τριγώνοις τÕ ¢πÕ τÁς τ¾ν Ñξε‹αν γωνίαν Øποτεινούσης πλευρ©ς τετράγωνον œλαττόν ™στι τîν ¢πÕ τîν τ¾ν Ñξε‹αν γωνίαν περιεχουσîν πλευρîν τετραγώνων τù περιεχοµένJ δˆς ØπÕ τε µι©ς τîν περˆ τ¾ν Ñξε‹αν γωνίαν, ™φ' ¿ν ¹ κάθετος πίπτει, κሠτÁς ¢πολαµβανοµένης ™ντÕς ØπÕ τÁς καθέτου πρÕς τÍ Ñξείv γωνίv· Óπερ œδει δε‹ξαι.

†

(is) equal to the (sum of the squares) on BD and DA. For the angle at (point) D is a right-angle [Prop. 1.47]. And the (square) on AC (is) equal to the (sum of the squares) on AD and DC [Prop. 1.47]. Thus, the (sum of the squares) on CB and BA is equal to the (square) on AC, and twice the (rectangle contained) by CB and BD. So the (square) on AC alone is less than the (sum of the) squares on CB and BA by twice the rectangle contained by CB and BD. Thus, in acute-angled triangles, the square on the side subtending the acute angle is less than the (sum of the) squares on the sides containing the acute angle by twice the (rectangle) contained by one of the sides around the acute angle, to which a perpendicular (straight-line) falls, and the (straight-line) cut off inside (the triangle) by the perpendicular (straight-line) towards the acute angle. (Which is) the very thing it was required to show.

This proposition is equivalent to the well-known cosine formula: AC 2 = AB 2 + BC 2 − 2 AB BC cos ABC, since cos ABC = BD/AB.

ιδ΄.

Proposition 14

Τù δοθέντι εÙθυγράµµJ ‡σον τετράγωνον συστήσασTo construct a square equal to a given rectilinear figθαι. ure.

Θ

H

Α

A

Β Γ

Η

Ε

Ζ

B

∆

C

”Εστω τÕ δοθν εÙθύγραµµον τÕ Α· δε‹ δ¾ τù Α εÙθυγράµµJ ‡σον τετράγωνον συστήσασθαι. Συνεστάτω γ¦ρ τù Α ™υθυγράµµJ ‡σον παραλληλόγραµµον Ñρθογώνιον τÕ Β∆· ε„ µν οâν ‡ση ™στˆν ¹ ΒΕ τÍ Ε∆, γεγονÕς ¨ν ε‡η τÕ ™πιταχθέν. συνέσταται γ¦ρ τù Α εÙθυγράµµJ ‡σον τετράγωνον τÕ Β∆· ε„ δ οÜ, µία τîν ΒΕ, Ε∆ µείζων ™στιν. œστω µείζων ¹ ΒΕ, κሠ™κβεβλήσθω ™πˆ τÕ Ζ, κሠκείσθω τÍ Ε∆ ‡ση ¹ ΕΖ, κሠτετµήσθω ¹ ΒΖ δίχα κατ¦ τÕ Η, κሠκέντρJ τù Η, διαστήµατι δ ˜νˆ τîν ΗΒ, ΗΖ ¹µικύκλιον γεγράφθω τÕ ΒΘΖ, κሠ™κβεβλήσθω ¹ ∆Ε ™πˆ τÕ Θ, κሠ™πεζεύχθω ¹ ΗΘ. 'Επεˆ οâν εÙθε‹α ¹ ΒΖ τέτµηται ε„ς µν ‡σα κατ¦ τÕ Η, ε„ς δ ¥νισα κατ¦ τÕ Ε, τÕ ¥ρα ØπÕ τîν ΒΕ, ΕΖ περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς ΕΗ

E G

F

D

Let A be the given rectilinear figure. So it is required to construct a square equal to the rectilinear figure A. For let the right-angled parallelogram BD have been constructed, equal to the rectilinear figure A [Prop. 1.45]. Therefore, if BE is equal to ED, then that (which) was prescribed has taken place. For the square BD has been constructed, equal to the rectilinear figure A. And if not, then one of BE or ED is greater (than the other). Let BE be greater, and let it have been produced to F , and let EF be made equal to ED [Prop. 1.3]. And let BF have been cut in half at (point) G [Prop. 1.10]. And, with center G, and radius one of GB or GF , let the semi-circle BHF have been drawn. And let DE have been produced to H, and let GH have been joined. Therefore, since the straight-line BF has been cut—

66

ΣΤΟΙΧΕΙΩΝ β΄.

ELEMENTS BOOK 2

τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς ΗΖ τετραγώνJ. ‡ση δ ¹ ΗΖ τÍ ΗΘ· τÕ ¥ρα ØπÕ τîν ΒΕ, ΕΖ µετ¦ τοà ¢πÕ τÁς ΗΕ ‡σον ™στˆ τù ¢πÕ τÁς ΗΘ. τù δ ¢πÕ τÁς ΗΘ ‡σα ™στˆ τ¦ ¢πÕ τîν ΘΕ, ΕΗ τετράγωνα· τÕ ¥ρα ØπÕ τîν ΒΕ, ΕΖ µετ¦ τοà ¢πÕ ΗΕ ‡σα ™στˆ το‹ς ¢πÕ τîν ΘΕ, ΕΗ. κοινÕν ¢φVρήσθω τÕ ¢πÕ τÁς ΗΕ τετράγωνον· λοιπÕν ¥ρα τÕ ØπÕ τîν ΒΕ, ΕΖ περιεχόµενον Ôρθογώνιον ‡σον ™στˆ τù ¢πÕ τÁς ΕΘ τετραγώνJ. ¢λλ¦ τÕ ØπÕ τîν ΒΕ, ΕΖ τÕ Β∆ ™στιν· ‡ση γ¦ρ ¹ ΕΖ τÍ Ε∆· τÕ ¥ρα Β∆ παραλληλόγραµµον ‡σον ™στˆ τù ¢πÕ τÁς ΘΕ τετραγώνJ. ‡σον δ τÕ Β∆ τù Α εÙθυγράµµJ. κሠτÕ Α ¥ρα εÙθύγραµµον ‡σον ™στˆ τù ¢πÕ τÁς ΕΘ ¢ναγραφησοµένJ τετραγώνJ. Τù ¥ρα δοθέντι εÙθυγράµµJ τù Α ‡σον τετράγωνον συνέσταται τÕ ¢πÕ τÁς ΕΘ ¢ναγραφησόµενον· Óπερ œδει ποιÁσαι.

equally at G, and unequally at E—the rectangle contained by BE and EF , plus the square on EG, is thus equal to the square on GF [Prop. 2.5]. And GF (is) equal to GH. Thus, the (rectangle contained) by BE and EF , plus the (square) on GE, is equal to the (square) on GH. And the (square) on GH is equal to the (sum of the) squares on HE and EG [Prop. 1.47]. Thus, the (rectangle contained) by BE and EF , plus the (square) on GE, is equal to the (sum of the squares) on HE and EG. Let the square on GE have been taken from both. Thus, the remaining rectangle contained by BE and EF is equal to the square on EH. But, BD is the (rectangle contained) by BE and EF . For EF (is) equal to ED. Thus, the parallelogram BD is equal to the square on HE. And BD (is) equal to the rectilinear figure A. Thus, the rectilinear figure A is also equal to the square (which) can be described on EH. Thus, a square—(namely), that (which) can be described on EH—has been constructed, equal to the given rectilinear figure A. (Which is) the very thing it was required to do.

67

68

ELEMENTS BOOK 3 Fundamentals of plane geometry involving circles

69

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

“Οροι.

Definitions

α΄. ”Ισοι κύκλοι ε„σίν, ïν αƒ διάµετροι ‡σαι ε„σίν, À ïν αƒ ™κ τîν κέντρων ‡σαι ε„σίν. β΄. ΕÙθε‹α κύκλου ™φάπτεσθαι λέγεται, ¼τις ¡πτοµένη τοà κύκλου κሠ™κβαλλοµένη οÙ τέµνει τÕν κύκλον. γ΄. Κύκλοι ™φάπτεσθαι ¢λλήλων λέγονται ο†τινες ¡πτόµενοι ¢λλήλων οÙ τέµνουσιν ¢λλήλους. δ΄. 'Εν κύκλJ ‡σον ¢πέχειν ¢πÕ τοà κέντρου εÙθε‹αι λέγονται, Óταν αƒ ¢πÕ τοà κέντρου ™π' αÙτ¦ς κάθετοι ¢γόµεναι ‡σαι ðσιν. ε΄. Με‹ζον δ ¢πέχειν λέγεται, ™φ' ¿ν ¹ µείζων κάθετος πίπτει. $΄. ΤµÁµα κύκλου ™στˆ τÕ περιεχόµενον σχÁµα Øπό τε εÙθείας κሠκύκλου περιφερείας. ζ΄. Τµήµατος δ γωνία ™στˆν ¹ περιεχοµένη Øπό τε εÙθείας κሠκύκλου περιφερείας. η΄. 'Εν τµήµατι δ γωνία ™στίν, Óταν ™πˆ τÁς περιφερείας τοà τµήµατος ληφθÍ τι σηµε‹ον κሠ¢π' αÙτοà ™πˆ τ¦ πέρατα τÁς εÙθείας, ¼ ™στι βάσις τοà τµήµατος, ™πιζευχθîσιν εÙθε‹αι, ¹ περιεχοµένη γωνία ØπÕ τîν ™πιζευχθεισîν εÙθειîν. θ΄. “Οταν δ αƒ περιέχουσαι τ¾ν γωνίαν εÙθε‹αι ¢πολαµβάνωσί τινα περιφέρειαν, ™π' ™κείνης λέγεται βεβηκέναι ¹ γωνία. ι΄. ΤοµεÝς δ κύκλου ™στίν, Óταν πρÕς τù κέντρù τοà κύκλου συσταθÍ γωνία, τÕ περιεχόµενον σχÁµα Øπό τε τîν τ¾ν γωνίαν περιεχουσîν εÙθειîν κሠτÁς ¢πολαµβανοµένης Øπ' αÙτîν περιφερείας. ια΄. “Οµοία τµήµατα κύκλων ™στˆ τ¦ δεχόµενα γωνίας ‡σας, ½ ™ν οŒς αƒ γωνίαι ‡σαι ¢λλήλαις ε„σίν.

1. Equal circles are (circles) whose diameters are equal, or whose (distances) from the centers (to the circumferences) are equal (i.e., whose radii are equal). 2. A straight-line said to touch a circle is any (straightline) which, meeting the circle and being produced, does not cut the circle. 3. Circles said to touch one another are any (circles) which, meeting one another, do not cut one another. 4. In a circle, straight-lines are said to be equally far from the center when the perpendiculars drawn to them from the center are equal. 5. And (that straight-line) is said to be further (from the center) on which the greater perpendicular falls (from the center). 6. A segment of a circle is the figure contained by a straight-line and a circumference of a circle. 7. And the angle of a segment is that contained by a straight-line and a circumference of a circle. 8. And the angle in a segment is the angle contained by the joined straight-lines, when any point is taken on the circumference of a segment, and straight-lines are joined from it to the ends of the straight-line which is the base of the segment. 9. And when the straight-lines containing an angle cut off some circumference, the angle is said to stand upon that (circumference). 10. And a sector of a circle is the figure contained by the straight-lines surrounding an angle, and the circumference cut off by them, when the angle is constructed at the center of a circle. 11. Similar segments of circles are those accepting equal angles, or in which the angles are equal to one another.

α΄.

Proposition 1

Τοà δοθέντος κύκλου τÕ κέντρον εØρε‹ν. ”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ· δε‹ δ¾ τοà ΑΒΓ κύκλου τÕ κέντρον εØρε‹ν. ∆ιήχθω τις ε„ς αÙτόν, æς œτυχεν, εÙθε‹α ¹ ΑΒ, κሠτετµήσθω δίχα κατ¦ τÕ ∆ σηµε‹ον, κሠ¢πÕ τοà ∆ τÍ ΑΒ πρÕς Ñρθ¦ς ½χθω ¹ ∆Γ κሠδιήχθω ™πˆ τÕ Ε, κሠτετµήσθω ¹ ΓΕ δίχα κατ¦ τÕ Ζ· λέγω, Óτι τÕ Ζ κέντρον ™στˆ τοà ΑΒΓ [κύκλου]. Μ¾ γάρ, ¢λλ' ε„ δυνατόν, œστω τÕ Η, κሠ™πεζεύχθωσαν αƒ ΗΑ, Η∆, ΗΒ. κሠ™πεˆ ‡ση ™στˆν ¹ Α∆ τÍ ∆Β, κοιν¾ δ ¹ ∆Η, δύο δ¾ αƒ Α∆, ∆Η δύο τα‹ς Η∆, ∆Β ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠβάσις ¹ ΗΑ βάσει τÍ ΗΒ ™στιν ‡ση· ™κ κέντρου γάρ· γωνία ¥ρα ¹ ØπÕ Α∆Η

To find the center of a given circle. Let ABC be the given circle. So it is required to find the center of circle ABC. Let some straight-line AB have been drawn through (ABC), at random, and let (AB) have been cut in half at point D [Prop. 1.9]. And let DC have been drawn from D, at right-angles to AB [Prop. 1.11]. And let (CD) have been drawn through to E. And let CE have been cut in half at F [Prop. 1.9]. I say that (point) F is the center of the [circle] ABC. For (if) not then, if possible, let G (be the center of the circle), and let GA, GD, and GB have been joined. And since AD is equal to DB, and DG (is) common, the two

70

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

γωνίv τÍ ØπÕ Η∆Β ‡ση ™στίν. Óταν δ εÙθε‹α ™π' εÙθε‹αν σταθε‹σα τ¦ς ™φεξÁς γωνίας ‡σας ¢λλήλαις ποιÍ, Ñρθ¾ ˜κατέρα τîν ‡σων γωνιîν ™στιν· Ñρθ¾ ¥ρα ™στˆν ¹ ØπÕ Η∆Β. ™στˆ δ κሠ¹ ØπÕ Ζ∆Β Ñρθή· ‡ση ¥ρα ¹ ØπÕ Ζ∆Β τÍ ØπÕ Η∆Β, ¹ µείζων τÍ ™λάττονι· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τÕ Η κέντρον ™στˆ τοà ΑΒΓ κύκλου. еοίως δ¾ δείξοµεν, Óτι οÙδ' ¥λλο τι πλ¾ν τοà Ζ.

(straight-lines) AD, DG are equal to the two (straightlines) BD, DG,† respectively. And the base GA is equal to the base GB. For (they are both) radii. Thus, the angle ADG is equal to GDB [Prop. 1.8]. And when a straight-line stood upon (another) straight-line make adjacent angles (which are) equal to one another, each of the equal angles is a right-angle [Def. 1.10]. Thus, GDB is a right-angle. And F DB is also a right-angle. Thus, F DB (is) equal to GDB, the greater to the lesser. The very thing is impossible. Thus, (point) G is not the center of the circle ABC. So, similarly, we can show that neither is any other (point) than F .

Γ

C

Ζ Α

Η

∆

F

Β

A

Ε

G

D

B

E

ΤÕ Ζ ¥ρα σηµε‹ον κέντρον ™στˆ τοà ΑΒΓ [κύκλου].

Thus, point F is the center of the [circle] ABC.

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι ™¦ν ™ν κύκλJ εÙθε‹ά So, from this, (it is) manifest that if any straight-line τις εÙθε‹άν τινα δίχα κሠπρÕς Ñρθ¦ς τέµνV, ™πˆ τÁς in a circle cuts any (other) straight-line in half, and at τεµνούσης ™στˆ τÕ κέντρον τοà κύκλου. — Óπερ œδει right-angles, then the center of the circle is on the forποιÁσαι. mer (straight-line). — (Which is) the very thing it was required to do. †

The Greek text has “GD, DB”, which is obviously a mistake.

β΄.

Proposition 2

'Ε¦ν κύκλου ™πˆ τÁς περιφερείας ληφθÍ δύο τυχόντα σηµε‹α, ¹ ™πˆ τ¦ σηµε‹α ™πιζευγνυµένη εÙθε‹α ™ντÕς πεσε‹ται τοà κύκλου. ”Εστω κύκλος Ð ΑΒΓ, κሠ™πˆ τÁς περιφερείας αÙτοà ε„λήφθω δύο τυχόντα σηµε‹α τ¦ Α, Β· λέγω, Óτι ¹ ¢πÕ τοà Α ™πˆ τÕ Β ™πιζευγνυµένη εÙθε‹α ™ντÕς πεσε‹ται τοà κύκλου. Μ¾ γάρ, ¢λλ' ε„ δυνατόν, πιπτέτω ™κτÕς æς ¹ ΑΕΒ, κሠε„λήφθω τÕ κέντρον τοà ΑΒΓ κύκλου, κሠœστω τÕ ∆, κሠ™πεζεύχθωσαν αƒ ∆Α, ∆Β, κሠδιήχθω ¹ ∆ΖΕ. 'Επεˆ οâν ‡ση ™στˆν ¹ ∆Α τÍ ∆Β, ‡ση ¥ρα κሠγωνία ¹ ØπÕ ∆ΑΕ τÍ ØπÕ ∆ΒΕ· κሠ™πεˆ τριγώνου τοà ∆ΑΕ

If two points are taken somewhere on the circumference of a circle then the straight-line joining the points will fall inside the circle. Let ABC be a circle, and let two points A and B have been taken somewhere on its circumference. I say that the straight-line joining A to B will fall inside the circle. For (if) not then otherwise, if possible, let it fall outside (the circle), like AEB (in the figure). And let the center of the circle ABC have been found [Prop. 3.1], and let it be (at point) D. And let DA and DB have been joined, and let DF E have been drawn through. Therefore, since DA is equal to DB, the angle DAE

71

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

µία πλευρ¦ προσεκβέβληται ¹ ΑΕΒ, µείζων ¥ρα ¹ ØπÕ ∆ΕΒ γωνία τÁς ØπÕ ∆ΑΕ. ‡ση δ ¹ ØπÕ ∆ΑΕ τÍ ØπÕ ∆ΒΕ· µείζων ¥ρα ¹ ØπÕ ∆ΕΒ τÁς ØπÕ ∆ΒΕ. ØπÕ δ τ¾ν µείζονα γωνίαν ¹ µείζων πλευρ¦ Øποτείνει· µείζων ¥ρα ¹ ∆Β τÁς ∆Ε. ‡ση δ ¹ ∆Β τÍ ∆Ζ. µείζων ¥ρα ¹ ∆Ζ τÁς ∆Ε ¹ ™λάττων τÁς µείζονος· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ¹ ¢πÕ τοà Α ™πˆ τÕ Β ™πιζευγνυµένη εÙθε‹α ™κτÕς πεσε‹ται τοà κύκλου. еοίως δ¾ δείξοµεν, Óτι οÙδ ™π' αÙτÁς τÁς περιφερείας· ™ντÕς ¥ρα.

(is) thus also equal to DBE [Prop. 1.5]. And since in triangle DAE the one side, AEB, has been produced, angle DEB (is) thus greater than DAE [Prop. 1.16]. And DAE (is) equal to DBE [Prop. 1.5]. Thus, DEB (is) greater than DBE. And the greater angle is subtended by the greater side [Prop. 1.19]. Thus, DB (is) greater than DE. And DB (is) equal to DF . Thus, DF (is) greater than DE, the lesser than the greater. The very thing is impossible. Thus, the straight-line joining A to B will not fall outside the circle. So, similarly, we can show that neither (will it fall) on the circumference itself. Thus, (it will fall) inside (the circle).

Γ

C

∆

Α

D A

Ζ Ε

F E

Β

B

'Ε¦ν ¥ρα κύκλου ™πˆ τÁς περιφερείας ληφθÍ δύο τυχόντα σηµε‹α, ¹ ™πˆ τ¦ σηµε‹α ™πιζευγνυµένη εÙθε‹α ™ντÕς πεσε‹ται τοà κύκλου· Óπερ œδει δε‹ξαι.

Thus, if two points are taken somewhere on the circumference of a circle then the straight-line joining the points will fall inside the circle. (Which is) the very thing it was required to show.

γ΄.

Proposition 3

'Ε¦ν ™ν κύκλJ εÙθε‹ά τις δι¦ τοà κέντρου εÙθε‹άν τινα µ¾ δι¦ τοà κέντρου δίχα τέµνV, κሠπρÕς Ñρθ¦ς αÙτ¾ν τέµνει· κሠ™¦ν πρÕς Ñρθ¦ς αÙτ¾ν τέµνV, κሠδίχα αÙτ¾ν τέµνει. ”Εστω κύκλος Ð ΑΒΓ, κሠ™ν αÙτù εÙθε‹ά τις δι¦ τοà κέντρου ¹ Γ∆ εÙθε‹άν τινα µ¾ δι¦ τοà κέντρου τ¾ν ΑΒ δίχα τεµνέτω κατ¦ τÕ Ζ σηµε‹ον· λέγω, Óτι κሠπρÕς Ñρθ¦ς αÙτ¾ν τέµνει. Ε„λήφθω γ¦ρ τÕ κέντρον τοà ΑΒΓ κύκλου, κሠœστω τÕ Ε, κሠ™πεζεύχθωσαν αƒ ΕΑ, ΕΒ. Κሠ™πεˆ ‡ση ™στˆν ¹ ΑΖ τÍ ΖΒ, κοιν¾ δ ¹ ΖΕ, δύο δυσˆν ‡σαι [ε„σίν]· κሠβάσις ¹ ΕΑ βάσει τÍ ΕΒ ‡ση· γωνία ¥ρα ¹ ØπÕ ΑΖΕ γωνίv τÍ ØπÕ ΒΖΕ ‡ση ™στίν. Óταν δ εÙθε‹α ™π' εÙθε‹αν σταθε‹σα τ¦ς ™φεξÁς γωνίας ‡σας ¢λλήλαις ποιÍ, Ñρθ¾ ˜κατέρα τîν ‡σων γωνιîν ™στιν· ˜κατέρα ¥ρα τîν ØπÕ ΑΖΕ, ΒΖΕ Ñρθή ™στιν. ¹ Γ∆ ¥ρα δι¦ τοà κέντρου οâσα τ¾ν ΑΒ µ¾ δι¦ τοà κέντρου οâσαν δίχα τέµνουσα κሠπρÕς Ñρθ¦ς τέµνει.

In a circle, if any straight-line through the center cuts in half any straight-line not through the center, then it also cuts it at right-angles. And (conversely) if it cuts it at right-angles, then it also cuts it in half. Let ABC be a circle, and within it, let some straightline through the center, CD, cut in half some straight-line not through the center, AB, at the point F . I say that (CD) also cuts (AB) at right-angles. For let the center of the circle ABC have been found [Prop. 3.1], and let it be (at point) E, and let EA and EB have been joined. And since AF is equal to F B, and F E (is) common, two (sides of triangle AF E) [are] equal to two (sides of triangle BF E). And the base EA (is) equal to the base EB. Thus, angle AF E is equal to angle BF E [Prop. 1.8]. And when a straight-line stood upon (another) straightline makes adjacent angles (which are) equal to one another, each of the equal angles is a right-angle [Def. 1.10]. Thus, AF E and BF E are each right-angles. Thus, the

72

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 (straight-line) CD, which is through the center and cuts in half the (straight-line) AB, which is not through the center, also cuts (AB) at right-angles.

Γ

C

Ε Α

Ζ

E Β

A

∆

F

B

D

'Αλλ¦ δ¾ ¹ Γ∆ τ¾ν ΑΒ πρÕς Ñρθ¦ς τεµνέτω· λέγω, Óτι κሠδίχα αÙτ¾ν τέµνει, τουτέστιν, Óτι ‡ση ™στˆν ¹ ΑΖ τÍ ΖΒ. Τîν γ¦ρ αÙτîν κατασκευασθέντων, ™πεˆ ‡ση ™στˆν ¹ ΕΑ τÍ ΕΒ, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ΕΑΖ τÍ ØπÕ ΕΒΖ. ™στˆ δ κሠÑρθ¾ ¹ ØπÕ ΑΖΕ ÑρθÍ τÍ ØπÕ ΒΖΕ ‡ση· δύο ¥ρα τρίγωνά ™στι ΕΑΖ, ΕΖΒ τ¦ς δύο γωνίας δυσˆ γωνίαις ‡σας œχοντα κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην κοιν¾ν αÙτîν τ¾ν ΕΖ Øποτείνουσαν ØπÕ µίαν τîν ‡σων γωνιîν· κሠτ¦ς λοιπ¦ς ¥ρα πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξει· ‡ση ¥ρα ¹ ΑΖ τÍ ΖΒ. 'Ε¦ν ¥ρα ™ν κύκλJ εÙθε‹ά τις δι¦ τοà κέντρου εÙθε‹άν τινα µ¾ δι¦ τοà κέντρου δίχα τέµνV, κሠπρÕς Ñρθ¦ς αÙτ¾ν τέµνει· κሠ™¦ν πρÕς Ñρθ¦ς αÙτ¾ν τέµνV, κሠδίχα αÙτ¾ν τέµνει· Óπερ œδει δε‹ξαι.

And so let CD cut AB at right-angles. I say that it also cuts (AB) in half. That is to say, that AF is equal to F B. For, with the same construction, since EA is equal to EB, angle EAF is also equal to EBF [Prop. 1.5]. And the right-angle AF E is also equal to the right-angle BF E. Thus, EAF and EF B are two triangles having two angles equal to two angles, and one side equal to one side—(namely), their common (side) EF , subtending one of the equal angles. Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26]. Thus, AF (is) equal to F B. Thus, in a circle, if any straight-line through the center cuts in half any straight-line not through the center, then it also cuts it at right-angles. And (conversely) if it cuts it at right-angles, then it also cuts it in half. (Which is) the very thing it was required to show.

δ΄.

Proposition 4

'Ε¦ν ™ν κύκλJ δύο εÙθε‹αι τέµνωσιν ¢λλήλας µ¾ δˆα τοà κέντρου οâσαι, οÙ τέµνουσιν ¢λλήλας δίχα. ”Εστω κύκλος Ð ΑΒΓ∆, κሠ™ν αÙτù δύο εÙθε‹αι αƒ ΑΓ, Β∆ τεµνέτωσαν ¢λλήλας κατ¦ τÕ Ε µ¾ δι¦ τοà κέντρου οâσαι· λέγω, Óτι οÙ τέµνουσιν ¢λλήλας δίχα. Ε„ γ¦ρ δυνατόν, τεµνέτωσαν ¢λλήλας δίχα éστε ‡σην εναι τ¾ν µν ΑΕ τÍ ΕΓ, τ¾ν δ ΒΕ τÍ Ε∆· κሠε„λήφθω τÕ κέντρον τοà ΑΒΓ∆ κύκλου, κሠœστω τÕ Ζ, κሠ™πεζεύχθω ¹ ΖΕ. 'Επεˆ οâν εÙθε‹ά τις δι¦ τοà κέντρου ¹ ΖΕ εÙθε‹άν τινα µ¾ δι¦ τοà κέντρου τ¾ν ΑΓ δίχα τέµνει, κሠπρÕς Ñρθ¦ς αÙτ¾ν τέµνει· Ñρθ¾ ¥ρα ™στˆν ¹ ØπÕ ΖΕΑ· πάλιν, ™πεˆ εÙθε‹ά τις ¹ ΖΕ εÙθε‹άν τινα τ¾ν Β∆ δίχα τέµνει, κሠπρÕς Ñρθ¦ς αÙτ¾ν τέµνει· Ñρθ¾ ¥ρα ¹ ØπÕ ΖΕΒ.

In a circle, if two straight-lines, which are not through the center, cut one another, then they do not cut one another in half. Let ABCD be a circle, and within it, let two straightlines, AC and BD, which are not through the center, cut one another at (point) E. I say that they do not cut one another in half. For, if possible, let them cut one another in half, such that AE is equal to EC, and BE to ED. And let the center of the circle ABCD have been found [Prop. 3.1], and let it be (at point) F , and let F E have been joined. Therefore, since some straight-line through the center, F E, cuts in half some straight-line not through the center, AC, it also cuts it at right-angles [Prop. 3.3]. Thus,

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ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

™δείχθη δ κሠ¹ ØπÕ ΖΕΑ Ñρθή· ‡ση ¥ρα ¹ ØπÕ ΖΕΑ τÍ ØπÕ ΖΕΒ ¹ ™λάττων τÍ µείζονι· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα αƒ ΑΓ, Β∆ τέµνουσιν ¢λλήλας δίχα.

∆

Ζ

Α

F EA is a right-angle. Again, since some straight-line F E cuts in half some straight-line BD, it also cuts it at rightangles [Prop. 3.3]. Thus, F EB (is) a right-angle. But F EA was also shown (to be) a right-angle. Thus, F EA (is) equal to F EB, the lesser to the greater. The very thing is impossible. Thus, AC and BD do not cut one another in half.

Ε

D

F

A

E

Γ

Β

C B

'Ε¦ν ¥ρα ™ν κύκλJ δύο εÙθε‹αι τέµνωσιν ¢λλήλας µ¾ δˆα τοà κέντρου οâσαι, οÙ τέµνουσιν ¢λλήλας δίχα· Óπερ œδει δε‹ξαι.

Thus, in a circle, if two straight-lines, which are not through the center, cut one another, then they do not cut one another in half. (Which is) the very thing it was required to show.

ε΄.

Proposition 5

'Ε¦ν δύο κύκλοι τέµνωσιν ¢λλήλους, οÙκ œσται αÙτîν τÕ αÙτÕ κέντρον.

If two circles cut one another then they will not have the same center.

Α

A

Γ ∆

D Ε

Β

C

E Ζ

B Η

F G

∆ύο γ¦ρ κύκλοι οƒ ΑΒΓ, Γ∆Η τεµνέτωσαν ¢λλήλους κατ¦ τ¦ Β, Γ σηµε‹α. λέγω, Óτι οÙκ œσται αÙτîν τÕ αÙτÕ κέντρον. Ε„ γ¦ρ δυνατόν, œστω τÕ Ε, κሠ™πεζεύχθω ¹ ΕΓ, κሠδιήχθω ¹ ΕΖΗ, æς œτυχεν. κሠ™πεˆ τÕ Ε σηµε‹ον κέντρον ™στˆ τοà ΑΒΓ κύκλου, †ση ™στˆν ¹ ΕΓ τÍ ΕΖ. πάλιν, ™πεˆ τÕ Ε σηµε‹ον κέντρον ™στˆ τοà Γ∆Η κύκλου, ‡ση ™στˆν ¹ ΕΓ τÍ ΕΗ· ™δείχθη δ ¹ ΕΓ κሠτÍ ΕΖ ‡ση· κሠ¹ ΕΖ ¥ρα τÍ ΕΗ ™στιν ‡ση ¹ ™λάσσων τÍ µείζονι· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τÕ Ε σηµε‹ον κέντρον

For let the two circles ABC and CDG cut one another at points B and C. I say that they will not have the same center. For, if possible, let E be (the common center), and let EC have been joined, and let EF G have been drawn through (the two circles), at random. And since point E is the center of the circle ABC, EC is equal to EF . Again, since point E is the center of the circle CDG, EC is equal to EG. But EC was also shown (to be) equal to EF . Thus, EF is also equal to EG, the lesser to the

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ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

™στˆ τîν ΑΒΓ, Γ∆Η κύκλων. greater. The very thing is impossible. Thus, point E is not 'Ε¦ν ¥ρα δύο κύκλοι τέµνωσιν ¢λλήλους, οÙκ œστιν the (common) center of the circles ABC and CDG. αÙτîν τÕ αÙτÕ κέντρον· Óπερ œδει δε‹ξαι. Thus, if two circles cut one another then they will not have the same center. (Which is) the very thing it was required to show.

$΄.

Proposition 6

'Ε¦ν δύο κύκλοι ™φάπτωνται ¢λλήλων, οÙκ œσται If two circles touch one another then they will not αÙτîν τÕ αÙτÕ κέντρον. have the same center.

Γ

C

Ζ

F

Ε

E

Β

∆

B

D

Α

A

∆ύο γ¦ρ κύκλοι οƒ ΑΒΓ, Γ∆Ε ™φαπτέσθωσαν ¢λλήλων κατ¦ τÕ Γ σηµε‹ον· λέγω, Óτι οÙκ œσται αÙτîν τÕ αÙτÕ κέντρον. Ε„ γ¦ρ δυνατόν, œστω τÕ Ζ, κሠ™πεζεύχθω ¹ ΖΓ, κሠδιήχθω, æς œτυχεν, ¹ ΖΕΒ. 'Επεˆ οâν τÕ Ζ σηµε‹ον κέντρον ™στˆ τοà ΑΒΓ κύκλου, ‡ση ™στˆν ¹ ΖΓ τÍ ΖΒ. πάλιν, ™πεˆ τÕ Ζ σηµε‹ον κέντρον ™στˆ τοà Γ∆Ε κύκλου, ‡ση ™στˆν ¹ ΖΓ τÍ ΖΕ. ™δείχθη δ ¹ ΖΓ τÍ ΖΒ ‡ση· κሠ¹ ΖΕ ¥ρα τÍ ΖΒ ™στιν ‡ση, ¹ ™λάττων τÍ µείζονι· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τÕ Ζ σηµε‹ον κέντρον ™στˆ τîν ΑΒΓ, Γ∆Ε κύκλων. 'Ε¦ν ¥ρα δύο κύκλοι ™φάπτωνται ¢λλήλων, οÙκ œσται αÙτîν τÕ αÙτÕ κέντρον· Óπερ œδει δε‹ξαι.

For let the two circles ABC and CDE touch one another at point C. I say that they will not have the same center. For, if possible, let F be (the common center), and let F C have been joined, and let F EB have been drawn through (the two circles), at random. Therefore, since point F is the center of the circle ABC, F C is equal to F B. Again, since point F is the center of the circle CDE, F C is equal to F E. But F C was shown (to be) equal to F B. Thus, F E is also equal to F B, the lesser to the greater. The very thing is impossible. Thus, point F is not the (common) center of the circles ABC and CDE. Thus, if two circles touch one another then they will not have the same center. (Which is) the very thing it was required to show.

ζ΄.

Proposition 7

'Ε¦ν κύκλου ™πˆ τÁς διαµέτρου ληφθÍ τι σηµε‹ον, Ö µή ™στι κέντρον τοà κύκλου, ¢πÕ δ τοà σηµείου πρÕς τÕν κύκλον προσπίπτωσιν εÙθε‹αί τινες, µεγίστη µν œσται, ™φ' Âς τÕ κέντρον, ™λαχίστη δ ¹ λοιπή, τîν δ ¥λλων ¢εˆ ¹ œγγιον τÁς δˆα τοà κέντρου τÁς ¢πώτερον µείζων ™στίν, δύο δ µόνον ‡σαι ¢πÕ τοà σηµείου προσπεσοàνται πρÕς τÕν κύκλον ™φ' ˜κάτερα τÁς ™λαχίστης.

If some point, which is not the center of the circle, is taken on the diameter of a circle, and some straightlines radiate from the point towards the (circumference of the) circle, then the greatest (straight-line) will be that on which the center (lies), and the least the remainder (of the same diameter). And for the others, a (straightline) nearer† to the (straight-line) through the center is always greater than a (straight-line) further away. And only two equal (straight-lines) will radiate from the point

75

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 towards the (circumference of the) circle, (one) on each (side) of the least (straight-line).

Γ

C

Η

Β

Α

G

B

Ε

Ζ

Κ

∆

A

Θ

F

D

E

H K

”Εστω κύκλος Ð ΑΒΓ∆, διάµετρος δ αÙτοà œστω ¹ Α∆, κሠ™πˆ τÁς Α∆ ε„λήφθω τι σηµε‹ον τÕ Ζ, Ö µή ™στι κέντρον τοà κύκλου, κέντρον δ τοà κύκλου œστω τÕ Ε, κሠ¢πÕ τοà Ζ πρÕς τÕν ΑΒΓ∆ κύκλον προσπιπτέτωσαν εÙθε‹αί τινες αƒ ΖΒ, ΖΓ, ΖΗ· λέγω, Óτι µεγίστη µέν ™στιν ¹ ΖΑ, ™λαχίστη δ ¹ Ζ∆, τîν δ ¥λλων ¹ µν ΖΒ τÁς ΖΓ µείζων, ¹ δ ΖΓ τÁς ΖΗ. 'Επεζεύχθωσαν γ¦ρ αƒ ΒΕ, ΓΕ, ΗΕ. κሠ™πεˆ παντÕς τριγώνου αƒ δύο πλευρሠτÁς λοιπÁς µείζονές ε„σιν, αƒ ¥ρα ΕΒ, ΕΖ τÁς ΒΖ µείζονές ε„σιν. ‡ση δ ¹ ΑΕ τÍ ΒΕ [αƒ ¥ρα ΒΕ, ΕΖ ‡σαι ε„σˆ τÍ ΑΖ]· µείζων ¥ρα ¹ ΑΖ τÁς ΒΖ. πάλιν, ™πεˆ ‡ση ™στˆν ¹ ΒΕ τÍ ΓΕ, κοιν¾ δ ¹ ΖΕ, δύο δ¾ αƒ ΒΕ, ΕΖ δυσˆ τα‹ς ΓΕ, ΕΖ ‡σαι ε„σίν. ¢λλ¦ κሠγωνία ¹ ØπÕ ΒΕΖ γωνίας τÁς ØπÕ ΓΕΖ µείζων· βάσις ¥ρα ¹ ΒΖ βάσεως τÁς ΓΖ µείζων ™στίν. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΓΖ τÁς ΖΗ µείζων ™στίν. Πάλιν, ™πεˆ αƒ ΗΖ, ΖΕ τÁς ΕΗ µείζονές ε„σιν, ‡ση δ ¹ ΕΗ τÍ Ε∆, αƒ ¥ρα ΗΖ, ΖΕ τÁς Ε∆ µείζονές ε„σιν. κοιν¾ ¢φVρήσθω ¹ ΕΖ· λοιπ¾ ¥ρα ¹ ΗΖ λοιπÁς τÁς Ζ∆ µείζων ™στίν. µεγίστη µν ¥ρα ¹ ΖΑ, ™λαχίστη δ ¹ Ζ∆, µείζων δ ¹ µν ΖΒ τÁς ΖΓ, ¹ δ ΖΓ τÁς ΖΗ. Λέγω, Óτι κሠ¢πÕ τοà Ζ σηµείου δύο µόνον ‡σαι προσπεσοàνται πρÕς τÕν ΑΒΓ∆ κύκλον ™φ' ˜κάτερα τÁς Ζ∆ ™λαχίστης. συνεστάτω γ¦ρ πρÕς τÍ ΕΖ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Ε τÍ ØπÕ ΗΕΖ γωνίv ‡ση ¹ ØπÕ ΖΕΘ, κሠ™πεζεύχθω ¹ ΖΘ. ™πεˆ οâν ‡ση ™στˆν ¹ ΗΕ τÍ ΕΘ, κοιν¾ δ ¹ ΕΖ, δύο δ¾ αƒ ΗΕ, ΕΖ δυσˆ τα‹ς ΘΕ, ΕΖ ‡σαι ε„σίν· κሠγωνία ¹ ØπÕ ΗΕΖ γωνίv τÍ ØπÕ ΘΕΖ ‡ση· βάσις ¥ρα ¹ ΖΗ βάσει τÍ ΖΘ ‡ση ™στίν. λέγω δή, Óτι τÍ ΖΗ ¥λλη ‡ση οÙ προσπεσε‹ται πρÕς τÕν κύκλον ¢πÕ τοà Ζ σηµείου. ε„ γ¦ρ δυνατόν, προσπιπτέτω ¹ ΖΚ. κሠ™πεˆ ¹ ΖΚ τÍ ΖΗ ‡ση ™στίν, ¢λλ¦ ¹ ΖΘ τÍ ΖΗ [‡ση ™στίν], κሠ¹ ΖΚ ¥ρα τÍ ΖΘ

Let ABCD be a circle, and let AD be its diameter, and let some point F , which is not the center of the circle, have been taken on AD. Let E be the center of the circle. And let some straight-lines, F B, F C, and F G, radiate from F towards (the circumference of) circle ABCD. I say that F A is the greatest (straight-line), F D the least, and of the others, F B (is) greater than F C, and F C than F G. For let BE, CE, and GE have been joined. And since for every triangle (any) two sides are greater than the remaining (side) [Prop. 1.20], EB and EF is thus greater than BF . And AE (is) equal to BE [thus, BE and EF is equal to AF ]. Thus, AF (is) greater than BF . Again, since BE is equal to CE, and F E (is) common, the two (straight-lines) BE, EF are equal to the two (straightlines) CE, EF (respectively). But, angle BEF (is) also greater than angle CEF .‡ Thus, the base BF is greater than the base CF Thus, the base BF is greater than the base CF [Prop. 1.24]. So, for the same (reasons), CF is greater than F G. Again, since GF and F E are greater than EG [Prop. 1.20], and EG (is) equal to ED, GF and F E are thus greater than ED. Let EF have been taken from both. Thus, the remainder GF is greater than the remainder F D. Thus, F A (is) the greatest (straight-line), F D the least, and F B (is) greater than F C, and F C than F G. I also say that from point F only two equal (straightlines) will radiate towards (the circumference of) circle ABCD, (one) on each (side) of the least (straight-line) F D. For let the (angle) F EH, equal to angle GEF , have been constructed at the point E on the straight-line EF [Prop. 1.23], and let F H have been joined. There-

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ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

™στιν ‡ση, ¹ œγγιον τÁς δι¦ τοà κέντρου τÍ ¢πώτερον ‡ση· Óπερ ¢δύνατον. οÙκ ¥ρα ¢πÕ τοà Ζ σηµείου ˜τέρα τις προσπεσε‹ται πρÕς τÕν κύκλον ‡ση τÍ ΗΖ· µία ¥ρα µόνη. 'Ε¦ν ¥ρα κύκλου ™πˆ τÁς διαµέτρου ληφθÍ τι σηµε‹ον, Ö µή ™στι κέντρον τοà κύκλου, ¢πÕ δ τοà σηµείου πρÕς τÕν κύκλον προσπίπτωσιν εÙθε‹αί τινες, µεγίστη µν œσται, ™φ' Âς τÕ κέντρον, ™λαχίστη δ ¹ λοιπή, τîν δ ¥λλων ¢εˆ ¹ œγγιον τÁς δˆα τοà κέντρου τÁς ¢πώτερον µείζων ™στίν, δύο δ µόνον ‡σαι ¢πÕ τοà αÙτοà σηµείου προσπεσοàνται πρÕς τÕν κύκλον ™φ' ˜κάτερα τÁς ™λαχίστης· Óπερ œδει δε‹ξαι.

† ‡

fore, since GE is equal to EH, and EF (is) common, the two (straight-lines) GE, EF are equal to the two (straight-lines) HE, EF (respectively). And angle GEF (is) equal to angle HEF . Thus, the base F G is equal to the base F H [Prop. 1.4]. So I say that another (straightline) equal to F G will not radiate towards (the circumference of) the circle from point F . For, if possible, let F K (so) radiate. And since F K is equal to F G, but F H [is equal] to F G, F K is thus also equal to F H, the nearer to the (straight-line) through the center equal to the further away. The very thing (is) impossible. Thus, another (straight-line) equal to GF will not radiate towards (the circumference of) the circle. Thus, (there is) only one (such straight-line). Thus, if some point, which is not the center of the circle, is taken on the diameter of a circle, and some straight-lines radiate from the point towards the (circumference of the) circle, then the greatest (straight-line) will be that on which the center (lies), and the least the remainder (of the same diameter). And for the others, a (straight-line) nearer to the (straight-line) through the center is always greater than a (straight-line) further away. And only two equal (straight-lines) will radiate from the same point towards the (circumference of the) circle, (one) on each (side) of the least (straight-line). (Which is) the very thing it was required to show.

Presumably, in an angular sense. This is not proved, except by reference to the figure.

η΄.

Proposition 8

'Ε¦ν κύκλου ληφθÍ τι σηµε‹ον ™κτός, ¢πÕ δ τοà σηµείου πρÕς τÕν κύκλον διαχθîσιν εÙθε‹αί τινες, ïν µία µν δι¦ τοà κέντρου, αƒ δ λοιπαί, æς œτυχεν, τîν µν πρÕς τ¾ν κοίλην περιφέρειαν προσπιπτουσîν εÙθειîν µεγίστη µέν ™στιν ¹ δι¦ τοà κέντρου, τîν δ ¥λλων ¢εˆ ¹ œγγιον τÁς δι¦ τοà κέντρου τÁς ¢πώτερον µείζων ™στίν, τîν δ πρÕς τ¾ν κυρτ¾ν περιφέρειαν προσπιπτουσîν εÙθειîν ™λαχίστη µέν ™στιν ¹ µεταξÝ τοà τε σηµείου κሠτÁς διαµέτρου, τîν δ ¥λλων ¢εˆ ¹ œγγιον τÁς ™λαχίστης τÁς ¢πώτερόν ™στιν ™λάττων, δύο δ µόνον ‡σαι ¢πÕ τοà σηµείου προσπεσοàνται πρÕς τÕν κύκλον ™φ' ˜κάτερα τÁς ™λαχίστης. ”Εστω κύκλος Ð ΑΒΓ, κሠτοà ΑΒΓ ε„λήφθω τι σηµε‹ον ™κτÕς τÕ ∆, κሠ¢π' αÙτοà διήχθωσαν εÙθε‹αί τινες αƒ ∆Α, ∆Ε, ∆Ζ, ∆Γ, œστω δ ¹ ∆Α δι¦ τοà κέντρου. λέγω, Óτι τîν µν πρÕς τ¾ν ΑΕΖΓ κοίλην περιφέρειαν προσπιπτουσîν εÙθειîν µεγίστη µέν ™στιν ¹ δι¦ τοà κέντρου ¹ ∆Α, µείζων δ ¹ µν ∆Ε τÁς ∆Ζ ¹ δ ∆Ζ τÁς ∆Γ, τîν δ πρÕς τ¾ν ΘΛΚΗ κυρτ¾ν πε-

If some point is taken outside a circle, and some straight-lines are drawn from the point to the (circumference of the) circle, one of which (passes) through the center, the remainder (being) random, then for the straight-lines radiating towards the concave (part of the) circumference, the greatest is that (passing) through the center. For the others, a (straight-line) nearer† to the (straight-line) through the center is always greater than one further away. For the straight-lines radiating towards the convex (part of the) circumference, the least is that between the point and the diameter. For the others, a (straight-line) nearer to the least (straight-line) is always less than one further away. And only two equal (straightlines) will radiate towards the (circumference of the) circle, (one) on each (side) of the least (straight-line). Let ABC be a circle, and let some point D have been taken outside ABC, and from it let some straight-lines, DA, DE, DF , and DC, have been drawn through (the circle), and let DA be through the center. I say that for

77

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

ριφέρειαν προσπιπτουσîν εÙθειîν ™λαχίστη µέν ™στιν ¹ ∆Η ¹ µεταξÝ τοà σηµείου κሠτÁς διαµέτρου τÁς ΑΗ, ¢εˆ δ ¹ œγγιον τÁς ∆Η ™λαχίστης ™λάττων ™στˆ τÁς ¢πώτερον, ¹ µν ∆Κ τÁς ∆Λ, ¹ δ ∆Λ τÁς ∆Θ.

the straight-lines radiating towards the concave (part of the) circumference, AEF C, the greatest is the one (passing) through the center, (namely) AD, and (that) DE (is) greater than DF , and DF than DC. For the straight-lines radiating towards the convex (part of the) circumference, HLKG, the least is the one between the point and the diameter AG, (namely) DG, and a (straight-line) nearer to the least (straight-line) DG is always less than one farther away, (so that) DK (is less) than DL, and DL than than DH.

∆

D

Θ Λ Κ Η Β

Γ

H L K

B

C Ν

Ζ

G N

F

Μ

M

Ε

E Α

A

Ε„λήφθω γ¦ρ τÕ κέντρον τοà ΑΒΓ κύκλου κሠœστω τÕ Μ· κሠ™πεζεύχθωσαν αƒ ΜΕ, ΜΖ, ΜΓ, ΜΚ, ΜΛ, ΜΘ. Κሠ™πεˆ ‡ση ™στˆν ¹ ΑΜ τÍ ΕΜ, κοιν¾ προσκείσθω ¹ Μ∆· ¹ ¥ρα Α∆ ‡ση ™στˆ τα‹ς ΕΜ, Μ∆. ¢λλ' αƒ ΕΜ, Μ∆ τÁς Ε∆ µείζονές ε„σιν· κሠ¹ Α∆ ¥ρα τÁς Ε∆ µείζων ™στίν. πάλιν, ™πεˆ ‡ση ™στˆν ¹ ΜΕ τÍ ΜΖ, κοιν¾ δ ¹ Μ∆, αƒ ΕΜ, Μ∆ ¥ρα τα‹ς ΖΜ, Μ∆ ‡σαι ε„σίν· κሠγωνία ¹ ØπÕ ΕΜ∆ γωνίας τÁς ØπÕ ΖΜ∆ µείζων ™στίν. βάσις ¥ρα ¹ Ε∆ βάσεως τÁς Ζ∆ µείζων ™στίν· еοίως δ¾ δείξοµεν, Óτι κሠ¹ Ζ∆ τÁς Γ∆ µείζων ™στίν· µεγίστη µν ¥ρα ¹ ∆Α, µείζων δ ¹ µν ∆Ε τÁς ∆Ζ, ¹ δ ∆Ζ τÁς ∆Γ. Κሠ™πεˆ αƒ ΜΚ, Κ∆ τÁς Μ∆ µείζονές ε„σιν, ‡ση δ ¹ ΜΗ τÍ ΜΚ, λοιπ¾ ¥ρα ¹ Κ∆ λοιπÁς τÁς Η∆ µείζων ™στίν· éστε ¹ Η∆ τÁς Κ∆ ™λάττων ™στίν· κሠ™πεˆ τριγώνου τοà ΜΛ∆ ™πˆ µι©ς τîν πλευρîν τÁς Μ∆ δύο εÙθε‹αι ™ντÕς συνεστάθησαν αƒ ΜΚ, Κ∆, αƒ ¥ρα ΜΚ, Κ∆ τîν ΜΛ, Λ∆ ™λάττονές ε„σιν· ‡ση δ ¹ ΜΚ τÍ ΜΛ· λοιπ¾ ¥ρα ¹ ∆Κ λοιπÁς τÁς ∆Λ ™λάττων ™στίν. еοίως δ¾ δείξοµεν, Óτι κሠ¹ ∆Λ τÁς ∆Θ ™λάττων ™στίν· ™λαχίστη µν ¥ρα ¹ ∆Η, ™λάττων δ ¹ µν ∆Κ τÁς ∆Λ ¹ δ ∆Λ τÁς ∆Θ. Λέγω, Óτι κሠδύο µόνον ‡σαι ¢πÕ τοà ∆ σηµείου προσπεσοàνται πρÕς τÕν κύκλον ™φ' ˜κάτερα τÁς ∆Η

For let the center of the circle have been found [Prop. 3.1], and let it be (at point) M [Prop. 3.1]. And let M E, M F , M C, M K, M L, and M H have been joined. And since AM is equal to EM , let M D have been added to both. Thus, AD is equal to EM and M D. But, EM and M D is greater than ED [Prop. 1.20]. Thus, AD is also greater than ED. Again, since M E is equal to M F , and M D (is) common, the (straight-lines) EM , M D are thus equal to F M , M D. And angle EM D is greater than angle F M D.‡ Thus, the base ED is greater than the base F D [Prop. 1.24]. So, similarly, we can show that F D is also greater than CD. Thus, AD (is) the greatest (straight-line), and DE (is) greater than DF , and DF than DC. And since M K and KD is greater than M D [Prop. 1.20], and M G (is) equal to M K, the remainder KD is thus greater than the remainder GD. So GD is less than KD. And since in triangle M LD, the two internal straight-lines M K and KD were constructed on one of the sides, M D, then M K and KD are thus less than M L and LD [Prop. 1.21]. And M K (is) equal to M L. Thus, the remainder DK is less than the remainder DL. So, similarly, we can show that DL is also less than DH. Thus, DG (is) the least (straight-line), and DK (is) less than DL, and DL than DH.

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ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

™λαχίστης· συνεστάτω πρÕς τÍ Μ∆ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Μ τÍ ØπÕ ΚΜ∆ γωνίv ‡ση γωνία ¹ ØπÕ ∆ΜΒ, κሠ™πεζεύχθω ¹ ∆Β. κሠ™πεˆ ‡ση ™στˆν ¹ ΜΚ τÍ ΜΒ, κοιν¾ δ ¹ Μ∆, δύο δ¾ αƒ ΚΜ, Μ∆ δύο τα‹ς ΒΜ, Μ∆ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΚΜ∆ γωνίv τÍ ØπÕ ΒΜ∆ ‡ση· βάσις ¥ρα ¹ ∆Κ βάσει τÍ ∆Β ‡ση ™στίν. λέγω [δή], Óτι τÍ ∆Κ εÙθείv ¥λλη ‡ση οÙ προσπεσε‹ται πρÕς τÕν κύκλον ¢πÕ τοà ∆ σηµείου. ε„ γ¦ρ δυνατόν, προσπιπτέτω κሠœστω ¹ ∆Ν. ™πεˆ οâν ¹ ∆Κ τÍ ∆Ν ™στιν ‡ση, ¢λλ' ¹ ∆Κ τÍ ∆Β ™στιν ‡ση, κሠ¹ ∆Β ¥ρα τÍ ∆Ν ™στιν ‡ση, ¹ œγγιον τÁς ∆Η ™λαχίστης τÍ ¢πώτερον [™στιν] ‡ση· Óπερ ¢δύνατον ™δείχθη. οÙκ ¥ρα πλείους À δύο ‡σαι πρÕς τÕν ΑΒΓ κύκλον ¢πÕ τοà ∆ σηµείου ™φ' ˜κάτερα τÁς ∆Η ™λαχίστης προσπεσοàνται. 'Ε¦ν ¥ρα κύκλου ληφθÍ τι σηµε‹ον ™κτός, ¢πÕ δ τοà σηµείου πρÕς τÕν κύκλον διαχθîσιν εÙθε‹αί τινες, ïν µία µν δι¦ τοà κέντρου αƒ δ λοιπαί, æς œτυχεν, τîν µν πρÕς τ¾ν κοίλην περιφέρειαν προσπιπτουσîν εÙθειîν µεγίστη µέν ™στιν ¹ δι¦ τοà κέντου, τîν δ ¥λλων ¢εˆ ¹ œγγιον τÁς δι¦ τοà κέντρου τÁς ¢πώτερον µείζων ™στίν, τîν δ πρÕς τ¾ν κυρτ¾ν περιφέρειαν προσπιπτουσîν εÙθειîν ™λαχίστη µέν ™στιν ¹ µεταξÝ τοà τε σηµείου κሠτÁς διαµέτρου, τîν δ ¥λλων ¢εˆ ¹ œγγιον τÁς ™λαχίστης τÁς ¢πώτερόν ™στιν ™λάττων, δύο δ µόνον ‡σαι ¢πÕ τοà σηµείου προσπεσοàνται πρÕς τÕν κύκλον ™φ' ˜κάτερα τÁς ™λαχίστης· Óπερ œδει δε‹ξαι.

†

I also say that only two equal (straight-lines) will radiate from point D towards (the circumference of) the circle, (one) on each (side) on the least (straight-line), DG. Let the angle DM B, equal to angle KM D, have been constructed at the point M on the straight-line M D [Prop. 1.23], and let DB have been joined. And since M K is equal to M B, and M D (is) common, the two (straight-lines) KM , M D are equal to the two (straightlines) BM , M D, respectively. And angle KM D (is) equal to angle BM D. Thus, the base DK is equal to the base DB [Prop. 1.4]. [So] I say that another (straightline) equal to DK will not radiate towards the (circumference of the) circle from point D. For, if possible, let (such a straight-line) radiate, and let it be DN . Therefore, since DK is equal to DN , but DK is equal to DB, then DB is thus also equal to DN , (so that) a (straightline) nearer to the least (straight-line) DG [is] equal to one further off. The very thing was shown (to be) impossible. Thus, not more than two equal (straight-lines) will radiate towards (the circumference of) circle ABC from point D, (one) on each side of the least (straight-line) DG. Thus, if some point is taken outside a circle, and some straight-lines are drawn from the point to the (circumference of the) circle, one of which (passes) through the center, the remainder (being) random, then for the straightlines radiating towards the concave (part of the) circumference, the greatest is that (passing) through the center. For the others, a (straight-line) nearer to the (straightline) through the center is always greater than one further away. For the straight-lines radiating towards the convex (part of the) circumference, the least is that between the point and the diameter. For the others, a (straight-line) nearer to the least (straight-line) is always less than one further away. And only two equal (straightlines) will radiate towards the (circumference of the) circle, (one) on each (side) of the least (straight-line). (Which is) the very thing it was required to show.

Presumably, in an angular sense.

‡ This is not proved, except by reference to the figure.

θ΄.

Proposition 9

'Ε¦ν κύκλου ληφθÍ τι σηµε‹ον ™ντός, ¢πο δ τοà σηµείου πρÕς τÕν κύκλον προσπίπτωσι πλείους À δύο ‡σαι εÙθε‹αι, τÕ ληφθν σηµε‹ον κέντρον ™στˆ τοà κύκλου. ”Εστω κύκλος Ð ΑΒΓ, ™ντÕς δ αÙτοà σηµε‹ον τÕ ∆, κሠ¢πÕ τοà ∆ πρÕς τÕν ΑΒΓ κύκλον προσπιπτέτωσαν πλείους À δύο ‡σαι εÙθε‹αι αƒ ∆Α, ∆Β, ∆Γ· λέγω, Óτι τÕ ∆ σηµε‹ον κέντρον ™στˆ τοà ΑΒΓ κύκλου.

If some point is taken inside a circle, and more than two equal straight-lines radiate from the point towards the (circumference of the) circle, then the point taken is the center of the circle. Let ABC be a circle, and D a point inside it, and let more than two equal straight-lines, DA, DB, and DC, radiate from D towards (the circumference of) circle ABC.

79

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 I say that point D is the center of circle ABC.

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'Επεζεύχθωσαν γ¦ρ αƒ ΑΒ, ΒΓ κሠτετµήσθωσαν δίχα κατ¦ τ¦ Ε, Ζ σηµε‹α, κሠ™πιζευχθε‹σαι αƒ Ε∆, Ζ∆ διήχθωσαν ™πˆ τ¦ Η, Κ, Θ, Λ σηµε‹α. 'Επεˆ οâν ‡ση ™στˆν ¹ ΑΕ τÍ ΕΒ, κοιν¾ δ ¹ Ε∆, δύο δ¾ αƒ ΑΕ, Ε∆ δύο τα‹ς ΒΕ, Ε∆ ‡σαι ε„σίν· κሠβάσις ¹ ∆Α βάσει τÍ ∆Β ‡ση· γωνία ¥ρα ¹ ØπÕ ΑΕ∆ γωνίv τÍ ØπÕ ΒΕ∆ ‡ση ™στίν· Ñρθ¾ ¥ρα ˜κατέρα τîν ØπÕ ΑΕ∆, ΒΕ∆ γωνιîν· ¹ ΗΚ ¥ρα τ¾ν ΑΒ τέµνει δίχα κሠπρÕς Ñρθάς. κሠ™πεί, ™¦ν ™ν κύκλJ εÙθε‹ά τις εÙθε‹άν τινα δίχα τε κሠπρÕς Ñρθ¦ς τέµνV, ™πˆ τÁς τεµνούσης ™στˆ τÕ κέντρον τοà κύκλου, ™πˆ τÁς ΗΚ ¥ρα ™στˆ τÕ κέντρον τοà κύκλου. δι¦ τ¦ αÙτ¦ δ¾ κሠ™πˆ τÁς ΘΛ ™στι τÕ κέντρον τοà ΑΒΓ κύκλου. κሠοÙδν ›τερον κοινÕν œχουσιν αƒ ΗΚ, ΘΛ εÙθε‹αι À τÕ ∆ σηµε‹ον· τÕ ∆ ¥ρα σηµε‹ον κέντρον ™στˆ τοà ΑΒΓ κύκλου. 'Ε¦ν ¥ρα κύκλου ληφθÍ τι σηµε‹ον ™ντός, ¢πÕ δ τοà σηµείου πρÕς τÕν κύκλον προσπίπτωσι πλείους À δύο ‡σαι εÙθε‹αι, τÕ ληφθν σηµε‹ον κέντρον ™στˆ τοà κύκλου· Óπερ œδει δε‹ξαι.

For let AB and BC have been joined, and (then) have been cut in half at points E and F (respectively) [Prop. 1.10]. And ED and F D being joined, let them have been drawn through to points G, K, H, and L. Therefore, since AE is equal to EB, and ED (is) common, the two (straight-lines) AE, ED are equal to the two (straight-lines) BE, ED (respectively). And the base DA (is) equal to the base DB. Thus, angle AED is equal to angle BED [Prop. 1.8]. Thus, angles AED and BED (are) each right-angles [Def. 1.10]. Thus, GK cuts AB in half, and at right-angles. And since, if some straight-line in a circle cuts some (other) straight-line in half, and at right-angles, then the center of the circle is on the former (straight-line) [Prop. 3.1 corr.], the center of the circle is thus on GK. So, for the same (reasons), the center of circle ABC is also on HL. And the straight-lines GK and HL have no common (point) other than point D. Thus, point D is the center of circle ABC. Thus, if some point is taken inside a circle, and more than two equal straight-lines radiate from the point towards the (circumference of the) circle, then the point taken is the center of the circle. (Which is) the very thing it was required to show.

ι΄.

Proposition 10

Κύκλος κύκλον οÙ τέµνει κατ¦ πλείονα σηµε‹α À δύο. Ε„ γ¦ρ δυνατόν, κύκλος Ð ΑΒΓ κύκλον τÕν ∆ΕΖ τεµνέτω κατ¦ πλείονα σηµε‹α À δύο τ¦ Β, Η, Ζ, Θ, κሠ™πιζευχθε‹σαι αƒ ΒΘ, ΒΗ δίχα τεµνέσθωσαν κατ¦ τ¦ Κ, Λ σηµε‹α· κሠ¢πÕ τîν Κ, Λ τα‹ς ΒΘ, ΒΗ πρÕς Ñρθ¦ς ¢χθε‹σαι αƒ ΚΓ, ΛΜ διήχθωσαν ™πˆ τ¦ Α, Ε σηµε‹α.

A circle does not cut a(nother) circle at more than two points. For, if possible, let the circle ABC cut the circle DEF at more than two points, B, G, F , and H. And BH and BG being joined, let them (then) have been cut in half at points K and L (respectively). And KC and LM being drawn at right-angles to BH and BG from K and L (respectively) [Prop. 1.11], let them (then) have been drawn through to points A and E (respectively).

80

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 Α

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'Επεˆ οâν ™ν κύκλJ τù ΑΒΓ εÙθε‹ά τις ¹ ΑΓ εÙθε‹άν τινα τ¾ν ΒΘ δίχα κሠπρÕς Ñρθ¦ς τέµνει, ™πˆ τÁς ΑΓ ¥ρα ™στˆ τÕ κέντρον τοà ΑΒΓ κύκλου. πάλιν, ™πεˆ ™ν κύκλJ τù αÙτù τù ΑΒΓ εÙθε‹ά τις ¹ ΝΞ εÙθε‹άν τινα τ¾ν ΒΗ δίχα κሠπρÕς Ñρθ¦ς τέµνει, ™πˆ τÁς ΝΞ ¥ρα ™στˆ τÕ κέντρον τοà ΑΒΓ κύκλου. ™δείχθη δ κሠ™πˆ τÁς ΑΓ, κሠκατ' οÙδν συµβάλλουσιν αƒ ΑΓ, ΝΞ εÙθε‹αι À κατ¦ τÕ Ο· τÕ Ο ¥ρα σηµε‹ον κέντρον ™στˆ τοà ΑΒΓ κύκλου. еοίως δ¾ δείξοµεν, Óτι κሠτοà ∆ΕΖ κύκλου κέντρον ™στˆ τÕ Ο· δύο ¥ρα κύκλων τεµνόντων ¢λλήλους τîν ΑΒΓ, ∆ΕΖ τÕ αÙτό ™στι κέντρον τÕ Ο· Óπερ ™στˆν ¢δύνατον. ΟÙκ ¥ρα κύκλος κύκλον τέµνει κατ¦ πλείονα σηµε‹α À δύο· Óπερ œδει δε‹ξαι.

Therefore, since in circle ABC some straight-line AC cuts some (other) straight-line BH in half, and at right-angles, the center of circle ABC is thus on AC [Prop. 3.1 corr.]. Again, since in the same circle ABC some straight-line N O cuts some (other straight-line) BG in half, and at right-angles, the center of circle ABC is thus on N O [Prop. 3.1 corr.]. And it was also shown (to be) on AC. And the straight-lines AC and N O meet at no other (point) than P . Thus, point P is the center of circle ABC. So, similarly, we can show that P is also the center of circle DEF . Thus, two circles cutting one another, ABC and DEF , have the same center P . The very thing is impossible [Prop. 3.5]. Thus, a circle does not cut a(nother) circle at more than two points. (Which is) the very thing it was required to show.

ια΄.

Proposition 11

'Ε¦ν δύο κύκλοι ™φάπτωνται ¢λλήλων ™ντός, κሠληφθÍ αÙτîν τ¦ κέντρα, ¹ ™πˆ τ¦ κέντρα αÙτîν ™πιζευγνυµένη εÙθε‹α κሠ™κβαλλοµένη ™πˆ τ¾ν συναφ¾ν πεσε‹ται τîν κύκλων. ∆ύο γ¦ρ κύκλοι οƒ ΑΒΓ, Α∆Ε ™φαπτέσθωσαν ¢λλήλων ™ντÕς κατ¦ τÕ Α σηµε‹ον, κሠε„λήφθω τοà µν ΑΒΓ κύκλου κέντρον τÕ Ζ, τοà δ Α∆Ε τÕ Η· λέγω, Óτι ¹ ¢πÕ τοà Η ™πˆ τÕ Ζ ™πιζευγνυµένη εÙθε‹α ™κβαλλοµένη ™πˆ τÕ Α πεσε‹ται. Μ¾ γάρ, ¢λλ' ε„ δυνατόν, πιπτέτω æς ¹ ΖΗΘ, κሠ™πεζεύχθωσαν αƒ ΑΖ, ΑΗ. 'Επεˆ οâν αƒ ΑΗ, ΗΖ τÁς ΖΑ, τουτέστι τÁς ΖΘ, µείζονές ε„σιν, κοιν¾ ¢φVρήσθω ¹ ΖΗ· λοιπ¾ ¥ρα ¹ ΑΗ λοιπÁς τÁς ΗΘ µείζων ™στίν. ‡ση δ ¹ ΑΗ τÍ Η∆· κሠ¹ Η∆ ¥ρα τÁς ΗΘ µείζων ™στˆν ¹ ™λάττων τÁς µείζονος· Óπερ ™στˆν ¢δύνατον· οÙκ ¥ρα ¹ ¢πÕ τοà Ζ ™πˆ τÕ Η ™πιζευγνυµένη εÙθεˆα ™κτÕς πεσε‹ται· κατ¦ τÕ Α ¥ρα ™πˆ τÁς συναφÁς πεσε‹ται.

If two circles touch one another internally, and their centers are found, then the straight-line joining their centers, being produced, will fall upon the point of union of the circles. For let two circles, ABC and ADE, touch one another internally at point A, and let the center F of circle ABC have been found [Prop. 3.1], and (the center) G of (circle) ADE [Prop. 3.1]. I say that the line joining G to F , being produced, will fall on A. For (if) not then, if possible, let it fall like F GH (in the figure), and let AF and AG have been joined. Therefore, since AG and GF is greater than F A, that is to say F H [Prop. 1.20], let F G have been taken from both. Thus, the remainder AG is greater than the remainder GH. And AG (is) equal to GD. Thus, GD is also greater than GH, the lesser than the greater. The very thing is impossible. Thus, the straight-line joining F to G will not fall outside (one circle but inside the other). Thus, it will fall upon the point of union (of the circles)

81

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 at point A.

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'Ε¦ν ¥ρα δύο κύκλοι ™φάπτωνται ¢λλήλων ™ντός, [κሠThus, if two circles touch one another internally, [and ληφθÍ αÙτîν τ¦ κέντρα], ¹ ™πˆ τ¦ κέντρα αÙτîν ™πι- their centers are found], then the straight-line joining ζευγνυµένη εÙθε‹α [κሠ™κβαλλοµένη] ™πˆ τ¾ν συναφ¾ν their centers, [being produced], will fall upon the point πεσε‹ται τîν κύκλων· Óπερ œδει δε‹ξαι. of union of the circles. (Which is) the very thing it was required to show.

ιβ΄.

Proposition 12

'Ε¦ν δύο κύκλοι ™φάπτωνται ¢λλήλων ™κτός, ¹ ™πˆ τ¦ If two circles touch one another externally then the κέντρα αÙτîν ™πιζευγνυµένη δι¦ τÁς ™παφÁς ™λεύσεται. (straight-line) joining their centers will go through the point of union.

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∆ύο γ¦ρ κύκλοι οƒ ΑΒΓ, Α∆Ε ™φαπτέσθωσαν ¢λλήλων ™κτÕς κατ¦ τÕ Α σηµε‹ον, κሠε„λήφθω τοà µν ΑΒΓ κέντρον τÕ Ζ, τοà δ Α∆Ε τÕ Η· λέγω, Óτι ¹ ¢πÕ τοà Ζ ™πˆ τÕ Η ™πιζευγνυµένη εÙθε‹α δι¦ τÁς κατ¦ τÕ Α ™παφÁς ™λεύσεται. Μ¾ γάρ, ¢λλ' ε„ δυνατόν, ™ρχέσθω æς ¹ ΖΓ∆Η, κሠ™πεζεύχθωσαν αƒ ΑΖ, ΑΗ. 'Επεˆ οâν τÕ Ζ σηµε‹ον κέντρον ™στˆ τοà ΑΒΓ κύκλου, ‡ση ™στˆν ¹ ΖΑ τÍ ΖΓ. πάλιν, ™πεˆ τÕ Η σηµε‹ον κέντρον ™στˆ τοà Α∆Ε κύκλου, ‡ση ™στˆν ¹ ΗΑ τÍ Η∆.

For let two circles, ABC and ADE, touch one another externally at point A, and let the center F of ABC have been found [Prop. 3.1], and (the center) G of ADE [Prop. 3.1]. I say that the straight-line joining F to G will go through the point of union at A. For (if) not then, if possible, let it go like F CDG (in the figure), and let AF and AG have been joined. Therefore, since point F is the center of circle ABC, F A is equal to F C. Again, since point G is the center of circle ADE, GA is equal to GD. And F A was also shown

82

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

™δείχθη δ κሠ¹ ΖΑ τÍ ΖΓ ‡ση· αƒ ¥ρα ΖΑ, ΑΗ τα‹ς ΖΓ, Η∆ ‡σαι ε„σίν· éστε Óλη ¹ ΖΗ τîν ΖΑ, ΑΗ µείζων ™στίν· ¢λλ¦ κሠ™λάττων· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ¹ ¢πÕ τοà Ζ ™πˆ τÕ Η ™πιζευγνυµένη εÙθε‹α δι¦ τÁς κατ¦ τÕ Α ™παφÁς οÙκ ™λεύσεται· δι' αÙτÁς ¥ρα. 'Ε¦ν ¥ρα δύο κύκλοι ™φάπτωνται ¢λλήλων ™κτός, ¹ ™πˆ τ¦ κέντρα αÙτîν ™πιζευγνυµένη [εÙθε‹α] δι¦ τÁς ™παφÁς ™λεύσεται· Óπερ œδει δε‹ξαι.

(to be) equal to F C. Thus, the (straight-lines) F A and AG are equal to the (straight-lines) F C and GD. So the whole of F G is greater than F A and AG. But, (it is) also less [Prop. 1.20]. The very thing is impossible. Thus, the straight-line joining F to G will not fail to go through the point of union at A. Thus, (it will go) through it. Thus, if two circles touch one another externally then the [straight-line] joining their centers will go through the point of union. (Which is) the very thing it was required to show.

ιγ΄.

Proposition 13

Κύκλος κύκλου οÙκ ™φάπτεται κατ¦ πλείονα σηµε‹α À καθ' ›ν, ™άν τε ™ντÕς ™άν τε ™κτÕς ™φάπτηται.

A circle does not touch a(nother) circle at more than one point, whether they touch internally or externally.

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Ε„ γ¦ρ δυνατόν, κύκλος Ð ΑΒΓ∆ κύκλου τοà ΕΒΖ∆ ™φαπτέσθω πρότερον ™ντÕς κατ¦ πλείονα σηµε‹α À žν τ¦ ∆, Β. Κሠε„λήφθω τοà µν ΑΒΓ∆ κύκλου κέντρον τÕ Η, τοà δ ΕΒΖ∆ τÕ Θ. `Η ¥ρα ¢πÕ τοà Η ™πˆ τÕ Θ ™πιζευγνυµένη ™πˆ τ¦ Β, ∆ πεσε‹ται. πιπτέτω æς ¹ ΒΗΘ∆. κሠ™πεˆ τÕ Η σηµε‹ον κέντρον ™στˆ τοà ΑΒΓ∆ κύκλου, ‡ση ™στˆν ¹ ΒΗ τÍ Η∆· µείζων ¥ρα ¹ ΒΗ τÁς Θ∆· πολλù ¥ρα µείζων ¹ ΒΘ τÁς Θ∆. πάλιν, ™πεˆ τÕ Θ σηµε‹ον κέντρον ™στˆ τοà ΕΒΖ∆ κύκλου, ‡ση ™στˆν ¹ ΒΘ τÍ Θ∆· ™δείχθη δ αÙτÁς κሠπολλù µείζων· Óπερ ¢δύνατον· οÙκ ¥ρα κύκλος κύκλου ™φάπτεται ™ντÕς κατ¦ πλείονα σηµε‹α À ›ν. Λέγω δή, Óτι οÙδ ™κτός. Ε„ γ¦ρ δυνατόν, κύκλος Ð ΑΓΚ κύκλου τοà ΑΒΓ∆ ™φαπτέσθω ™κτÕς κατ¦ πλείονα σηµε‹α À žν τ¦ Α, Γ, κሠ™πεζεύχθω ¹ ΑΓ. ”Επεˆ οâν κύκλων τîν ΑΒΓ∆, ΑΓΚ ε‡ληπται ™πˆ τÁς περιφερείας ˜κατέρου δύο τυχόντα σηµε‹α τ¦ Α, Γ, ¹ ™πˆ τ¦ σηµε‹α ™πιζευγνυµένη εÙθε‹α ™ντÕς ˜κατέρου πεσε‹ται· ¢λλ¦ τοà µν ΑΒΓ∆ ™ντÕς œπεσεν, τοà δ ΑΓΚ ™κτός· Óπερ ¥τοπον· οÙκ ¥ρα κύκλος κύκλου ™φάπτεται ™κτÕς κατ¦ πλείονα σηµε‹α À ›ν. ™δείχθη δέ, Óτι οÙδ ™ντός.

For, if possible, let circle ABDC † touch circle EBF D— first of all, internally—at more than one point, D and B. And let the center G of circle ABDC have been found [Prop. 3.1], and (the center) H of EBF D [Prop. 3.1]. Thus, the (straight-line) joining G and H will fall on B and D [Prop. 3.11]. Let it fall like BGHD (in the figure). And since point G is the center of circle ABDC, BG is equal to GD. Thus, BG (is) greater than HD. Thus, BH (is) much greater than HD. Again, since point H is the center of circle EBF D, BH is equal to HD. But it was also shown (to be) much greater than the same. The very thing (is) impossible. Thus, a circle does not touch a(nother) circle internally at more than one point. So, I say that neither (does it touch) externally (at more than one point). For, if possible, let circle ACK touch circle ABDC externally at more than one point, A and C. And let AC have been joined. Therefore, since two points, A and C, have been taken somewhere on the circumference of each of the circles ABDC and ACK, the straight-line joining the points will fall inside each (circle) [Prop. 3.2]. But, it fell inside ABDC, and outside ACK [Def. 3.3]. The very thing

83

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

Κύκλος ¥ρα κύκλου οÙκ ™φάπτεται κατ¦ πλείονα σηµε‹α À [καθ'] ›ν, ™άν τε ™ντÕς ™άν τε ™κτÕς ™φάπτηται· Óπερ œδει δε‹ξαι.

†

(is) absurd. Thus, a circle does not touch a(nother) circle externally at more than one point. And it was shown that neither (does it) internally. Thus, a circle does not touch a(nother) circle at more than one point, whether they touch internally or externally. (Which is) the very thing it was required to show.

The Greek text has “ABCD”, which is obviously a mistake.

ιδ΄.

Proposition 14

'Εν κύκλJ αƒ ‡σαι εÙθε‹αι ‡σον ¢πέχουσιν ¢πÕ τοà In a circle, equal straight-lines are equally far from the κέντρου, καˆ αƒ ‡σον ¢πέχουσαι ¢πÕ τοà κέντρου ‡σαι center, and (straight-lines) which are equally far from the ¢λλήλαις ε„σίν. center are equal to one another.

∆

Β

D

B Η

Ε

G

E

Ζ

F Γ

C

Α

A

”Εστω κύκλος Ð ΑΒΓ∆, κሠ™ν αÙτù ‡σαι εÙθε‹αι œστωσαν αƒ ΑΒ, Γ∆· λέγω, Óτι αƒ ΑΒ, Γ∆ ‡σον ¢πέχουσιν ¢πÕ τοà κέντρου. Ε„λήφθω γ¦ρ τÕ κέντον τοà ΑΒΓ∆ κύκλου κሠœστω τÕ Ε, κሠ¢πÕ τοà Ε ™πˆ τ¦ς ΑΒ, Γ∆ κάθετοι ½χθωσαν αƒ ΕΖ, ΕΗ, κሠ™πεζεύχθωσαν αƒ ΑΕ, ΕΓ. 'Επεˆ οâν εÙθε‹ά τις δˆα τοà κέντρου ¹ ΕΖ εÙθε‹άν τινα µ¾ δι¦ τοà κέντρου τ¾ν ΑΒ πρÕς Ñρθ¦ς τέµνει, κሠδίχα αÙτ¾ν τέµνει. ‡ση ¥ρα ¹ ΑΖ τÍ ΖΒ· διπλÁ ¥ρα ¹ ΑΒ τÁς ΑΖ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ Γ∆ τÁς ΓΗ ™στι διπλÁ· καί ™στιν ‡ση ¹ ΑΒ τÍ Γ∆· ‡ση ¥ρα κሠ¹ ΑΖ τÍ ΓΗ. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΕ τÍ ΕΓ, ‡σον κሠτÕ ¢πÕ τÁς ΑΕ τù ¢πÕ τÁς ΕΓ. ¢λλ¦ τù µν ¢πÕ τÁς ΑΕ ‡σα τ¦ ¢πÕ τîν ΑΖ, ΕΖ· Ñρθ¾ γ¦ρ ¹ πρÕς τù Ζ γωνία· τù δ ¢πÕ τÁς ΕΓ ‡σα τ¦ ¢πÕ τîν ΕΗ, ΗΓ· Ñρθ¾ γ¦ρ ¹ πρÕς τù Η γωνία· τ¦ ¥ρα ¢πÕ τîν ΑΖ, ΖΕ ‡σα ™στˆ το‹ς ¢πÕ τîν ΓΗ, ΗΕ, ïν τÕ ¢πÕ τÁς ΑΖ ‡σον ™στˆ τù ¢πÕ τÁς ΓΗ· ‡ση γάρ ™στιν ¹ ΑΖ τÍ ΓΗ· λοιπÕν ¥ρα τÕ ¢πÕ τÁς ΖΕ τù ¢πÕ τÁς ΕΗ ‡σον ™στίν· ‡ση ¥ρα ¹ ΕΖ τÍ ΕΗ. ™ν δ κύκλJ ‡σον ¢πέχειν ¢πÕ τοà κέντρου εÙθε‹αι λέγονται, Óταν αƒ ¢πÕ τοà κέντρου ™π' αÙτ¦ς κάθετοι ¢γόµεναι ‡σαι ðσιν· αƒ ¥ρα ΑΒ, Γ∆ ‡σον ¢πέχουσιν ¢πÕ τοà κέντρου. 'Αλλ¦ δ¾ αƒ ΑΒ, Γ∆ εÙθε‹αι ‡σον ¢πεχέτωσαν ¢πÕ

Let ABDC † be a circle, and let AB and CD be equal straight-lines within it. I say that AB and CD are equally far from the center. For let the center of circle ABDC have been found [Prop. 3.1], and let it be (at) E. And let EF and EG have been drawn from (point) E, perpendicular to AB and CD (respectively) [Prop. 1.12]. And let AE and EC have been joined. Therefore, since some straight-line, EF , through the center (of the circle), cuts some (other) straight-line, AB, not through the center, at right-angles, it also cuts it in half [Prop. 3.3]. Thus, AF (is) equal to F B. Thus, AB (is) double AF . So, for the same (reasons), CD is also double CG. And AB is equal to CD. Thus, AF (is) also equal to CG. And since AE is equal to EC, the (square) on AE (is) also equal to the (square) on EC. But, the (sum of the squares) on AF and EF (is) equal to the (square) on AE. For the angle at F (is) a rightangle [Prop. 1.47]. And the (sum of the squares) on EG and GC (is) equal to the (square) on EC. For the angle at G (is) a right-angle [Prop. 1.47]. Thus, the (sum of the squares) on AF and F E is equal to the (sum of the squares) on CG and GE, of which the (square) on AF is equal to the (square) on CG. For AF is equal to CG.

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ELEMENTS BOOK 3

τοà κέντρου, τουτέστιν ‡ση œστω ¹ ΕΖ τÍ ΕΗ. λέγω, Óτι ‡ση ™στˆ κሠ¹ ΑΒ τÍ Γ∆. Τîν γ¦ρ αÙτîν κατασκευασθέντων еοίως δείξοµεν, Óτι διπλÁ ™στιν ¹ µν ΑΒ τÁς ΑΖ, ¹ δ Γ∆ τÁς ΓΗ· κሠ™πεˆ ‡ση ™στˆν ¹ ΑΕ τÍ ΓΕ, ‡σον ™στˆ τÕ ¢πÕ τÁς ΑΕ τù ¢πÕ τÁς ΓΕ· ¢λλ¦ τù µν ¢πÕ τÁς ΑΕ ‡σα ™στˆ τ¦ ¢πÕ τîν ΕΖ, ΖΑ, τù δ ¢πÕ τÁς ΓΕ ‡σα τ¦ ¢πÕ τîν ΕΗ, ΗΓ. τ¦ ¥ρα ¢πÕ τîν ΕΖ, ΖΑ ‡σα ™στˆ το‹ς ¢πÕ τîν ΕΗ, ΗΓ· ïν τÕ ¢πÕ τÁς ΕΖ τù ¢πÕ τÁς ΕΗ ™στιν ‡σον· ‡ση γ¦ρ ¹ ΕΖ τÍ ΕΗ· λοιπÕν ¥ρα τÕ ¢πÕ τÁς ΑΖ ‡σον ™στˆ τù ¢πÕ τÁς ΓΗ· ‡ση ¥ρα ¹ ΑΖ τÍ ΓΗ· καί ™στι τÁς µν ΑΖ διπλÁ ¹ ΑΒ, τÁς δ ΓΗ διπλÁ ¹ Γ∆· ‡ση ¥ρα ¹ ΑΒ τÍ Γ∆. 'Εν κύκλJ ¥ρα αƒ ‡σαι εÙθε‹αι ‡σον ¢πέχουσιν ¢πÕ τοà κέντρου, καˆ αƒ ‡σον ¢πέχουσαι ¢πÕ τοà κέντρου ‡σαι ¢λλήλαις ε„σίν· Óπερ œδει δε‹ξαι.

†

Thus, the remaining (square) on F E is equal to the (remaining square) on EG. Thus, EF (is) equal to EG. And straight-lines in a circle are said to be equally far from the center when perpendicular (straight-lines) which are drawn to them from the center are equal [Def. 3.4]. Thus, AB and CD are equally far from the center. So, let the straight-lines AB and CD be equally far from the center. That is to say, let EF be equal to EG. I say that AB is also equal to CD. For, with the same construction, we can, similarly, show that AB is double AF , and CD (double) CG. And since AE is equal to CE, the (square) on AE is equal to the (square) on CE. But, the (sum of the squares) on EF and F A is equal to the (square) on AE [Prop. 1.47]. And the (sum of the squares) on EG and GC (is) equal to the (square) on CE [Prop. 1.47]. Thus, the (sum of the squares) on EF and F A is equal to the (sum of the squares) on EG and GC, of which the (square) on EF is equal to the (square) on EG. For EF (is) equal to EG. Thus, the remaining (square) on AF is equal to the (remaining square) on CG. Thus, AF (is) equal to CG. And AB is double AF , and CD double CG. Thus, AB (is) equal to CD. Thus, in a circle, equal straight-lines are equally far from the center, and (straight-lines) which are equally far from the center are equal to one another. (Which is) the very thing it was required to show.

The Greek text has “ABCD”, which is obviously a mistake.

ιε΄.

Proposition 15

'Εν κύκλJ µεγίστη µν ¹ διάµετρος, τîν δ ¥λλων ¢εˆ ¹ œγγιον τοà κέντρου τÁς ¢πώτερον µείζων ™στίν. ”Εστω κύκλος Ð ΑΒΓ∆, διάµετρος δ αÙτοà œστω ¹ Α∆, κέντρον δ τÕ Ε, κሠœγγιον µν τÁς Α∆ διαµέτρου œστω ¹ ΒΓ, ¢πώτερον δ ¹ ΖΗ· λέγω, Óτι µεγίστη µέν ™στιν ¹ Α∆, µείζων δ ¹ ΒΓ τÁς ΖΗ. ”Ηχθωσαν γ¦ρ ¢πÕ τοà Ε κέντρου ™πˆ τ¦ς ΒΓ, ΖΗ κάθετοι αƒ ΕΘ, ΕΚ. κሠ™πεˆ œγγιον µν τοà κέντρου ™στˆν ¹ ΒΓ, ¢πώτερον δ ¹ ΖΗ, µείζων ¥ρα ¹ ΕΚ τÁς ΕΘ. κείσθω τÍ ΕΘ ‡ση ¹ ΕΛ, κሠδι¦ τοà Λ τÍ ΕΚ πρÕς Ñρθ¦ς ¢χθε‹σα ¹ ΛΜ διήχθω ™πˆ τÕ Ν, κሠ™πεζεύχθωσαν αƒ ΜΕ, ΕΝ, ΖΕ, ΕΗ. Κሠ™πεˆ ‡ση ™στˆν ¹ ΕΘ τÍ ΕΛ, ‡ση ™στˆ κሠ¹ ΒΓ τÍ ΜΝ. πάλιν, ™πεˆ ‡ση ™στˆν ¹ µν ΑΕ τÍ ΕΜ, ¹ δ Ε∆ τÍ ΕΝ, ¹ ¥ρα Α∆ τα‹ς ΜΕ, ΕΝ ‡ση ™στίν. ¢λλ' αƒ µν ΜΕ, ΕΝ τÁς ΜΝ µείζονές ε„σιν [κሠ¹ Α∆ τÁς ΜΝ µείζων ™στίν], ‡ση δ ¹ ΜΝ τÍ ΒΓ· ¹ Α∆ ¥ρα τÁς ΒΓ µείζων ™στίν. κሠ™πεˆ δύο αƒ ΜΕ, ΕΝ δύο τα‹ς ΖΕ, ΕΗ ‡σαι ε„σίν, κሠγωνία ¹ ØπÕ ΜΕΝ γωνίας τÁς ØπÕ ΖΕΗ

In a circle, a diameter (is) the greatest (straight-line), and for the others, a (straight-line) nearer to the center is always greater than one further away. Let ABCD be a circle, and let AD be its diameter, and E (its) center. And let BC be nearer to the diameter AD,† and F G further away. I say that AD is the greatest (straight-line), and BC (is) greater than F G. For let EH and EK have been drawn from the center E, at right-angles to BC and F G (respectively) [Prop. 1.12]. And since BC is nearer to the center, and F G further away, EK (is) thus greater than EH [Def. 3.5]. Let EL be made equal to EH [Prop. 1.3]. And LM being drawn through L, at right-angles to EK [Prop. 1.11], let it have been drawn through to N . And let M E, EN , F E, and EG have been joined. And since EH is equal to EL, BC is also equal to M N [Prop. 3.14]. Again, since AE is equal to EM , and ED to EN , AD is thus equal to M E and EN . But, M E and EN is greater than M N [Prop. 1.20] [also AD is

85

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

µείζων [™στίν], βάσις ¥ρα ¹ ΜΝ βάσεως τÁς ΖΗ µείζων ™στίν. ¢λλ¦ ¹ ΜΝ τÍ ΒΓ ™δείχθη ‡ση [κሠ¹ ΒΓ τÁς ΖΗ µείζων ™στίν]. µεγίστη µν ¥ρα ¹ Α∆ διάµετρος, µείζων δ ¹ ΒΓ τÁς ΖΗ.

Μ

Α Β

greater than M N ], and M N (is) equal to BC. Thus, AD is greater than BC. And since the two (straight-lines) M E, EN are equal to the two (straight-lines) F E, EG (respectively), and angle M EN [is] greater than angle F EG,‡ the base M N is thus greater than the base F G [Prop. 1.24]. But, M N was shown (to be) equal to BC [(so) BC is also greater than F G]. Thus, the diameter AD (is) the greatest (straight-line), and BC (is) greater than F G. A M B

Ζ

Κ

F

Λ

Ε

K

L

E H

Θ

G

Η

C D 'Εν κύκλJ ¥ρα µεγίστη µν έστιν ¹ διάµετρος, τîν Thus, in a circle, a diameter (is) the greatest (straightδ ¥λλων ¢εˆ ¹ œγγιον τοà κέντρου τÁς ¢πώτερον µείζων line), and for the others, a (straight-line) nearer to the ™στίν· Óπερ œδει δε‹ξαι. center is always greater than one further away. (Which is) the very thing it was required to show. Ν

† ‡

∆

Γ

N

Euclid should have said “to the center”, rather than ”to the diameter AD”, since BC, AD and F G are not necessarily parallel. This is not proved, except by reference to the figure.

ι$΄.

Proposition 16

`Η τÍ διαµέτρJ τοà κύκλου πρÕς Ñρθ¦ς ¢π' ¥κρας ¢γοµένη ™κτÕς πεσε‹ται τοà κύκλου, κሠε„ς τÕν µεταξÝ τόπον τÁς τε εÙθείας κሠτÁς περιφερείας ˜τέρα εÙθε‹α οÙ παρεµπεσε‹ται, κሠ¹ µν τοà ¹µικυκλίου γωνία ¡πάσης γωνίας Ñξείας εÙθυγράµµου µείζων ™στίν, ¹ δ λοιπ¾ ™λάττων. ”Εστω κύκλος Ð ΑΒΓ περˆ κέντρον τÕ ∆ κሠδιάµετρον τ¾ν ΑΒ· λέγω, Óτι ¹ ¢πÕ τοà Α τÍ ΑΒ πρÕς Ñρθ¦ς ¢π' ¥κρας ¢γοµένη ™κτÕς πεσε‹ται τοà κύκλου. Μ¾ γάρ, ¢λλ' ε„ δυνατόν, πιπτέτω ™ντÕς æς ¹ ΓΑ, κሠ™πεζεύχθω ¹ ∆Γ. 'Επεˆ ‡ση ™στˆν ¹ ∆Α τÍ ∆Γ, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ∆ΑΓ γωνίv τÍ ØπÕ ΑΓ∆. Ñρθ¾ δ ¹ ØπÕ ∆ΑΓ· Ñρθ¾ ¥ρα κሠ¹ ØπÕ ΑΓ∆· τριγώνου δ¾ τοà ΑΓ∆ αƒ δύο γωνίαι αƒ ØπÕ ∆ΑΓ, ΑΓ∆ δύο Ñρθα‹ς ‡σαι ε„σίν· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ¹ ¢πÕ τοà Α σηµείου τÍ ΒΑ πρÕς Ñρθ¦ς ¢γοµένη ™ντÕς πεσε‹ται τοà κύκλου. еοίως δ¾ δε‹ξοµεν, Óτι οÙδ' ™πˆ τÁς περιφερείας· ™κτÕς ¥ρα.

A (straight-line) drawn at right-angles to the diameter of a circle, from its end, will fall outside the circle. And another straight-line cannot be inserted into the space between the (aforementioned) straight-line and the circumference. And the angle of the semi-circle is greater than any acute rectilinear angle whatsoever, and the remaining (angle is) less (than any acute rectilinear angle). Let ABC be a circle around the center D and the diameter AB. I say that the (straight-line) drawn from A, at right-angles to AB [Prop 1.11], from its end, will fall outside the circle. For (if) not then, if possible, let it fall inside, like CA (in the figure), and let DC have been joined. Since DA is equal to DC, angle DAC is also equal to angle ACD [Prop. 1.5]. And DAC (is) a right-angle. Thus, ACD (is) also a right-angle. So, in triangle ACD, the two angles DAC and ACD are equal to two rightangles. The very thing is impossible [Prop. 1.17]. Thus, the (straight-line) drawn from point A, at right-angles

86

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 to BA, will not fall inside the circle. So, similarly, we can show that neither (will it fall) on the circumference. Thus, (it will fall) outside (the circle).

Β

Γ

Ζ Ε

Η

B

C

∆

D

Θ

H F E

Α

Πιπτέτω æς ¹ ΑΕ· λέγω δή, Óτι ε„ς τÕν µεταξÝ τόπον τÁς τε ΑΕ εÙθείας κሠτÁς ΓΘΑ περιφερείας ˜τέρα εÙθε‹α οÙ παρεµπεσε‹ται. Ε„ γ¦ρ δυνατόν, παρεµπιπτέτω æς ¹ ΖΑ, κሠ½χθω ¢πÕ τοà ∆ σηµείου ™πˆ τÁν ΖΑ κάθετος ¹ ∆Η. κሠ™πεˆ Ñρθή ™στιν ¹ ØπÕ ΑΗ∆, ™λάττων δ ÑρθÁς ¹ ØπÕ ∆ΑΗ, µείζων ¥ρα ¹ Α∆ τÁς ∆Η. ‡ση δ ¹ ∆Α τÍ ∆Θ· µείζων ¥ρα ¹ ∆Θ τÁς ∆Η, ¹ ™λάττων τÁς µείζονος· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ε„ς τÕν µεταξÝ τόπον τÁς τε εÙθείας κሠτÁς περιφερείας ˜τέρα εÙθε‹α παρεµπεσε‹ται. Λέγω, Óτι κሠ¹ µν τοà ¹µικυκλίου γωνία ¹ περιεχοµένη Øπό τε τÁς ΒΑ εÙθείας κሠτÁς ΓΘΑ περιφερείας ¡πάσης γωνίας Ñξείας εÙθυγράµµου µείζων ™στίν, ¹ δ λοιπ¾ ¹ περιεχοµένη Øπό τε τÁς ΓΘΑ περιφερείας κሠτÁς ΑΕ εÙθείας ¡πάσης γωνίας Ñξείας εÙθυγράµµου ™λάττων ™στίν. Ε„ γ¦ρ ™στί τις γωνία εÙθύγραµµος µείζων µν τÁς περιεχοµένης Øπό τε τÁς ΒΑ εÙθείας κሠτÁς ΓΘΑ περιφερείας, ™λάττων δ τÁς περιεχοµένης Øπό τε τÁς ΓΘΑ περιφερείας κሠτ¾ς ΑΕ εÙθείας, ε„ς τÕν µεταξÝ τόπον τÁς τε ΓΘΑ περιφερείας κሠτÁς ΑΕ εÙθείας εÙθε‹α παρεµπεσε‹ται, ¼τις ποιήσει µείζονα µν τÁς περιεχοµένης ØπÕ τε τÁς ΒΑ εÙθείας κሠτÁς ΓΘΑ περιφερείας ØπÕ εÙθειîν περιεχοµένην, ™λάττονα δ τÁς περιεχοµένης Øπό τε τÁς ΓΘΑ περιφερείας κሠτÁς ΑΕ εÙθείας. οÙ παρεµπίπτει δέ· οÙκ ¥ρα τÁς περιεχοµένης γωνίας Øπό τε τÁς ΒΑ εÙθείας κሠτÁς ΓΘΑ περιφερείας œσται µείζων Ñξε‹α ØπÕ εÙθειîν περιεχοµένη, οÙδ µ¾ν ™λάττων τÁς περιεχοµένης Øπό τε τÁς ΓΘΑ περιφερείας κሠτÁς ΑΕ εÙθείας.

G A

Let it fall like AE (in the figure). So, I say that another straight-line cannot be inserted into the space between the straight-line AE and the circumference CHA. For, if possible, let it be inserted like F A (in the figure), and let DG have been drawn from point D, perpendicular to F A [Prop. 1.12]. And since AGD is a rightangle, and DAG (is) less than a right-angle, AD (is) thus greater than DG [Prop. 1.19]. And DA (is) equal to DH. Thus, DH (is) greater than DG, the lesser than the greater. The very thing is impossible. Thus, another straight-line cannot be inserted into the space between the straight-line (AE) and the circumference. And I also say that the semi-circular angle contained by the straight-line BA and the circumference CHA is greater than any acute rectilinear angle whatsoever, and the remaining (angle) contained by the circumference CHA and the straight-line AE is less than any acute rectilinear angle whatsoever. For if any rectilinear angle is greater than the (angle) contained by the straight-line BA and the circumference CHA, or less than the (angle) contained by the circumference CHA and the straight-line AE, then a straight-line can be inserted into the space between the circumference CHA and the straight-line AE—anything which will make (an angle) contained by straight-lines greater than the angle contained by the straight-line BA and the circumference CHA, or less than the (angle) contained by the circumference CHA and the straightline AE. But (such a straight-line) cannot be inserted. Thus, an acute (angle) contained by straight-lines cannot be greater than the angle contained by the straight-line BA and the circumference CHA, neither (can it be) less than the (angle) contained by the circumference CHA and the straight-line AE.

87

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι ¹ τÍ διαµέτρJ τοà κύκλου πρÕς Ñρθ¦ς ¢π' ¥κρας ¢γοµένη ™φάπτεται τοà κύκλου [κሠÓτι εÙθε‹α κύκλου καθ' žν µόνον ™φάπτεται σηµε‹ον, ™πειδήπερ κሠ¹ κατ¦ δύο αÙτù συµβάλλουσα ™ντÕς αÙτοà πίπτουσα ™δείχθη]· Óπερ œδει δε‹ξαι.

So, from this, (it is) manifest that a (straight-line) drawn at right-angles to the diameter of a circle, from its end, touches the circle [and that the straight-line touches the circle at a single point, inasmuch as it was also shown that a (straight-line) meeting (the circle) at two (points) falls inside it [Prop. 3.2] ]. (Which is) the very thing it was required to show.

ιζ΄.

Proposition 17

'ΑπÕ τοà δοθέντος σηµείου τοà δοθέντος κύκλου ™φαπτοµένην εÙθε‹αν γραµµ¾ν ¢γαγε‹ν.

To draw a straight-line touching a given circle from a given point.

Α Ζ

A F

∆ Β

D B

Ε

Γ

Η

C

G

E

”Εστω τÕ µν δοθν σηµε‹ον τÕ Α, Ð δ δοθεˆς κύκλος Ð ΒΓ∆· δε‹ δ¾ ¢πÕ τοà Α σηµείου τοà ΒΓ∆ κύκλου ™φαπτοµένην εÙθε‹αν γραµµ¾ν ¢γαγε‹ν. Ε„λήφθω γ¦ρ τÕ κέντρον τοà κύκλου τÕ Ε, κሠ™πεζεύχθω ¹ ΑΕ, κሠκέντρJ µν τù Ε διαστήµατι δ τù ΕΑ κύκλος γεγράφθω Ð ΑΖΗ, κሠ¢πÕ τοà ∆ τÍ ΕΑ πρÕς Ñρθ¦ς ½χθω ¹ ∆Ζ, κሠ™πεζεύχθωσαν αƒ ΕΖ, ΑΒ· λέγω, Óτι ¢πÕ τοà Α σηµείου τοà ΒΓ∆ κύκλου ™φαπτοµένη Ãκται ¹ ΑΒ. 'Επεˆ γ¦ρ τÕ Ε κέντρον ™στˆ τîν ΒΓ∆, ΑΖΗ κύκλων, ‡ση ¥ρα ™στˆν ¹ µν ΕΑ τÍ ΕΖ, ¹ δ Ε∆ τÍ ΕΒ· δύο δ¾ αƒ ΑΕ, ΕΒ δύο τα‹ς ΖΕ, Ε∆ ‡σαι ε„σίν· κሠγωνίαν κοιν¾ν περιέχουσι τ¾ν πρÕς τù Ε· βάσις ¥ρα ¹ ∆Ζ βάσει τÍ ΑΒ ‡ση ™στίν, κሠτÕ ∆ΕΖ τρίγωνον τù ΕΒΑ τριγώνJ ‡σον ™στίν, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις· ‡ση ¥ρα ¹ ØπÕ Ε∆Ζ τÍ ØπÕ ΕΒΑ. Ñρθ¾ δ ¹ ØπÕ Ε∆Ζ· Ñρθ¾ ¥ρα κሠ¹ ØπÕ ΕΒΑ. καί ™στιν ¹ ΕΒ ™κ τοà κέντρου· ¹ δ τÍ διαµέτρJ τοà κύκλου πρÕς Ñρθ¦ς ¢π' ¥κρας ¢γοµένη ™φάπτεται τοà κύκλου· ¹ ΑΒ ¥ρα ™φάπτεται τοà ΒΓ∆ κύκλου. 'ΑπÕ τοà ¥ρα δοθέντος σηµείου τοà Α τοà δοθέντος κύκλου τοà ΒΓ∆ ™φαπτοµένη εÙθε‹α γραµµ¾ Ãκται ¹ ΑΒ· Óπερ œδει ποιÁσαι.

Let A be the given point, and BCD the given circle. So it is required to draw a straight-line touching circle BCD from point A. For let the center E of the circle have been found [Prop. 3.1], and let AE have been joined. And let (the circle) AF G have been drawn with center E and radius EA. And let DF have been drawn from from (point) D, at right-angles to EA [Prop. 1.11]. And let EF and AB have been joined. I say that the (straight-line) AB has been drawn from point A touching circle BCD. For since E is the center of circles BCD and AF G, EA is thus equal to EF , and ED to EB. So the two (straight-lines) AE, EB are equal to the two (straightlines) F E, ED (respectively). And they contain a common angle at E. Thus, the base DF is equal to the base AB, and triangle DEF is equal to triangle EBA, and the remaining angles (are equal) to the (corresponding) remaining angles [Prop. 1.4]. Thus, (angle) EDF (is) equal to EBA. And EDF (is) a right-angle. Thus, EBA (is) also a right-angle. And EB is a radius. And a (straight-line) drawn at right-angles to the diameter of a circle, from its end, touches the circle [Prop. 3.16 corr.]. Thus, AB touches circle BCD. Thus, the straight-line AB has been drawn touching

88

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 the given circle BCD from the given point A. (Which is) the very thing it was required to do.

ιη΄.

Proposition 18

'Ε¦ν κύκλου ™φάπτηταί τις εÙθε‹α, ¢πÕ δ τοà If some straight-line touches a circle, and some κέντρου ™πˆ τ¾ν ¡φ¾ν ™πιζευχθÍ τις εÙθε‹α, ¹ ™πιζευ- (other) straight-line is joined from the center (of the cirχθε‹σα κάθετος œσται ™πˆ τ¾ν ™φαπτοµένην. cle) to the point of contact, then the (straight-line) so joined will be perpendicular to the tangent.

A

Α ∆ Β

Ζ

D

Η

B

F

Γ

G

C

Ε

E

Κύκλου γ¦ρ τοà ΑΒΓ ™φαπτέσθω τις εÙθε‹α ¹ ∆Ε κατ¦ τÕ Γ σηµε‹ον, κሠε„λήφθω τÕ κέντρον τοà ΑΒΓ κύκλου τÕ Ζ, κሠ¢πÕ τοà Ζ ™πˆ τÕ Γ ™πεζεύχθω ¹ ΖΓ· λέγω, Óτι ¹ ΖΓ κάθετός ™στιν ™πˆ τ¾ν ∆Ε. Ε„ γ¦ρ µή, ½χθω ¢πÕ τοà Ζ ™πˆ τ¾ν ∆Ε κάθετος ¹ ΖΗ. 'Επεˆ οâν ¹ ØπÕ ΖΗΓ γωνία Ñρθή ™στιν, Ñξε‹α ¥ρα ™στˆν ¹ ØπÕ ΖΓΗ· ØπÕ δ τ¾ν µείζονα γωνίαν ¹ µείζων πλευρ¦ Øποτείνει· µείζων ¥ρα ¹ ΖΓ τÁς ΖΗ· ‡ση δ ¹ ΖΓ τÍ ΖΒ· µείζων ¥ρα κሠ¹ ΖΒ τÁς ΖΗ ¹ ™λάττων τÁς µείζονος· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ¹ ΖΗ κάθετός ™στιν ™πˆ τ¾ν ∆Ε. еοίως δ¾ δε‹ξοµεν, Óτι οÙδ' ¥λλη τις πλ¾ν τÁς ΖΓ· ¹ ΖΓ ¥ρα κάθετός ™στιν ™πˆ τ¾ν ∆Ε. 'Ε¦ν ¥ρα κύκλου ™φάπτηταί τις εÙθε‹α, ¢πÕ δ τοà κέντρου ™πˆ τ¾ν ¡φ¾ν ™πιζευχθÍ τις εÙθε‹α, ¹ ™πιζευχθε‹σα κάθετος œσται ™πˆ τ¾ν ™φαπτοµένην· Óπερ œδει δε‹ξαι.

For let some straight-line DE touch the circle ABC at point C, and let the center F of circle ABC have been found [Prop. 3.1], and let F C have been joined from F to C. I say that F C is perpendicular to DE. For if not, let F G have been drawn from F , perpendicular to DE [Prop. 1.12]. Therefore, since angle F GC is a right-angle, (angle) F CG is thus acute [Prop. 1.17]. And the greater angle subtends the greater side [Prop. 1.19]. Thus, F C (is) greater than F G. And F C (is) equal to F B. Thus, F B (is) also greater than F G, the lesser than the greater. The very thing is impossible. Thus, F G is not perpendicular to DE. So, similarly, we can show that neither (is) any other (straight-line) than F C. Thus, F C is perpendicular to DE. Thus, if some straight-line touches a circle, and some (other) straight-line is joined from the center (of the circle) to the point of contact, then the (straight-line) so joined will be perpendicular to the tangent. (Which is) the very thing it was required to show.

ιθ΄.

Proposition 19

'Ε¦ν κύκλου ™φάπτηταί τις εÙθε‹α, ¢πÕ δ τÁς ¡φÁς τÍ ™φαπτοµένV πρÕς Ñρθ¦ς [γωνίας] εÙθε‹α γραµµ¾ ¢χθÍ, ™πˆ τÁς ¢χθείσης œσται τÕ κέντρον τοà κύκλου. Κύκλου γ¦ρ τοà ΑΒΓ ™φαπτέσθω τις εÙθε‹α ¹ ∆Ε κατ¦ τÕ Γ σηµε‹ον, κሠ¢πÕ τοà Γ τÍ ∆Ε πρÕς Ñρθ¦ς ½χθω ¹ ΓΑ· λέγω, Óτι ™πˆ τÁς ΑΓ ™στι τÕ κέντρον τοà

If some straight-line touches a circle, and a straightline is drawn from the point of contact, at right-[angles] to the tangent, then the center (of the circle) will be on the (straight-line) so drawn. For let some straight-line DE touch the circle ABC at point C. And let CA have been drawn from C, at right-

89

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

κύκλου.

angles to DE [Prop. 1.11]. I say that the center of the circle is on AC.

Α

B

Ζ

Β

∆

A

Γ

Ε

D

F

C

E

Μ¾ γάρ, ¢λλ' ε„ δυνατόν, œστω τÕ Ζ, κሠ™πεζεύχθω ¹ ΓΖ. 'Επεˆ [οâν] κύκλου τοà ΑΒΓ ™φάπτεταί τις εÙθε‹α ¹ ∆Ε, ¢πÕ δ τοà κέντρου ™πˆ τ¾ν ¡φ¾ν ™πέζευκται ¹ ΖΓ, ¹ ΖΓ ¥ρα κάθετός ™στιν ™πˆ τ¾ν ∆Ε· Ñρθ¾ ¥ρα ™στˆν ¹ ØπÕ ΖΓΕ. ™στˆ δ κሠ¹ ØπÕ ΑΓΕ Ñρθή· ‡ση ¥ρα ™στˆν ¹ ØπÕ ΖΓΕ τÍ ØπÕ ΑΓΕ ¹ ™λάττων τÍ µείζονι· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τÕ Ζ κέντρον ™στˆ τοà ΑΒΓ κύκλου. еοίως δ¾ δείξοµεν, Óτι οÙδ' ¥λλο τι πλ¾ν ™πˆ τÁς ΑΓ. 'Ε¦ν ¥ρα κύκλου ™φάπτηταί τις εÙθε‹α, ¢πÕ δ τÁς ¡φÁς τÍ ™φαπτοµένV πρÕς Ñρθ¦ς εÙθε‹α γραµµ¾ ¢χθÍ, ™πˆ τÁς ¢χθείσης œσται τÕ κέντρον τοà κύκλου· Óπερ œδει δε‹ξαι.

For (if) not, if possible, let F be (the center of the circle), and let CF have been joined. [Therefore], since some straight-line DE touches the circle ABC, and F C has been joined from the center to the point of contact, F C is thus perpendicular to DE [Prop. 3.18]. Thus, F CE is a right-angle. And ACE is also a right-angle. Thus, F CE is equal to ACE, the lesser to the greater. The very thing is impossible. Thus, F is not the center of circle ABC. So, similarly, we can show that neither is any (point) other (than one) on AC.

κ΄.

Proposition 20

'Εν κύκλJ ¹ πρÕς τù κέντρJ γωνία διπλασίων ™στˆ τÁς πρÕς τÍ περιφερείv, Óταν τ¾ν αÙτ¾ν περιφέρειαν βάσιν œχωσιν αƒ γωνίαι. ”Εστω κύκλος Ð ΑΒΓ, κሠπρÕς µν τù κέντρJ αÙτοà γωνία œστω ¹ ØπÕ ΒΕΓ, πρÕς δ τÍ περιφερείv ¹ ØπÕ ΒΑΓ, ™χέτωσαν δ τ¾ν αÙτ¾ν περιφέρειαν βάσιν τ¾ν ΒΓ· λέγω, Óτι διπλασίων ™στˆν ¹ ØπÕ ΒΕΓ γωνία τÁς ØπÕ ΒΑΓ. 'Επιζευχθε‹σα γ¦ρ ¹ ΑΕ διήχθω ™πˆ τÕ Ζ. 'Επεˆ οâν ‡ση ™στˆν ¹ ΕΑ τÍ ΕΒ, ‡ση κሠγωνία ¹ ØπÕ ΕΑΒ τÍ ØπÕ ΕΒΑ· αƒ ¥ρα ØπÕ ΕΑΒ, ΕΒΑ γωνίαι τÁς ØπÕ ΕΑΒ διπλασίους ε„σίν. ‡ση δ ¹ ØπÕ ΒΕΖ τα‹ς ØπÕ ΕΑΒ, ΕΒΑ· κሠ¹ ØπÕ ΒΕΖ ¥ρα τÁς ØπÕ ΕΑΒ ™στι διπλÁ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ ΖΕΓ τÁς ØπÕ ΕΑΓ ™στι διπλÁ. Óλη ¥ρα ¹ ØπÕ ΒΕΓ Óλης τÁς ØπÕ ΒΑΓ ™στι διπλÁ.

In a circle, the angle at the center is double that at the circumference, when the angles have the same circumference base. Let ABC be a circle, and let BEC be an angle at its center, and BAC (one) at (its) circumference. And let them have the same circumference base BC. I say that angle BEC is double (angle) BAC. For being joined, let AE have been drawn through to F. Therefore, since EA is equal to EB, angle EAB (is) also equal to EBA [Prop. 1.5]. Thus, angle EAB and EBA is double (angle) EAB. And BEF (is) equal to EAB and EBA [Prop. 1.32]. Thus, BEF is also double EAB. So, for the same (reasons), F EC is also double EAC. Thus, the whole (angle) BEC is double the whole (angle) BAC.

Thus, if some straight-line touches a circle, and a straightline is drawn from the point of contact, at right-angles to the tangent, then the center (of the circle) will be on the (straight-line) so drawn. (Which is) the very thing it was required to show.

90

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

Α

A

∆ Ε

D

E Γ

Η

C

Ζ

G

F

Β

B

Κεκλάσθω δ¾ πάλιν, κሠœστω ˜τέρα γωνία ¹ ØπÕ Β∆Γ, κሠ™πιζευχθε‹σα ¹ ∆Ε ™κβεβλήσθω ™πˆ τÕ Η. еοίως δ¾ δείξοµεν, Óτι διπλÁ ™στιν ¹ ØπÕ ΗΕΓ γωνία τÁς ØπÕ Ε∆Γ, ïν ¹ ØπÕ ΗΕΒ διπλÁ ™στι τÁς ØπÕ Ε∆Β· λοιπ¾ ¥ρα ¹ ØπÕ ΒΕΓ διπλÁ ™στι τÁς ØπÕ Β∆Γ. 'Εν κύκλJ ¥ρα ¹ πρÕς τù κέντρJ γωνία διπλασίων ™στˆ τÁς πρÕς τÍ περιφερείv, Óταν τ¾ν αÙτ¾ν περιφέρειαν βάσιν œχωσιν [αƒ γωνίαι]· Óπερ œδει δε‹ξαι.

So let a (straight-line) have been inflected again, and let there be another angle, BDC. And DE being joined, let it have been produced to G. So, similarly, we can show that angle GEC is double EDC, of which GEB is double EDB. Thus, the remaining (angle) BEC is double the (remaining angle) BDC. Thus, in a circle, the angle at the center is double that at the circumference, when [the angles] have the same circumference base. (Which is) the very thing it was required to show.

κα΄.

Proposition 21

'Εν κύκλJ αƒ ™ν τù αÙτù τµήµατι γωνίαι ‡σαι In a circle, angles in the same segment are equal to ¢λλήλαις ε„σίν. one another.

Α

A

Ε

Ζ Β

E

F

∆

B

Γ

D C

”Εστω κύκλος Ð ΑΒΓ∆, κሠ™ν τù αÙτù τµήµατι τù ΒΑΕ∆ γωνίαι œστωσαν αƒ ØπÕ ΒΑ∆, ΒΕ∆· λέγω, Óτι αƒ ØπÕ ΒΑ∆, ΒΕ∆ γωνίαι ‡σαι ¢λλήλαις ε„σίν. Ε„λήφθω γ¦ρ τοà ΑΒΓ∆ κύκλου τÕ κέντρον, κሠœστω τÕ Ζ, κሠ™πεζεύχθωσαν αƒ ΒΖ, Ζ∆. Κሠ™πεˆ ¹ µν ØπÕ ΒΖ∆ γωνία πρÕς τù κέντρJ ™στίν, ¹ δ ØπÕ ΒΑ∆ πρÕς τÍ περιφερείv, κሠœχουσι τ¾ν αÙτ¾ν περιφέρειαν βάσιν τ¾ν ΒΓ∆, ¹ ¥ρα ØπÕ ΒΖ∆ γωνία διπλασίων ™στˆ τÁς ØπÕ ΒΑ∆. δι¦ τ¦ αÙτ¦ δ¾ ¹

Let ABCD be a circle, and let BAD and BED be angles in the same segment BAED. I say that angles BAD and BED are equal to one another. For let the center of circle ABCD have been found [Prop. 3.1], and let it be (at point) F . And let BF and F D have been joined. And since angle BF D is at the center, and BAD at the circumference, and they have the same circumference base BCD, angle BF D is thus double BAD [Prop. 3.20].

91

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

ØπÕ ΒΖ∆ κሠτÁς ØπÕ ΒΕ∆ ™στι διπλσίων· ‡ση ¥ρα ¹ So, for the same (reasons), BF D is also double BED. ØπÕ ΒΑ∆ τÍ ØπÕ ΒΕ∆. Thus, BAD (is) equal to BED. 'Εν κύκλJ ¥ρα αƒ ™ν τù αÙτù τµήµατι γωνίαι ‡σαι Thus, in a circle, angles in the same segment are equal ¢λλήλαις ε„σίν· Óπερ œδει δε‹ξαι. to one another. (Which is) the very thing it was required to show.

κβ΄.

Proposition 22

Τîν ™ν το‹ς κύκλοις τετραπλεύρων αƒ ¢πεναντίον γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν.

For quadrilaterals within circles, the (sum of the) opposite angles is equal to two right-angles.

Β

B

Α

A

Γ

C

∆

D

”Εστω κύκλος Ð ΑΒΓ∆, κሠ™ν αÙτù τετράπλευρον œστω τÕ ΑΒΓ∆· λέγω, Óτι αƒ ¢πεναντίον γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν. 'Επεζεύχθωσαν αƒ ΑΓ, Β∆. 'Επεˆ οâν παντÕς τριγώνου αƒ τρε‹ς γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν, τοà ΑΒΓ ¥ρα τριγώνου αƒ τρε‹ς γωνίαι αƒ ØπÕ ΓΑΒ, ΑΒΓ, ΒΓΑ δυσˆν Ñρθα‹ς ‡σαι ε„σίν. ‡ση δ ¹ µν ØπÕ ΓΑΒ τÍ ØπÕ Β∆Γ· ™ν γ¦ρ τù αÙτù τµήµατί ε„σι τù ΒΑ∆Γ· ¹ δ ØπÕ ΑΓΒ τÍ ØπÕ Α∆Β· ™ν γ¦ρ τù αÙτù τµήµατί ε„σι τù Α∆ΓΒ· Óλη ¥ρα ¹ ØπÕ Α∆Γ τα‹ς ØπÕ ΒΑΓ, ΑΓΒ ‡ση ™στίν. κοιν¾ προσκείσθω ¹ ØπÕ ΑΒΓ· αƒ ¥ρα ØπÕ ΑΒΓ, ΒΑΓ, ΑΓΒ τα‹ς ØπÕ ΑΒΓ, Α∆Γ ‡σαι ε„σίν. ¢λλ' αƒ ØπÕ ΑΒΓ, ΒΑΓ, ΑΓΒ δυσˆν Ñρθα‹ς ‡σαι ε„σίν. καˆ αƒ ØπÕ ΑΒΓ, Α∆Γ ¥ρα δυσˆν Ñρθα‹ς ‡σαι ε„σίν. еοίως δ¾ δείξοµεν, Óτι καˆ αƒ ØπÕ ΒΑ∆, ∆ΓΒ γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν. Τîν ¥ρα ™ν το‹ς κύκλοις τετραπλεύρων αƒ ¢πεναντίον γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν· Óπερ œδει δε‹ξαι.

Let ABCD be a circle, and let ABCD be a quadrilateral within it. I say that the (sum of the) opposite angles is equal to two right-angles. Let AC and BD have been joined. Therefore, since the three angles of every triangle are equal to two right-angles [Prop. 1.32], the three angles CAB, ABC, and BCA of triangle ABC are thus equal to two right-angles. And CAB (is) equal to BDC. For they are in the same segment BADC [Prop. 3.21]. And ACB (is equal) to ADB. For they are in the same segment ADCB [Prop. 3.21]. Thus, the whole of ADC is equal to BAC and ACB. Let ABC have been added to both. Thus, ABC, BAC, and ACB are equal to ABC and ADC. But, ABC, BAC, and ACB are equal to two right-angles. Thus, ABC and ADC are also equal to two right-angles. Similarly, we can show that angles BAD and DCB are also equal to two right-angles. Thus, for quadrilaterals within circles, the (sum of the) opposite angles is equal to two right-angles. (Which is) the very thing it was required to show.

κγ΄.

Proposition 23

'Επˆ τÁς αÙτÁς εÙθείας δύο τµήµατα κύκλων Óµοια Two similar and unequal segments of circles cannot be κሠ¥νισα οÙ συσταθήσεται ™πˆ τ¦ αÙτ¦ µέρη. constructed on the same side of the same straight-line. Ε„ γ¦ρ δυνατόν, ™πˆ τÁς αÙτÁς εÙθείας τÁς ΑΒ For, if possible, let the two similar and unequal segδύο τµήµατα κύκλων Óµοια κሠ¥νισα συνεστάτω ™πˆ ments of circles, ACB and ADB, have been constructed τ¦ αÙτ¦ µέρη τ¦ ΑΓΒ, Α∆Β, κሠδιήχθω ¹ ΑΓ∆, κሠon the same side of the same straight-line AB. And let

92

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

™πεζεύχθωσαν αƒ ΓΒ, ∆Β.

ACD have been drawn through (the segments), and let CB and DB have been joined.

∆

Γ Α

D

C

Β A

B

'Επεˆ οâν Óµοιόν ™στι τÕ ΑΓΒ τµÁµα τù Α∆Β τµήµατι, Óµοια δ τµήµατα κύκλων ™στˆ τ¦ δεχόµενα γωνίας ‡σας, ‡ση ¥ρα ™στˆν ¹ ØπÕ ΑΓΒ γωνία τÍ ØπÕ Α∆Β ¹ ™κτÕς τÍ ™ντός· Óπερ ™στˆν ¢δύνατον. ΟÙκ ¥ρα ™πˆ τÁς αÙτÁς εÙθείας δύο τµήµατα κύκλων Óµοια κሠ¥νισα συσταθήσεται ™πˆ τ¦ αÙτ¦ µέρη· Óπερ œδει δε‹ξαι.

Therefore, since segment ACB is similar to segment ADB, and similar segments of circles are those accepting equal angles [Def. 3.11], angle ACB is thus equal to ADB, the external to the internal. The very thing is impossible [Prop. 1.16]. Thus, two similar and unequal segments of circles cannot be constructed on the same side of the same straight-line.

κδ΄.

Proposition 24

Τ¦ ™πˆ ‡σων εÙθειîν Óµοια τµήµατα κύλων ‡σα Similar segments of circles on equal straight-lines are ¢λλήλοις ™στίν. equal to one another.

Ε

E

Α

Β Ζ

Γ

A

Η

B G F

∆

C

”Εστωσαν γ¦ρ ™πˆ ‡σων εÙθειîν τîν ΑΒ, Γ∆ Óµοια τµήµατα κύκλων τ¦ ΑΕΒ, ΓΖ∆· λέγω, Óτι ‡σον ™στˆ τÕ ΑΕΒ τµÁµα τù ΓΖ∆ τµήµατι. 'Εφαρµοζοµένου γ¦ρ τοà ΑΕΒ τµήµατος ™πˆ τÕ ΓΖ∆ κሠτιθεµένου τοà µν Α σηµείου ™πˆ τÕ Γ τÁς δ ΑΒ εÙθείας ™πˆ τ¾ν Γ∆, ™φαρµόσει κሠτÕ Β σηµε‹ον ™πˆ τÕ ∆ σηµε‹ον δι¦ τÕ ‡σην εναι τ¾ν ΑΒ τÍ Γ∆· τÁς δ ΑΒ ™πˆ τ¾ν Γ∆ ™φαρµοσάσης ™φαρµόσει κሠτÕ ΑΕΒ τµÁµα ™πˆ τÕ ΓΖ∆. ε„ γ¦ρ ¹ ΑΒ εÙθε‹α ™πˆ τ¾ν Γ∆ ™φαρµόσει, τÕ δ ΑΕΒ τµÁµα ™πˆ τÕ ΓΖ∆ µ¾ ™φαρµόσει, ½τοι ™ντÕς αÙτοà πεσε‹ται À ™κτÕς À παραλλάξει, æς τÕ ΓΗ∆, κሠκύκλος κύκλον τέµνει κατ¦ πλείονα σηµε‹α À δύο· Óπερ ™στίν ¢δύνατον. οÙκ ¥ρα ™φαρµοζοµένης τÁς ΑΒ εÙθείας

D

For let AEB and CF D be similar segments of circles on the equal straight-lines AB and CD (respectively). I say that segment AEB is equal to segment CF D. For let the segment AEB be applied to the segment CF D, the point A being placed on (point) C, and the straight-line AB on CD. The point B will also coincide with point D, on account of AB being equal to CD. And if AB coincides with CD, the segment AEB will also coincide with CF D. For if the straight-line AB coincides with CD, and the segment AEB does not coincide with CF D, then it will surely either fall inside it, outside (it),† or it will miss like CGD (in the figure), and a circle (will) cut (another) circle at more than two points. The very

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ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

™πˆ τ¾ν Γ∆ οÙκ ™φαρµόσει κሠτÕ ΑΕΒ τµÁµα ™πˆ τÕ ΓΖ∆· ™φαρµόσει ¥ρα, κሠ‡σον αÙτù œσται. Τ¦ ¥ρα ™πˆ ‡σων εÙθειîν Óµοια τµήµατα κύκλων ‡σα ¢λλήλοις ™στίν· Óπερ œδει δε‹ξαι.

†

thing is impossible [Prop. 3.10]. Thus, if the straight-line AB is applied to CD, the segment AEB cannot fail to also coincide with CF D. Thus, it will coincide, and will be equal to it [C.N. 4]. Thus, similar segments of circles on equal straightlines are equal to one another. (Which is) the very thing it was required to show.

Both this possibility, and the previous one, are precluded by Prop. 3.23.

κε΄.

Proposition 25

Κύκλου τµήµατος δοθέντος προσαναγράψαι τÕν κύκλον, οáπέρ ™στι τµÁµα.

To complete the circle for a given segment of a circle, the very one of which it is a segment.

Α

Β

∆

Γ

Α

Ε

Β

∆

Γ

A

Α

Β

Ε

∆

B

D

A

E

B

D

A

B

D E

Γ

C

”Εστω τÕ δοθν τµÁµα κύκλου τÕ ΑΒΓ· δε‹ δ¾ τοà ΑΒΓ τµήµατος προσαναγράψαι τÕν κύκλον, οâπέρ ™στι τµÁµα. Τετµήσθω γ¦ρ ¹ ΑΓ δίχα κατ¦ τÕ ∆, κሠ½χθω ¢πÕ τοà ∆ σηµείου τÍ ΑΓ πρÕς Ñρθ¦ς ¹ ∆Β, κሠ™πεζεύχθω ¹ ΑΒ· ¹ ØπÕ ΑΒ∆ γωνία ¥ρα τÁς ØπÕ ΒΑ∆ ½τοι µείζων ™στˆν À ‡ση À ™λάττων. ”Εστω πρότερον µείζων, κሠσυνεστάτω πρÕς τÍ ΒΑ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ ØπÕ ΑΒ∆ γωνίv ‡ση ¹ ØπÕ ΒΑΕ, κሠδιήχθω ¹ ∆Β ™πˆ τÕ Ε, κሠ™πεζεύχθω ¹ ΕΓ. ™πεˆ οâν ‡ση ™στˆν ¹ ØπÕ ΑΒΕ γωνία τÍ ØπÕ ΒΑΕ, ‡ση ¥ρα ™στˆ κሠ¹ ΕΒ εÙθε‹α τÍ ΕΑ. κሠ™πεˆ ‡ση ™στˆν ¹ Α∆ τÍ ∆Γ, κοιν¾ δ ¹ ∆Ε, δύο δ¾ αƒ Α∆, ∆Ε δύο τα‹ς Γ∆, ∆Ε ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ Α∆Ε γωνίv τÍ ØπÕ Γ∆Ε ™στιν ‡ση· Ñρθ¾ γ¦ρ ˜κατέρα· βάσις ¥ρα ¹ ΑΕ βάσει τÍ ΓΕ ™στιν ‡ση. ¢λλ¦ ¹ ΑΕ τÍ ΒΕ ™δείχθη ‡ση· κሠ¹ ΒΕ ¥ρα τÍ ΓΕ ™στιν ‡ση· αƒ τρε‹ς ¥ρα αƒ ΑΕ, ΕΒ, ΕΓ ‡σαι ¢λλήλαις ε„σίν· Ð ¥ρα κέντρù τù Ε διαστήµατι δ ˜νˆ τîν ΑΕ, ΕΒ, ΕΓ κύκλος γραφόµενος ¼ξει κሠδι¦ τîν λοιπîν σηµείων κሠœσται προσαναγεγραµµένος. κύκλου ¥ρα τµήµατος δοθέντος προσαναγέγραπται Ð κύκλος. κሠδÁλον, æς τÕ ΑΒΓ τµÁµα œλαττόν ™στιν ¹µικυκλίου δι¦ τÕ τÕ Ε κέντρον ™κτÕς αÙτοà τυγχάνειν. `Οµοίως [δ] κ¨ν Ï ¹ ØπÕ ΑΒ∆ γωνία ‡ση τÍ ØπÕ ΒΑ∆, τÁς Α∆ ‡σης γενοµένης ˜κατέρv τîν Β∆, ∆Γ αƒ τρε‹ς αƒ ∆Α, ∆Β, ∆Γ ‡σαι ¢λλήλαις œσονται, κሠœσται τÕ ∆ κέντρον τοà προσαναπεπληρωµένου κύκλου, κሠδηλαδ¾ œσται τÕ ΑΒΓ ¹µικύκλιον.

C

C

Let ABC be the given segment of a circle. So it is required to complete the circle for segment ABC, the very one of which it is a segment. For let AC have been cut in half at (point) D [Prop. 1.10], and let DB have been drawn from point D, at right-angles to AC [Prop. 1.11]. And let AB have been joined. Thus, angle ABD is surely either greater than, equal to, or less than (angle) BAD. First of all, let it be greater. And let (angle) BAE have been constructed, equal to angle ABD, at the point A on the straight-line BA [Prop. 1.23]. And let DB have been drawn through to E, and let EC have been joined. Therefore, since angle ABE is equal to BAE, the straight-line EB is thus also equal to EA [Prop. 1.6]. And since AD is equal to DC, and DE (is) common, the two (straight-lines) AD, DE are equal to the two (straightlines) CD, DE, respectively. And angle ADE is equal to angle CDE. For each (is) a right-angle. Thus, the base AE is equal to the base CE [Prop. 1.4]. But, AE was shown (to be) equal to BE. Thus, BE is also equal to CE. Thus, the three (straight-lines) AE, EB, and EC are equal to one another. Thus, if a circle is drawn with center E, and radius one of AE, EB, or EC, it will also go through the remaining points (of the segment), and the (associated circle) will be completed [Prop. 3.9]. Thus, a circle has been completed from the given segment of a circle. And (it is) clear that the segment ABC is less than a semi-circle, on account of the center E lying outside it. [And], similarly, even if angle ABD is equal to BAD,

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ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

'Ε¦ν δ ¹ ØπÕ ΑΒ∆ ™λάττων Ï τÁς ØπÕ ΒΑ∆, κሠσυστησώµεθα πρÕς τÍ ΒΑ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ ØπÕ ΑΒ∆ γωνίv ‡σην, ™ντÕς τοà ΑΒΓ τµήµατος πεσε‹ται τÕ κέντρον ™πˆ τÁς ∆Β, κሠœσται δηλαδ¾ τÕ ΑΒΓ τµÁµα µε‹ζον ¹µικυκλίου. Κύκλου ¥ρα τµήµατος δοθέντος προσαναγέγραπται Ð κύκλος· Óπερ œδει ποιÁσαι.

(since) AD becomes equal to each of BD [Prop. 1.6] and DC, the three (straight-lines) DA, DB, and DC will be equal to one another. And point D will be the center of the completed circle. And ABC will manifestly be a semi-circle. And if ABD is less than BAD, and we construct (angle BAE), equal to angle ABD, at the point A on the straight-line BA [Prop. 1.23], then the center will fall on DB, inside the segment ABC. And segment ABC will manifestly be greater than a semi-circle. Thus, a circle has been completed from the given segment of a circle. (Which is) the very thing it was required to do.

κ$΄.

Proposition 26

'Εν το‹ς ‡σοις κύκλοις αƒ ‡σαι γωνίαι ™πˆ ‡σων περιφεEqual angles stand upon equal circumferences in ρειîν βεβήκασιν, ™άν τε πρÕς το‹ς κέντροις ™άν τε πρÕς equal circles, whether they are standing at the center τα‹ς περιφερείαις ðσι βεβηκυ‹αι. or at the circumference.

Α

A ∆ Θ

Η Β

Γ Κ

D

Ε

H

G Ζ

C

B

Λ

K

”Εστωσαν ‡σοι κύκλοι οƒ ΑΒΓ, ∆ΕΖ κሠ™ν αÙτο‹ς ‡σαι γωνίαι œστωσαν πρÕς µν το‹ς κέντροις αƒ ØπÕ ΒΗΓ, ΕΘΖ, πρÕς δ τα‹ς περιφερείαις αƒ ØπÕ ΒΑΓ, Ε∆Ζ· λέγω, Óτι ‡ση ™στˆν ¹ ΒΚΓ περιφέρεια τÍ ΕΛΖ περιφερείv. 'Επεζεύχθωσαν γ¦ρ αƒ ΒΓ, ΕΖ. Κሠ™πεˆ ‡σοι ε„σˆν οƒ ΑΒΓ, ∆ΕΖ κύκλοι, ‡σαι ε„σˆν αƒ ™κ τîν κέντρων· δύο δ¾ αƒ ΒΗ, ΗΓ δύο τα‹ς ΕΘ, ΘΖ ‡σαι· κሠγωνία ¹ πρÕς τù Η γωνίv τÍ πρÕς τù Θ ‡ση· βάσις ¥ρα ¹ ΒΓ βάσει τÍ ΕΖ ™στιν ‡ση. κሠ™πεˆ ‡ση ™στˆν ¹ πρÕς τù Α γωνία τÍ πρÕς τù ∆, Óµοιον ¥ρα ™στˆ τÕ ΒΑΓ τµÁµα τù Ε∆Ζ τµήµατι· καί ε„σιν ™πˆ ‡σων εÙθειîν [τîν ΒΓ, ΕΖ]· τ¦ δ ™πˆ ‡σων εÙθειîν Óµοια τµήµατα κύκλων ‡σα ¢λλήλοις ™στίν· ‡σον ¥ρα τÕ ΒΑΓ τµÁµα τù Ε∆Ζ. œστι δ κሠÓλος Ð ΑΒΓ κύκλος ÓλJ τù ∆ΕΖ κύκλJ ‡σος· λοιπ¾ ¥ρα ¹ ΒΚΓ περιφέρεια τÍ ΕΛΖ περιφερείv ™στˆν ‡ση. 'Εν ¥ρα το‹ς ‡σοις κύκλοις αƒ ‡σαι γωνίαι ™πˆ ‡σων περιφερειîν βεβήκασιν, ™άν τε πρÕς το‹ς κέντροις ™άν τε πρÕς τα‹ς περιφερείας ðσι βεβηκυ‹αι· Óπερ œδει δε‹ξαι.

F

E L

Let ABC and DEF be equal circles, and within them let BGC and EHF be equal angles at the center, and BAC and EDF (equal angles) at the circumference. I say that circumference BKC is equal to circumference ELF . For let BC and EF have been joined. And since circles ABC and DEF are equal, their radii are equal. So the two (straight-lines) BG, GC (are) equal to the two (straight-lines) EH, HF (respectively). And the angle at G (is) equal to the angle at H. Thus, the base BC is equal to the base EF [Prop. 1.4]. And since the angle at A is equal to the (angle) at D, the segment BAC is thus similar to the segment EDF [Def. 3.11]. And they are on equal straight-lines [BC and EF ]. And similar segments of circles on equal straight-lines are equal to one another [Prop. 3.24]. Thus, segment BAC is equal to (segment) EDF . And the whole circle ABC is also equal to the whole circle DEF . Thus, the remaining circumference BKC is equal to the (remaining) circumference ELF . Thus, equal angles stand upon equal circumferences in equal circles, whether they are standing at the center or at the circumference. (Which is) the very thing which

95

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 it was required to show.

κζ΄.

Proposition 27

'Εν το‹ς ‡σοις κύκλοις αƒ ™πˆ ‡σων περιφερειîν βεβηκυ‹αι γωνίαι ‡σαι ¢λλήλαις ε„σίν, ™άν τε πρÕς το‹ς κέντροις ™άν τε πρÕς τα‹ς περιφερείαις ðσι βεβηκυ‹αι.

Angles standing upon equal circumferences in equal circles are equal to one another, whether they are standing at the center or at the circumference.

Α

∆

A

Θ

Η Β

Γ

Ε

D

H

G Ζ

B

Κ

C

E

F

K

'Εν γ¦ρ ‡σοις κύκλοις το‹ς ΑΒΓ, ∆ΕΖ ™πˆ ‡σων περιφερειîν τîν ΒΓ, ΕΖ πρÕς µν το‹ς Η, Θ κέντροις γωνίαι βεβηκέτωσαν αƒ ØπÕ ΒΗΓ, ΕΘΖ, πρÕς δ τα‹ς περιφερείαις αƒ ØπÕ ΒΑΓ, Ε∆Ζ· λέγω, Óτι ¹ µν ØπÕ ΒΗΓ γωνία τÍ ØπÕ ΕΘΖ ™στιν ‡ση, ¹ δ ØπÕ ΒΑΓ τÍ ØπÕ Ε∆Ζ ™στιν ‡ση. Ε„ γ¦ρ ¥νισός ™στιν ¹ ØπÕ ΒΗΓ τÍ ØπÕ ΕΘΖ, µία αÙτîν µείζων ™στίν. œστω µείζων ¹ ØπÕ ΒΗΓ, κሠσυνεστάτω πρÕς τÍ ΒΗ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Η τÍ ØπÕ ΕΘΖ γωνίv ‡ση ¹ ØπÕ ΒΗΚ· αƒ δ ‡σαι γωνίαι ™πˆ ‡σων περιφερειîν βεβήκασιν, Óταν πρÕς το‹ς κέντροις ðσιν· ‡ση ¥ρα ¹ ΒΚ περιφέρεια τÍ ΕΖ περιφερείv. ¢λλ¦ ¹ ΕΖ τÍ ΒΓ ™στιν ‡ση· κሠ¹ ΒΚ ¥ρα τÍ ΒΓ ™στιν ‡ση ¹ ™λάττων τÍ µείζονι· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ¥νισός ™στιν ¹ ØπÕ ΒΗΓ γωνία τÍ ØπÕ ΕΘΖ· ‡ση ¥ρα. καί ™στι τÁς µν ØπÕ ΒΗΓ ¹µίσεια ¹ πρÕς τù Α, τÁς δ ØπÕ ΕΘΖ ¹µίσεια ¹ πρÕς τù ∆· ‡ση ¥ρα κሠ¹ πρÕς τù Α γωνία τÍ πρÕς τù ∆. 'Εν ¥ρα το‹ς ‡σοις κύκλοις αƒ ™πˆ ‡σων περιφερειîν βεβηκυ‹αι γωνίαι ‡σαι ¢λλήλαις ε„σίν, ™άν τε πρÕς το‹ς κέντροις ™άν τε πρÕς τα‹ς περιφερείαις ðσι βεβηκυ‹αι· Óπερ œδει δε‹ξαι.

For let the angles BGC and EHF at the centers G and H, and the (angles) BAC and EDF at the circumferences, stand upon the equal circumferences BC and EF , in the equal circles ABC and DEF (respectively). I say that angle BGC is equal to (angle) EHF , and BAC is equal to EDF . For if BGC is unequal to EHF , one of them is greater. Let BGC be greater, and let the (angle) BGK, equal to the angle EHF , have been constructed at the point G on the straight-line BG [Prop. 1.23]. But equal angles (in equal circles) stand upon equal circumferences, when they are at the centers [Prop. 3.26]. Thus, circumference BK (is) equal to circumference EF . But, EF is equal to BC. Thus, BK is also equal to BC, the lesser to the greater. The very thing is impossible. Thus, angle BGC is not unequal to EHF . Thus, (it is) equal. And the (angle) at A is half BGC, and the (angle) at D half EHF [Prop. 3.20]. Thus, the angle at A (is) also equal to the (angle) at D. Thus, angles standing upon equal circumferences in equal circles are equal to one another, whether they are standing at the center or at the circumference. (Which is) the very thing it was required to show.

κη΄.

Proposition 28

'Εν το‹ς ‡σοις κύκλοις αƒ ‡σαι εÙθε‹αι ‡σας περιφερείας ¢φαιροàσι τ¾ν µν µείζονα τÍ µείζονι τ¾ν δ ™λάττονα τÍ ™λάττονι. ”Εστωσαν ‡σοι κύκλοι οƒ ΑΒΓ, ∆ΕΖ, κሠ™ν το‹ς κύκλοις ‡σαι εÙθε‹αι œστωσαν αƒ ΑΒ, ∆Ε τ¦ς µν ΑΓΒ, ΑΖΕ περιφερείας µείζονας ¢φαιροàσαι τ¦ς δ ΑΗΒ, ∆ΘΕ ™λάττονας· λέγω, Óτι ¹ µν ΑΓΒ µείζων περιφέρεια ‡ση ™στˆ τÍ ∆ΖΕ µείζονι περιφερείv ¹ δ ΑΗΒ ™λάττων περιφέρεια τÍ ∆ΘΕ.

Equal straight-lines cut off equal circumferences in equal circles, the greater (circumference being equal) to the greater, and the lesser to the lesser. Let ABC and DEF be equal circles, and let AB and DE be equal straight-lines in these circles, cutting off the greater circumferences ACB and DF E, and the lesser (circumferences) AGB and DHE (respectively). I say that the greater circumference ACB is equal to the greater circumference DF E, and the lesser circumfer-

96

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 ence AGB to (the lesser) DHE.

Γ

Ζ

C

F

Κ

Λ

K

L

Β

Α

∆

Η

Ε

B

A

Θ

D

G

E H

Ε„λήφθω γ¦ρ τ¦ κέντρα τîν κύκλων τ¦ Κ, Λ, κሠ™πεζεύχθωσαν αƒ ΑΚ, ΚΒ, ∆Λ, ΛΕ. Κሠ™πεˆ ‡σοι κύκλοι ε„σίν, ‡σαι ε„σˆ καˆ αƒ ™κ τîν κέντρων· δύο δ¾ αƒ ΑΚ, ΚΒ δυσˆ τα‹ς ∆Λ, ΛΕ ‡σαι ε„σίν· κሠβάσις ¹ ΑΒ βάσει τÍ ∆Ε ‡ση· γωνία ¥ρα ¹ ØπÕ ΑΚΒ γωνίv τÍ ØπÕ ∆ΛΕ ‡ση ™στίν. αƒ δ ‡σαι γωνίαι ™πˆ ‡σων περιφερειîν βεβήκασιν, Óταν πρÕς το‹ς κέντροις ðσιν· ‡ση ¥ρα ¹ ΑΗΒ περιφέρεια τÍ ∆ΘΕ. ™στˆ δ κሠÓλος Ð ΑΒΓ κύκλος ÓλJ τù ∆ΕΖ κύκλJ ‡σος· κሠλοιπ¾ ¥ρα ¹ ΑΓΒ περιφέρεια λοιπÍ τÍ ∆ΖΕ περιφερείv ‡ση ™στίν. 'Εν ¥ρα το‹ς ‡σοις κύκλοις αƒ ‡σαι εÙθε‹αι ‡σας περιφερείας ¢φαιροàσι τ¾ν µν µείζονα τÍ µείζονι τ¾ν δ ™λάττονα τÍ ™λάττονι· Óπερ œδει δε‹ξαι.

For let the centers of the circles, K and L, have been found [Prop. 3.1], and let AK, KB, DL, and LE have been joined. And since (ABC and DEF ) are equal circles, their radii are also equal [Def. 3.1]. So the two (straightlines) AK, KB are equal to the two (straight-lines) DL, LE (respectively). And the base AB (is) equal to the base DE. Thus, angle AKB is equal to angle DLE [Prop. 1.8]. And equal angles stand upon equal circumferences, when they are at the centers [Prop. 3.26]. Thus, circumference AGB (is) equal to DHE. And the whole circle ABC is also equal to the whole circle DEF . Thus, the remaining circumference ACB is also equal to the remaining circumference DF E. Thus, equal straight-lines cut off equal circumferences in equal circles, the greater (circumference being equal) to the greater, and the lesser to the lesser. (Which is) the very thing it was required to show.

κθ΄.

Proposition 29

'Εν το‹ς ‡σοις κύκλοις τ¦ς ‡σας περιφερείας ‡σαι Equal straight-lines subtend equal circumferences in εÙθε‹αι Øποτείνουσιν. equal circles.

Α

∆

A

D

Κ

Λ

K

L

Β

Γ Η

Ε

Ζ

B

Θ

C G

”Εστωσαν ‡σοι κύκλοι οƒ ΑΒΓ, ∆ΕΖ, κሠ™ν αÙτο‹ς ‡σαι περιφέρειαι ¢πειλήφθωσαν αƒ ΒΗΓ, ΕΘΖ, κሠ™πεζεύχθωσαν αƒ ΒΓ, ΕΖ εÙθε‹αι· λέγω, Óτι ‡ση ™στˆν ¹ ΒΓ τÍ ΕΖ. Ε„λήφθω γ¦ρ τ¦ κέντρα τîν κύκλων, κሠœστω τ¦ Κ, Λ, κሠ™πεζεύχθωσαν αƒ ΒΚ, ΚΓ, ΕΛ, ΛΖ.

E

F H

Let ABC and DEF be equal circles, and within them let the equal circumferences BGC and EHF have been cut off. And let the straight-lines BC and EF have been joined. I say that BC is equal to EF . For let the centers of the circles have been found [Prop. 3.1], and let them be (at) K and L. And let BK,

97

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

Κሠ™πεˆ ‡ση ™στˆν ¹ ΒΗΓ περιφέρεια τÍ ΕΘΖ περιφερείv, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ΒΚΓ τÍ ØπÕ ΕΛΖ. κሠ™πεˆ ‡σοι ε„σˆν οƒ ΑΒΓ, ∆ΕΖ κύκλοι, ‡σαι ε„σˆ καˆ αƒ ™κ τîν κέντρων· δύο δ¾ αƒ ΒΚ, ΚΓ δυσˆ τα‹ς ΕΛ, ΛΖ ‡σαι ε„σίν· κሠγωνίας ‡σας περιέχουσιν· βάσις ¥ρα ¹ ΒΓ βάσει τÍ ΕΖ ‡ση ™στίν· 'Εν ¥ρα το‹ς ‡σοις κύκλοις τ¦ς ‡σας περιφερείας ‡σαι εÙθε‹αι Øποτείνουσιν· Óπερ œδει δε‹ξαι.

KC, EL, and LF have been joined. And since the circumference BGC is equal to the circumference EHF , the angle BKC is also equal to (angle) ELF [Prop. 3.27]. And since the circles ABC and DEF are equal, their radii are also equal [Def. 3.1]. So the two (straight-lines) BK, KC are equal to the two (straight-lines) EL, LF (respectively). And they contain equal angles. Thus, the base BC is equal to the base EF [Prop. 1.4]. Thus, equal straight-lines subtend equal circumferences in equal circles. (Which is) the very thing it was required to show.

λ΄.

Proposition 30

Τ¾ν δοθε‹σαν περιφέρειαν δίχα τεµε‹ν.

To cut a given circumference in half.

∆

Α

Γ

D

Β

A

C

B

”Εστω ¹ δοθε‹σα περιφέρεια ¹ Α∆Β· δε‹ δ¾ τ¾ν Α∆Β περιφέρειαν δίχα τεµε‹ν. 'Επεζεύχθω ¹ ΑΒ, κሠτετµήσθω δίχα κατ¦ τÕ Γ, κሠ¢πÕ τοà Γ σηµείου τÍ ΑΒ εÙθείv πρÕς Ñρθ¦ς ½χθω ¹ Γ∆, κሠ™πεζεύχθωσαν αƒ Α∆, ∆Β. Κሠ™πεˆ ‡ση ™στˆν ¹ ΑΓ τÍ ΓΒ, κοιν¾ δ ¹ Γ∆, δύο δ¾ αƒ ΑΓ, Γ∆ δυσˆ τα‹ς ΒΓ, Γ∆ ‡σαι ε„σίν· κሠγωνία ¹ ØπÕ ΑΓ∆ γωνίv τÍ ØπÕ ΒΓ∆ ‡ση· Ñρθ¾ γ¦ρ ˜κατέρα· βάσις ¥ρα ¹ Α∆ βάσει τÍ ∆Β ‡ση ™στίν. αƒ δ ‡σαι εÙθε‹αι ‡σας περιφερείας ¢φαιροàσι τ¾ν µν µείζονα τÍ µείζονι τ¾ν δ ™λάττονα τÍ ™λάττονι· κάι ™στιν ˜κατέρα τîν Α∆, ∆Β περιφερειîν ™λάττων ¹µικυκλίου· ‡ση ¥ρα ¹ Α∆ περιφέρεια τÍ ∆Β περιφερείv. `Η ¥ρα δοθε‹σα περιφέρεια δίχα τέτµηται κατ¦ τÕ ∆ σηµε‹ον· Óπερ œδει ποιÁσαι.

Let ADB be the given circumference. So it is required to cut circumference ADB in half. Let AB have been joined, and let it have been cut in half at (point) C [Prop. 1.10]. And let CD have been drawn from point C, at right-angles to AB [Prop. 1.11]. And let AD, and DB have been joined. And since AC is equal to CB, and CD (is) common, the two (straight-lines) AC, CD are equal to the two (straight-lines) BC, CD (respectively). And angle ACD (is) equal to angle BCD. For (they are) each rightangles. Thus, the base AD is equal to the base DB [Prop. 1.4]. And equal straight-lines cut off equal circumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser [Prop. 1.28]. And the circumferences AD and DB are each less than a semicircle. Thus, circumference AD (is) equal to circumference DB. Thus, the given circumference has been cut in half at point D. (Which is) the very thing it was required to do.

λα΄.

Proposition 31

'Εν κύκλJ ¹ µν ™ν τù ¹µικυκλίJ γωνία Ñρθή ™στιν, In a circle, the angle in a semi-circle is a right-angle, ¹ δ ™ν τù µείζονι τµήµατι ™λάττων ÑρθÁς, ¹ δ ™ν and that in a greater segment (is) less than a right-angle, τù ™λάττονι τµήµατι µείζων ÑρθÁς· κሠœπι ¹ µν τοà and that in a lesser segment (is) greater than a rightµείζονος τµήµατος γωνία µείζων ™στˆν ÑρθÁς, ¹ δ τοà angle. And, further, the angle of a segment greater (than 98

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

™λάττονος τµήµατος γωνία ™λάττων ÑρθÁς.

a semi-circle) is greater than a right-angle, and the angle of a segment less (than a semi-circle) is less than a right-angle.

Ζ

F ∆

D Γ

C

A

Α Ε

E

Β

B

”Εστω κύκλος Ð ΑΒΓ∆, διάµετρος δ αÙτοà œστω ¹ ΒΓ, κέντρον δ τÕ Ε, κሠ™πεζεύχθωσαν αƒ ΒΑ, ΑΓ, Α∆, ∆Γ· λέγω, Óτι ¹ µν ™ν τù ΒΑΓ ¹µικυκλίJ γωνία ¹ ØπÕ ΒΑΓ Ñρθή ™στιν, ¹ δ ™ν τù ΑΒΓ µείζονι τοà ¹µικυκλίου τµήµατι γωνία ¹ ØπÕ ΑΒΓ ™λάττων ™στˆν ÑρθÁς, ¹ δ ™ν τù Α∆Γ ™λάττονι τοà ¹µικυκλίου τµήµατι γωνία ¹ ØπÕ Α∆Γ µείζων ™στˆν ÑρθÁς. 'Επεζεύχθω ¹ ΑΕ, κሠδιήχθω ¹ ΒΑ ™πˆ τÕ Ζ. Κሠ™πεˆ ‡ση ™στˆν ¹ ΒΕ τÍ ΕΑ, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ΑΒΕ τÍ ØπÕ ΒΑΕ. πάλιν, ™πεˆ ‡ση ™στˆν ¹ ΓΕ τÍ ΕΑ, ‡ση ™στˆ κሠ¹ ØπÕ ΑΓΕ τÍ ØπÕ ΓΑΕ· Óλη ¥ρα ¹ ØπÕ ΒΑΓ δυσˆ τα‹ς ØπÕ ΑΒΓ, ΑΓΒ ‡ση ™στίν. ™στˆ δ κሠ¹ ØπÕ ΖΑΓ ™κτÕς τοà ΑΒΓ τριγώνου δυσˆ τα‹ς ØπÕ ΑΒΓ, ΑΓΒ γωνίαις ‡ση· ‡ση ¥ρα κሠ¹ ØπÕ ΒΑΓ γωνία τÍ ØπÕ ΖΑΓ· Ñρθ¾ ¥ρα ˜κατέρα· ¹ ¥ρα ™ν τù ΒΑΓ ¹µικυκλίJ γωνία ¹ ØπÕ ΒΑΓ Ñρθή ™στιν. Κሠ™πεˆ τοà ΑΒΓ τρίγωνου δύο γωνίαι αƒ ØπÕ ΑΒΓ, ΒΑΓ δύο Ñρθîν ™λάττονές ε„σιν, Ñρθ¾ δ ¹ ØπÕ ΒΑΓ, ™λάττων ¥ρα ÑρθÁς ™στιν ¹ ØπÕ ΑΒΓ γωνία· καί ™στιν ™ν τù ΑΒΓ µείζονι τοà ¹µικυκλίου τµήµατι. Κሠ™πεˆ ™ν κύκλJ τετράπλευρόν ™στι τÕ ΑΒΓ∆, τîν δ ™ν το‹ς κύκλοις τετραπλεύρων αƒ ¢πεναντίον γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν [αƒ ¥ρα ØπÕ ΑΒΓ, Α∆Γ γωνίαι δυσˆν Ñρθα‹ς ‡σας ε„σίν], καί ™στιν ¹ ØπÕ ΑΒΓ ™λάττων ÑρθÁς· λοιπ¾ ¥ρα ¹ ØπÕ Α∆Γ γωνία µείζων ÑρθÁς ™στιν· καί ™στιν ™ν τù Α∆Γ ™λάττονι τοà ¹µικυκλίου τµήµατι. Λέγω, Óτι κሠ¹ µν τοà µείζονος τµήµατος γωνία ¹ περιεχοµένη Øπό [τε] τÁς ΑΒΓ περιφερείας κሠτÁς ΑΓ εÙθείας µείζων ™στˆν ÑρθÁς, ¹ δ τοà ™λάττονος τµήµατος γωνία ¹ περιεχοµένη Øπό [τε] τÁς Α∆[Γ] περιφερείας κሠτÁς ΑΓ εÙθείας ™λάττων ™στˆν ÑρθÁς. καί ™στιν αÙτόθεν φανερόν. ™πεˆ γ¦ρ ¹ ØπÕ τîν ΒΑ, ΑΓ εÙθειîν Ñρθή ™στιν, ¹ ¥ρα ØπÕ τÁς ΑΒΓ περιφερείας κሠτÁς ΑΓ εÙθείας περιεχοµένη µείζων ™στˆν ÑρθÁς. πάλιν,

Let ABCD be a circle, and let BC be its diameter, and E its center. And let BA, AC, AD, and DC have been joined. I say that the angle BAC in the semi-circle BAC is a right-angle, and the angle ABC in the segment ABC, (which is) greater than a semi-circle, is less than a rightangle, and the angle ADC in the segment ADC, (which is) less than a semi-circle, is greater than a right-angle. Let AE have been joined, and let BA have been drawn through to F . And since BE is equal to EA, angle ABE is also equal to BAE [Prop. 1.5]. Again, since CE is equal to EA, ACE is also equal to CAE [Prop. 1.5]. Thus, the whole (angle) BAC is equal to the two (angles) ABC and ACB. And F AC, (which is) external to triangle ABC, is also equal to the two angles ABC and ACB [Prop. 1.32]. Thus, angle BAC (is) also equal to F AC. Thus, (they are) each right-angles. [Def. 1.10]. Thus, the angle BAC in the semi-circle BAC is a right-angle. And since the two angles ABC and BAC of triangle ABC are less than two right-angles [Prop. 1.17], and BAC is a right-angle, angle ABC is thus less than a rightangle. And it is in segment ABC, (which is) greater than a semi-circle. And since ABCD is a quadrilateral within a circle, and for quadrilaterals within circles the (sum of the) opposite angles is equal to two right-angles [Prop. 3.22] [angles ABC and ADC are thus equal to two rightangles], and (angle) ABC is less than a right-angle. The remaining angle ADC is thus greater than a right-angle. And it is in segment ADC, (which is) less than a semicircle. I also say that the angle of the greater segment, (namely) that contained by the circumference ABC and the straight-line AC, is greater than a right-angle. And

99

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

™πεˆ ¹ ØπÕ τîν ΑΓ, ΑΖ εÙθειîν Ñρθή ™στιν, ¹ ¥ρα ØπÕ τÁς ΓΑ εÙθείας κሠτÁς Α∆[Γ] περιφερείας περιεχοµένη ™λάττων ™στˆν ÑρθÁς. 'Εν κύκλJ ¥ρα ¹ µν ™ν τù ¹µικυκλίJ γωνία Ñρθή ™στιν, ¹ δ ™ν τù µείζονι τµήµατι ™λάττων ÑρθÁς, ¹ δ ™ν τù ™λάττονι [τµήµατι] µείζων ÑρθÁς· κሠœπι ¹ µν τοà µείζονος τµήµατος [γωνία] µείζων [™στˆν] ÑρθÁς, ¹ δ τοà ™λάττονος τµήµατος [γωνία] ™λάττων ÑρθÁς· Óπερ œδει δε‹ξαι.

the angle of the lesser segment, (namely) that contained by the circumference AD[C] and the straight-line AC, is less than a right-angle. And this is immediately apparent. For since the (angle contained by) the two straight-lines BA and AC is a right-angle, the (angle) contained by the circumference ABC and the straight-line AC is thus greater than a right-angle. Again, since the (angle contained by) the straight-lines AC and AF is a right-angle, the (angle) contained by the circumference AD[C] and the straight-line CA is thus less than a right-angle. Thus, in a circle, the angle in a semi-circle is a rightangle, and that in a greater segment (is) less than a right-angle, and that in a lesser [segment] (is) greater than a right-angle. And, further, the [angle] of a segment greater (than a semi-circle) [is] greater than a rightangle, and the [angle] of a segment less (than a semicircle) is less than a right-angle. (Which is) the very thing it was required to show.

λβ΄.

Proposition 32

'Ε¦ν κύκλου ™φάπτηταί τις εÙθε‹α, ¢πÕ δ τÁς ¡φÁς ε„ς τÕν κύκλον διαχθÍ τις εÙθε‹α τέµνουσα τÕν κύκλον, §ς ποιε‹ γωνίας πρÕς τÍ ™φαπτοµένV, ‡σαι œσονται τα‹ς ™ν το‹ς ™ναλλ¦ξ τοà κύκλου τµήµασι γωνίαις.

If some straight-line touches a circle, and some (other) straight-line is drawn across, from the point of contact into the circle, cutting the circle (in two), then those angles the (straight-line) makes with the tangent will be equal to the angles in the alternate segments of the circle.

Α

A ∆

D

Γ Ε

Β

C Ζ

E

F B

Κύκλου γ¦ρ τοà ΑΒΓ∆ ™φαπτέσθω τις εÙθε‹α ¹ ΕΖ κατ¦ τÕ Β σηµε‹ον, κሠ¢πÕ τοà Β σηµείου διήχθω τις εÙθε‹α ε„ς τÕν ΑΒΓ∆ κύκλον τέµνουσα αÙτÕν ¹ Β∆. λέγω, Óτι §ς ποιε‹ γωνίας ¹ Β∆ µετ¦ τÁς ΕΖ ™φαπτοµένης, ‡σας œσονται τα‹ς ™ν το‹ς ™ναλλ¦ξ τµήµασι τοà κύκλου γωνίαις, τουτέστιν, Óτι ¹ µν ØπÕ ΖΒ∆ γωνία ‡ση ™στˆ τÍ ™ν τù ΒΑ∆ τµήµατι συνισταµένV γωνίv, ¹ δ ØπÕ ΕΒ∆ γωνία ‡ση ™στˆ τÍ ™ν τù ∆ΓΒ τµήµατι συνισταµένV γωνίv. ”Ηχθω γ¦ρ ¢πÕ τοà Β τÍ ΕΖ πρÕς Ñρθ¦ς ¹ ΒΑ, κሠε„λήφθω ™πˆ τÁς Β∆ περιφερείας τυχÕν σηµε‹ον τÕ

For let some straight-line EF touch the circle ABCD at the point B, and let some (other) straight-line BD have been drawn from point B into the circle ABCD, cutting it (in two). I say that the angles BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle. That is to say, that angle F BD is equal to one (of the) angle(s) constructed in segment BAD, and angle EBD is equal to one (of the) angle(s) constructed in segment DCB. For let BA have been drawn from B, at right-angles to EF [Prop. 1.11]. And let the point C have been taken

100

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

Γ, κሠ™πεζεύχθωσαν αƒ Α∆, ∆Γ, ΓΒ. Κሠ™πεˆ κύκλου τοà ΑΒΓ∆ ™φάπτεταί τις εÙθε‹α ¹ ΕΖ κατ¦ τÕ Β, κሠ¢πÕ τÁς ¡φÁς Ãκται τÍ ™φαπτοµένV πρÕς Ñρθ¦ς ¹ ΒΑ, ™πˆ τÁς ΒΑ ¥ρα τÕ κέντρον ™στˆ τοà ΑΒΓ∆ κύκλου. ¹ ΒΑ ¥ρα διάµετός ™στι τοà ΑΒΓ∆ κύκλου· ¹ ¥ρα ØπÕ Α∆Β γωνία ™ν ¹µικυκλίJ οâσα Ñρθή ™στιν. λοιπሠ¥ρα αƒ ØπÕ ΒΑ∆, ΑΒ∆ µι´ ÑρθÍ ‡σαι ε„σίν. ™στˆ δ κሠ¹ ØπÕ ΑΒΖ Ñρθή· ¹ ¥ρα ØπÕ ΑΒΖ ‡ση ™στˆ τα‹ς ØπÕ ΒΑ∆, ΑΒ∆. κοιν¾ ¢φVρήσθω ¹ ØπÕ ΑΒ∆· λοιπ¾ ¥ρα ¹ ØπÕ ∆ΒΖ γωνία ‡ση ™στˆ τÍ ™ν τù ™ναλλ¦ξ τµήµατι τοà κύκλου γωνίv τÍ ØπÕ ΒΑ∆. κሠ™πεˆ ™ν κύκλJ τετράπλευρόν ™στι τÕ ΑΒΓ∆, αƒ ¢πεναντίον αÙτοà γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν. ε„σˆ δ καˆ αƒ ØπÕ ∆ΒΖ, ∆ΒΕ δυσˆν Ñρθα‹ς ‡σαι· αƒ ¥ρα ØπÕ ∆ΒΖ, ∆ΒΕ τα‹ς ØπÕ ΒΑ∆, ΒΓ∆ ‡σαι ε„σίν, ïν ¹ ØπÕ ΒΑ∆ τÍ ØπÕ ∆ΒΖ ™δείχθη ‡ση· λοιπ¾ ¥ρα ¹ ØπÕ ∆ΒΕ τÍ ™ν τù ™ναλλ¦ξ τοà κύκλου τµήµατι τù ∆ΓΒ τÍ ØπÕ ∆ΓΒ γωνίv ™στˆν ‡ση. 'Ε¦ν ¥ρα κύκλου ™φάπτηταί τις εÙθε‹α, ¢πÕ δ τÁς ¡φÁς ε„ς τÕν κύκλον διαχθÍ τις εÙθε‹α τέµνουσα τÕν κύκλον, §ς ποιε‹ γωνίας πρÕς τÍ ™φαπτοµένV, ‡σαι œσονται τα‹ς ™ν το‹ς ™ναλλ¦ξ τοà κύκλου τµήµασι γωνίαις· Óπερ œδει δε‹ξαι.

somewhere on the circumference BD. And let AD, DC, and CB have been joined. And since some straight-line EF touches the circle ABCD at point B, and BA has been drawn from the point of contact, at right-angles to the tangent, the center of circle ABCD is thus on BA [Prop. 3.19]. Thus, BA is a diameter of circle ABCD. Thus, angle ADB, being in a semi-circle, is a right-angle [Prop. 3.31]. Thus, the remaining angles (of triangle ADB) BAD and ABD are equal to one right-angle [Prop. 1.32]. And ABF is also a right-angle. Thus, ABF is equal to BAD and ABD. Let ABD have been subtracted from both. Thus, the remaining angle DBF is equal to the angle BAD in the alternate segment of the circle. And since ABCD is a quadrilateral in a circle, (the sum of) its opposite angles is equal to two right-angles [Prop. 3.22]. And DBF and DBE is also equal to two right-angles [Prop. 1.13]. Thus, DBF and DBE is equal to BAD and BCD, of which BAD was shown (to be) equal to DBF . Thus, the remaining angle DBE is equal to the angle DCB in the alternate segment DCB of the circle. Thus, if some straight-line touches a circle, and some (other) straight-line is drawn across, from the point of contact into the circle, cutting the circle (in two), then those angles the (straight-line) makes with the tangent will be equal to the angles in the alternate segments of the circle. (Which is) the very thing it was required to show.

λγ΄.

Proposition 33

'Επˆ τÁς δοθείσης εÙθείας γράψαι τµÁµα κύκλου δεχόµενον γωνίαν ‡σην τÍ δοθείσV γωνίv εÙθυγράµµJ.

To draw a segment of a circle, accepting an angle equal to a given rectilinear angle, on a given straight-line.

Γ Α

∆ Ζ

Η

Α

∆

C A

Γ

Γ

Α

D

∆

C

C A

D

A H

Θ Ε

F

Ζ Ζ

Β

F G

Η

E

G

F B

B

Β Ε

Β

Ε

E

”Εστω ¹ δοθε‹σα εÙθε‹α ¹ ΑΒ, ¹ δ δοθε‹σα γωνία εÙθύγραµµος ¹ πρÕς τù Γ· δε‹ δ¾ ™πˆ τÁς δοθείσης εÙθείας τÁς ΑΒ γράψαι τµÁµα κύκλου δεχόµενον γωνίαν ‡σην τÍ πρÕς τù Γ. `Η δ¾ πρÕς τù Γ [γωνία] ½τοι Ñξε‹ά ™στιν À Ñρθ¾ À ¢µβλε‹α· œστω πρότερον Ñξε‹α, κሠæς ™πˆ τÁς πρώτης καταγραφÁς συνεστάτω πρÕς τÍ ΑΒ εÙθείv κሠτù Α σηµείJ τÍ πρÕς τù Γ γωνίv ‡ση ¹ ØπÕ ΒΑ∆· Ñξε‹α ¥ρα ™στˆ κሠ¹ ØπÕ ΒΑ∆. ½χθω τÍ ∆Α πρÕς Ñρθ¦ς ¹ ΑΕ,

B

E

Let AB be the given straight-line, and C the given rectilinear angle. So it is required to draw a segment of a circle, accepting an angle equal to C, on the given straight-line AB. So the [angle] C is surely either acute, a right-angle, or obtuse. First of all, let it be acute. And, as in the first diagram (from the left), let (angle) BAD, equal to angle C, have been constructed at the point A on the straightline AB [Prop. 1.23]. Thus, BAD is also acute. Let AE

101

D

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

κሠτετµήσθω ¹ ΑΒ δίχα κατ¦ τÕ Ζ, κሠ½χθω ¢πÕ τοà Ζ σηµείου τÍ ΑΒ πρÕς Ñρθ¦ς ¹ ΖΗ, κሠ™πεζεύχθω ¹ ΗΒ. Κሠ™πεˆ ‡ση ™στˆν ¹ ΑΖ τÍ ΖΒ, κοιν¾ δ ¹ ΖΗ, δύο δ¾ αƒ ΑΖ, ΖΗ δύο τα‹ς ΒΖ, ΖΗ ‡σαι ε„σίν· κሠγωνία ¹ ØπÕ ΑΖΗ [γωνίv] τÍ ØπÕ ΒΖΗ ‡ση· βάσις ¥ρα ¹ ΑΗ βάσει τÍ ΒΗ ‡ση ™στίν. Ð ¥ρα κέντρJ µν τù Η διαστήµατι δ τù ΗΑ κύκλος γραφόµενος ¼ξει κሠδι¦ τοà Β. γεγράφθω κሠœστω Ð ΑΒΕ, κሠ™πεζεύχθω ¹ ΕΒ. ™πεˆ οâν ¢π' ¥κρας τÁς ΑΕ διαµέτρου ¢πÕ τοà Α τÍ ΑΕ πρÕς Ñρθάς ™στιν ¹ Α∆, ¹ Α∆ ¥ρα ™φάπτεται τοà ΑΒΕ κύκλου· ™πεˆ οâν κύκλου τοà ΑΒΕ ™φάπτεταί τις εÙθε‹α ¹ Α∆, κሠ¢πÕ τÁς κατ¦ τÕ Α ¡φÁς ε„ς τÕν ΑΒΕ κύκλον διÁκταί τις εÙθε‹α ¹ ΑΒ, ¹ ¥ρα ØπÕ ∆ΑΒ γωνία ‡ση ™στˆ τÍ ™ν τù ™ναλλ¦ξ τοà κύκλου τµήµατι γωνίv τÍ ØπÕ ΑΕΒ. ¢λλ' ¹ ØπÕ ∆ΑΒ τÍ πρÕς τù Γ ™στιν ‡ση· κሠ¹ πρÕς τù Γ ¥ρα γωνία ‡ση ™στˆ τÍ ØπÕ ΑΕΒ. 'Επˆ τÁς δοθείσης ¥ρα εÙθείας τÁς ΑΒ τµÁµα κύκλου γέγραπται τÕ ΑΕΒ δεχόµενον γωνίαν τ¾ν ØπÕ ΑΕΒ ‡σην τÍ δοθείσV τÍ πρÕς τù Γ. 'Αλλ¦ δ¾ Ñρθ¾ œστω ¹ πρÕς τù Γ· κሠδέον πάλιν œστω ™πˆ τÁς ΑΒ γράψαι τµÁµα κύκλου δεχόµενον γωνίαν ‡σην τÍ πρÕς τù Γ ÑρθÍ [γωνίv]. συνεστάτω [πάλιν] τÍ πρÕς τù Γ ÑρθÍ γωνίv ‡ση ¹ ØπÕ ΒΑ∆, æς œχει ™πˆ τÁς δευτέρας καταγραφÁς, κሠτετµήσθω ¹ ΑΒ δίχα κατ¦ τÕ Ζ, κሠκέντρJ τù Ζ, διαστήµατι δ ÐποτέρJ τîν ΖΑ, ΖΒ, κύκλος γεγράφθω Ð ΑΕΒ. 'Εφάπτεται ¥ρα ¹ Α∆ εÙθε‹α τοà ΑΒΕ κύκλου δι¦ τÕ Ñρθ¾ν εναι τ¾ν πρÕς τù Α γωνίαν. κሠ‡ση ™στˆν ¹ ØπÕ ΒΑ∆ γωνία τÍ ™ν τù ΑΕΒ τµήµατι· Ñρθ¾ γ¦ρ κሠαÙτ¾ ™ν ¹µικυκλίJ οâσα. ¢λλ¦ κሠ¹ ØπÕ ΒΑ∆ τÍ πρÕς τù Γ ‡ση ™στίν. κሠ¹ ™ν τù ΑΕΒ ¥ρα ‡ση ™στˆ τÍ πρÕς τù Γ. Γέγραπται ¥ρα πάλιν ™πˆ τÁς ΑΒ τµÁµα κύκλου τÕ ΑΕΒ δεχόµενον γωνίαν ‡σην τÍ πρÕς τù Γ. 'Αλλ¦ δ¾ ¹ πρÕς τù Γ ¢µβλε‹α œστω· κሠσυνεστάτω αÙτÍ ‡ση πρÕς τÍ ΑΒ εÙθείv κሠτù Α σηµείJ ¹ ØπÕ ΒΑ∆, æς œχει ™πˆ τÁς τρίτης καταγραφÁς, κሠτÍ Α∆ πρÕς Ñρθ¦ς ½χθω ¹ ΑΕ, κሠτετµήσθω πάλιν ¹ ΑΒ δίχα κατ¦ τÕ Ζ, κሠτÍ ΑΒ πρÕς Ñρθ¦ς ½χθω ¹ ΖΗ, κሠ™πεζεύχθω ¹ ΗΒ. Κሠ™πεˆ πάλιν ‡ση ™στˆν ¹ ΑΖ τÍ ΖΒ, κሠκοιν¾ ¹ ΖΗ, δύο δ¾ αƒ ΑΖ, ΖΗ δύο τα‹ς ΒΖ, ΖΗ ‡σαι ε„σίν· κሠγωνία ¹ ØπÕ ΑΖΗ γωνίv τÍ ØπÕ ΒΖΗ ‡ση· βάσις ¥ρα ¹ ΑΗ βάσει τÍ ΒΗ ‡ση ™στίν· Ð ¥ρα κέντρJ µν τù Η διαστήµατι δ τù ΗΑ κύκλος γραφόµενος ¼ξει κሠδι¦ τοà Β. ™ρχέσθω æς Ð ΑΕΒ. κሠ™πεˆ τÍ ΑΕ διαµέτρJ ¢π' ¥κρας πρÕς Ñρθάς ™στιν ¹ Α∆, ¹ Α∆ ¥ρα ™φάπτεται τοà ΑΕΒ κύκλου. κሠ¢πÕ τÁς κατ¦ τÕ Α ™παφÁς διÁκται ¹ ΑΒ· ¹ ¥ρα ØπÕ ΒΑ∆ γωνία ‡ση ™στˆ τÍ ™ν τù ™ναλλ¦ξ τοà κύκλου τµήµατι τù ΑΘΒ συνισταµένV γωνίv. ¢λλ'

have been drawn, at right-angles to DA [Prop. 1.11]. And let AB have been cut in half at F [Prop. 1.10]. And let F G have been drawn from point F , at right-angles to AB [Prop. 1.11]. And let GB have been joined. And since AF is equal to F B, and F G (is) common, the two (straight-lines) AF , F G are equal to the two (straight-lines) BF , F G (respectively). And angle AF G (is) equal to [angle] BF G. Thus, the base AG is equal to the base BG [Prop. 1.4]. Thus, the circle drawn with center G, and radius GA, will also go through B (as well as A). Let it have been drawn, and let it be (denoted) ABE. And let EB have been joined. Therefore, since AD is at the end of diameter AE, at (point) A, at right-angles to AE, the (straight-line) AD thus touches the circle ABE [Prop. 3.16 corr.]. Therefore, since some straight-line AD touches the circle ABE, and some (other) straight-line AB has been drawn across from the point of contact A into circle ABE, angle DAB is thus equal to the angle AEB in the alternate segment of the circle [Prop. 3.32]. But, DAB is equal to C. Thus, angle C is also equal to AEB. Thus, a segment AEB of a circle, accepting the angle AEB (which is) equal to the given (angle) C, has been drawn on the given straight-line AB. And so let C be a right-angle. And let it again be necessary to draw a segment of a circle on AB, accepting an angle equal to the right-[angle] C. Let the (angle) BAD [again] have been constructed, equal to the rightangle C [Prop. 1.23], as in the second diagram (from the left). And let AB have been cut in half at F [Prop. 1.10]. And let the circle AEB have been drawn with center F , and radius either F A or F B. Thus, the straight-line AD touches the circle ABE, on account of the angle at A being a right-angle [Prop. 3.16 corr.]. And angle BAD is equal to the angle in segment AEB. For (the latter angle), being in a semi-circle, is also a right-angle [Prop. 3.31]. But, BAD is also equal to C. Thus, the (angle) in (segment) AEB is also equal to C. Thus, a segment AEB of a circle, accepting an angle equal to C, has again been drawn on AB. And so let (angle) C be obtuse. And let (angle) BAD, equal to (C), have been constructed at the point A on the straight-line AB [Prop. 1.23], as in the third diagram (from the left). And let AE have been drawn, at right-angles to AD [Prop. 1.11]. And let AB have again been cut in half at F [Prop. 1.10]. And let F G have been drawn, at right-angles to AB [Prop. 1.10]. And let GB have been joined. And again, since AF is equal to F B, and F G (is) common, the two (straight-lines) AF , F G are equal to the two (straight-lines) BF , F G (respectively). And an-

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ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

¹ ØπÕ ΒΑ∆ γωνία τÍ πρÕς τù Γ ‡ση ™στίν. κሠ¹ ™ν τù ΑΘΒ ¥ρα τµήµατι γωνία ‡ση ™στˆ τÍ πρÕς τù Γ. 'Επˆ τÁς ¥ρα δοθείσης εÙθείας τÁς ΑΒ γέγραπται τµÁµα κύκλου τÕ ΑΘΒ δεχόµενον γωνίαν ‡σην τÍ πρÕς τù Γ· Óπερ œδει ποιÁσαι.

gle AF G (is) equal to angle BF G. Thus, the base AG is equal to the base BG [Prop. 1.4]. Thus, a circle of center G, and radius GA, being drawn, will also go through B (as well as A). Let it go like AEB (in the third diagram from the left). And since AD is at right-angles to the diameter AE, at the end, AD thus touches circle AEB [Prop. 3.16 corr.]. And AB has been drawn across (the circle) from the point of contact A. Thus, angle BAD is equal to the angle constructed in the alternate segment AHB of the circle [Prop. 3.32]. But, angle BAD is equal to C. Thus, the angle in segment AHB is also equal to C. Thus, a segment AHB of a circle, accepting an angle equal to C, has been drawn on the given straight-line AB. (Which is) the very thing it was required to do.

λδ΄.

Proposition 34

'ΑπÕ τοà δοθέντος κύκλου τµÁµα ¢φελε‹ν δεχόµενον γωνίαν ‡σην τÍ δοθείσV γωνίv εÙθυγράµµJ.

To cut off a segment, accepting an angle equal to a given rectilinear angle, from a given circle.

Ζ

Γ

F

C

B

Β ∆ Ε

D

Α

E

”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ, ¹ δ δοθε‹σα γωνία εÙθύγραµµος ¹ πρÕς τù ∆· δε‹ δ¾ ¢πÕ τοà ΑΒΓ κύκλου τµÁµα ¢φελε‹ν δεχόµενον γωνίαν ‡σην τÍ δοθείσV γωνίv εÙθυγράµµJ τÍ πρÕς τù ∆. ”Ηχθω τοà ΑΒΓ ™φαπτοµένη ¹ ΕΖ κατ¦ τÕ Β σηµε‹ον, κሠσυνεστάτω πρÕς τÍ ΖΒ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Β τÍ πρÕς τù ∆ γωνίv ‡ση ¹ ØπÕ ΖΒΓ. 'Επεˆ οâν κύκλου τοà ΑΒΓ ™φάπτεταί τις εÙθε‹α ¹ ΕΖ, κሠ¢πÕ τÁς κατ¦ τÕ Β ™παφÁς διÁκται ¹ ΒΓ, ¹ ØπÕ ΖΒΓ ¥ρα γωνία ‡ση ™στˆ τÍ ™ν τù ΒΑΓ ™ναλλ¦ξ τµήµατι συνισταµένV γωνίv. ¢λλ' ¹ ØπÕ ΖΒΓ τÍ πρÕς τù ∆ ™στιν ‡ση· κሠ¹ ™ν τù ΒΑΓ ¥ρα τµήµατι ‡ση ™στˆ τÍ πρÕς τù ∆ [γωνίv]. 'ΑπÕ τοà δοθέντος ¥ρα κύκλου τοà ΑΒΓ τµÁµα ¢φÇρηται τÕ ΒΑΓ δεχόµενον γωνίαν ‡σην τÍ δοθείσV γωνίv εÙθυγράµµJ τÍ πρÕς τù ∆· Óπερ œδει ποιÁσαι.

†

A

Let ABC be the given circle, and D the given rectilinear angle. So it is required to cut off a segment, accepting an angle equal to the given rectilinear angle D, from the given circle ABC. Let EF have been drawn touching ABC at point B.† And let (angle) F BC, equal to angle D, have been constructed at the point B on the straight-line F B [Prop. 1.23]. Therefore, since some straight-line EF touches the circle ABC, and BC has been drawn across (the circle) from the point of contact B, angle F BC is thus equal to the angle constructed in the alternate segment BAC [Prop. 1.32]. But, F BC is equal to D. Thus, the (angle) in the segment BAC is also equal to [angle] D. Thus, the segment BAC, accepting an angle equal to the given rectilinear angle D, has been cut off from the given circle ABC. (Which is) the very thing it was required to do.

Presumably, by finding the center of ABC [Prop. 3.1], drawing a straight-line between the center and point B, and then drawing EF through

103

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

point B, at right-angles to the aforementioned straight-line [Prop. 1.11].

λε΄.

Proposition 35

'Ε¦ν ™ν κύκλJ δύο εÙθε‹αι τέµνωσιν ¢λλήλας, τÕ If two straight-lines in a circle cut one another then ØπÕ τîν τÁς µι©ς τµηµάτων περιεχόµενον Ñρθογώνιον the rectangle contained by the pieces of one is equal to ‡σον ™στˆ τù ØπÕ τîν τÁς ˜τέρας τµηµάτων περιεχοµένJ the rectangle contained by the pieces of the other. ÑρθογωνίJ.

Α

A ∆

Α Β

Ε

Ζ

∆

B

E

F

D

Θ

Η Γ

D A

Ε Β

H

G E C

Γ

'Εν γ¦ρ κύκλJ τù ΑΒΓ∆ δύο εÙθε‹αι αƒ ΑΓ, Β∆ τεµνέτωσαν ¢λλήλας κατ¦ τÕ Ε σηµε‹ον· λέγω, Óτι τÕ ØπÕ τîν ΑΕ, ΕΓ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ØπÕ τîν ∆Ε, ΕΒ περιεχοµένJ ÑρθογωνίJ. Ε„ µν οâν αƒ ΑΓ, Β∆ δι¦ τοà κέντρου ε„σˆν éστε τÕ Ε κέντρον εναι τοà ΑΒΓ∆ κύκλου, φανερόν, Óτι ‡σων οÙσîν τîν ΑΕ, ΕΓ, ∆Ε, ΕΒ κሠτÕ ØπÕ τîν ΑΕ, ΕΓ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ØπÕ τîν ∆Ε, ΕΒ περιεχοµένJ ÑρθογωνίJ. Μ¾ œστωσαν δ¾ αƒ ΑΓ, ∆Β δι¦ τοà κέντρου, κሠε„λήφθω τÕ κέντρον τοà ΑΒΓ∆, κሠœστω τÕ Ζ, κሠ¢πÕ τοà Ζ ™πˆ τ¦ς ΑΓ, ∆Β εÙθείας κάθετοι ½χθωσαν αƒ ΖΗ, ΖΘ, κሠ™πεζεύχθωσαν αƒ ΖΒ, ΖΓ, ΖΕ. Κሠ™πεˆ εÙθε‹ά τις δι¦ τοà κέντρου ¹ ΗΖ εÙθε‹άν τινα µ¾ δι¦ τοà κέντρου τ¾ν ΑΓ πρÕς Ñρθ¦ς τέµνει, κሠδίχα αÙτ¾ν τέµνει· ‡ση ¥ρα ¹ ΑΗ τÍ ΗΓ. ™πεˆ οâν εÙθε‹α ¹ ΑΓ τέτµηται ε„ς µν ‡σα κατ¦ τÕ Η, ε„ς δ ¥νισα κατ¦ τÕ Ε, τÕ ¥ρα ØπÕ τîν ΑΕ, ΕΓ περιεχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς ΕΗ τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς ΗΓ· [κοινÕν] προσκείσθω τÕ ¢πÕ τÁς ΗΖ· τÕ ¥ρα ØπÕ τîν ΑΕ, ΕΓ µετ¦ τîν ¢πÕ τîν ΗΕ, ΗΖ ‡σον ™στˆ το‹ς ¢πÕ τîν ΓΗ, ΗΖ. ¢λλ¦ το‹ς µν ¢πÕ τîν ΕΗ, ΗΖ ‡σον ™στˆ τÕ ¢πÕ τÁς ΖΕ, τοˆς δ ¢πÕ τîν ΓΗ, ΗΖ ‡σον ™στˆ τÕ ¢πÕ τÁς ΖΓ· τÕ ¥ρα ØπÕ τîν ΑΕ, ΕΓ µετ¦ τοà ¢πÕ τÁς ΖΕ ‡σον ™στˆ τù ¢πÕ τÁς ΖΓ. ‡ση δ ¹ ΖΓ τÍ ΖΒ· τÕ ¥ρα ØπÕ τîν ΑΕ, ΕΓ µετ¦ τοà ¢πÕ τÁς ΕΖ ‡σον ™στˆ τù ¢πÕ τÁς ΖΒ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ØπÕ τîν ∆Ε, ΕΒ µετ¦ τοà ¢πÕ τÁς ΖΕ „σον ™στˆ τù ¢πÕ τÁς ΖΒ. ™δείχθη δ κሠτÕ ØπÕ τîν ΑΕ, ΕΓ µετ¦ τοà ¢πÕ τÁς ΖΕ ‡σον τù ¢πÕ τÁς ΖΒ· τÕ ¥ρα ØπÕ τîν ΑΕ, ΕΓ µετ¦ τοà ¢πÕ τÁς ΖΕ ‡σον ™στˆ τù ØπÕ τîν ∆Ε, ΕΒ µετ¦ τοà ¢πÕ τÁς ΖΕ. κοινÕν ¢φÍρήσθω τÕ ¢πÕ τÁς ΖΕ· λοιπÕν ¥ρα τÕ ØπÕ τîν ΑΕ, ΕΓ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ØπÕ τîν ∆Ε, ΕΒ περιεχοµένJ ÑρθογωνίJ.

B

C

For let the two straight-lines AC and BD, in the circle ABCD, cut one another at point E. I say that the rectangle contained by AE and EC is equal to the rectangle contained by DE and EB. In fact, if AC and BD are through the center (as in the first diagram from the left), so that E is the center of circle ABCD, then (it is) clear that, AE, EC, DE, and EB being equal, the rectangle contained by AE and EC is also equal to the rectangle contained by DE and EB. So let AC and DB not be though the center (as in the second diagram from the left), and let the center of ABCD have been found [Prop. 3.1], and let it be (at) F . And let F G and F H have been drawn from F , perpendicular to the straight-lines AC and DB (respectively) [Prop. 1.12]. And let F B, F C, and F E have been joined. And since some straight-line, GF , through the center cuts at right-angles some (other) straight-line, AC, not through the center, then it also cuts it in half [Prop. 3.3]. Thus, AG (is) equal to GC. Therefore, since the straightline AC is cut equally at G, and unequally at E, the rectangle contained by AE and EC plus the square on EG is thus equal to the (square) on GC [Prop. 2.5]. Let the (square) on GF have been added [to both]. Thus, the (rectangle contained) by AE and EC plus the (sum of the squares) on GE and GF is equal to the (sum of the squares) on CG and GF . But, the (sum of the squares) on EG and GF is equal to the (square) on F E [Prop. 1.47], and the (sum of the squares) on CG and GF is equal to the (square) on F C [Prop. 1.47]. Thus, the (rectangle contained) by AE and EC plus the (square) on F E is equal to the (square) on F C. And F C (is) equal to F B. Thus, the (rectangle contained) by AE and EC plus the (square) on F E is equal to the (square) on F B. So, for the same (reasons), the (rectangle contained) by DE and

104

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

'Ε¦ν ¥ρα ™ν κύκλJ εÙθε‹αι δύο τέµνωσιν ¢λλήλας, τÕ ØπÕ τîν τÁς µι©ς τµηµάτων περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ØπÕ τîν τÁς ˜τέρας τµηµάτων περιεχοµένJ ÑρθογωνίJ· Óπερ œδει δε‹ξαι.

EB plus the (square) on F E is equal to the (square) on F B. And the (rectangle contained) by AE and EC plus the (square) on F E was also shown (to be) equal to the (square) on F B. Thus, the (rectangle contained) by AE and EC plus the (square) on F E is equal to the (rectangle contained) by DE and EB plus the (square) on F E. Let the (square) on F E have been taken from both. Thus, the remaining rectangle contained by AE and EC is equal to the rectangle contained by DE and EB. Thus, if two straight-lines in a circle cut one another then the rectangle contained by the pieces of one is equal to the rectangle contained by the pieces of the other. (Which is) the very thing it was required to show.

λ$΄.

Proposition 36

'Ε¦ν κύκλου ληφθÍ τι σηµε‹ον ™κτός, κሠ¢π' αÙτοà πρÕς τÕν κύκλον προσπίπτωσι δύο εÙθε‹αι, κሠ¹ µν αÙτîν τέµνV τÕν κύκλον, ¹ δ ™φάπτηται, œσται τÕ ØπÕ Óλης τÁς τεµνούσης κሠτÁς ™κτÕς ¢πολαµβανοµένης µεταξÝ τοà τε σηµείου κሠτÁς κυρτÁς περιφερείας ‡σον τù ¢πÕ τÁς ™φαπτοµένης τετραγώνJ.

If some point is taken outside a circle, and two straight-lines radiate from it towards the circle, and (one) of them cuts the circle, and the (other) touches (it), then the (rectangle contained) by the whole (straight-line) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumference, will be equal to the square on the tangent (line).

Α

Β

A

Ε

Ζ

∆

E

F

Α

Γ Γ

B

A C

Ζ

C

Β

∆

D

F B

D

Κύκλου γ¦ρ τοà ΑΒΓ ε„λήφθω τι σηµε‹ον ™κτÕς τÕ ∆, κሠ¢πÕ τοà ∆ πρÕς τÕν ΑΒΓ κύκλον προσπιπτέτωσαν δύο εÙθε‹αι αƒ ∆Γ[Α], ∆Β· κሠ¹ µν ∆ΓΑ τεµνέτω τÕν ΑΒΓ κύκλον, ¹ δ Β∆ ™φαπτέσθω· λέγω, Óτι τÕ ØπÕ τîν Α∆, ∆Γ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ¢πÕ τÁς ∆Β τετραγώνJ. `Η ¥ρα [∆]ΓΑ ½τοι δι¦ τοà κέντρου ™στˆν À οÜ. œστω πρότερον δι¦ τοà κέντρου, κሠœστω τÕ Ζ κέντρον τοà ΑΒΓ κύκλου, κሠ™πεζεύχθω ¹ ΖΒ· Ñρθ¾ ¥ρα ™στˆν ¹ ØπÕ ΖΒ∆. κሠ™πεˆ εÙθε‹α ¹ ΑΓ δίχα τέτµηται κατ¦ τÕ Ζ, πρόσκειται δ αÙτÍ ¹ Γ∆, τÕ ¥ρα ØπÕ τîν Α∆, ∆Γ µετ¦ τοà ¢πÕ τÁς ΖΓ ‡σον ™στˆ τù ¢πÕ τÁς Ζ∆. ‡ση δ ¹ ΖΓ τÍ ΖΒ· τÕ ¥ρα ØπÕ τîν Α∆, ∆Γ µετ¦ τοà ¢πÕ τÁς ΖΒ ‡σον ™στˆ τù ¢πÕ τ¾ς Ζ∆. τù δ ¢πÕ τÁς Ζ∆ ‡σα ™στˆ τ¦ ¢πÕ τîν ΖΒ, Β∆· τÕ ¥ρα ØπÕ τîν Α∆, ∆Γ µετ¦ τοà ¢πÕ τÁς ΖΒ ‡σον ™στˆ το‹ς ¢πÕ τîν ΖΒ, Β∆. κοινÕν ¢φVρήσθω τÕ ¢πÕ τÁς ΖΒ· λοιπÕν ¥ρα τÕ ØπÕ τîν Α∆,

For let some point D have been taken outside circle ABC, and let two straight-lines, DC[A] and DB, radiate from D towards circle ABC. And let DCA cut circle ABC, and let BD touch (it). I say that the rectangle contained by AD and DC is equal to the square on DB. [D]CA is surely either through the center, or not. Let it first of all be through the center, and let F be the center of circle ABC, and let F B have been joined. Thus, (angle) F BD is a right-angle [Prop. 3.18]. And since straight-line AC is cut in half at F , let CD have been added to it. Thus, the (rectangle contained) by AD and DC plus the (square) on F C is equal to the (square) on F D [Prop. 2.6]. And F C (is) equal to F B. Thus, the (rectangle contained) by AD and DC plus the (square) on F B is equal to the (square) on F D. And the (square) on F D is equal to the (sum of the squares) on F B and BD [Prop. 1.47]. Thus, the (rectangle contained) by AD

105

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

∆Γ ‡σον ™στˆ τù ¢πÕ τÁς ∆Β ™φαπτοµένης. 'Αλλ¦ δ¾ ¹ ∆ΓΑ µ¾ œστω δι¦ τοà κέντρου τοà ΑΒΓ κύκλου, κሠε„λήφθω τÕ κέντρον τÕ Ε, κሠ¢πÕ τοà Ε ™πˆ τ¾ν ΑΓ κάθετος ½χθω ¹ ΕΖ, κሠ™πεζεύχθωσαν αƒ ΕΒ, ΕΓ, Ε∆· Ñρθ¾ ¥ρα ™στˆν ¹ ØπÕ ΕΒ∆. κሠ™πεˆ εÙθε‹ά τις δι¦ τοà κέντρου ¹ ΕΖ εÙθε‹άν τινα µ¾ δι¦ τοà κέντρου τ¾ν ΑΓ πρÕς Ñρθ¦ς τέµνει, κሠδίχα αÙτ¾ν τέµνει· ¹ ΑΖ ¥ρα τÍ ΖΓ ™στιν ‡ση. κሠ™πεˆ εÙθε‹α ¹ ΑΓ τέτµηται δίχα κατ¦ τÕ Ζ σηµε‹ον, πρόσκειται δ αÙτÍ ¹ Γ∆, τÕ ¥ρα ØπÕ τîν Α∆, ∆Γ µετ¦ τοà ¢πÕ τÁς ΖΓ ‡σον ™στˆ τù ¢πÕ τÁς Ζ∆. κοινÕν προσκείσθω τÕ ¢πÕ τÁς ΖΕ· τÕ ¥ρα ØπÕ τîν Α∆, ∆Γ µετ¦ τîν ¢πÕ τîν ΓΖ, ΖΕ ‡σον ™στˆ το‹ς ¢πÕ τîν Ζ∆, ΖΕ. το‹ς δ ¢πÕ τîν ΓΖ, ΖΕ ‡σον ™στˆ τÕ ¢πÕ τÁς ΕΓ· Ñρθ¾ γ¦ρ [™στιν] ¹ ØπÕ ΕΖΓ [γωνία]· το‹ς δ ¢πÕ τîν ∆Ζ, ΖΕ ‡σον ™στˆ τÕ ¢πÕ τÁς Ε∆· τÕ ¥ρα ØπÕ τîν Α∆, ∆Γ µετ¦ τοà ¢πÕ τÁς ΕΓ ‡σον ™στˆ τù ¢πÕ τÁς Ε∆. ‡ση δ ¹ ΕΓ τÊ ΕΒ· τÕ ¥ρα ØπÕ τîν Α∆, ∆Γ µετ¦ τοà ¢πÕ τÁς ΕΒ ‡σον ™στˆ τù ¥πÕ τÁς Ε∆. τù δ ¢πÕ τÁς Ε∆ ‡σα ™στˆ τ¦ ¢πÕ τîν ΕΒ, Β∆· Ñρθ¾ γ¦ρ ¹ ØπÕ ΕΒ∆ γωνία· τÕ ¥ρα ØπÕ τîν Α∆, ∆Γ µετ¦ τοà ¢πÕ τÁς ΕΒ ‡σον ™στˆ το‹ς ¢πÕ τîν ΕΒ, Β∆. κοινÕν ¢φVρήσθω τÕ ¢πÕ τÁς ΕΒ· λοιπÕν ¥ρα τÕ ØπÕ τîν Α∆, ∆Γ ‡σον ™στˆ τù ¢πÕ τÁς ∆Β. 'Ε¦ν ¥ρα κύκλου ληφθÍ τι σηµε‹ον ™κτός, κሠ¢π' αÙτοà πρÕς τÕν κύκλον προσπίπτωσι δύο εÙθε‹αι, κሠ¹ µν αÙτîν τέµνV τÕν κύκλον, ¹ δ ™φάπτηται, œσται τÕ ØπÕ Óλης τÁς τεµνούσης κሠτÁς ™κτÕς ¢πολαµβανοµένης µεταξÝ τοà τε σηµείου κሠτÁς κυρτÁς περιφερείας ‡σον τù ¢πÕ τÁς ™φαπτοµένης τετραγώνJ· Óπερ œδει δε‹ξαι.

and DC plus the (square) on F B is equal to the (sum of the squares) on F B and BD. Let the (square) on F B have been subtracted from both. Thus, the remaining (rectangle contained) by AD and DC is equal to the (square) on the tangent DB. And so let DCA not be through the center of circle ABC, and let the center E have been found, and let EF have been drawn from E, perpendicular to AC [Prop. 1.12]. And let EB, EC, and ED have been joined. (Angle) EBD (is) thus a right-angle [Prop. 3.18]. And since some straight-line, EF , through the center cuts some (other) straight-line, AC, not through the center, at right-angles, it also cuts it in half [Prop. 3.3]. Thus, AF is equal to F C. And since the straight-line AC is cut in half at point F , let CD have been added to it. Thus, the (rectangle contained) by AD and DC plus the (square) on F C is equal to the (square) on F D [Prop. 2.6]. Let the (square) on F E have been added to both. Thus, the (rectangle contained) by AD and DC plus the (sum of the squares) on CF and F E is equal to the (sum of the squares) on F D and F E. But the (sum of the squares) on CF and F E is equal to the (square) on EC. For [angle] EF C [is] a right-angle [Prop. 1.47]. And the (sum of the squares) on DF and F E is equal to the (square) on ED [Prop. 1.47]. Thus, the (rectangle contained) by AD and DC plus the (square) on EC is equal to the (square) on ED. And EC (is) equal to EB. Thus, the (rectangle contained) by AD and DC plus the (square) on EB is equal to the (square) on ED. And the (square) on ED is equal to the (sum of the squares) on EB and BD. For EBD (is) a right-angle [Prop. 1.47]. Thus, the (rectangle contained) by AD and DC plus the (square) on EB is equal to the (sum of the squares) on EB and BD. Let the (square) on EB have been subtracted from both. Thus, the remaining (rectangle contained) by AD and DC is equal to the (square) on BD. Thus, if some point is taken outside a circle, and two straight-lines radiate from it towards the circle, and (one) of them cuts the circle, and (the other) touches (it), then the (rectangle contained) by the whole (straight-line) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumference, will be equal to the square on the tangent (line). (Which is) the very thing it was required to show.

λζ΄.

Proposition 37

'Ε¦ν κύκλου ληφθÍ τι σηµε‹ον ™κτός, ¢πÕ δ τοà If some point is taken outside a circle, and two σηµείου πρÕς τÕν κύκλον προσπίπτωσι δύο εÙθε‹αι, straight-lines radiate from the point towards the circle, κሠ¹ µν αÙτîν τέµνV τÕν κύκλον, ¹ δ προσπίπτV, and one of them cuts the circle, and the (other) meets Ï δ τÕ ØπÕ [τÁς] Óλης τÁς τεµνούσης κሠτÁς ™κτÕς (it), and the (rectangle contained) by the whole (straight-

106

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3

¢πολαµβανοµένης µεταξÝ τοà τε σηµείου κሠτÁς line) cutting (the circle), and the (part of it) cut off outκυρτÁς περιφερείας ‡σον τù ¢πÕ τÁς προσπιπτούσης, side (the circle), between the point and the convex cir¹ προσπίπτουσα ™φάψεται τοà κύκλου. cumference, is equal to the (square) on the (straight-line) meeting (the circle), then the (straight-line) meeting (the circle) will touch the circle.

∆

D Ε Γ

E

C

Ζ Β

F B

Α

Κύκλου γ¦ρ τοà ΑΒΓ ε„λήφθω τι σηµε‹ον ™κτÕς τÕ ∆, κሠ¢πÕ τοà ∆ πρÕς τÕν ΑΒΓ κύκλον προσπιπτέτωσαν δύο εÙθε‹αι αƒ ∆ΓΑ, ∆Β, κሠ¹ µν ∆ΓΑ τεµνέτω τÕν κύκλον, ¹ δ ∆Β προσπιπτέτω, œστω δ τÕ ØπÕ τîν Α∆, ∆Γ ‡σον τù ¢πÕ τÁς ∆Β. λέγω, Óτι ¹ ∆Β ™φάπτεται τοà ΑΒΓ κύκλου. ”Ηχθω γ¦ρ τοà ΑΒΓ ™φαπτοµένη ¹ ∆Ε, κሠε„λήφθω τÕ κέντρον τοà ΑΒΓ κύκλου, κሠœστω τÕ Ζ, κሠ™πεζεύχθωσαν αƒ ΖΕ, ΖΒ, Ζ∆. ¹ ¥ρα ØπÕ ΖΕ∆ Ñρθή ™στιν. κሠ™πεˆ ¹ ∆Ε ™φάπτεται τοà ΑΒΓ κύκλου, τέµνει δ ¹ ∆ΓΑ, τÕ ¥ρα ØπÕ τîν Α∆, ∆Γ ‡σον ™στˆ τù ¢πÕ τÁς ∆Ε. Ãν δ κሠτÕ ØπÕ τîν Α∆, ∆Γ ‡σον τù ¢πÕ τÁς ∆Β· τÕ ¥ρα ¢πÕ τÁς ∆Ε ‡σον ™στˆ τù ¢πÕ τÁς ∆Β· ‡ση ¥ρα ¹ ∆Ε τÍ ∆Β. ™στˆ δ κሠ¹ ΖΕ τÍ ΖΒ ‡ση· δύο δ¾ αƒ ∆Ε, ΕΖ δύο τα‹ς ∆Β, ΒΖ ‡σαι ε„σίν· κሠβάσις αÙτîν κοιν¾ ¹ Ζ∆· γωνία ¥ρα ¹ ØπÕ ∆ΕΖ γωνίv τÍ ØπÕ ∆ΒΖ ™στιν ‡ση. Ñρθ¾ δ ¹ ØπÕ ∆ΕΖ· Ñρθ¾ ¥ρα κሠ¹ ØπÕ ∆ΒΖ. καί ™στιν ¹ ΖΒ ™κβαλλοµένη διάµετρος· ¹ δ τÍ διαµέτρJ τοà κύκλου πρÕς Ñρθ¦ς ¢π' ¥κρας ¢γοµένη ™φάπτεται τοà κύκλου· ¹ ∆Β ¥ρα ™φάπτεται τοà ΑΒΓ κύκλου. еοίως δ¾ δειχθήσεται, κ¨ν τÕ κέντρον ™πˆ τÁς ΑΓ τυγχάνV. 'Ε¦ν ¥ρα κύκλου ληφθÍ τι σηµε‹ον ™κτός, ¢πÕ δ τοà σηµείου πρÕς τÕν κύκλον προσπίπτωσι δύο εÙθε‹αι, κሠ¹ µν αÙτîν τέµνV τÕν κύκλον, ¹ δ προσπίπτV, Ï δ τÕ ØπÕ Óλης τÁς τεµνούσης κሠτÁς ™κτÕς ¢πολαµβανοµένης µεταξÝ τοà τε σηµείου κሠτÁς κυρτÁς περιφερείας ‡σον τù ¢πÕ τÁς προσπιπτούσης, ¹ προσπίπτουσα ™φάψεται τοà κύκλου· Óπερ œδει δε‹ξαι.

A

For let some point D have been taken outside circle ABC, and let two straight-lines, DCA and DB, radiate from D towards circle ABC, and let DCA cut the circle, and let DB meet (the circle). And let the (rectangle contained) by AD and DC be equal to the (square) on DB. I say that DB touches circle ABC. For let DE have been drawn touching ABC [Prop. 3.17], and let the center of the circle ABC have been found, and let it be (at) F . And let F E, F B, and F D have been joined. (Angle) F ED is thus a right-angle [Prop. 3.18]. And since DE touches circle ABC, and DCA cuts (it), the (rectangle contained) by AD and DC is thus equal to the (square) on DE [Prop. 3.36]. And the (rectangle contained) by AD and DC was also equal to the (square) on DB. Thus, the (square) on DE is equal to the (square) on DB. Thus, DE (is) equal to DB. And F E is also equal to F B. So the two (straight-lines) DE, EF are equal to the two (straight-lines) DB, BF (respectively). And their base, F D, is common. Thus, angle DEF is equal to angle DBF [Prop. 1.8]. And DEF (is) a right-angle. Thus, DBF (is) also a right-angle. And F B produced is a diameter, And a (straight-line) drawn at right-angles to a diameter of a circle, at its end, touches the circle [Prop. 3.16 corr.]. Thus, DB touches circle ABC. Similarly, (the same thing) can be shown, even if the center is somewhere on AC. Thus, if some point is taken outside a circle, and two straight-lines radiate from the point towards the circle, and one of them cuts the circle, and the (other) meets (it), and the (rectangle contained) by the whole (straightline) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumference, is equal to the (square) on the (straight-line)

107

ΣΤΟΙΧΕΙΩΝ γ΄.

ELEMENTS BOOK 3 meeting (the circle), then the (straight-line) meeting (the circle) will touch the circle. (Which is) the very thing it was required to show.

108

ELEMENTS BOOK 4 Construction of rectilinear figures in and around circles

109

ΣΤΟΙΧΕΙΩΝ δ΄.

ELEMENTS BOOK 4

“Οροι.

Definitions

α΄. ΣχÁµα εÙθύγραµµον ε„ς σχÁµα εÙθύγραµµον ™γγράφεσθαι λέγεται, Óταν ˜κάστη τîν τοà ™γγραφοµένου σχήµατος γωνιîν ˜κάστης πλευρ©ς τοà, ε„ς Ö ™γγράφεται, ¤πτηται. β΄. ΣχÁµα δ Ðµοίως περˆ σχÁµα περιγράφεσθαι λέγεται, Óταν ˜κάστη πλευρ¦ τοà περιγραφοµένου ˜κάστης γωνίας τοà, περˆ Ö περιγράφεται, ¤πτηται. γ΄. ΣχÁµα εÙθύγραµµον ε„ς κύκλον ™γγράφεσθαι λέγεται, Óταν ˜κάστη γωνία τοà ™γγραφοµένου ¤πτηται τÁς τοà κύκλου περιφερείας. δ΄. ΣχÁµα δ εÙθύγραµµον περˆ κύκλον περιγράφεθαι λέγεται, Óταν ˜κάστη πλευρ¦ τοà περιγραφοµένου ™φάπτηται τÁς τοà κύκλου περιφερείας. ε΄. Κύκλος δ ε„ς σχÁµα еοίως ™γγράφεσθαι λέγεται, Óταν ¹ τοà κύκλου περιφέρεια ˜κάστης πλευρ©ς τοà, ε„ς Ö ™γγράφεται, ¤πτηται. $΄. Κύκλος δ περˆ σχÁµα περιγράφεσθαι λέγεται, Óταν ¹ τοà κύκλου περιφέρεια ˜κάστης γωνίας τοà, περˆ Ö περιγράφεται, ¤πτηται. ζ΄. ΕÙθε‹α ε„ς κύκλον ™ναρµόζεσθαι λέγεται, Óταν τ¦ πέρατα αÙτÁς ™πˆ τÁς περιφερείας Ï τοà κύκλου.

1. A rectilinear figure is said to be inscribed in a(nother) rectilinear figure when each of the angles of the inscribed figure touches each (respective) side of the (figure) in which it is inscribed. 2. And, similarly, a (rectilinear) figure is said to be circumscribed about a(nother rectilinear) figure when each side of the circumscribed (figure) touches each (respective) angle of the (figure) about which it is circumscribed. 3. A rectilinear figure is said to be inscribed in a circle when each angle of the inscribed (figure) touches the circumference of the circle. 4. And a rectilinear figure is said to be circumscribed about a circle when each side of the circumscribed (figure) touches the circumference of the circle. 5. And, similarly, a circle is said to be inscribed in a (rectilinear) figure when the circumference of the circle touches each side of the (figure) in which it is inscribed. 6. And a circle is said to be circumscribed about a rectilinear (figure) when the circumference of the circle touches each angle of the (figure) about which it is circumscribed. 7. A straight-line is said to be inserted into a circle when its ends are on the circumference of the circle.

α΄.

Proposition 1

Ε„ς τÕν δοθέντα κύκλον τÍ δοθείσV εÙθείv µ¾ µείζονι οÜσV τÁς τοà κύκλου διαµέτρου ‡σην εÙθε‹αν ™ναρµόσαι.

To insert a straight-line equal to a given straight-line into a circle, (the latter straight-line) not being greater than the diameter of the circle.

∆

D

Α

Β

Ε

A

Γ

B

Ζ

E

C

F

”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ, ¹ δ δοθε‹σα εÙθε‹α µ¾ µείζων τÁς τοà κύκλου διαµέτρου ¹ ∆. δε‹ δ¾ ε„ς τÕν ΑΒΓ κύκλον τÍ ∆ εÙθείv ‡σην εÙθε‹αν ™ναρµόσαι. ”Ηχθω τοà ΑΒΓ κύκλου διάµετρος ¹ ΒΓ. ε„ µν οâν ‡ση ™στˆν ¹ ΒΓ τÍ ∆, γεγονÕς ¨ν ε‡η τÕ ™πιταχθέν· ™νήρµοσται γ¦ρ ε„ς τÕν ΑΒΓ κύκλον τÍ ∆ εÙθείv ‡ση

Let ABC be the given circle, and D the given straightline (which is) not greater than the diameter of the circle. So it is required to insert a straight-line, equal to the straight-line D, into the circle ABC. Let a diameter BC of circle ABC have been drawn.† Therefore, if BC is equal to D, then that (which) was

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ELEMENTS BOOK 4

¹ ΒΓ. ε„ δ µείζων ™στˆν ¹ ΒΓ τÁς ∆, κείσθω τÍ ∆ ‡ση ¹ ΓΕ, κሠκέντρJ τù Γ διαστήµατι δ τù ΓΕ κύκλος γεγράφθω Ð ΕΑΖ, κሠ™πεζεύχθω ¹ ΓΑ. 'Επεˆ οâν το Γ σηµε‹ον κέντρον ™στˆ τοà ΕΑΖ κύκλου, ‡ση ™στˆν ¹ ΓΑ τÍ ΓΕ. ¢λλ¦ τÍ ∆ ¹ ΓΕ ™στιν ‡ση· κሠ¹ ∆ ¥ρα τÍ ΓΑ ™στιν ‡ση. Ε„ς ¥ρα τÕν δοθέντα κύκλον τÕν ΑΒΓ τÍ δοθείσV εÙθείv τÍ ∆ ‡ση ™νήρµοσται ¹ ΓΑ· Óπερ œδει ποιÁσαι.

†

prescribed has taken place. For the (straight-line) BC, equal to the straight-line D, has been inserted into the circle ABC. And if BC is greater than D, then let CE be made equal to D [Prop. 1.3], and let the circle EAF have been drawn with center C and radius CE. And let CA have been joined. Therefore, since the point C is the center of circle EAF , CA is equal to CE. But, CE is equal to D. Thus, D is also equal to CA. Thus, CA, equal to the given straight-line D, has been inserted into the given circle ABC. (Which is) the very thing it was required to do.

Presumably, by finding the center of the circle [Prop. 3.1], and then drawing a line through it.

β΄.

Proposition 2

Ε„ς τÕν δοθέντα κύκλον τù δοθέντι τριγώνJ „σογώνιον To inscribe a triangle, equiangular to a given triangle, τρίγωνον ™γγράψαι. in a given circle.

Β

Ε

B

E

Ζ

F

Γ

Η

C G

∆ Α

D A

Θ

H

”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ, τÕ δ δοθν τριγωνον τÕ ∆ΕΖ· δε‹ δ¾ ε„ς τÕν ΑΒΓ κύκλον τù ∆ΕΖ τριγώνJ „σογώνιον τρίγωνον ™γγράψαι. ”Ηχθω τοà ΑΒΓ κύκλου ™φαπτοµένη ¹ ΗΘ κατ¦ τÕ Α, κሠσυνεστάτω πρÕς τÍ ΑΘ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ ØπÕ ∆ΕΖ γωνίv ‡ση ¹ ØπÕ ΘΑΓ, πρÕς δ τÍ ΑΗ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ ØπÕ ∆ΖΕ [γωνίv] ‡ση ¹ ØπÕ ΗΑΒ, κሠ™πεζεύχθω ¹ ΒΓ. 'Επεˆ οâν κύκλου τοà ΑΒΓ ™φάπτεταί τις εÙθε‹α ¹ ΑΘ, κሠ¢πÕ τÁς κατ¦ τÕ Α ™παφÁς ε„ς τÕν κύκλον διÁκται εÙθε‹α ¹ ΑΓ, ¹ ¥ρα ØπÕ ΘΑΓ ‡ση ™στˆ τÍ ™ν τù ™ναλλ¦ξ τοà κύκλου τµήµατι γωνίv τÍ ØπÕ ΑΒΓ. ¢λλ' ¹ ØπÕ ΘΑΓ τÍ ØπÕ ∆ΕΖ ™στιν ‡ση· κሠ¹ ØπÕ ΑΒΓ ¥ρα γωνία τÍ ØπÕ ∆ΕΖ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ ΑΓΒ τÍ ØπÕ ∆ΖΕ ™στιν ‡ση· κሠλοιπ¾ ¥ρα ¹ ØπÕ ΒΑΓ λοιπÍ τÍ ØπÕ Ε∆Ζ ™στιν ‡ση [„σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ, κሠ™γγέγραπται ε„ς τÕν ΑΒΓ κύκλον]. Ε„ς τÕν δοθέντα ¥ρα κύκλον τù δοθέντι τριγώνJ „σογώνιον τρίγωνον ™γγέγραπται· Óπερ œδει ποιÁσαι.

Let ABC be the given circle, and DEF the given triangle. So it is required to inscribe a triangle, equiangular to triangle DEF , in circle ABC. Let GH have been drawn touching circle ABC at A.† And let (angle) HAC, equal to angle DEF , have been constructed at the point A on the straight-line AH, and (angle) GAB, equal to [angle] DF E, at the point A on the straight-line AG [Prop. 1.23]. And let BC have been joined. Therefore, since some straight-line AH touches the circle ABC, and the straight-line AC has been drawn across (the circle) from the point of contact A, (angle) HAC is thus equal to the angle ABC in the alternate segment of the circle [Prop. 3.32]. But, HAC is equal to DEF . Thus, angle ABC is also equal to DEF . So, for the same (reasons), ACB is also equal to DF E. Thus, the remaining (angle) BAC is equal to the remaining (angle) EDF [Prop. 1.32]. [Thus, triangle ABC is equiangular to triangle DEF , and has been inscribed in circle ABC]. Thus, a triangle, equiangular to the given triangle, has

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ELEMENTS BOOK 4 been inscribed in the given circle. (Which is) the very thing it was required to do.

†

See the footnote to Prop. 3.34.

γ΄.

Proposition 3

Περˆ τÕν δοθέντα κύκλον τù δοθέντι τριγώνJ To circumscribe a triangle, equiangular to a given tri„σογώνιον τρίγωνον περιγράψαι. angle, about a given circle.

Μ Α

Θ

∆

Ζ

Γ

A Ε Ν

D

F

Β Κ

Λ

M H B E

K Η

L

”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ, τÕ δ δοθν τρίγωνον τÕ ∆ΕΖ· δε‹ δ¾ περˆ τÕν ΑΒΓ κύκλον τù ∆ΕΖ τριγώνJ „σογώνιον τρίγωνον περιγράψαι. 'Εκβεβλήσθω ¹ ΕΖ ™φ' ˜κάτερα τ¦ µέρη κατ¦ τ¦ Η, Θ σηµε‹α, κሠε„λήφθω τοà ΑΒΓ κύκλου κέντρον τÕ Κ, κሠδιήχθω, æς œτυχεν, εÙθε‹α ¹ ΚΒ, κሠσυνεστάτω πρÕς τÍ ΚΒ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Κ τÍ µν ØπÕ ∆ΕΗ γωνίv ‡ση ¹ ØπÕ ΒΚΑ, τÍ δ ØπÕ ∆ΖΘ ‡ση ¹ ØπÕ ΒΚΓ, κሠδι¦ τîν Α, Β, Γ σηµείων ½χθωσαν ™φαπτόµεναι τοà ΑΒΓ κύκλου αƒ ΛΑΜ, ΜΒΝ, ΝΓΛ. Κሠ™πεˆ ™φάπτονται τοà ΑΒΓ κύκλου αƒ ΛΜ, ΜΝ, ΝΛ κατ¦ τ¦ Α, Β, Γ σηµε‹α, ¢πÕ δ τοà Κ κέντρου ™πˆ τ¦ Α, Β, Γ σηµε‹α ™πεζευγµέναι ε„σˆν αƒ ΚΑ, ΚΒ, ΚΓ, Ñρθሠ¥ρα ε„σˆν αƒ πρÕς το‹ς Α, Β, Γ σηµείοις γωνίαι. κሠ™πεˆ τοà ΑΜΒΚ τετραπλεύρου αƒ τέσσαρες γωνίαι τέτρασιν Ñρθα‹ς ‡σαι ε„σίν, ™πειδήπερ κሠε„ς δύο τρίγωνα διαιρε‹ται τÕ ΑΜΒΚ, καί ε„σιν Ñρθαˆ αƒ ØπÕ ΚΑΜ, ΚΒΜ γωνίαι, λοιπሠ¥ρα αƒ ØπÕ ΑΚΒ, ΑΜΒ δυσˆν Ñρθα‹ς ‡σαι ε„σίν. ε„σˆ δ καˆ αƒ ØπÕ ∆ΕΗ, ∆ΕΖ δυσˆν Ñρθα‹ς ‡σαι· αƒ ¥ρα ØπÕ ΑΚΒ, ΑΜΒ τα‹ς ØπÕ ∆ΕΗ, ∆ΕΖ ‡σαι ε„σίν, ïν ¹ ØπÕ ΑΚΒ τÍ ØπÕ ∆ΕΗ ™στιν ‡ση· λοιπ¾ ¥ρα ¹ ØπÕ ΑΜΒ λοιπÍ τÍ ØπÕ ∆ΕΖ ™στιν ‡ση. еοίως δ¾ δειχθήσεται, Óτι κሠ¹ ØπÕ ΛΝΒ τÍ ØπÕ ∆ΖΕ ™στιν ‡ση· κሠλοιπ¾ ¥ρα ¹ ØπÕ ΜΛΝ [λοιπÍ] τÍ ØπÕ Ε∆Ζ ™στιν ‡ση. „σογώνιον ¥ρα ™στˆ τÕ ΛΜΝ τρίγωνον τù ∆ΕΖ τριγώνJ· κሠπεριγέγραπται περˆ τÕν ΑΒΓ κύκλον. Περˆ τÕν δοθέντα ¥ρα κύκλον τù δοθέντι τριγώνJ „σογώνιον τρίγωνον περιγέγραπται· Óπερ œδει ποιÁσαι.

C

N

G

Let ABC be the given circle, and DEF the given triangle. So it is required to circumscribe a triangle, equiangular to triangle DEF , about circle ABC. Let EF have been produced in each direction to points G and H. And let the center K of circle ABC have been found [Prop. 3.1]. And let the straight-line KB have been drawn across (ABC), at random. And let (angle) BKA, equal to angle DEG, have been constructed at the point K on the straight-line KB, and (angle) BKC, equal to DF H [Prop. 1.23]. And let the (straightlines) LAM , M BN , and N CL have been drawn through the points A, B, and C (respectively), touching the circle ABC.† And since LM , M N , and N L touch circle ABC at points A, B, and C (respectively), and KA, KB, and KC are joined from the center K to points A, B, and C (respectively), the angles at points A, B, and C are thus right-angles [Prop. 3.18]. And since the (sum of the) four angles of quadrilateral AM BK is equal to four rightangles, inasmuch as AM BK (can) also (be) divided into two triangles [Prop. 1.32], and angles KAM and KBM are (both) right-angles, the (sum of the) remaining (angles), AKB and AM B, is thus equal to two right-angles. And DEG and DEF is also equal to two right-angles [Prop. 1.13]. Thus, AKB and AM B is equal to DEG and DEF , of which AKB is equal to DEG. Thus, the remainder AM B is equal to the remainder DEF . So, similarly, it can be shown that LN B is also equal to DF E. Thus, the remaining (angle) M LN is also equal to the [remaining] (angle) EDF [Prop. 1.32]. Thus, triangle LM N is equiangular to triangle DEF . And it has been

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ELEMENTS BOOK 4 drawn around circle ABC. Thus, a triangle, equiangular to the given triangle, has been circumscribed about the given circle. (Which is) the very thing it was required to do.

†

See the footnote to Prop. 3.34.

δ΄.

Proposition 4

Ε„ς τÕ δοθν τρίγωνον κύκλον ™γγράψαι.

To inscribe a circle in a given triangle.

Α Ε

A E

Η ∆

Β

Ζ

G D

Γ

B

”Εστω τÕ δοθν τρίγωνον τÕ ΑΒΓ· δε‹ δ¾ ε„ς τÕ ΑΒΓ τρίγωνον κύκλον ™γγράψαι. Τετµήσθωσαν αƒ ØπÕ ΑΒΓ, ΑΓΒ γωνίαι δίχα τα‹ς Β∆, Γ∆ εÙθείαις, κሠσυµβαλλέτωσαν ¢λλήλαις κατ¦ τÕ ∆ σηµε‹ον, κሠ½χθωσαν ¢πÕ τοà ∆ ™πˆ τ¦ς ΑΒ, ΒΓ, ΓΑ εÙθείας κάθετοι αƒ ∆Ε, ∆Ζ, ∆Η. Κሠ™πεˆ ‡ση ™στˆν ¹ ØπÕ ΑΒ∆ γωνία τÍ ØπÕ ΓΒΑ, ™στˆ δ κሠÑρθ¾ ¹ ØπÕ ΒΕ∆ ÑρθÍ τÍ ØπÕ ΒΖ∆ ‡ση, δύο δ¾ τρίγωνά ™στι τ¦ ΕΒ∆, ΖΒ∆ τ¦ς δύο γωνίας τα‹ς δυσˆ γωνίαις ‡σας œχοντα κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην τ¾ν Øποτείνουσαν ØπÕ µίαν τîν ‡σων γωνιîν κοιν¾ν αÙτîν τ¾ν Β∆· κሠτ¦ς λοιπ¦ς ¥ρα πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξουσιν· ‡ση ¥ρα ¹ ∆Ε τÍ ∆Ζ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ∆Η τÍ ∆Ζ ™στιν ‡ση. αƒ τρε‹ς ¥ρα εÙθε‹αι αƒ ∆Ε, ∆Ζ, ∆Η ‡σαι ¢λλήλαις ε„σίν· Ð ¥ρα κέντρù τù ∆ κሠδιαστήµατι ˜νˆ τîν Ε, Ζ, Η κύκλος γραφόµενος ¼ξει κሠδι¦ τîν λοιπîν σηµείων κሠ™φάψεται τîν ΑΒ, ΒΓ, ΓΑ εÙθειîν δι¦ τÕ Ñρθ¦ς εναι τ¦ς πρÕς το‹ς Ε, Ζ, Η σηµείοις γωνίας. ε„ γ¦ρ τεµε‹ αÙτάς, œσται ¹ τÍ διαµέτρJ τοà κύκλου πρÕς Ñρθ¦ς ¢π' ¥κρας ¢γοµένη ™ντÕς πίπτουσα τοà κύκλου· Óπερ ¥τοπον ™δείχθη· οÙκ ¥ρα Ð κέντρJ τù ∆ διαστήµατι δ ˜νˆ τîν Ε, Ζ, Η γραφόµενος κύκλος τεµε‹ τ¦ς ΑΒ, ΒΓ, ΓΑ εÙθείας· ™φάψεται ¥ρα αÙτîν, κሠœσται Ð κύκλος ™γγεγραµµένος ε„ς τÕ ΑΒΓ τρίγωνον. ™γγεγράφθω æς Ð ΖΗΕ. Ε„ς ¥ρα τÕ δοθν τρίγωνον τÕ ΑΒΓ κύκλος ™γγέγραπται Ð ΕΖΗ· Óπερ œδει ποιÁσαι.

F

C

Let ABC be the given triangle. So it is required to inscribe a circle in triangle ABC. Let the angles ABC and ACB have been cut in half by the straight-lines BD and CD (respectively) [Prop. 1.9], and let them meet one another at point D, and let DE, DF , and DG have been drawn from point D, perpendicular to the straight-lines AB, BC, and CA (respectively) [Prop. 1.12]. And since angle ABD is equal to CBD, and the rightangle BED is also equal to the right-angle BF D, EBD and F BD are thus two triangles having two angles equal to two angles, and one side equal to one side—the (one) subtending one of the equal angles (which is) common to the (triangles)—(namely), BD. Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26]. Thus, DE (is) equal to DF . So, for the same (reasons), DG is also equal to DF . Thus, the three straight-lines DE, DF , and DG are equal to one another. Thus, the circle drawn with center D, and radius one of E, F , or G,† will also go through the remaining points, and will touch the straight-lines AB, BC, and CA, on account of the angles at E, F , and G being right-angles. For if it cuts (one of) them then it will be a (straight-line) drawn at right-angles to a diameter of the circle, from its end, falling inside the circle. The very thing was shown (to be) absurd [Prop. 3.16]. Thus, the circle drawn with center D, and radius one of E, F , or G, does not cut the straight-lines AB, BC, and CA. Thus, it will touch them. And the circle will have been inscribed

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ΣΤΟΙΧΕΙΩΝ δ΄.

ELEMENTS BOOK 4 in triangle ABC. Let it have been (so) inscribed, like F GE (in the figure). Thus, the circle EF G has been inscribed in the given triangle ABC. (Which is) the very thing it was required to do.

†

Here, and in the following propositions, it is understood that the radius is actually one of DE, DF , or DG.

ε΄.

Proposition 5

Περˆ τÕ δοθν τρίγωνον κύκλον περιγράψαι. Α

Α ∆

∆

Ε

Β

Ζ Β

Β

Ε Ζ

Γ

To circumscribe a circle about a given triangle. A

Α

∆

D B

D

Ε Ζ

A

A

D Γ

B

F B

Γ

”Εστω τÕ δοθν τρίγωνον τÕ ΑΒΓ· δε‹ δ περˆ τÕ δοθν τρίγωνον τÕ ΑΒΓ κύκλον περιγράψαι. Τετµήσθωσαν αƒ ΑΒ, ΑΓ εÙθε‹αι δίχα κατ¦ τ¦ ∆, Ε σηµε‹α, κሠ¢πÕ τîν ∆, Ε σηµείων τα‹ς ΑΒ, ΑΓ πρÕς Ñρθ¦ς ½χθωσαν αƒ ∆Ζ, ΕΖ· συµπεσοàνται δ¾ ½τοι ™ντÕς τοà ΑΒΓ τριγώνου À ™πˆ τÁς ΒΓ εÙθείας À ™κτÕς τÁς ΒΓ. Συµπιπτέτωσαν πρότερον ™ντÕς κατ¦ τÕ Ζ, κሠ™πεζεύχθωσαν αƒ ΖΒ, ΖΓ, ΖΑ. κሠ™πεˆ ‡ση ™στˆν ¹ Α∆ τÍ ∆Β, κοιν¾ δ κሠπρÕς Ñρθ¦ς ¹ ∆Ζ, βάσις ¥ρα ¹ ΑΖ βάσει τÍ ΖΒ ™στιν ‡ση. еοίως δ¾ δείξοµεν, Óτι κሠ¹ ΓΖ τÍ ΑΖ ™στιν ‡ση· éστε κሠ¹ ΖΒ τÍ ΖΓ ™στιν ‡ση· αƒ τρε‹ς ¥ρα αƒ ΖΑ, ΖΒ, ΖΓ ‡σαι ¢λλήλαις ε„σίν. Ð ¥ρα κέντρJ τù Ζ διαστήµατι δ ˜νˆ τîν Α, Β, Γ κύκλος γραφόµενος ¼ξει κሠδι¦ τîν λοιπîν σηµείων, κሠœσται περιγεγραµµένος Ð κύκλος περˆ τÕ ΑΒΓ τρίγωνον. περιγεγράφθω æς Ð ΑΒΓ. 'Αλλ¦ δ¾ αƒ ∆Ζ, ΕΖ συµπιπτέτωσαν ™πˆ τÁς ΒΓ εÙθείας κατ¦ τÕ Ζ, æς œχει ™πˆ τÁς δευτέρας καταγραφÁς, κሠ™πεζεύχθω ¹ ΑΖ. еοίως δ¾ δείξοµεν, Óτι τÕ Ζ σηµε‹ον κέντρον ™στˆ τοà περˆ τÕ ΑΒΓ τρίγωνον περιγραφοµένου κύκλου. 'Αλλ¦ δ¾ αƒ ∆Ζ, ΕΖ συµπιπτέτωσαν ™κτÕς τοà ΑΒΓ τριγώνου κατ¦ τÕ Ζ πάλιν, æς œχει ™πˆ τÁς τρίτης καταγραφÁς, καί ™πεζεύχθωσαν αƒ ΑΖ, ΒΖ, ΓΖ. κሠ™πεˆ πάλιν ‡ση ™στˆν ¹ Α∆ τÍ ∆Β, κοιν¾ δ κሠπρÕς Ñρθ¦ς ¹ ∆Ζ, βάσις ¥ρα ¹ ΑΖ βάσει τÍ ΒΖ ™στιν ‡ση. еοίως δ¾ δείξοµεν, Óτι κሠ¹ ΓΖ τÍ ΑΖ ™στιν ‡ση· éστε κሠ¹ ΒΖ τÍ ΖΓ ™στιν ‡ση· Ð ¥ρα [πάλιν] κέντρJ τù Ζ διαστήµατι δ ˜νˆ τîν ΖΑ, ΖΒ, ΖΓ κύκλος γραφόµενος ¼ξει κሠδι¦ τîν λοιπîν σηµείων, κሠœσται περιγεγραµµένος περˆ τÕ ΑΒΓ τρίγωνον. Περˆ τÕ δοθν ¥ρα τρίγωνον κύκλος περιγέγραπται· Óπερ œδει ποιÁσαι.

E

E

E F

C

F

C

C

Let ABC be the given triangle. So it is required to circumscribe a circle about the given triangle ABC. Let the straight-lines AB and AC have been cut in half at points D and E (respectively) [Prop. 1.10]. And let DF and EF have been drawn from points D and E, at right-angles to AB and AC (respectively) [Prop. 1.11]. So (DF and EF ) will surely either meet inside triangle ABC, on the straight-line BC, or beyond BC. Let them, first of all, meet inside (triangle ABC) at (point) F , and let F B, F C, and F A have been joined. And since AD is equal to DB, and DF is common and at right-angles, the base AF is thus equal to the base F B [Prop. 1.4]. So, similarly, we can show that CF is also equal to AF . So that F B is also equal to F C. Thus, the three (straight-lines) F A, F B, and F C are equal to one another. Thus, the circle drawn with center F , and radius one of A, B, or C, will also go through the remaining points. And the circle will have been circumscribed about triangle ABC. Let it have been (so) circumscribed, like ABC (in the first diagram from the left). And so, let DF and EF meet on the straight-line BC at (point) F , like in the second diagram (from the left). And let AF have been joined. So, similarly, we can show that point F is the center of the circle circumscribed about triangle ABC. And so, let DF and EF meet outside triangle ABC, again at (point) F , like in the third diagram (from the left). And let AF , BF , and CF have been joined. And, again, since AD is equal to DB, and DF is common and at right-angles, the base AF is thus equal to the base BF [Prop. 1.4]. So, similarly, we can show that CF is also equal to AF . So that BF is also equal to F C. Thus, [again] the circle drawn with center F , and radius one of F A, F B, and F C, will also go through the remaining

114

ΣΤΟΙΧΕΙΩΝ δ΄.

ELEMENTS BOOK 4 points. And it will have been circumscribed about triangle ABC. Thus, a circle has been circumscribed about the given triangle. (Which is) the very thing it was required to do.

$΄.

Proposition 6

Ε„ς τÕν δοθέντα κύκλον τετράγωνον ™γγράψαι.

To inscribe a square in a given circle.

Α

Β

Ε

A

∆

B

E

D

Γ

C

”Εστω ¹ δοθεˆς κύκλος Ð ΑΒΓ∆· δε‹ δ¾ ε„ς τÕν ΑΒΓ∆ κύκλον τετράγωνον ™γγράψαι. ”Ηχθωσαν τοà ΑΒΓ∆ κύκλου δύο διάµετροι πρÕς Ñρθ¦ς ¢λλήλαις αƒ ΑΓ, Β∆, κሠ™πεζεύχθωσαν αƒ ΑΒ, ΒΓ, Γ∆, ∆Α. Κሠ™πεˆ ‡ση ™στˆν ¹ ΒΕ τÍ Ε∆· κέντρον γ¦ρ τÕ Ε· κοιν¾ δ κሠπρÕς Ñρθ¦ς ¹ ΕΑ, βάσις ¥ρα ¹ ΑΒ βάσει τÍ Α∆ ‡ση ™στίν. δι¦ τ¦ αÙτ¦ δ¾ κሠ˜κατέρα τîν ΒΓ, Γ∆ ˜κατέρv τîν ΑΒ, Α∆ ‡ση ™στίν· „σόπλευρον ¥ρα ™στˆ τÕ ΑΒΓ∆ τετράπλευρον. λέγω δή, Óτι κሠÑρθογώνιον. ™πεˆ γ¦ρ ¹ Β∆ εÙθε‹α διάµετρός ™στι τοà ΑΒΓ∆ κύκλου, ¹µικύκλιον ¥ρα ™στˆ τÕ ΒΑ∆· Ñρθ¾ ¥ρα ¹ ØπÕ ΒΑ∆ γωνία. δι¦ τ¦ αÙτ¦ δ¾ κሠ˜κάστη τîν ØπÕ ΑΒΓ, ΒΓ∆, Γ∆Α Ñρθή ™στιν· Ñρθογώνιον ¥ρα ™στˆ τÕ ΑΒΓ∆ τετράπλευρον. ™δείχθη δ κሠ„σόπλευρον· τετράγωνον ¥ρα ™στίν. κሠ™γγέγραπται ε„ς τÕν ΑΒΓ∆ κύκλον. Ε„ς ¥ρα τÕν δοθέντα κύκλον τετράγωνον ™γγέγραπται τÕ ΑΒΓ∆· Óπερ œδει ποιÁσαι.

Let ABCD be the given circle. So it is required to inscribe a square in circle ABCD. Let two diameters of circle ABCD, AC and BD, have been drawn at right-angles to one another.† And let AB, BC, CD, and DA have been joined. And since BE is equal to ED, for E (is) the center (of the circle), and EA is common and at right-angles, the base AB is thus equal to the base AD [Prop. 1.4]. So, for the same (reasons), each of BC and CD is equal to each of AB and AD. Thus, the quadrilateral ABCD is equilateral. So I say that (it is) also right-angled. For since the straight-line BD is a diameter of circle ABCD, BAD is thus a semi-circle. Thus, angle BAD (is) a right-angle [Prop. 3.31]. So, for the same (reasons), (angles) ABC, BCD, and CDA are each right-angles. Thus, the quadrilateral ABCD is right-angled. And it was also shown (to be) equilateral. Thus, it is a square [Def. 1.22]. And it has been inscribed in circle ABCD. Thus, the square ABCD has been inscribed in the given circle. (Which is) the very thing it was required to do.

†

Presumably, by finding the center of the circle [Prop. 3.1], drawing a line through it, and then drawing a second line through it, at right-angles

to the first [Prop. 1.11].

ζ΄.

Proposition 7

Περˆ τÕν δοθέντα κύκλον τετράγωνον περιγράψαι. ”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ∆· δε‹ δ¾ περˆ τÕν ΑΒΓ∆ κύκλον τετράγωνον περιγράψαι.

To circumscribe a square about a given circle. Let ABCD be the given circle. So it is required to circumscribe a square about circle ABCD.

115

ΣΤΟΙΧΕΙΩΝ δ΄.

Η

Β

Θ

ELEMENTS BOOK 4

Α

Ε

Γ

Ζ

G

∆

B

Κ

H

”Ηχθωσαν τοà ΑΒΓ∆ κύκλου δύο διάµετροι πρÕς Ñρθ¦ς ¢λλήλαις αƒ ΑΓ, Β∆, κሠδι¦ τîν Α, Β, Γ, ∆ σηµείων ½χθωσαν ™φαπτόµεναι τοà ΑΒΓ∆ κύκλου αƒ ΖΗ, ΗΘ, ΘΚ, ΚΖ. 'Επεˆ οâν ™φάπτεται ¹ ΖΗ τοà ΑΒΓ∆ κύκλου, ¢πÕ δ τοà Ε κέντρου ™πˆ τ¾ν κατ¦ τÕ Α ™παφ¾ν ™πέζευκται ¹ ΕΑ, αƒ ¥ρα πρÕς τù Α γωνίαι Ñρθαί ε„σιν. δι¦ τ¦ αÙτ¦ δ¾ καˆ αƒ πρÕς το‹ς Β, Γ, ∆ σηµείοις γωνίαι Ñρθαί ε„σιν. κሠ™πεˆ Ñρθή ™στιν ¹ ØπÕ ΑΕΒ γωνία, ™στˆ δ Ñρθ¾ κሠ¹ ØπÕ ΕΒΗ, παράλληλος ¥ρα ™στˆν ¹ ΗΘ τÍ ΑΓ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΑΓ τÍ ΖΚ ™στι παράλληλος. éστε κሠ¹ ΗΘ τÍ ΖΚ ™στι παράλληλος. еοίως δ¾ δείξοµεν, Óτι κሠ˜κατέρα τîν ΗΖ, ΘΚ τÍ ΒΕ∆ ™στι παράλληλος. παραλληλόγραµµα ¥ρα ™στˆ τ¦ ΗΚ, ΗΓ, ΑΚ, ΖΒ, ΒΚ· ‡ση ¥ρα ™στˆν ¹ µν ΗΖ τÍ ΘΚ, ¹ δ ΗΘ τÍ ΖΚ. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΓ τÍ Β∆, ¢λλ¦ κሠ¹ µν ΑΓ ˜κατέρv τîν ΗΘ, ΖΚ, ¹ δ Β∆ ˜κατέρv τîν ΗΖ, ΘΚ ™στιν ‡ση [κሠ˜κατέρα ¥ρα τîν ΗΘ, ΖΚ ˜κατέρv τîν ΗΖ, ΘΚ ™στιν ‡ση], „σόπλευρον ¥ρα ™στˆ τÕ ΖΗΘΚ τετράπλευρον. λέγω δή, Óτι κሠÑρθογώνιον. ™πεˆ γ¦ρ παραλληλόγραµµόν ™στι τÕ ΗΒΕΑ, καί ™στιν Ñρθ¾ ¹ ØπÕ ΑΕΒ, Ñρθ¾ ¥ρα κሠ¹ ØπÕ ΑΗΒ. еοίως δ¾ δείξοµεν, Óτι καˆ αƒ πρÕς το‹ς Θ, Κ, Ζ γωνίαι Ñρθαί ε„σιν. Ñρθογώνιον ¥ρα ™στˆ τÕ ΖΗΘΚ. ™δείχθη δ κሠ„σόπλευρον· τετράγωνον ¥ρα ™στίν. κሠπεριγέγραπται περˆ τÕν ΑΒΓ∆ κύκλον. Περˆ τÕν δοθέντα ¥ρα κύκλον τετράγωνον περιγέγραπται· Óπερ œδει ποιÁσαι.

† ‡

A

E

C

F

D

K

Let two diameters of circle ABCD, AC and BD, have been drawn at right-angles to one another.† And let F G, GH, HK, and KF have been drawn through points A, B, C, and D (respectively), touching circle ABCD.‡ Therefore, since F G touches circle ABCD, and EA has been joined from the center E to the point of contact A, the angle at A is thus a right-angle [Prop. 3.18]. So, for the same (reasons), the angles at points B, C, and D are also right-angles. And since angle AEB is a rightangle, and EBG is also a right-angle, GH is thus parallel to AC [Prop. 1.29]. So, for the same (reasons), AC is also parallel to F K. So that GH is also parallel to F K [Prop. 1.30]. So, similarly, we can show that GF and HK are each parallel to BED. Thus, GK, GC, AK, F B, and BK are (all) parallelograms. Thus, GF is equal to HK, and GH to F K [Prop. 1.34]. And since AC is equal to BD, but AC (is) also (equal) to each of GH and F K, and BD is equal to each of GF and HK [Prop. 1.34] [and each of GH and F K is thus equal to each of GF and HK], the quadrilateral F GHK is thus equilateral. So I say that (it is) also right-angled. For since GBEA is a parallelogram, and AEB is a right-angle, AGB is thus also a right-angle [Prop. 1.34]. So, similarly, we can show that the angles at H, K, and F are also right-angles. Thus, F GHK is right-angled. And it was also shown (to be) equilateral. Thus, it is a square [Def. 1.22]. And it has been circumscribed about circle ABCD. Thus, a square has been circumscribed about the given circle. (Which is) the very thing it was required to do.

See the footnote to the previous proposition. See the footnote to Prop. 3.34.

η΄.

Proposition 8

Ε„ς τÕ δοθν τετράγωνον κύκλον ™γγράψαι. ”Εστω τÕ δοθν τετράγωνον τÕ ΑΒΓ∆. δε‹ δ¾ ε„ς τÕ 116

To inscribe a circle in a given square. Let the given square be ABCD. So it is required to

ΣΤΟΙΧΕΙΩΝ δ΄.

ELEMENTS BOOK 4

ΑΒΓ∆ τετράγωνον κύκλον ™γγράψαι.

Α

Ζ

Β

Ε

Η

Θ

inscribe a circle in square ABCD.

∆

A

Κ

F

Γ

B

E

G

H

D

K

C

Τετµήσθω ˜κατέρα τîν Α∆, ΑΒ δίχα κατ¦ τ¦ Ε, Ζ σηµε‹α, κሠδι¦ µν τοà Ε Ðποτέρv τîν ΑΒ, Γ∆ παράλληλος ½χθω Ð ΕΘ, δι¦ δ τοà Ζ Ðποτέρv τîν Α∆, ΒΓ παράλληλος ½χθω ¹ ΖΚ· παραλληλόγραµµον ¥ρα ™στˆν ›καστον τîν ΑΚ, ΚΒ, ΑΘ, Θ∆, ΑΗ, ΗΓ, ΒΗ, Η∆, καˆ αƒ ¢πεναντίον αÙτîν πλευρሠδηλονότι ‡σαι [ε„σίν]. κሠ™πεˆ ‡ση ™στˆν ¹ Α∆ τÍ ΑΒ, καί ™στι τÁς µν Α∆ ¹µίσεια ¹ ΑΕ, τÁς δ ΑΒ ¹µίσεια ¹ ΑΖ, ‡ση ¥ρα κሠ¹ ΑΕ τÍ ΑΖ· éστε καˆ αƒ ¢πεναντίον· ‡ση ¥ρα κሠ¹ ΖΗ τÍ ΗΕ. еοίως δ¾ δείξοµεν, Óτι κሠ˜κατέρα τîν ΗΘ, ΗΚ ˜κατέρv τîν ΖΗ, ΗΕ ™στιν ‡ση· αƒ τέσσαρες ¥ρα αƒ ΗΕ, ΗΖ, ΗΘ, ΗΚ ‡σαι ¢λλήλαις [ε„σίν]. Ð ¥ρα κέντρJ µν τù Η διαστήµατι δ ˜νˆ τîν Ε, Ζ, Θ, Κ κύκλος γραφόµενος ¼ξει κሠδι¦ τîν λοιπîν σηµείων· κሠ™φάψεται τîν ΑΒ, ΒΓ, Γ∆, ∆Α εÙθειîν δι¦ τÕ Ñρθ¦ς εναι τ¦ς πρÕς το‹ς Ε, Ζ, Θ, Κ γωνίας· ε„ γ¦ρ τεµε‹ Ð κύκλος τ¦ς ΑΒ, ΒΓ, Γ∆, ∆Α, ¹ τÍ διαµέτρJ τοà κύκλου πρÕς Ñρθ¦ς ¢π' ¥κρας ¢γοµένη ™ντÕς πεσε‹ται τοà κύκλου· Óπερ ¥τοπον ™δείχθη. οÙκ ¥ρα Ð κέντρJ τù Η διαστήµατι δ ˜νˆ τîν Ε, Ζ, Θ, Κ κύκλος γραφόµενος τεµε‹ τ¦ς ΑΒ, ΒΓ, Γ∆, ∆Α εÙθείας. ™φάψεται ¥ρα αÙτîν κሠœσται ™γγεγραµµένος ε„ς τÕ ΑΒΓ∆ τετράγωνον. Ε„ς ¥ρα τÕ δοθν τετράγωνον κύκλος ™γγέγραπται· Óπερ œδει ποιÁσαι.

Let AD and AB each have been cut in half at points E and F (respectively) [Prop. 1.10]. And let EH have been drawn through E, parallel to either of AB or CD, and let F K have been drawn through F , parallel to either of AD or BC [Prop. 1.31]. Thus, AK, KB, AH, HD, AG, GC, BG, and GD are each parallelograms, and their opposite sides [are] manifestly equal [Prop. 1.34]. And since AD is equal to AB, and AE is half of AD, and AF half of AB, AE (is) thus also equal to AF . So that the opposite (sides are) also (equal). Thus, F G (is) also equal to GE. So, similarly, we can also show that each of GH and GK is equal to each of F G and GE. Thus, the four (straight-lines) GE, GF , GH, and GK [are] equal to one another. Thus, the circle drawn with center G, and radius one of E, F , H, or K, will also go through the remaining points. And it will touch the straight-lines AB, BC, CD, and DA, on account of the angles at E, F , H, and K being right-angles. For if the circle cuts AB, BC, CD, or DA, then a (straight-line) drawn at right-angles to a diameter of the circle, from its end, will fall inside the circle. The very thing was shown (to be) absurd [Prop. 3.16]. Thus, the circle drawn with center G, and radius one of E, F , H, or K, does not cut the straight-lines AB, BC, CD, or DA. Thus, it will touch them, and will have been inscribed in the square ABCD. Thus, a circle has been inscribed in the given square. (Which is) the very thing it was required to do.

θ΄.

Proposition 9

Περˆ τÕ δοθν τετράγωνον κύκλον περιγράψαι. To circumscribe a circle about a given square. ”Εστω τÕ δοθν τετράγωνον τÕ ΑΒΓ∆· δε‹ δ¾ περˆ Let ABCD be the given square. So it is required to τÕ ΑΒΓ∆ τετράγωνον κύκλον περιγράψαι. circumscribe a circle about square ABCD. 'Επιζευχθε‹σαι γ¦ρ αƒ ΑΓ, Β∆ τεµνέτωσαν ¢λλήλας AC and BD being joined, let them cut one another at κατ¦ τÕ Ε. E.

117

ΣΤΟΙΧΕΙΩΝ δ΄.

ELEMENTS BOOK 4

Α

Β

Ε

A

∆

B

E

D

Γ

C

Κሠ™πεˆ ‡ση ™στˆν ¹ ∆Α τÍ ΑΒ, κοιν¾ δ ¹ ΑΓ, δύο δ¾ αƒ ∆Α, ΑΓ δυσˆ τα‹ς ΒΑ, ΑΓ ‡σαι ε„σίν· κሠβάσις ¹ ∆Γ βάσει τÍ ΒΓ ‡ση· γωνία ¥ρα ¹ ØπÕ ∆ΑΓ γωνίv τÍ ØπÕ ΒΑΓ ‡ση ™στίν· ¹ ¥ρα ØπÕ ∆ΑΒ γωνία δίχα τέτµηται ØπÕ τÁς ΑΓ. еοίως δ¾ δείξοµεν, Óτι κሠ˜κάστη τîν ØπÕ ΑΒΓ, ΒΓ∆, Γ∆Α δίχα τέτµηται ØπÕ τîν ΑΓ, ∆Β εÙθειîν. κሠ™πεˆ ‡ση ™στˆν ¹ ØπÕ ∆ΑΒ γωνία τÍ ØπÕ ΑΒΓ, καί ™στι τÁς µν ØπÕ ∆ΑΒ ¹µίσεια ¹ ØπÕ ΕΑΒ, τÁς δ ØπÕ ΑΒΓ ¹µίσεια ¹ ØπÕ ΕΒΑ, κሠ¹ ØπÕ ΕΑΒ ¥ρα τÍ ØπÕ ΕΒΑ ™στιν ‡ση· éστε κሠπλευρ¦ ¹ ΕΑ τÍ ΕΒ ™στιν ‡ση. еοίως δ¾ δείξοµεν, Óτι κሠ˜κατέρα τîν ΕΑ, ΕΒ [εÙθειîν] ˜κατέρv τîν ΕΓ, Ε∆ ‡ση ™στίν. αƒ τέσσαρες ¥ρα αƒ ΕΑ, ΕΒ, ΕΓ, Ε∆ ‡σαι ¢λλήλαις ε„σίν. Ð ¥ρα κέντρJ τù Ε κሠδιαστήµατι ˜νˆ τîν Α, Β, Γ, ∆ κύκλος γραφόµενος ¼ξει κሠδι¦ τîν λοιπîν σηµείων κሠœσται περιγεγραµµένος περˆ τÕ ΑΒΓ∆ τετράγωνον. περιγεγράφθω æς Ð ΑΒΓ∆. Περˆ τÕ δοθν ¥ρα τετράγωνον κύκλος περιγέγραπται· Óπερ œδει ποιÁσαι.

And since DA is equal to AB, and AC (is) common, the two (straight-lines) DA, AC are thus equal to the two (straight-lines) BA, AC. And the base DC (is) equal to the base BC. Thus, angle DAC is equal to angle BAC [Prop. 1.8]. Thus, the angle DAB has been cut in half by AC. So, similarly, we can show that ABC, BCD, and CDA have each been cut in half by the straight-lines AC and DB. And since angle DAB is equal to ABC, and EAB is half of DAB, and EBA half of ABC, EAB is thus also equal to EBA. So that side EA is also equal to EB [Prop. 1.6]. So, similarly, we can show that each of the [straight-lines] EA and EB are also equal to each of EC and ED. Thus, the four (straight-lines) EA, EB, EC, and ED are equal to one another. Thus, the circle drawn with center E, and radius one of A, B, C, or D, will also go through the remaining points, and will have been circumscribed about the square ABCD. Let it have been (so) circumscribed, like ABCD (in the figure). Thus, a circle has been circumscribed about the given square. (Which is) the very thing it was required to do.

ι΄.

Proposition 10

'Ισοσκελς τρίγωνον συστήσασθαι œχον ˜κατέραν τîν πρÕς τÍ βάσει γωνιîν διπλασίονα τÁς λοιπÁς. 'Εκκείσθω τις εÙθε‹α ¹ ΑΒ, κሠτετµήσθω κατ¦ τÕ Γ σηµε‹ον, éστε τÕ ØπÕ τîν ΑΒ, ΒΓ περιεχόµενον Ñρθογώνιον ‡σον εναι τù ¢πÕ τÁς ΓΑ τετραγώνJ· κሠκέντρJ τù Α κሠδιαστήµατι τù ΑΒ κύκλος γεγράγθω Ð Β∆Ε, κሠ™νηρµόσθω ε„ς τÕν Β∆Ε κύκλον τÍ ΑΓ εÙθείv µ¾ µείζονι οÜσV τÁς τοà Β∆Ε κύκλου διαµέτρου ‡ση εÙθε‹α ¹ Β∆· κሠ™πεζεύχθωσαν αƒ Α∆, ∆Γ, κሠπεριγεγράφθω περˆ τÕ ΑΓ∆ τρίγωνον κύκλος Ð ΑΓ∆.

To construct an isosceles triangle having each of the angles at the base double the remaining (angle). Let some straight-line AB be taken, and let it have been cut at point C so that the rectangle contained by AB and BC is equal to the square on CA [Prop. 2.11]. And let the circle BDE have been drawn with center A, and radius AB. And let the straight-line BD, equal to the straight-line AC, being not greater than the diameter of circle BDE, have been inserted into circle BDE [Prop. 4.1]. And let AD and DC have been joined. And let the circle ACD have been circumscribed about triangle ACD [Prop. 4.5].

118

ΣΤΟΙΧΕΙΩΝ δ΄.

ELEMENTS BOOK 4

Β

B

Γ

C ∆

D

Α

A Ε

E

Κሠ™πεˆ τÕ ØπÕ τîν ΑΒ, ΒΓ ‡σον ™στˆ τù ¢πÕ τÁς ΑΓ, ‡ση δ ¹ ΑΓ τÍ Β∆, τÕ ¥ρα ØπÕ τîν ΑΒ, ΒΓ ‡σον ™στˆ τù ¢πÕ τÁς Β∆. κሠ™πεˆ κύκλου τοà ΑΓ∆ ε‡ληπταί τι σηµε‹ον ™κτÕς τÕ Β, κሠ¢πÕ τοà Β πρÕς τÕν ΑΓ∆ κύκλον προσπεπτώκασι δύο εÙθε‹αι αƒ ΒΑ, Β∆, κሠ¹ µν αÙτîν τέµνει, ¹ δ προσπίπτει, καί ™στι τÕ ØπÕ τîν ΑΒ, ΒΓ ‡σον τù ¢πÕ τÁς Β∆, ¹ Β∆ ¥ρα ™φάπτεται τοà ΑΓ∆ κύκλου. ™πεˆ οâν ™φάπτεται µν ¹ Β∆, ¢πÕ δ τÁς κατ¦ τÕ ∆ ™παφÁς διÁκται ¹ ∆Γ, ¹ ¥ρα ØπÕ Β∆Γ γωνιά ‡ση ™στˆ τÍ ™ν τù ™ναλλ¦ξ τοà κύκλου τµήµατι γωνίv τÍ ØπÕ ∆ΑΓ. ™πεˆ οâν ‡ση ™στˆν ¹ ØπÕ Β∆Γ τÍ ØπÕ ∆ΑΓ, κοιν¾ προσκείσθω ¹ ØπÕ Γ∆Α· Óλη ¥ρα ¹ ØπÕ Β∆Α ‡ση ™στˆ δυσˆ τα‹ς ØπÕ Γ∆Α, ∆ΑΓ. ¢λλ¦ τα‹ς ØπÕ Γ∆Α, ∆ΑΓ ‡ση ™στˆν ¹ ™κτÕς ¹ ØπÕ ΒΓ∆· κሠ¹ ØπÕ Β∆Α ¥ρα ‡ση ™στˆ τÍ ØπÕ ΒΓ∆. ¢λλ¦ ¹ ØπÕ Β∆Α τÍ ØπÕ ΓΒ∆ ™στιν ‡ση, ™πεˆ κሠπλευρ¦ ¹ Α∆ τÍ ΑΒ ™στιν ‡ση· éστε κሠ¹ ØπÕ ∆ΒΑ τÍ ØπÕ ΒΓ∆ ™στιν ‡ση. αƒ τρε‹ς ¥ρα αƒ ØπÕ Β∆Α, ∆ΒΑ, ΒΓΑ ‡σαι ¢λλήλαις ε„σίν. κሠ™πεˆ ‡ση ™στˆν ¹ ØπÕ ∆ΒΓ γωνία τÍ ØπÕ ΒΓ∆, ‡ση ™στˆ κሠπλευρ¦ ¹ Β∆ πλευρ´ τÍ ∆Γ. ¢λλ¦ ¹ Β∆ τÍ ΓΑ Øπόκειται ‡ση· κሠ¹ ΓΑ ¥ρα τÍ Γ∆ ™στιν ‡ση· éστε κሠγωνία ¹ ØπÕ Γ∆Α γωνίv τÍ ØπÕ ∆ΑΓ ™στιν ‡ση· αƒ ¥ρα ØπÕ Γ∆Α, ∆ΑΓ τÁς ØπÕ ∆ΑΓ ε„σι διπλασίους. ‡ση δ ¹ ØπÕ ΒΓ∆ τα‹ς ØπÕ Γ∆Α, ∆ΑΓ· κሠ¹ ØπÕ ΒΓ∆ ¥ρα τÁς ØπÕ ΓΑ∆ ™στι διπλÁ. ‡ση δ ¹ ØπÕ ΒΓ∆ ˜κατέρv τîν ØπÕ Β∆Α, ∆ΒΑ· κሠ˜κατέρα ¥ρα τîν ØπÕ Β∆Α, ∆ΒΑ τÁς ØπÕ ∆ΑΒ ™στι διπλÁ. 'Ισοσκελς ¥ρα τρίγωνον συνέσταται τÕ ΑΒ∆ œχον ˜κατέραν τîν πρÕς τÍ ∆Β βάσει γωνιîν διπλασίονα τÁς λοιπÁς· Óπερ œδει ποιÁσαι.

And since the (rectangle contained) by AB and BC is equal to the (square) on AC, and AC (is) equal to BD, the (rectangle contained) by AB and BC is thus equal to the (square) on BD. And since some point B has been taken outside of circle ACD, and two straightlines BA and BD have radiated from B towards the circle ABC, and (one) of them cuts (the circle), and (the other) meets (the circle), and the (rectangle contained) by AB and BC is equal to the (square) on BD, BD thus touches circle ABC [Prop. 3.37]. Therefore, since BD touches (the circle), and DC has been drawn across (the circle) from the point of contact D, the angle BDC is thus equal to the angle DAC in the alternate segment of the circle [Prop. 3.32]. Therefore, since BDC is equal to DAC, let CDA have been added to both. Thus, the whole of BDA is equal to the two (angles) CDA and DAC. But, CDA and DAC is equal to the external (angle) BCD [Prop. 1.32]. Thus, BDA is also equal to BCD. But, BDA is equal to CBD, since the side AD is also equal to AB [Prop. 1.5]. So that DBA is also equal to BCD. Thus, the three (angles) BDA, DBA, and BCD are equal to one another. And since angle DBC is equal to BCD, side BD is also equal to side DC [Prop. 1.6]. But, BD was assumed (to be) equal to CA. Thus, CA is also equal to CD. So that angle CDA is also equal to angle DAC [Prop. 1.5]. Thus, CDA and DAC is double DAC. But BCD (is) equal to CDA and DAC. Thus, BCD is also double CAD. And BCD (is) equal to to each of BDA and DBA. Thus, BDA and DBA are each double DAB. Thus, the isosceles triangle ABD has been constructed having each of the angles at the base BD double the remaining (angle). (Which is) the very thing it was required to do.

ια΄.

Proposition 11

Ε„ς τÕν δοθέντα κύκλον πεντάγωνον „σόπλευρόν τε 119

To inscribe an equilateral and equiangular pentagon

ΣΤΟΙΧΕΙΩΝ δ΄.

ELEMENTS BOOK 4

κሠ„σογώνιον ™γγράψαι.

in a given circle.

Α

A Ζ

F

Ε

Β

Γ

∆

E

B

Η

Θ

C

D

G

H

”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ∆Ε· δε‹ δ¾ ε„ς τÕν ΑΒΓ∆Ε κύκλον πεντάγωνον „σόπλευρόν τε κሠ„σογώνιον ™γγράψαι. 'Εκκείσθω τρίγωνον „σοσκελς τÕ ΖΗΘ διπλασίονα œχον ˜κατέραν τîν πρÕς το‹ς Η, Θ γωνιîν τÁς πρÕς τù Ζ, κሠ™γγεγράφθω ε„ς τÕν ΑΒΓ∆Ε κύκλον τù ΖΗΘ τριγώνJ „σογώνιον τρίγωνον τÕ ΑΓ∆, éστε τÍ µν πρÕς τù Ζ γωνίv ‡σην εναι τ¾ν ØπÕ ΓΑ∆, ˜κατέραν δ τîν πρÕς το‹ς Η, Θ ‡σην ˜κατέρv τîν ØπÕ ΑΓ∆, Γ∆Α· κሠ˜κατέρα ¥ρα τîν ØπÕ ΑΓ∆, Γ∆Α τÁς ØπÕ ΓΑ∆ ™στι διπλÁ. τετµήσθω δ¾ ˜κατέρα τîν ØπÕ ΑΓ∆, Γ∆Α δίχα ØπÕ ˜κατέρας τîν ΓΕ, ∆Β εÙθειîν, κሠ™πεζεύχθωσαν αƒ ΑΒ, ΒΓ, [Γ∆], ∆Ε, ΕΑ. 'Επεˆ οâν ˜κατέρα τîν ØπÕ ΑΓ∆, Γ∆Α γωνιîν διπλασίων ™στˆ τÁς ØπÕ ΓΑ∆, κሠτετµηµέναι ε„σˆ δίχα ØπÕ τîν ΓΕ, ∆Β εÙθειîν, αƒ πέντε ¥ρα γωνίαι αƒ ØπÕ ∆ΑΓ, ΑΓΕ, ΕΓ∆, Γ∆Β, Β∆Α ‡σαι ¢λλήλαις ε„σίν. αƒ δ ‡σαι γωνίαι ™πˆ ‡σων περιφερειîν βεβήκασιν· αƒ πέντε ¥ρα περιφέρειαι αƒ ΑΒ, ΒΓ, Γ∆, ∆Ε, ΕΑ ‡σαι ¢λλήλαις ε„σίν. ØπÕ δ τ¦ς ‡σας περιφερείας ‡σαι εÙθε‹αι Øποτείνουσιν· αƒ πέντε ¥ρα εÙθε‹αι αƒ ΑΒ, ΒΓ, Γ∆, ∆Ε, ΕΑ ‡σαι ¢λλήλαις ε„σίν· „σόπλευρον ¥ρα ™στˆ τÕ ΑΒΓ∆Ε πεντάγωνον. λέγω δή, Óτι κሠ„σογώνιον. ™πεˆ γ¦ρ ¹ ΑΒ περιφέρεια τÍ ∆Ε περιφερείv ™στˆν ‡ση, κοιν¾ προσκείσθω ¹ ΒΓ∆· Óλη ¥ρα ¹ ΑΒΓ∆ περιφέρια ÓλV τÍ Ε∆ΓΒ περιφερείv ™στˆν ‡ση. κሠβέβηκεν ™πˆ µν τÁς ΑΒΓ∆ περιφερείας γωνία ¹ ØπÕ ΑΕ∆, ™πˆ δ τÁς Ε∆ΓΒ περιφερείας γωνία ¹ ØπÕ ΒΑΕ· κሠ¹ ØπÕ ΒΑΕ ¥ρα γωνία τÍ ØπÕ ΑΕ∆ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ˜κάστη τîν ØπÕ ΑΒΓ, ΒΓ∆, Γ∆Ε γωνιîν ˜κατέρv τîν ØπÕ ΒΑΕ, ΑΕ∆ ™στιν ‡ση· „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ∆Ε πεντάγωνον. ™δείχθη δ κሠ„σόπλευρον. Ε„ς ¥ρα τÕν δοθέντα κύκλον πεντάγωνον „σόπλευρόν τε κሠ„σογώνιον ™γγέγραπται· Óπερ œδει ποιÁσαι.

Let ABCDE be the given circle. So it is required to inscribed an equilateral and equiangular pentagon in circle ABCDE. Let the the isosceles triangle F GH be set up having each of the angles at G and H double the (angle) at F [Prop. 4.10]. And let triangle ACD, equiangular to F GH, have been inscribed in circle ABCDE, so that CAD is equal to the angle at F , and the (angles) at G and H (are) equal to ACD and CDA, respectively [Prop. 4.2]. Thus, ACD and CDA are each double CAD. So let ACD and CDA have been cut in half by the straight-lines CE and DB, respectively [Prop. 1.9]. And let AB, BC, [CD], DE and EA have been joined. Therefore, since angles ACD and CDA are each double CAD, and are cut in half by the straight-lines CE and DB, the five angles DAC, ACE, ECD, CDB, and BDA are thus equal to one another. And equal angles stand upon equal circumferences [Prop. 3.26]. Thus, the five circumferences AB, BC, CD, DE, and EA are equal to one another [Prop. 3.29]. Thus, the pentagon ABCDE is equilateral. So I say that (it is) also equiangular. For since the circumference AB is equal to the circumference DE, let BCD have been added to both. Thus, the whole circumference ABCD is equal to the whole circumference EDCB. And the angle AED stands upon circumference ABCD, and angle BAE upon circumference EDCB. Thus, angle BAE is also equal to AED [Prop. 3.27]. So, for the same (reasons), each of the angles ABC, BCD, and CDE are also equal to each of BAE and AED. Thus, pentagon ABCDE is equiangular. And it was also shown (to be) equilateral. Thus, an equilateral and equiangular pentagon has been inscribed in the given circle. (Which is) the very thing it was required to do.

ιβ΄.

Proposition 12

Περˆ τÕν δοθέντα κύκλον πεντάγωνον „σόπλευρόν τε To circumscribe an equilateral and equiangular penκሠ„σογώνιον περιγράψαι. tagon about a given circle.

120

ΣΤΟΙΧΕΙΩΝ δ΄.

ELEMENTS BOOK 4

Η Α

G Ε

A Μ

Θ

H

M

Ζ

F

Β

∆ Κ

Γ

E

B

Λ

D

K

”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ∆Ε· δε‹ δ περˆ τÕν ΑΒΓ∆Ε κύκλον πεντάγωνον „σόπλευρόν τε κሠ„σογώνιον περιγράψαι. Νενοήσθω τοà ™γγεγραµµένου πενταγώνου τîν γωνιîν σηµε‹α τ¦ Α, Β, Γ, ∆, Ε, éστε ‡σας εναι τ¦ς ΑΒ, ΒΓ, Γ∆, ∆Ε, ΕΑ περιφερείας· κሠδι¦ τîν Α, Β, Γ, ∆, Ε ½χθωσαν τοà κύκλου ™φαπτόµεναι αƒ ΗΘ, ΘΚ, ΚΛ, ΛΜ, ΜΗ, κሠε„λήφθω τοà ΑΒΓ∆Ε κύκλου κέντρον τÕ Ζ, κሠ™πεζεύχθωσαν αƒ ΖΒ, ΖΚ, ΖΓ, ΖΛ, Ζ∆. Κሠ™πεˆ ¹ µν ΚΛ εÙθε‹α ™φάπτεται τοà ΑΒΓ∆Ε κατ¦ τÕ Γ, ¢πÕ δ τοà Ζ κέντρου ™πˆ τ¾ν κατ¦ τÕ Γ ™παφ¾ν ™πέζευκται ¹ ΖΓ, ¹ ΖΓ ¥ρα κάθετός ™στιν ™πˆ τ¾ν ΚΛ· Ñρθ¾ ¥ρα ™στˆν ˜κατέρα τîν πρÕς τù Γ γωνιîν. δι¦ τ¦ αÙτ¦ δ¾ καˆ αƒ πρÕς το‹ς Β, ∆ σηµείοις γωνίαι Ñρθαί ε„σιν. κሠ™πεˆ Ñρθή ™στιν ¹ ØπÕ ΖΓΚ γωνία, τÕ ¥ρα ¢πÕ τÁς ΖΚ ‡σον ™στˆ το‹ς ¢πÕ τîν ΖΓ, ΓΚ. δι¦ τ¦ αÙτ¦ δ¾ κሠτο‹ς ¢πÕ τîν ΖΒ, ΒΚ ‡σον ™στˆ τÕ ¢πÕ τÁς ΖΚ· éστε τ¦ ¢πÕ τîν ΖΓ, ΓΚ το‹ς ¢πÕ τîν ΖΒ, ΒΚ ™στιν ‡σα, ïν τÕ ¢πÕ τÁς ΖΓ τù ¢πÕ τÁς ΖΒ ™στιν ‡σον· λοιπÕν ¥ρα τÕ ¢πÕ τÁς ΓΚ τù ¢πÕ τÁς ΒΚ ™στιν ‡σον. ‡ση ¥ρα ¹ ΒΚ τÍ ΓΚ. κሠ™πεˆ ‡ση ™στˆν ¹ ΖΒ τÍ ΖΓ, κሠκοιν¾ ¹ ΖΚ, δύο δ¾ αƒ ΒΖ, ΖΚ δυσˆ τα‹ς ΓΖ, ΖΚ ‡σαι ε„σίν· κሠβάσις ¹ ΒΚ βάσει τÍ ΓΚ [™στιν] ‡ση· γωνία ¥ρα ¹ µν ØπÕ ΒΖΚ [γωνίv] τÍ ØπÕ ΚΖΓ ™στιν ‡ση· ¹ δ ØπÕ ΒΚΖ τÍ ØπÕ ΖΚΓ· διπλÁ ¥ρα ¹ µν ØπÕ ΒΖΓ τÁς ØπÕ ΚΖΓ, ¹ δ ØπÕ ΒΚΓ τÁς ØπÕ ΖΚΓ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ µν ØπÕ ΓΖ∆ τÁς ØπÕ ΓΖΛ ™στι διπλÁ, ¹ δ ØπÕ ∆ΛΓ τÁς ØπÕ ΖΛΓ. κሠ™πεˆ ‡ση ™στˆν ¹ ΒΓ περιφέρεια τÍ Γ∆, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ΒΖΓ τÍ ØπÕ ΓΖ∆. καί ™στιν ¹ µν ØπÕ ΒΖΓ τÁς ØπÕ ΚΖΓ διπλÁ, ¹ δ ØπÕ ∆ΖΓ τÁς ØπÕ ΛΖΓ· ‡ση ¥ρα κሠ¹ ØπÕ ΚΖΓ τÍ ØπÕ ΛΖΓ· ™στˆ δ κሠ¹ ØπÕ ΖΓΚ γωνία τÍ ØπÕ ΖΓΛ ‡ση. δύο δ¾ τρίγωνά ™στι τ¦ ΖΚΓ, ΖΛΓ τ¦ς δύο γωνίας τα‹ς δυσˆ γωνίαις ‡σας œχοντα κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην κοιν¾ν αÙτîν τ¾ν ΖΓ· κሠτ¦ς λοιπ¦ς ¥ρα πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξει κሠτ¾ν λοιπ¾ν γωνίαν τÍ λοιπÍ γωνίv· ‡ση ¥ρα ¹ µν ΚΓ εÙθε‹α τÍ ΓΛ, ¹ δ ØπÕ ΖΚΓ γωνία τÍ ØπÕ ΖΛΓ. κሠ™πεˆ ‡ση ™στˆν ¹ ΚΓ τÍ ΓΛ, διπλÁ ¥ρα ¹ ΚΛ τÁς ΚΓ. δι¦ τ¦ αÙτα δ¾

C

L

Let ABCDE be the given circle. So it is required to circumscribe an equilateral and equiangular pentagon about circle ABCDE. Let A, B, C, D, and E have been conceived as the angular points of a pentagon having been inscribed (in circle ABCDE) [Prop. 3.11], such that the circumferences AB, BC, CD, DE, and EA are equal. And let GH, HK, KL, LM , and M G have been drawn through (points) A, B, C, D, and E (respectively), touching the circle.† And let the center F of the circle ABCDE have been found [Prop. 3.1]. And let F B, F K, F C, F L, and F D have been joined. And since the straight-line KL touches (circle) ABCDE at C, and F C has been joined from the center F to the point of contact C, F C is thus perpendicular to KL [Prop. 3.18]. Thus, each of the angles at C is a rightangle. So, for the same (reasons), the angles at B and D are also right-angles. And since angle F CK is a rightangle, the (square) on F K is thus equal to the (sum of the squares) on F C and CK [Prop. 1.47]. So, for the same (reasons), the (square) on F K is also equal to the (sum of the squares) on F B and BK. So that the (sum of the squares) on F C and CK is equal to the (sum of the squares) on F B and BK, of which the (square) on F C is equal to the (square) on F B. Thus, the remaining (square) on CK is equal to the remaining (square) on BK. Thus, BK (is) equal to CK. And since F B is equal to F C, and F K (is) common, the two (straightlines) BF , F K are equal to the two (straight-lines) CF , F K. And the base BK [is] equal to the base CK. Thus, angle BF K is equal to [angle] KF C [Prop. 1.8]. And BKF (is equal) to F KC [Prop. 1.8]. Thus, BF C (is) double KF C, and BKC (is double) F KC. So, for the same (reasons), CF D is also double CF L, and DLC (is also double) F LC. And since circumference BC is equal to CD, angle BF C is also equal to CF D [Prop. 3.27]. And BF C is double KF C, and DF C (is double) LF C. Thus, KF C is also equal to LF C. And angle F CK is also equal to F CL. So, F KC and F LC are two triangles hav-

121

ΣΤΟΙΧΕΙΩΝ δ΄.

ELEMENTS BOOK 4

δειχθήσεται κሠ¹ ΘΚ τÁς ΒΚ διπλÁ. καί ™στιν ¹ ΒΚ τÍ ΚΓ ‡ση· κሠ¹ ΘΚ ¥ρα τÍ ΚΛ ™στιν ‡ση. еοίως δ¾ δειχθήσεται κሠ˜κάστη τîν ΘΗ, ΗΜ, ΜΛ ˜κατέρv τîν ΘΚ, ΚΛ ‡ση· „σόπλευρον ¥ρα ™στˆ τÕ ΗΘΚΛΜ πεντάγωνον. λέγω δή, Óτι κሠ„σογώνιον. ™πεˆ γ¦ρ ‡ση ™στˆν ¹ ØπÕ ΖΚΓ γωνία τÍ ØπÕ ΖΛΓ, κሠ™δείχθη τÁς µν ØπÕ ΖΚΓ διπλÁ ¹ ØπÕ ΘΚΛ, τÁς δ ØπÕ ΖΛΓ διπλÁ ¹ ØπÕ ΚΛΜ, κሠ¹ ØπÕ ΘΚΛ ¥ρα τÍ ØπÕ ΚΛΜ ™στιν ‡ση. еοίως δ¾ δειχθήσεται κሠ˜κάστη τîν ØπÕ ΚΘΗ, ΘΗΜ, ΗΜΛ ˜κατέρv τîν ØπÕ ΘΚΛ, ΚΛΜ ‡ση· αƒ πέντε ¥ρα γωνίαι αƒ ØπÕ ΗΘΚ, ΘΚΛ, ΚΛΜ, ΛΜΗ, ΜΗΘ ‡σαι ¢λλήλαις ε„σίν. „σογώνιον ¥ρα ™στˆ τÕ ΗΘΚΛΜ πεντάγωνον. ™δείχθη δ κሠ„σόπλευρον, κሠπεριγέγραπται περˆ τÕν ΑΒΓ∆Ε κύκλον. [Περˆ τÕν δοθέντα ¥ρα κύκλον πεντάγωνον „σόπλευρόν τε κሠ„σογώνιον περιγέγραπται]· Óπερ œδει ποιÁσαι.

†

ing two angles equal to two angles, and one side equal to one side, (namely) their common (side) F C. Thus, they will also have the remaining sides equal to the (corresponding) remaining sides, and the remaining angle to the remaining angle [Prop. 1.26]. Thus, the straight-line KC (is) equal to CL, and the angle F KC to F LC. And since KC is equal to LC, KL (is) thus double KC. So, for the same (reasons), it can be shown that HK (is) also double BK. And BK is equal to KC. Thus, HK is also equal to KL. So, similarly, each of HG, GM , and M L can also be shown (to be) equal to each of HK and KL. Thus, pentagon GHKLM is equilateral. So I say that (it is) also equiangular. For since angle F KC is equal to F LC, and HKL was shown (to be) double F KC, and KLM double F LC, HKL is thus also equal to KLM . So, similarly, each of KHG, HGM , and GM L can also be shown (to be) equal to each of HKL and KLM . Thus, the five angles GHK, HKL, KLM , LM G, and M GH are equal to one another. Thus, the pentagon GHKLM is equiangular. And it was also shown (to be) equilateral, and has been circumscribed about circle ABCDE. [Thus, an equilateral and equiangular pentagon has been circumscribed about the given circle]. (Which is) the very thing it was required to do.

See the footnote to Prop. 3.34.

ιγ΄.

Proposition 13

Ε„ς τÕ δοθν πεντάγωνον, Ó ™στιν „σόπλευρόν τε κሠ„σογώνιον, κύκλον ™γγράψαι.

To inscribe a circle in a given pentagon, which is equilateral and equiangular.

Α

A

Η

Μ

Β

G Ε

Ζ

B

Γ

Κ

E F

Λ

Θ

M

L

H

∆

C

”Εστω τÕ δοθν πεντάγωνον „σόπλευρόν τε κሠ„σογώνιον τÕ ΑΒΓ∆Ε· δε‹ δ¾ ε„ς τÕ ΑΒΓ∆Ε πεντάγωνον κύκλον ™γγράψαι. Τετµήσθω γ¦ρ ˜κατέρα τîν ØπÕ ΒΓ∆, Γ∆Ε γωνιîν δίχα ØπÕ ˜κατέρας τîν ΓΖ, ∆Ζ εÙθειîν· κሠ¢πÕ τοà Ζ σηµείου, καθ' Ö συµβάλλουσιν ¢λλήλαις αƒ ΓΖ, ∆Ζ εÙθε‹αι, ™πεζεύχθωσαν αƒ ΖΒ, ΖΑ, ΖΕ εÙθε‹αι. κሠ™πεˆ

K

D

Let ABCDE be the given equilateral and equiangular pentagon. So it is required to inscribe a circle in pentagon ABCDE. For let angles BCD and CDE have each been cut in half by each of the straight-lines CF and DF (respectively) [Prop. 1.9]. And from the point F , at which the straight-lines CF and DF meet one another, let the

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ELEMENTS BOOK 4

‡ση ™στˆν ¹ ΒΓ τÍ Γ∆, κοιν¾ δ ¹ ΓΖ, δύο δ¾ αƒ ΒΓ, ΓΖ δυσˆ τα‹ς ∆Γ, ΓΖ ‡σαι ε„σίν· κሠγωνία ¹ ØπÕ ΒΓΖ γωνίv τÍ ØπÕ ∆ΓΖ [™στιν] ‡ση· βάσις ¥ρα ¹ ΒΖ βάσει τÍ ∆Ζ ™στιν ‡ση, κሠτÕ ΒΓΖ τρίγωνον τù ∆ΓΖ τριγώνJ ™στιν ‡σον, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν· ‡ση ¥ρα ¹ ØπÕ ΓΒΖ γωνία τÍ ØπÕ Γ∆Ζ. κሠ™πεˆ διπλÁ ™στιν ¹ ØπÕ Γ∆Ε τÁς ØπÕ Γ∆Ζ, ‡ση δ ¹ µν ØπÕ Γ∆Ε τÍ ØπÕ ΑΒΓ, ¹ δ ØπÕ Γ∆Ζ τÍ ØπÕ ΓΒΖ, κሠ¹ ØπÕ ΓΒΑ ¥ρα τÁς ØπÕ ΓΒΖ ™στι διπλÁ· ‡ση ¥ρα ¹ ØπÕ ΑΒΖ γωνία τÍ ØπÕ ΖΒΓ· ¹ ¥ρα ØπÕ ΑΒΓ γωνία δίχα τέτµηται ØπÕ τÁς ΒΖ εÙθείας. еοίως δ¾ δειχθήσεται, Óτι κሠ˜κατέρα τîν ØπÕ ΒΑΕ, ΑΕ∆ δίχα τέτµηται ØπÕ ˜κατέρας τîν ΖΑ, ΖΕ εÙθειîν. ½χθωσαν δ¾ ¢πÕ τοà Ζ σηµείου ™πˆ τ¦ς ΑΒ, ΒΓ, Γ∆, ∆Ε, ΕΑ εÙθείας κάθετοι αƒ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ. κሠ™πεˆ ‡ση ™στˆν ¹ ØπÕ ΘΓΖ γωνία τÍ ØπÕ ΚΓΖ, ™στˆ δ κሠÑρθ¾ ¹ ØπÕ ΖΘΓ [ÑρθÍ] τÍ ØπÕ ΖΚΓ ‡ση, δύο δ¾ τρίγωνά ™στι τ¦ ΖΘΓ, ΖΚΓ τ¦ς δύο γωνίας δυσˆ γωνίαις ‡σας œχοντα κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην κοιν¾ν αÙτîν τ¾ν ΖΓ Øποτείνουσαν ØπÕ µίαν τîν ‡σων γωνιîν· κሠτ¦ς λοιπ¦ς ¥ρα πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξει· ‡ση ¥ρα ¹ ΖΘ κάθετος τÊ ΖΚ καθέτJ. еοίως δ¾ δειχθήσεται, Óτι κሠ˜κάστη τîν ΖΛ, ΖΜ, ΖΗ ˜κατέρv τîν ΖΘ, ΖΚ ‡ση ™στίν· αƒ πέντε ¥ρα εÙθε‹αι αƒ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ ‡σαι ¢λλήλαις ε„σίν. Ð ¥ρα κέντρJ τù Ζ διαστήµατι δ ˜νˆ τîν Η, Θ, Κ, Λ, Μ κύκλος γραφόµενος ¼ξει κሠδι¦ τîν λοιπîν σηµείων κሠ™φάψεται τîν ΑΒ, ΒΓ, Γ∆, ∆Ε, ΕΑ εÙθειîν δι¦ τÕ Ñρθ¦ς εναι τ¦ς πρÕς το‹ς Η, Θ, Κ, Λ, Μ σηµείοις γωνίας. ε„ γ¦ρ οÙκ ™φάψεται αÙτîν, ¢λλ¦ τεµε‹ αÙτάς, συµβήσεται τ¾ν τÍ διαµέτρJ τοà κύκλου πρÕς Ñρθ¦ς ¢π' ¥κρας ¢γοµένην ™ντÕς πίπτειν τοà κύκλου· Óπερ ¥τοπον ™δείχθη. οÙκ ¥ρα Ð κέντρJ τù Ζ διαστήµατι δ ˜νˆ τîν Η, Θ, Κ, Λ, Μ σηµείων γραφόµενος κύκλος τεµε‹ τ¦ς ΑΒ, ΒΓ, Γ∆, ∆Ε, ΕΑ εÙθείας· ™φάψεται ¥ρα αÙτîν. γεγράφθω æς Ð ΗΘΚΛΜ. Ε„ς ¥ρα τÕ δοθν πεντάγωνον, Ó ™στιν „σόπλευρόν τε κሠ„σογώνιον, κύκλος ™γγέγραπται· Óπερ œδει ποιÁσαι.

straight-lines F B, F A, and F E have been joined. And since BC is equal to CD, and CF (is) common, the two (straight-lines) BC, CF are equal to the two (straightlines) DC, CF . And angle BCF [is] equal to angle DCF . Thus, the base BF is equal to the base DF , and triangle BCF is equal to triangle DCF , and the remaining angles will be equal to the (corresponding) remaining angles, which the equal sides subtend [Prop. 1.4]. Thus, angle CBF (is) equal to CDF . And since CDE is double CDF , and CDE (is) equal to ABC, and CDF to CBF , CBA is thus also double CBF . Thus, angle ABF is equal to F BC. Thus, angle ABC has been cut in half by the straight-line BF . So, similarly, it can be shown that BAE and AED have been cut in half by the straight-lines F A and F E, respectively. So let F G, F H, F K, F L, and F M have been drawn from point F , perpendicular to the straight-lines AB, BC, CD, DE, and EA (respectively) [Prop. 1.12]. And since angle HCF is equal to KCF , and the right-angle F HC is also equal to the [right-angle] F KC, F HC and F KC are two triangles having two angles equal to two angles, and one side equal to one side, (namely) their common (side) F C, subtending one of the equal angles. Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26]. Thus, the perpendicular F H (is) equal to the perpendicular F K. So, similarly, it can be shown that F L, F M , and F G are each equal to each of F H and F K. Thus, the five straight-lines F G, F H, F K, F L, and F M are equal to one another. Thus, the circle drawn with center F , and radius one of G, H, K, L, or M , will also go through the remaining points, and will touch the straight-lines AB, BC, CD, DE, and EA, on account of the angles at points G, H, K, L, and M being right-angles. For if it does not touch them, but cuts them, it follows that a (straight-line) drawn at rightangles to the diameter of the circle, from the end, falls inside the circle. The very thing was shown (to be) absurd [Prop. 3.16]. Thus, the circle drawn with center F , and radius one of G, H, K, L, or M , does not cut the straight-lines AB, BC, CD, DE, or EA. Thus, it will touch them. Let it have been drawn, like GHKLM (in the figure). Thus, a circle has been inscribed in the given pentagon, which is equilateral and equiangular. (Which is) the very thing it was required to do.

ιδ΄.

Proposition 14

Περˆ τÕ δοθν πεντάγωνον, Ó ™στιν „σόπλευρόν τε To circumscribe a circle about a given pentagon, κሠ„σογώνιον, κύκλον περιγράψαι. which is equilateral and equiangular. ”Εστω τÕ δοθν πεντάγωνον, Ó ™στιν „σόπλευρόν τε Let ABCDE be the given pentagon, which is equilat-

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ELEMENTS BOOK 4

κሠ„σογώνιον, τÕ ΑΒΓ∆Ε· δε‹ δ¾ περˆ τÕ ΑΒΓ∆Ε πεντάγωνον κύκλον περιγράψαι.

eral and equiangular. So it is required to circumscribe a circle about the pentagon ABCDE.

Α

A

Β

Ε

Ζ

Γ

B

E F

∆

C

D

Τετµήσθω δ¾ ˜κατέρα τîν ØπÕ ΒΓ∆, Γ∆Ε γωνιîν δίχα ØπÕ ˜κατέρας τîν ΓΖ, ∆Ζ, κሠ¢πÕ τοà Ζ σηµείου, καθ' Ö συµβάλλουσιν αƒ εÙθε‹αι, ™πˆ τ¦ Β, Α, Ε σηµε‹α ™πεζεύχθωσαν εÙθε‹αι αƒ ΖΒ, ΖΑ, ΖΕ. еοίως δ¾ τù πρÕ τούτου δειχθήσεται, Óτι κሠ˜κάστη τîν ØπÕ ΓΒΑ, ΒΑΕ, ΑΕ∆ γωνιîν δίχα τέτµηται ØπÕ ˜κάστης τîν ΖΒ, ΖΑ, ΖΕ εÙθειîν. κሠ™πεˆ ‡ση ™στˆν ¹ ØπÕ ΒΓ∆ γωνία τÍ ØπÕ Γ∆Ε, καί ™στι τÁς µν ØπÕ ΒΓ∆ ¹µίσεια ¹ ØπÕ ΖΓ∆, τÁς δ ØπÕ Γ∆Ε ¹µίσεια ¹ ØπÕ Γ∆Ζ, κሠ¹ ØπÕ ΖΓ∆ ¥ρα τÍ ØπÕ Ζ∆Γ ™στιν ‡ση· éστε κሠπλευρ¦ ¹ ΖΓ πλευρ´ τÍ Ζ∆ ™στιν ‡ση. еοίως δ¾ δειχθήσεται, Óτι κሠ˜κάστη τîν ΖΒ, ΖΑ, ΖΕ ˜κατέρv τîν ΖΓ, Ζ∆ ™στιν ‡ση· αƒ πέντε ¥ρα εÙθε‹αι αƒ ΖΑ, ΖΒ, ΖΓ, Ζ∆, ΖΕ ‡σαι ¢λλήλαις ε„σίν. Ñ ¥ρα κέντρJ τù Ζ κሠδιαστήµατι ˜νˆ τîν ΖΑ, ΖΒ, ΖΓ, Ζ∆, ΖΕ κύκλος γραφόµενος ¼ξει κሠδι¦ τîν λοιπîν σηµείων κሠœσται περιγεγραµµένος. περιγεγράφθω κሠœστω Ð ΑΒΓ∆Ε. Περˆ ¥ρα τÕ δοθν πεντάγωνον, Ó ™στιν „σόπλευρόν τε κሠ„σογώνιον, κύκλος περιγέγραπται· Óπερ œδει ποιÁσαι.

So let angles BCD and CDE have been cut in half by the (straight-lines) CF and DF , respectively [Prop. 1.9]. And let the straight-lines F B, F A, and F E have been joined from point F , at which the straight-lines meet, to the points B, A, and E (respectively). So, similarly, to the (proposition) before this (one), it can be shown that angles CBA, BAE, and AED have also been cut in half by the straight-lines F B, F A, and F E, respectively. And since angle BCD is equal to CDE, and F CD is half of BCD, and CDF half of CDE, F CD is thus also equal to F DC. So that side F C is also equal to side F D [Prop. 1.6]. So, similarly, it can be shown that F B, F A, and F E are also each equal to each of F C and F D. Thus, the five straight-lines F A, F B, F C, F D, and F E are equal to one another. Thus, the circle drawn with center F , and radius one of F A, F B, F C, F D, or F E, will also go through the remaining points, and will have been circumscribed. Let it have been (so) circumscribed, and let it be ABCDE. Thus, a circle has been circumscribed about the given pentagon, which is equilateral and equiangular. (Which is) the very thing it was required to do.

ιε΄.

Proposition 15

Ε„ς τÕν δοθέντα κύκλον ˜ξάγωνον „σόπλευρόν τε κሠ„σογώνιον ™γγράψαι. ”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ∆ΕΖ· δε‹ δ¾ ε„ς τÕν ΑΒΓ∆ΕΖ κύκλον ˜ξάγωνον „σόπλευρόν τε κሠ„σογώνιον ™γγράψαι. ”Ηχθω τοà ΑΒΓ∆ΕΖ κύκλου διάµετρος ¹ Α∆, κሠε„λήφθω τÕ κέντρον τοà κύκλου τÕ Η, κሠκέντρJ µν τù ∆ διαστήµατι δ τù ∆Η κύκλος γεγράφθω Ð ΕΗΓΘ, κሠ™πιζευχθε‹σαι αƒ ΕΗ, ΓΗ διήχθωσαν ™πˆ τ¦ Β, Ζ σηµε‹α, κሠ™πεζεύχθωσαν αƒ ΑΒ, ΒΓ, Γ∆, ∆Ε, ΕΖ, ΖΑ·

To inscribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle. So it is required to inscribe an equilateral and equiangular hexagon in circle ABCDEF . Let the diameter AD of circle ABCDEF have been drawn,† and let the center G of the circle have been found [Prop. 3.1]. And let the circle EGCH have been drawn, with center D, and radius DG. And EG and CG being joined, let them have been drawn across (the cir-

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λέγω, Óτι τÕ ΑΒΓ∆ΕΖ ˜ξάγωνον „σόπλευρόν τέ ™στι κሠ„σογώνιον.

Θ

cle) to points B and F (respectively). And let AB, BC, CD, DE, EF , and F A have been joined. I say that the hexagon ABCDEF is equilateral and equiangular. H

∆

D Ε

Γ Η

Β

E

C G

Ζ

B

Α 'Επεˆ γ¦ρ τÕ Η σηµε‹ον κέντρον ™στˆ τοà ΑΒΓ∆ΕΖ κύκλου, ‡ση ™στˆν ¹ ΗΕ τÍ Η∆. πάλιν, ™πεˆ τÕ ∆ σηµε‹ον κέντρον ™στˆ τοà ΗΓΘ κύκλου, ‡ση ™στˆν ¹ ∆Ε τÍ ∆Η. ¢λλ' ¹ ΗΕ τÍ Η∆ ™δείχθη ‡ση· κሠ¹ ΗΕ ¥ρα τÍ Ε∆ ‡ση ™στίν· „σόπλευρον ¥ρα ™στˆ τÕ ΕΗ∆ τρίγωνον· καˆ αƒ τρε‹ς ¥ρα αÙτοà γωνίαι αƒ ØπÕ ΕΗ∆, Η∆Ε, ∆ΕΗ ‡σαι ¢λλήλαις ε„σίν, ™πειδήπερ τîν „σοσκελîν τριγώνων αƒ πρÕς τÍ βάσει γωνίαι ‡σαι ¢λλήλαις ε„σίν· καί ε„σιν αƒ τρε‹ς τοà τριγώνου γωνίαι δυσˆν Ñρθα‹ς ‡σαι· ¹ ¥ρα ØπÕ ΕΗ∆ γωνία τρίτον ™στˆ δύο Ñρθîν. еοίως δ¾ δειχθήσεται κሠ¹ ØπÕ ∆ΗΓ τρίτον δύο Ñρθîν. κሠ™πεˆ ¹ ΓΗ εÙθε‹α ™πˆ τ¾ν ΕΒ σταθε‹σα τ¦ς ™φεξÁς γωνίας τ¦ς ØπÕ ΕΗΓ, ΓΗΒ δυσˆν Ñρθα‹ς ‡σας ποιε‹, κሠλοιπ¾ ¥ρα ¹ ØπÕ ΓΗΒ τρίτον ™στˆ δύο Ñρθîν· αƒ ¥ρα ØπÕ ΕΗ∆, ∆ΗΓ, ΓΗΒ γωνίαι ‡σαι ¢λλήλαις ε„σίν· éστε καˆ αƒ κατ¦ κορυφ¾ν αÙτα‹ς αƒ ØπÕ ΒΗΑ, ΑΗΖ, ΖΗΕ ‡σαι ε„σˆν [τα‹ς ØπÕ ΕΗ∆, ∆ΗΓ, ΓΗΒ]. αƒ žξ ¥ρα γωνίαι αƒ ØπÕ ΕΗ∆, ∆ΗΓ, ΓΗΒ, ΒΗΑ, ΑΗΖ, ΖΗΕ ‡σαι ¢λλήλαις ε„σίν. αƒ δ ‡σαι γωνίαι ™πˆ ‡σων περιφερειîν βεβήκασιν· αƒ žξ ¥ρα περιφέρειαι αƒ ΑΒ, ΒΓ, Γ∆, ∆Ε, ΕΖ, ΖΑ ‡σαι ¢λλήλαις ε„σίν. ØπÕ δ τ¦ς ‡σας περιφερείας αƒ ‡σαι εÙθε‹αι Øποτείνουσιν· αƒ žξ ¥ρα εÙθε‹αι ‡σαι ¢λλήλαις ε„σίν· „σόπλευρον ¥ρα ™στˆ το ΑΒΓ∆ΕΖ ˜ξάγωνον. λέγω δή, Óτι κሠ„σογώνιον. ™πεˆ γ¦ρ ‡ση ™στˆν ¹ ΖΑ περιφέρεια τÍ Ε∆ περιφερείv, κοιν¾ προσκείσθω ¹ ΑΒΓ∆ περιφέρεια· Óλη ¥ρα ¹ ΖΑΒΓ∆ ÓλV τÍ Ε∆ΓΒΑ ™στιν ‡ση· κሠβέβηκεν ™πˆ µν τÁς ΖΑΒΓ∆ περιφερείας ¹ ØπÕ ΖΕ∆ γωνία, ™πˆ δ τÁς Ε∆ΓΒΑ περιφερείας ¹ ØπÕ ΑΖΕ γωνία· ‡ση ¥ρα ¹ ØπÕ ΑΖΕ γωνία τÍ ØπÕ ∆ΕΖ. еοίως δ¾ δειχθήσεται, Óτι καˆ αƒ λοιπሠγωνίαι τοà ΑΒΓ∆ΕΖ ˜ξαγώνου κατ¦ µίαν ‡σαι ε„σˆν ˜κατέρv τîν ØπÕ ΑΖΕ, ΖΕ∆ γωνιîν· „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ∆ΕΖ ˜ξάγωνον. ™δείχθη δ κሠ„σόπλευρον· καˆ

F

A For since point G is the center of circle ABCDEF , GE is equal to GD. Again, since point D is the center of circle GCH, DE is equal to DG. But, GE was shown (to be) equal to GD. Thus, GE is also equal to ED. Thus, triangle EGD is equilateral. Thus, its three angles EGD, GDE, and DEG are also equal to one another, inasmuch as the angles at the base of isosceles triangles are equal to one another [Prop. 1.5]. And the three angles of the triangle are equal to two right-angles [Prop. 1.32]. Thus, angle EGD is one third of two right-angles. So, similarly, DGC can also be shown (to be) one third of two right-angles. And since the straight-line CG, standing on EB, makes adjacent angles EGC and CGB equal to two right-angles [Prop. 1.13], the remaining angle CGB is thus also equal to one third of two right-angles. Thus, angles EGD, DGC, and CGB are equal to one another. And hence the (angles) opposite to them BGA, AGF , and F GE are also equal [to EGD, DGC, and CGB (respectively)] [Prop. 1.15]. Thus, the six angles EGD, DGC, CGB, BGA, AGF , and F GE are equal to one another. And equal angles stand on equal [Prop. 3.26]. Thus, the six circumferences AB, BC, CD, DE, EF , and F A are equal to one another. And equal straight-lines subtend equal circumferences [Prop. 3.29]. Thus, the six straight-lines (AB, BC, CD, DE, EF , and F A) are equal to one another. Thus, hexagon ABCDEF is equilateral. So, I say that (it is) also equiangular. For since circumference F A is equal to circumference ED, let circumference ABCD have been added to both. Thus, the whole of F ABCD is equal to the whole of EDCBA. And angle F ED stands on circumference F ABCD, and angle AF E on circumference EDCBA. Thus, angle AF E is equal to DEF [Prop. 3.27]. Similarly, it can also be

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™γγέγραπται ε„ς τÕν ΑΒΓ∆ΕΖ κύκλον. shown that the remaining angles of hexagon ABCDEF Ε„ς ¥ρα τÕν δοθέντα κύκλον ˜ξάγωνον „σόπλευρόν are individually equal to each of angles AF E and F ED. τε κሠ„σογώνιον ™γγέγραπται· Óπερ œδει ποιÁσαι. Thus, hexagon ABCDEF is equiangular. And it was also shown (to be) equilateral. And it has been inscribed in circle ABCDE. Thus, an equilateral and equiangular hexagon has been inscribed in the given circle. (Which is) the very thing it was required to do.

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι ¹ τοà ˜ξαγώνου πλευρ¦ ‡ση ™στˆ τÍ ™κ τοà κέντρου τοà κύκλου. `Οµοίως δ το‹ς ™πˆ τοà πενταγώνου ™¦ν δι¦ τîν κατ¦ τÕν κύκλον διαιρέσεων ™φαπτοµένας τοà κύκλου ¢γάγωµεν, περιγραφήσεται περˆ τÕν κύκλον ˜ξάγωνον „σόπλευρόν τε κሠ„σογώνιον ¢κολούθως το‹ς ™πˆ τοà πενταγώνου ε„ρηµένοις. κሠœτι δι¦ τîν еοίων το‹ς ™πˆ τοà πενταγώνου ε„ρηµένοις ε„ς τÕ δοθν ˜ξάγωνον κύκλον ™γγράψοµέν τε κሠπεριγράψοµεν· Óπερ œδει ποιÁσαι.

So, from this, (it is) manifest that a side of the hexagon is equal to the radius of the circle. And similarly to a pentagon, if we draw tangents to the circle through the (sixfold) divisions of the (circumference of the) circle, an equilateral and equiangular hexagon can be circumscribed about the circle, analogously to the aforementioned pentagon. And, further, by (means) similar to the aforementioned pentagon, we can inscribe and circumscribe a circle in (and about) a given hexagon. (Which is) the very thing it was required to do.

†

See the footnote to Prop. 4.6.

ι$΄.

Proposition 16

Ε„ς τÕν δοθέντα κύκλον πεντεκαιδεκάγωνον „σόπλευρTo inscribe an equilateral and equiangular fifteenόν τε κሠ„σογώνιον ™γγράψαι. sided figure in a given circle.

Α

A

Β

B

Ε

E

Γ

∆

C

”Εστω Ð δοθεˆς κύκλος Ð ΑΒΓ∆· δε‹ δ¾ ε„ς τÕν ΑΒΓ∆ κύκλον πεντεκαιδεκάγωνον „σόπλευρόν τε κሠ„σογώνιον ™γγράψαι. 'Εγγεγράφθω ε„ς τÕν ΑΒΓ∆ κύκλον τριγώνου µν „σοπλεύρου τοà ε„ς αÙτÕν ™γγραφοµένου πλευρ¦ ¹ ΑΓ, πενταγώνου δ „σοπλεύρου ¹ ΑΒ· ο†ων ¥ρα ™στˆν Ð ΑΒΓ∆ κύκλος ‡σων τµήµατων δεκαπέντε, τοιούτων ¹ µν ΑΒΓ περιφέρεια τρίτον οâσα τοà κύκλου œσται πέντε, ¹ δ ΑΒ περιφέρεια πέµτον οâσα τοà κύκλου œσται τριîν· λοιπ¾ ¥ρα ¹ ΒΓ τîν ‡σων δύο. τετµήσθω ¹ ΒΓ δίχα

D

Let ABCD be the given circle. So it is required to inscribe an equilateral and equiangular fifteen-sided figure in circle ABCD. Let the side AC of an equilateral triangle inscribed in (the circle) [Prop. 4.2], and (the side) AB of an (inscribed) equilateral pentagon [Prop. 4.11], have been inscribed in circle ABCD. Thus, just as the circle ABCD is (made up) of fifteen equal pieces, the circumference ABC, being a third of the circle, will be (made up) of five such (pieces), and the circumference AB, being a fifth of

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ELEMENTS BOOK 4

κατ¦ τÕ Ε· ˜κατέρα ¥ρα τîν ΒΕ, ΕΓ περιφερειîν πεντεκαιδέκατόν ™στι τοà ΑΒΓ∆ κύκλου. 'Ε¦ν ¥ρα ™πιζεύξαντες τ¦ς ΒΕ, ΕΓ ‡σας αÙτα‹ς κατ¦ τÕ συνεχς εÙθείας ™ναρµόσωµεν ε„ς τÕν ΑΒΓ∆[Ε] κύκλον, œσται ε„ς αÙτÕν ™γγεγραµµένον πεντεκαιδεκάγωνον „σόπλευρόν τε κሠ„σογώνιον· Óπερ œδει ποιÁσαι. `Οµοίως δ το‹ς ™πˆ τοà πενταγώνου ™¦ν δι¦ τîν κατ¦ τÕν κύκλον διαιρέσεων ™φαπτοµένας τοà κύκλου ¢γάγωµεν, περιγραφήσεται περˆ τÕν κύκλον πεντεκαιδεκάγωνον „σόπλευρόν τε κሠ„σογώνιον. œτι δ δι¦ τîν еοίων το‹ς ™πˆ τοà πενταγώνου δείξεων κሠε„ς τÕ δοθν πεντεκαιδεκάγωνον κύκλον ™γγράψοµέν τε κሠπεριγράψοµεν· Óπερ œδει ποιÁσαι.

the circle, will be (made up) of three. Thus, the remainder BC (will be made up) of two equal (pieces). Let (circumference) BC have been cut in half at E [Prop. 3.30]. Thus, each of the circumferences BE and EC is one fifteenth of the circle ABCDE. Thus, if, joining BE and EC, we continuously insert straight-lines equal to them into circle ABCD[E] [Prop. 4.1], then an equilateral and equiangular fifteensided figure will have been inserted into (the circle). (Which is) the very thing it was required to do. And similarly to the pentagon, if we draw tangents to the circle through the (fifteenfold) divisions of the (circumference of the) circle, we can circumscribe an equilateral and equiangular fifteen-sided figure about the circle. And, further, through similar proofs to the pentagon, we can also inscribe and circumscribe a circle in (and about) a given fifteen-sided figure. (Which is) the very thing it was required to do.

127

128

ELEMENTS BOOK 5 Proportion†

† The theory of proportion set out in this book is generally attributed to Eudoxus of Cnidus. The novel feature of this theory is its ability to deal with irrational magnitudes, which had hitherto been a major stumbling block for Greek mathematicians. Throughout the footnotes in this book, α, β, γ, etc., denote general (possibly irrational) magnitudes, whereas m, n, l, etc., denote positive integers.

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“Οροι.

Definitions

α΄. Μέρος ™στˆ µέγεθος µεγέθους τÕ œλασσον τοà µείζονος, Óταν καταµετρÍ τÕ µε‹ζον. β΄. Πολλαπλάσιον δ τÕ µε‹ζον τοà ™λάττονος, Óταν καταµετρÁται ØπÕ τοà ™λάττονος. γ΄. Λόγος ™στˆ δύο µεγεθîν еογενîν ¹ κατ¦ πηλικότητά ποια σχέσις. δ΄. Λόγον œχειν πρÕς ¥λληλα µεγέθη λέγεται, § δύναται πολλαπλασιαζόµενα ¢λλήλων Øπερέχειν. ε΄. 'Εν τù αÙτù λόγJ µεγέθη λέγεται εναι πρîτον πρÕς δεύτερον κሠτρίτον πρÕς τέταρτον, Óταν τ¦ τοà πρώτου καί τρίτου „σάκις πολλαπλάσια τîν τοà δευτέρου κሠτετάρτου „σάκις πολλαπλασίων καθ' Ðποιονοàν πολλαπλασιασµÕν ˜κάτερον ˜κατέρου À ¤µα ØπερέχV À ¤µα ‡σα Ï À ¤µα ™λλείπÍ ληφθέντα κατάλληλα. $΄. Τ¦ δ τÕν αÙτÕν œχοντα λόγον µεγέθη ¢νάλογον καλείσθω. ζ΄. “Οταν δ τîν „σάκις πολλαπλασίων τÕ µν τοà πρώτου πολλαπλάσιον ØπερέχV τοà τοà δευτέρου πολλαπλασίου, τÕ δ τοà τρίτου πολλαπλάσιον µ¾ ØπερέχV τοà τοà τετάρτου πολλαπλασίου, τότε τÕ πρîτον πρÕς τÕ δεύτερον µείζονα λόγον œχειν λέγεται, ½περ τÕ τρίτον πρÕς τÕ τέταρτον. η΄. 'Αναλογία δ ™ν τρισˆν Óροις ™λαχίστη ™στίν. θ΄. “Οταν δ τρία µεγέθη ¢νάλογον Ï, τÕ πρîτον πρÕς τÕ τρίτον διπλασίονα λόγον œχειν λέγεται ½περ πρÕς τÕ δεύτερον. ι΄. “Οταν δ τέσσαρα µεγέθη ¢νάλογον Ï, τÕ πρîτον πρÕς τÕ τέταρτον τριπλασίονα λόγον œχειν λέγεται ½περ πρÕς τÕ δεύτερον, κሠ¢εˆ ˜ξÁς еοίως, æς ¨ν ¹ ¢ναλογία ØπάρχV. ια΄. `Οµόλογα µεγέθη λέγεται τ¦ µν ¹γούµενα το‹ς ¹γουµένοις τ¦ δ ˜πόµενα το‹ς ˜ποµένοις. ιβ΄. 'Εναλλ¦ξ λόγος ™στˆ λÁψις τοà ¹γουµένου πρÕς τÕ ¹γούµενον κሠτοà ˜ποµένου πρÕς τÕ ˜πόµενον. ιγ΄. 'Ανάπαλιν λόγος ™στˆ λÁψις τοà ˜ποµένου æς ¹γουµένου πρÕς τÕ ¹γούµενον æς ˜πόµενον. ιδ΄. Σύνθεσις λόγου ™στˆ λÁψις τοà ¹γουµένου µετ¦ τοà ˜ποµένου æς ˜νÕς πρÕς αÙτÕ τÕ ˜πόµενον. ιε΄. ∆ιαίρεσις λόγου ™στˆ λÁψις τÁς ØπεροχÁς, Î Øπερέχει τÕ ¹γούµενον τοà ˜ποµένου, πρÕς αÙτÕ τÕ ˜πόµενον. ι$΄. 'Αναστροφ¾ λόγου ™στˆ λÁψις τοà ¹γουµένου πρÕς τ¾ν Øπεροχήν, Î Øπερέχει τÕ ¹γούµενον τοà ˜ποµένου. ιζ΄. ∆ι' ‡σου λόγος ™στˆ πλειόνων Ôντων µεγεθîν κሠ¥λλων αÙτο‹ς ‡σων τÕ πλÁθος σύνδυο λαµβανοµένων κሠ™ν τù αÙτù λόγJ, Óταν Ï æς ™ν το‹ς πρώτοις µεγέθεσι

1. A magnitude is a part of a(nother) magnitude, the lesser of the greater, when it measures the greater.† 2. And the greater (magnitude is) a multiple of the lesser when it is measured by the lesser. 3. A ratio is a certain type of condition with respect to size of two magnitudes of the same kind.‡ 4. (Those) magnitudes are said to have a ratio with respect to one another which, being multiplied, are capable of exceeding one another.§ 5. Magnitudes are said to be in the same ratio, the first to the second, and the third to the fourth, when equal multiples of the first and the third either both exceed, are both equal to, or are both less than, equal multiples of the second and the fourth, respectively, being taken in corresponding order, according to any kind of multiplication whatever.¶ 6. And let magnitudes having the same ratio be called proportional.∗ 7. And when for equal multiples (as in Def. 5), the multiple of the first (magnitude) exceeds the multiple of the second, and the multiple of the third (magnitude) does not exceed the multiple of the fourth, then the first (magnitude) is said to have a greater ratio to the second than the third (magnitude has) to the fourth. 8. And a proportion in three terms is the smallest (possible).$ 9. And when three magnitudes are proportional, the first is said to have a squaredk ratio to the third with respect to the second.†† 10. And when four magnitudes are (continuously) proportional, the first is said to have a cubed‡‡ ratio to the fourth with respect to the second.§§ And so on, similarly, in successive order, whatever the (continuous) proportion might be. 11. These magnitudes are said to be corresponding (magnitudes): the leading to the leading (of two ratios), and the following to the following. 12. An alternate ratio is a taking of the (ratio of the) leading (magnitude) to the leading (of two equal ratios), and (setting it equal to) the (ratio of the) following (magnitude) to the following.¶¶ 13. An inverse ratio is a taking of the (ratio of the) following (magnitude) as the leading and the leading (magnitude) as the following.∗∗ 14. A composition of a ratio is a taking of the (ratio of the) leading plus the following (magnitudes), as one, to the same following (magnitude).$$

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ELEMENTS BOOK 5

τÕ πρîτον πρÕς τÕ œσχατον, οÛτως ™ν το‹ς δευτέροις µεγέθεσι τÕ πρîτον πρÕς τÕ œσχατον· À ¥λλως· λÁψις τîν ¥κρων καθ' Øπεξαίρεσιν τîν µέσων. ιη΄. Τεταραγµένη δ ¢ναλογία ™στίν, Óταν τριîν Ôντων µεγεθîν κሠ¥λλων αÙτο‹ς ‡σων τÕ πλÁθος γίνηται æς µν ™ν το‹ς πρώτοις µεγέθεσιν ¹γούµενον πρÕς ™πόµενον, οÛτως ™ν το‹ς δευτέροις µεγέθεσιν ¹γούµενον πρÕς ˜πόµενον, æς δ ™ν το‹ς πρώτοις µεγέθεσιν ˜πόµενον πρÕς ¥λλο τι, οÛτως ™ν το‹ς δευτέροις ¥λλο τι πρÕς ¹γούµενον.

15. A separation of a ratio is a taking of the (ratio of the) excess by which the leading (magnitude) exceeds the following to the same following (magnitude).kk 16. A conversion of a ratio is a taking of the (ratio of the) leading (magnitude) to the excess by which the leading (magnitude) exceeds the following.††† 17. There being several magnitudes, and other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, a ratio via equality (or ex aequali) occurs when as the first is to the last in the first (set of) magnitudes, so the first (is) to the last in the second (set of) magnitudes. Or alternately, (it is) a taking of the (ratio of the) outer (magnitudes) by the removal of the inner (magnitudes).‡‡‡ 18. There being three magnitudes, and other (magnitudes) of equal number to them, a perturbed proportion occurs when as the leading is to the following in the first (set of) magnitudes, so the leading (is) to the following in the second (set of) magnitudes, and as the following (is) to some other (i.e., the remaining magnitude) in the first (set of) magnitudes, so some other (is) to the leading in the second (set of) magnitudes.§§§

†

In other words, α is said to be a part of β if β = m α.

‡

In modern notation, the ratio of two magnitudes, α and β, is denoted α : β.

§

In other words, α has a ratio with respect to β if m α > β and n β > α, for some m and n.

¶

In other words, α : β :: γ : δ if and only if m α > n β whenever m γ > n δ, and m α = n β whenever m γ = n δ, and m α < n β whenever

m γ < n δ, for all m and n. This definition is the kernel of Eudoxus’ theory of proportion, and is valid even if α, β, etc., are irrational. ∗

Thus if α and β have the same ratio as γ and δ then they are proportional. In modern notation, α : β :: γ : δ.

$

In modern notation, a proportion in three terms—α, β, and γ—is written: α : β :: β : γ.

k

Literally, “double”.

††

In other words, if α : β :: β : γ then α : γ :: α 2 : β 2 .

‡‡

Literally, “triple”.

§§

In other words, if α : β :: β : γ :: γ : δ then α : δ :: α 3 : β 3 .

¶¶

In other words, if α : β :: γ : δ then the alternate ratio corresponds to α : γ :: β : δ.

∗∗

In other words, if α : β then the inverse ratio corresponds to β : α.

$$

In other words, if α : β then the composed ratio corresponds to α + β : β.

kk

In other words, if α : β then the separated ratio corresponds to α − β : β.

††† ‡‡‡

In other words, if α : β then the converted ratio corresponds to α : α − β.

In other words, if α, β, γ are the first set of magnitudes, and δ, ǫ, ζ the second set, and α : β : γ :: δ : ǫ : ζ, then the ratio via equality (or ex aequali) corresponds to α : γ :: δ : ζ. §§§

In other words, if α, β, γ are the first set of magnitudes, and δ, ǫ, ζ the second set, and α : β :: δ : ǫ as well as β : γ :: ζ : δ, then the proportion

is said to be perturbed.

α΄.

Proposition 1†

'Ε¦ν Ï Ðποσαοàν µεγέθη Ðποσωνοàν µεγεθîν ‡σων If there are any number of magnitudes whatsoever τÕ πλÁθος ›καστον ˜κάστου „σάκις πολλαπλάσιον, (which are) equal multiples, respectively, of some (other)

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Ðσαπλάσιόν ™στιν žν τîν µεγεθîν ˜νός, τοσαυταπλάσια œσται κሠτ¦ πάντα τîν πάντων.

Α Ε

Η

Β Γ

Θ

A

∆

E

Ζ

”Εστω Ðποσαοàν µεγέθη τ¦ ΑΒ, Γ∆ Ðποσωνοàν µεγεθîν τîν Ε, Ζ ‡σων τÕ πλÁθος ›καστον ˜κάστου „σάκις πολλαπλάσιον· λέγω, Óτι Ðσαπλάσιόν ™στι τÕ ΑΒ τοà Ε, τοσαυταπλάσια œσται κሠτ¦ ΑΒ, Γ∆ τîν Ε, Ζ. 'Επεˆ γ¦ρ „σάκις ™στˆ πολλαπλάσιον τÕ ΑΒ τοà Ε κሠτÕ Γ∆ τοà Ζ, Óσα ¥ρα ™στˆν ™ν τù ΑΒ µεγέθη ‡σα τù Ε, τοσαàτα κሠ™ν τù Γ∆ ‡σα τù Ζ. διVρήσθω τÕ µν ΑΒ ε„ς τ¦ τù Ε µεγέθη ‡σα τ¦ ΑΗ, ΗΒ, τÕ δ Γ∆ ε„ς τ¦ τù Ζ ‡σα τ¦ ΓΘ, Θ∆· œσται δ¾ ‡σον τÕ πλÁθος τîν ΑΗ, ΗΒ τù πλήθει τîν ΓΘ, Θ∆. κሠ™πεˆ ‡σον ™στˆ τÕ µν ΑΗ τù Ε, τÕ δ ΓΘ τù Ζ, ‡σον ¥ρα τÕ ΑΗ τù Ε, κሠτ¦ ΑΗ, ΓΘ το‹ς Ε, Ζ. δι¦ τ¦ αÙτ¦ δ¾ ‡σον ™στˆ τÕ ΗΒ τù Ε, κሠτ¦ ΗΒ, Θ∆ το‹ς Ε, Ζ· Óσα ¥ρα ™στˆν ™ν τù ΑΒ ‡σα τù Ε, τοσαàτα κሠ™ν το‹ς ΑΒ, Γ∆ ‡σα το‹ς Ε, Ζ· Ðσαπλάσιον ¥ρα ™στˆ τÕ ΑΒ τοà Ε, τοσαυταπλάσια œσται κሠτ¦ ΑΒ, Γ∆ τîν Ε, Ζ. 'Ε¦ν ¥ρα Ï Ðποσαοàν µεγέθη Ðποσωνοàν µεγεθîν ‡σων τÕ πλÁθος ›καστον ˜κάστου „σάκις πολλαπλάσιον, Ðσαπλάσιόν ™στιν žν τîν µεγεθîν ˜νός, τοσαυταπλάσια œσται κሠτ¦ πάντα τîν πάντων· Óπερ œδει δε‹ξαι.

†

magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second).

G

B C

H

D

F

Let there be any number of magnitudes whatsoever, AB, CD, (which are) equal multiples, respectively, of some (other) magnitudes, E, F , of equal number (to them). I say that as many times as AB is (divisible) by E, so many times will AB, CD also be (divisible) by E, F . For since AB, CD are equal multiples of E, F , thus as many magnitudes as (there) are in AB equal to E, so many (are there) also in CD equal to F . Let AB have been divided into magnitudes AG, GB, equal to E, and CD into (magnitudes) CH, HD, equal to F . So, the number of (divisions) AG, GB will be equal to the number of (divisions) CH, HD. And since AG is equal to E, and CH to F , AG (is) thus equal to E, and AG, CH to E, F . So, for the same (reasons), GB is equal to E, and GB, HD to E, F . Thus, as many (magnitudes) as (there) are in AB equal to E, so many (are there) also in AB, CD equal to E, F . Thus, as many times as AB is (divisible) by E, so many times will AB, CD also be (divisible) by E, F . Thus, if there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second). (Which is) the very thing it was required to show.

In modern notation, this proposition reads m α + m β + · · · = m (α + β + · · · ).

β΄.

Proposition 2†

'Ε¦ν πρîτον δευτέρου „σάκις Ï πολλαπλάσιον κሠτρίτον τετάρτου, Ï δ κሠπέµπτον δευτέρου „σάκις πολλαπλάσιον κሠ›κτον τετάρτου, κሠσυντεθν πρîτον κሠπέµπτον δευτέρου „σάκις œσται πολλαπλάσιον κሠτρίτον κሠ›κτον τετάρτου. Πρîτον γ¦ρ τÕ ΑΒ δευτέρου τοà Γ „σάκις œστω πολλαπλάσιον κሠτρίτον τÕ ∆Ε τετάρτου τοà Ζ, œστω δ κሠπέµπτον τÕ ΒΗ δευτέρου τοà Γ „σάκις πολλαπλάσιον κሠ›κτον τÕ ΕΘ τετάρτου τοà Ζ· λέγω, Óτι κሠσυντεθν πρîτον κሠπέµπτον τÕ ΑΗ δευτέρου τοà Γ „σάκις œσται πολλαπλάσιον κሠτρίτον κሠ›κτον τÕ ∆Θ τετάρτου τοà Ζ.

If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and a fifth (magnitude) and a sixth (are) also equal multiples of the second and fourth (respectively), then the first (magnitude) and the fifth, being added together, and the third and the sixth, (being added together), will also be equal multiples of the second (magnitude) and the fourth (respectively). For let a first (magnitude) AB and a third DE be equal multiples of a second C and a fourth F (respectively). And let a fifth (magnitude) BG and a sixth EH also be (other) equal multiples of the second C and the fourth F (respectively). I say that the first (magnitude) and the fifth, being added together, (to give) AG, and the

132

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ELEMENTS BOOK 5 third (magnitude) and the sixth, (being added together, to give) DH, will also be equal multiples of the second (magnitude) C and the fourth F (respectively).

Α

Β

A

Η

G

C

Γ ∆

Ε

D

Θ

E

H

F

Ζ 'Επεˆ γ¦ρ „σάκις ™στˆ πολλαπλάσιον τÕ ΑΒ τοà Γ κሠτÕ ∆Ε τοà Ζ, Óσα ¥ρα ™στˆν ™ν τù ΑΒ ‡σα τù Γ, τοσαàτα κሠ™ν τù ∆Ε ‡σα τù Ζ. δι¦ τ¦ αÙτ¦ δ¾ κሠÓσα ™στˆν ™ν τù ΒΗ ‡σα τù Γ, τοσαàτα κሠ™ν τù ΕΘ ‡σα τù Ζ· Óσα ¥ρα ™στˆν ™ν ÓλJ τù ΑΗ ‡σα τù Γ, τοσαàτα κሠ™ν ÓλJ τù ∆Θ ‡σα τù Ζ· Ðσαπλάσιον ¥ρα ™στˆ τÕ ΑΗ τοà Γ, τοσαυταπλάσιον œσται κሠτÕ ∆Θ τοà Ζ. κሠσυντεθν ¥ρα πρîτον κሠπέµπτον τÕ ΑΗ δευτέρου τοà Γ „σάκις œσται πολλαπλάσιον κሠτρίτον κሠ›κτον τÕ ∆Θ τετάρτου τοà Ζ. 'Ε¦ν ¥ρα πρîτον δευτέρου „σάκις Ï πολλαπλάσιον κሠτρίτον τετάρτου, Ï δ κሠπέµπτον δευτέρου „σάκις πολλαπλάσιον κሠ›κτον τετάρτου, κሠσυντεθν πρîτον κሠπέµπτον δευτέρου „σάκις œσται πολλαπλάσιον κሠτρίτον κሠ›κτον τετάρτου· Óπερ œδει δε‹ξαι.

†

B

For since AB and DE are equal multiples of C and F (respectively), thus as many (magnitudes) as (there) are in AB equal to C, so many (are there) also in DE equal to F . And so, for the same (reasons), as many (magnitudes) as (there) are in BG equal to C, so many (are there) also in EH equal to F . Thus, as many (magnitudes) as (there) are in the whole of AG equal to C, so many (are there) also in the whole of DH equal to F . Thus, as many times as AG is (divisible) by C, so many times will DH also be divisible by F . Thus, the first (magnitude) and the fifth, being added together, (to give) AG, and the third (magnitude) and the sixth, (being added together, to give) DH, will also be equal multiples of the second (magnitude) C and the fourth F (respectively). Thus, if a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and a fifth (magnitude) and a sixth (are) also equal multiples of the second and fourth (respectively), then the first (magnitude) and the fifth, being added together, and the third and sixth, (being added together), will also be equal multiples of the second (magnitude) and the fourth (respectively). (Which is) the very thing it was required to show.

In modern notation, this propostion reads m α + n α = (m + n) α.

γ΄.

Proposition 3†

'Ε¦ν πρîτον δευτέρου „σάκις Ï πολλαπλάσιον κሠτρίτον τετάρτου, ληφθÍ δ „σάκις πολλαπλάσια τοà τε πρώτου κሠτρίτου, κሠδι' ‡σου τîν ληφθέντων ˜κάτερον ˜κατέρου „σάκις œσται πολλαπλάσιον τÕ µν τοà δευτέρου τÕ δ τοà τετάρτου. Πρîτον γ¦ρ τÕ Α δευτέρου τοà Β „σάκις œστω πολλαπλάσιον κሠτρίτον τÕ Γ τετάρτου τοà ∆, κሠε„λήφθω τîν Α, Γ „σάκις πολλαπλάσια τ¦ ΕΖ, ΗΘ· λέγω, Óτι „σάκις ™στˆ πολλαπλάσιον τÕ ΕΖ τοà Β κሠτÕ ΗΘ τοà ∆. 'Επεˆ γ¦ρ „σάκις ™στˆ πολλαπλάσιον τÕ ΕΖ τοà Α κሠτÕ ΗΘ τοà Γ, Óσα ¥ρα ™στˆν ™ν τù ΕΖ ‡σα τù

If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and equal multiples are taken of the first and the third, then, via equality, the (magnitudes) taken will also be equal multiples of the second (magnitude) and the fourth, respectively. For let a first (magnitude) A and a third C be equal multiples of a second B and a fourth D (respectively), and let the equal multiples EF and GH have been taken of A and C (respectively). I say that EF and GH are equal multiples of B and D (respectively). For since EF and GH are equal multiples of A and C (respectively), thus as many (magnitudes) as (there)

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ELEMENTS BOOK 5

Α, τοσαàτα κሠ™ν τù ΗΘ ‡σα τù Γ. διVρήσθω τÕ µν ΕΖ ε„ς τ¦ τù Α µεγέθη ‡σα τ¦ ΕΚ, ΚΖ, τÕ δ ΗΘ ε„ς τ¦ τù Γ ‡σα τ¦ ΗΛ, ΛΘ· œσται δ¾ ‡σον τÕ πλÁθος τîν ΕΚ, ΚΖ τù πλήθει τîν ΗΛ, ΛΘ. κሠ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ Α τοà Β κሠτÕ Γ τοà ∆, ‡σον δ τÕ µν ΕΚ τù Α, τÕ δ ΗΛ τù Γ, „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΕΚ τοà Β κሠτÕ ΗΛ τοà ∆. δι¦ τ¦ αÙτ¦ δ¾ „σάκις ™στˆ πολλαπλάσιον τÕ ΚΖ τοà Β κሠτÕ ΛΘ τÕà ∆. ™πεˆ οâν πρîτον τÕ ΕΚ δευτέρου τοà Β ‡σάκις ™στˆ πολλαπλάσιον κሠτρίτον τÕ ΗΛ τετάρτου τοà ∆, œστι δ κሠπέµπτον τÕ ΚΖ δευτέρου τοà Β „σάκις πολλαπλάσιον κሠ›κτον τÕ ΛΘ τετάρτου τοà ∆, κሠσυντεθν ¥ρα πρîτον κሠπέµπτον τÕ ΕΖ δευτέρου τοà Β „σάκις ™στˆ πολλαπλάσιον κሠτρίτον κሠ›κτον τÕ ΗΘ τετάρτου τοà ∆.

Α Β Ε Γ ∆ Η

Κ

Λ

A B E

Ζ

C D G

Θ

'Ε¦ν ¥ρα πρîτον δευτέρου „σάκις Ï πολλαπλάσιον κሠτρίτον τετάρτου, ληφθÍ δ τοà πρώτου κሠτρίτου „σάκις πολλαπλάσια, κሠδι' ‡σου τîν ληφθέντων ˜κάτερον ˜κατέρου „σάκις œσται πολλαπλάσιον τÕ µν τοà δευτέρου τÕ δ τοà τετάρτου· Óπερ œδει δε‹ξαι. †

are in EF equal to A, so many (are there) also in GH equal to C. Let EF have been divided into magnitudes EK, KF equal to A, and GH into (magnitudes) GL, LH equal to C. So, the number of (magnitudes) EK, KF will be equal to the number of (magnitudes) GL, LH. And since A and C are equal multiples of B and D (respectively), and EK (is) equal to A, and GL to C, EK and GL are thus equal multiples of B and D (respectively). So, for the same (reasons), KF and LH are equal multiples of B and D (respectively). Therefore, since the first (magnitude) EK and the third GL are equal multiples of the second B and the fourth D (respectively), and the fifth (magnitude) KF and the sixth LH are also equal multiples of the second B and the fourth D (respectively), then the first (magnitude) and fifth, being added together, (to give) EF , and the third (magnitude) and sixth, (being added together, to give) GH, are thus also equal multiples of the second (magnitude) B and the fourth D (respectively) [Prop. 5.2].

K

L

F

H

Thus, if a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and equal multiples are taken of the first and the third, then, via equality, the (magnitudes) taken will also be equal multiples of the second (magnitude) and the fourth, respectively. (Which is) the very thing it was required to show.

In modern notation, this proposition reads m(n α) = (m n) α.

δ΄.

Proposition 4†

'Ε¦ν πρîτον πρÕς δεύτερον τÕν αÙτÕν œχV λόγον κሠτρίτον πρÕς τέταρτον, κሠτ¦ „σάκις πολλαπλάσια τοà τε πρώτου κሠτρίτου πρÕς τ¦ „σάκις πολλαπλάσια τοà δευτέρου κሠτετάρτου καθ' Ðποιονοàν πολλαπλασιασµÕν τÕν αÙτÕν ›ξει λόγον ληφθέντα κατάλληλα. Πρîτον γ¦ρ τÕ Α πρÕς δεύτερον τÕ Β τÕν αÙτÕν ™χέτω λόγον κሠτρίτον τÕ Γ πρÕς τέταρτον τÕ ∆, κሠε„λήφθω τîν µν Α, Γ „σάκις πολλαπλάσια τ¦ Ε, Ζ, τîν δ Β, ∆ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Η, Θ·

If a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multiples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever. For let a first (magnitude) A have the same ratio to a second B that a third C (has) to a fourth D. And let equal multiples E and F have been taken of A and C

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ELEMENTS BOOK 5

λέγω, Óτι ™στˆν æς τÕ Ε πρÕς τÕ Η, οÛτως τÕ Ζ πρÕς (respectively), and other random equal multiples G and τÕ Θ. H of B and D (respectively). I say that as E (is) to G, so F (is) to H.

Α Β Ε Η Κ Μ Γ ∆ Ζ Θ Λ Ν

A B E G K M C D F H L N

Ε„λήφθω γ¦ρ τîν µν Ε, Ζ „σάκις πολλαπλάσια τ¦ Κ, Λ, τîν δ Η, Θ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Μ, Ν. [Καˆ] ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ µν Ε τοà Α, τÕ δ Ζ τοà Γ, κሠε‡ληπται τîν Ε, Ζ ‡σάκις πολλαπλάσια τ¦ Κ, Λ, ‡σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ Κ τοà Α κሠτÕ Λ τοà Γ. δι¦ τ¦ αÙτ¦ δ¾ „σάκις ™στˆ πολλαπλάσιον τÕ Μ τοà Β κሠτÕ Ν τοà ∆. κሠ™πεί ™στιν æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆, κሠε‡ληπται τîν µν Α, Γ „σάκις πολλαπλάσια τ¦ Κ, Λ, τîν δ Β, ∆ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Μ, Ν, ε„ ¥ρα Øπερέχει τÕ Κ τοà Μ, Øπερέχει κሠτÕ Λ τοà Ν, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον. καί ™στι τ¦ µν Κ, Λ τîν Ε, Ζ „σάκις πολλαπλάσια, τ¦ δ Μ, Ν τîν Η, Θ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια· œστιν ¥ρα æς τÕ Ε πρÕς τÕ Η, οÛτως τÕ Ζ πρÕς τÕ Θ. 'Ε¦ν ¥ρα πρîτον πρÕς δεύτερον τÕν αÙτÕν œχV λόγον κሠτρίτον πρÕς τέταρτον, κሠτ¦ „σάκις πολλαπλάσια τοà τε πρώτου κሠτρίτου πρÕς τ¦ „σάκις πολλαπλάσια τοà δευτέρου κሠτετάρτου τÕν αÙτÕν ›ξει λόγον καθ' Ðποιονοàν πολλαπλασιασµÕν ληφθέντα κατάλληλα· Óπερ œδει δε‹ξαι.

†

For let equal multiples K and L have been taken of E and F (respectively), and other random equal multiples M and N of G and H (respectively). [And] since E and F are equal multiples of A and C (respectively), and the equal multiples K and L have been taken of E and F (respectively), K and L are thus equal multiples of A and C (respectively) [Prop. 5.3]. So, for the same (reasons), M and N are equal multiples of B and D (respectively). And since as A is to B, so C (is) to D, and the equal multiples K and L have been taken of A and C (respectively), and the other random equal multiples M and N of B and D (respectively), then if K exceeds M then L also exceeds N , and if (K is) equal (to M then L is also) equal (to N ), and if (K is) less (than M then L is also) less (than N ) [Def. 5.5]. And K and L are equal multiples of E and F (respectively), and M and N other random equal multiples of G and H (respectively). Thus, as E (is) to G, so F (is) to H [Def. 5.5]. Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multiples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: γ : δ then m α : n β :: m γ : n δ, for all m and n.

135

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ELEMENTS BOOK 5 ε΄.

Proposition 5†

'Ε¦ν µέγεθος µεγέθους „σάκις Ï πολλαπλάσιον, Óπερ If a magnitude is the same multiple of a magnitude ¢φαιρεθν ¢φαιρεθέντος, κሠτÕ λοιπÕν τοà λοιποà that a (part) taken away (is) of a (part) taken away (re„σάκις œσται πολλαπλάσιον, Ðσαπλάσιόν ™στι τÕ Óλον spectively) then the remainder will also be the same mulτοà Óλου. tiple of the remainder as that which the whole (is) of the whole (respectively).

Α Η Γ

Ε

Β

A

Ζ ∆

G C

Μέγεθος γ¦ρ τÕ ΑΒ µεγέθους τοà Γ∆ „σάκις œστω πολλαπλάσιον, Óπερ ¢φαιρεθν τÕ ΑΕ ¢φαιρεθέντος τοà ΓΖ· λέγω, Óτι κሠλοιπÕν τÕ ΕΒ λοιποà τοà Ζ∆ „σάκις œσται πολλαπλάσιον, Ðσαπλάσιόν ™στιν Óλον τÕ ΑΒ Óλου τοà Γ∆. `Οσαπλάσιον γάρ ™στι τÕ ΑΕ τοà ΓΖ, τοσαυταπλάσιον γεγονέτω κሠτÕ ΕΒ τοà ΓΗ. Κሠ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ ΑΕ τοà ΓΖ κሠτÕ ΕΒ τοà ΗΓ, „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΑΕ τοà ΓΖ κሠτÕ ΑΒ τοà ΗΖ. κε‹ται δ „σάκις πολλαπλάσιον τÕ ΑΕ τοà ΓΖ κሠτÕ ΑΒ τοà Γ∆. „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΑΒ ˜κατέρου τîν ΗΖ, Γ∆· ‡σον ¥ρα τÕ ΗΖ τù Γ∆. κοινÕν ¢φVρήσθω τÕ ΓΖ· λοιπÕν ¥ρα τÕ ΗΓ λοιπù τù Ζ∆ ‡σον ™στίν. κሠ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ ΑΕ τοà ΓΖ κሠτÕ ΕΒ τοà ΗΓ, ‡σον δ τÕ ΗΓ τù ∆Ζ, „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΑΕ τοà ΓΖ κሠτÕ ΕΒ τοà Ζ∆. „σάκις δ Øπόκειται πολλαπλάσιον τÕ ΑΕ τοà ΓΖ κሠτÕ ΑΒ τοà Γ∆· „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΕΒ τοà Ζ∆ κሠτÕ ΑΒ τοà Γ∆. κሠλοιπÕν ¥ρα τÕ ΕΒ λοιποà τοà Ζ∆ „σάκις œσται πολλαπλάσιον, Ðσαπλάσιόν ™στιν Óλον τÕ ΑΒ Óλου τοà Γ∆. 'Ε¦ν ¥ρα µέγεθος µεγέθους „σάκις Ï πολλαπλάσιον, Óπερ ¢φαιρεθν ¢φαιρεθέντος, κሠτÕ λοιπÕν τοà λοιποà „σάκις œσται πολλαπλάσιον, Ðσαπλάσιόν ™στι κሠτÕ Óλον τοà Óλου· Óπερ œδει δε‹ξαι.

†

E

B

F D

For let the magnitude AB be the same multiple of the magnitude CD that the (part) taken away AE (is) of the (part) taken away CF (respectively). I say that the remainder EB will also be the same multiple of the remainder F D as that which the whole AB (is) of the whole CD (respectively). For as many times as AE is (divisible) by CF , so many times let EB also have been made (divisible) by CG. And since AE and EB are equal multiples of CF and GC (respectively), AE and AB are thus equal multiples of CF and GF (respectively) [Prop. 5.1]. And AE and AB are assumed (to be) equal multiples of CF and CD (respectively). Thus, AB is an equal multiple of each of GF and CD. Thus, GF (is) equal to CD. Let CF have been subtracted from both. Thus, the remainder GC is equal to the remainder F D. And since AE and EB are equal multiples of CF and GC (respectively), and GC (is) equal to DF , AE and EB are thus equal multiples of CF and F D (respectively). And AE and AB are assumed (to be) equal multiples of CF and CD (respectively). Thus, EB and AB are equal multiples of F D and CD (respectively). Thus, the remainder EB will also be the same multiple of the remainder F D as that which the whole AB (is) of the whole CD (respectively). Thus, if a magnitude is the same multiple of a magnitude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively). (Which is) the very thing it was required to show.

In modern notation, this proposition reads m α − m β = m (α − β).

$΄.

Proposition 6†

'Ε¦ν δύο µεγέθη δύο µεγεθîν „σάκις Ï πολλαπλάσια, κሠ¢φαιρεθέντα τιν¦ τîν αÙτîν „σάκις Ï πολλαπλάσια, κሠτ¦ λοιπ¦ το‹ς αÙτο‹ς ½τοι ‡σα ™στˆν À „σάκις αÙτîν πολλαπλάσια. ∆ύο γ¦ρ µεγέθη τ¦ ΑΒ, Γ∆ δύο µεγεθîν τîν Ε, Ζ

If two magnitudes are equal multiples of two (other) magnitudes, and some (parts) taken away (from the former magnitudes) are equal multiples of the latter (magnitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples

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ELEMENTS BOOK 5

„σάκις œστω πολλαπλάσια, κሠ¢φαιρεθέντα τ¦ ΑΗ, ΓΘ of them (respectively). τîν αÙτîν τîν Ε, Ζ „σάκις œστω πολλαπλάσια· λέγω, Óτι For let two magnitudes AB and CD be equal multiκሠλοιπ¦ τ¦ ΗΒ, Θ∆ το‹ς Ε, Ζ ½τοι ‡σα ™στˆν À „σάκις ples of two magnitudes E and F (respectively). And let αÙτîν πολλαπλάσια. the (parts) taken away (from the former) AG and CH be equal multiples of E and F (respectively). I say that the remainders GB and HD are also either equal to E and F (respectively), or (are) equal multiples of them.

Α

Η

Ε Κ Γ

Β

A E K

Θ ∆

Ζ

C

B

H D

F

”Εστω γ¦ρ πρότερον τÕ ΗΒ τù Ε ‡σον· λέγω, Óτι κሠτÕ Θ∆ τù Ζ ‡σον ™στίν. Κείσθω γ¦ρ τù Ζ ‡σον τÕ ΓΚ. ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ ΑΗ τοà Ε κሠτÕ ΓΘ τοà Ζ, ‡σον δ τÕ µν ΗΒ τù Ε, τÕ δ ΚΓ τù Ζ, „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΑΒ τοà Ε κሠτÕ ΚΘ τοà Ζ. „σάκις δ Øπόκειται πολλαπλάσιον τÕ ΑΒ τοà Ε κሠτÕ Γ∆ τοà Ζ· ‡σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΚΘ τοà Ζ κሠτÕ Γ∆ τοà Ζ. ™πεˆ οâν ˜κάτερον τîν ΚΘ, Γ∆ τοà Ζ „σάκις ™στˆ πολλαπλάσιον, ‡σον ¥ρα ™στˆ τÕ ΚΘ τù Γ∆. κοινÕν ¢φVρήσθω τÕ ΓΘ· λοιπÕν ¥ρα τÕ ΚΓ λοιπù τù Θ∆ ‡σον ™στίν. ¢λλ¦ τÕ Ζ τù ΚΓ ™στιν ‡σον· κሠτÕ Θ∆ ¥ρα τù Ζ ‡σον ™στίν. éστε ε„ τÕ ΗΒ τù Ε ‡σον ™στίν, κሠτÕ Θ∆ ‡σον œσται τù Ζ. `Οµοίως δ¾ δείξοµεν, Óτι, κ³ν πολλαπλάσιον Ï τÕ ΗΒ τοà Ε, τοσαυταπλάσιον œσται κሠτÕ Θ∆ τοà Ζ. 'Ε¦ν ¥ρα δύο µεγέθη δύο µεγεθîν „σάκις Ï πολλαπλάσια, κሠ¢φαιρεθέντα τιν¦ τîν αÙτîν „σάκις Ï πολλαπλάσια, κሠτ¦ λοιπ¦ το‹ς αÙτο‹ς ½τοι ‡σα ™στˆν À „σάκις αÙτîν πολλαπλάσια· Óπερ œδει δε‹ξαι.

†

G

For let GB be, first of all, equal to E. I say that HD is also equal to F . For let CK be made equal to F . Since AG and CH are equal multiples of E and F (respectively), and GB (is) equal to E, and KC to F , AB and KH are thus equal multiples of E and F (respectively) [Prop. 5.2]. And AB and CD are assumed (to be) equal multiples of E and F (respectively). Thus, KH and CD are equal multiples of F and F (respectively). Therefore, KH and CD are each equal multiples of F . Thus, KH is equal to CD. Let CH have be taken away from both. Thus, the remainder KC is equal to the remainder HD. But, F is equal to KC. Thus, HD is also equal to F . Hence, if GB is equal to E then HD will also be equal to F . So, similarly, we can show that even if GB is a multiple of E then HD will be the same multiple of F . Thus, if two magnitudes are equal multiples of two (other) magnitudes, and some (parts) taken away (from the former magnitudes) are equal multiples of the latter (magnitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples of them (respectively). (Which is) the very thing it was required to show.

In modern notation, this proposition reads m α − n α = (m − n) α.

ζ΄.

Proposition 7†

Τ¦ ‡σα πρÕς τÕ αÙτÕ τÕν αÙτÕν œχει λόγον κሠτÕ Equal (magnitudes) have the same ratio to the same αÙτÕ πρÕς τ¦ ‡σα. (magnitude), and the latter (magnitude has the same ra”Εστω ‡σα µεγέθη τ¦ Α, Β, ¥λλο δέ τι, Ö œτυχεν, tio) to the equal (magnitudes). µέγεθος τÕ Γ· λέγω, Óτι ˜κάτερον τîν Α, Β πρÕς τÕ Γ Let A and B be equal magnitudes, and C some other τÕν αÙτÕν œχει λόγον, κሠτÕ Γ πρÕς ˜κάτερον τîν Α, random magnitude. I say that A and B each have the

137

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ELEMENTS BOOK 5

Β.

same ratio to C, and (that) C (has the same ratio) to each of A and B.

Α Β Γ

∆ Ε Ζ

A B C

D E F

Ε„λήφθω γ¦ρ τîν µν Α, Β „σάκις πολλαπλάσια τ¦ ∆, Ε, τοà δ Γ ¥λλο, Ö œτυχεν, πολλαπλάσιον τÕ Ζ. 'Επεˆ οâν „σάκις ™στˆ πολλαπλάσιον τÕ ∆ τοà Α κሠτÕ Ε τοà Β, ‡σον δ τÕ Α τù Β, ‡σον ¥ρα κሠτÕ ∆ τù Ε. ¥λλο δέ, Ó œτυχεν, τÕ Ζ. Ε„ ¥ρα Øπερέχει τÕ ∆ τοà Ζ, Øπερέχει κሠτÕ Ε τοà Ζ, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον. καί ™στι τ¦ µν ∆, Ε τîν Α, Β „σάκις πολλαπλάσια, τÕ δ Ζ τοà Γ ¥λλο, Ö œτυχεν, πολλαπλάσιον· œστιν ¥ρα æς τÕ Α πρÕς τÕ Γ, οÛτως τÕ Β πρÕς τÕ Γ. Λέγω [δή], Óτι κሠτÕ Γ πρÕς ˜κάτερον τîν Α, Β τÕν αÙτÕν œχει λόγον. Τîν γ¦ρ αÙτîν κατασκευασθέντων еοίως δείξοµεν, Óτι ‡σον ™στˆ τÕ ∆ τù Ε· ¥λλο δέ τι τÕ Ζ· ε„ ¥ρα Øπερέχει τÕ Ζ τοà ∆, Øπερέχει κሠτοà Ε, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον. καί ™στι τÕ µν Ζ τοà Γ πολλαπλάσιον, τ¦ δ ∆, Ε τîν Α, Β ¥λλα, § œτυχεν, „σάκις πολλαπλάσια· œστιν ¥ρα æς τÕ Γ πρÕς τÕ Α, οÛτως τÕ Γ πρÕς τÕ Β. Τ¦ ‡σα ¥ρα πρÕς τÕ αÙτÕ τÕν αÙτÕν œχει λόγον κሠτÕ αÙτÕ πρÕς τ¦ ‡σα.

For let the equal multiples D and E have been taken of A and B (respectively), and the other random multiple F of C. Therefore, since D and E are equal multiples of A and B (respectively), and A (is) equal to B, D (is) thus also equal to E. And F (is) different, at random. Thus, if D exceeds F then E also exceeds F , and if (D is) equal (to F then E is also) equal (to F ), and if (D is) less (than F then E is also) less (than F ). And D and E are equal multiples of A and B (respectively), and F another random multiple of C. Thus, as A (is) to C, so B (is) to C [Def. 5.5]. [So] I say that C † also has the same ratio to each of A and B. For, similarly, we can show, by the same construction, that D is equal to E. And F (has) some other (value). Thus, if F exceeds D then it also exceeds E, and if (F is) equal (to D then it is also) equal (to E), and if (F is) less (than D then it is also) less (than E). And F is a multiple of C, and D and E other random equal multiples of A and B. Thus, as C (is) to A, so C (is) to B [Def. 5.5]. Thus, equal (magnitudes) have the same ratio to the same (magnitude), and the latter (magnitude has the same ratio) to the equal (magnitudes).

Πόρισµα.

Corollary‡

'Εκ δ¾ τούτου φανερόν, Óτι ™¦ν µεγέθη τιν¦ ¢νάλογον Ï, κሠ¢νάπαλιν ¢νάλογον œσται. Óπερ œδει δε‹ξαι.

So (it is) clear, from this, that if some magnitudes are proportional then they will also be proportional inversely. (Which is) the very thing it was required to show.

†

The Greek text has “E,” which is obviously a mistake.

‡

In modern notation, this corollary reads that if α : β :: γ : δ then β : α :: δ : γ.

η΄.

Proposition 8

Τîν ¢νίσων µεγεθîν τÕ µε‹ζον πρÕς τÕ αÙτÕ µείζονα λόγον œχει ½περ τÕ œλαττον. κሠτÕ αÙτÕ πρÕς τÕ œλαττον µείζονα λόγον œχει ½περ πρÕς τÕ µε‹ζον. ”Εστω ¥νισα µεγέθη τ¦ ΑΒ, Γ, κሠœστω µε‹ζον τÕ ΑΒ, ¥λλο δέ, Ö œτυχεν, τÕ ∆· λέγω, Óτι τÕ ΑΒ πρÕς τÕ ∆ µείζονα λόγον œχει ½περ τÕ Γ πρÕς τÕ ∆, κሠτÕ ∆ πρÕς τÕ Γ µείζονα λόγον œχει ½περ πρÕς τÕ ΑΒ.

For unequal magnitudes, the greater (magnitude) has a greater ratio than the lesser to the same (magnitude). And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater. Let AB and C be unequal magnitudes, and let AB be the greater (of the two), and D another random magnitude. I say that AB has a greater ratio to D than C (has) to D, and (that) D has a greater ratio to C than (it has) to AB.

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ΣΤΟΙΧΕΙΩΝ ε΄.

A E G K D L M N

Z

ELEMENTS BOOK 5

B H

A J

G K D L M N

E B

A E

B

A C

C

Z

H

J

F K D L M N

'Επεˆ γ¦ρ µε‹ζόν ™στι τÕ ΑΒ τοà Γ, κείσθω τù Γ ‡σον τÕ ΒΕ· τÕ δ¾ œλασσον τîν ΑΕ, ΕΒ πολλαπλασιαζόµενον œσται ποτ τοà ∆ µε‹ζον. œστω πρότερον τÕ ΑΕ œλαττον τοà ΕΒ, κሠπεπολλαπλασιάσθω τÕ ΑΕ, κሠœστω αÙτοà πολλαπλάσιον τÕ ΖΗ µε‹ζον ×ν τοà ∆, κሠÐσαπλάσιόν ™στι τÕ ΖΗ τοà ΑΕ, τοσαυταπλάσιον γεγονέτω κሠτÕ µν ΗΘ τοà ΕΒ τÕ δ Κ τοà Γ· κሠε„λήφθω τοà ∆ διπλάσιον µν τÕ Λ, τριπλάσιον δ τÕ Μ, κሠ˜ξÁς ˜νˆ πλε‹ον, ›ως ¨ν τÕ λαµβανόµενον πολλαπλάσιον µν γένηται τοà ∆, πρώτως δ µε‹ζον τοà Κ. ε„λήφθω, κሠœστω τÕ Ν τετραπλάσιον µν τοà ∆, πρώτως δ µε‹ζον τοà Κ. 'Επεˆ οâν τÕ Κ τοà Ν πρώτως ™στˆν œλαττον, τÕ Κ ¥ρα τοà Μ οÜκ ™στιν œλαττον. κሠ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ ΖΗ τοà ΑΕ κሠτÕ ΗΘ τοà ΕΒ, „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΖΗ τοà ΑΕ κሠτÕ ΖΘ τοà ΑΒ. „σάκις δέ ™στι πολλαπλάσιον τÕ ΖΗ τοà ΑΕ κሠτÕ Κ τοà Γ· „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΖΘ τοà ΑΒ κሠτÕ Κ τοà Γ. τ¦ ΖΘ, Κ ¥ρα τîν ΑΒ, Γ „σάκις ™στˆ πολλαπλάσια. πάλιν, ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ ΗΘ τοà ΕΒ κሠτÕ Κ τοà Γ, ‡σον δ τÕ ΕΒ τù Γ, ‡σον ¥ρα κሠτÕ ΗΘ τù Κ. τÕ δ Κ τοà Μ οÜκ ™στιν œλαττον· οÙδ' ¥ρα τÕ ΗΘ τοà Μ œλαττόν ™στιν. µε‹ζον δ τÕ ΖΗ τοà ∆· Óλον ¥ρα τÕ ΖΘ συναµφοτέρων τîν ∆, Μ µε‹ζόν ™στιν. ¢λλ¦ συναµφότερα τ¦ ∆, Μ τù Ν ™στιν ‡σα, ™πειδήπερ τÕ Μ τοà ∆ τριπλάσιόν ™στιν, συναµφότερα δ τ¦ Μ, ∆ τοà ∆ ™στι τετραπλάσια, œστι δ κሠτÕ Ν τοà ∆ τετραπλάσιον· συναµφότερα ¥ρα τ¦ Μ, ∆ τù Ν ‡σα ™στίν. ¢λλ¦ τÕ ΖΘ τîν Μ, ∆ µε‹ζόν ™στιν· τÕ ΖΘ ¥ρα τοà Ν Øπερέχει· τÕ δ Κ τοà Ν οÙχ Øπερέχει. καί ™στι τ¦ µν ΖΘ, Κ τîν ΑΒ, Γ „σάκις πολλαπλάσια, τÕ δ Ν τοà ∆ ¥λλο, Ö œτυχεν, πολλαπλάσιον· τÕ ΑΒ ¥ρα πρÕς τÕ ∆ µείζονα λόγον œχει ½περ τÕ Γ πρÕς τÕ ∆. Λέγω δή, Óτι κሠτÕ ∆ πρÕς τÕ Γ µείζονα λόγον œχει ½περ τÕ ∆ πρÕς τÕ ΑΒ. Τîν γ¦ρ αÙτîν κατασκευασθέντων еοίως δείξοµεν, Óτι τÕ µν Ν τοà Κ Øπερέχει, τÕ δ Ν τοà ΖΘ οÙχ Øπερέχει. καί ™στι τÕ µν Ν τοà ∆ πολλαπλάσιον, τ¦ δ ΖΘ, Κ τîν ΑΒ, Γ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια· τÕ ∆ ¥ρα πρÕς τÕ Γ µείζονα λόγον œχει ½περ τÕ ∆ πρÕς

E B

G

H

F

G

H

K D L M N

For since AB is greater than C, let BE be made equal to C. So, the lesser of AE and EB, being multiplied, will sometimes be greater than D [Def. 5.4]. First of all, let AE be less than EB, and let AE have been multiplied, and let F G be a multiple of it which (is) greater than D. And as many times as F G is (divisible) by AE, so many times let GH also have become (divisible) by EB, and K by C. And let the double multiple L of D have been taken, and the triple multiple M , and several more, (each increasing) in order by one, until the (multiple) taken becomes the first multiple of D (which is) greater than K. Let it have been taken, and let it also be the quadruple multiple N of D—the first (multiple) greater than K. Therefore, since K is less than N first, K is thus not less than M . And since F G and GH are equal multiples of AE and EB (respectively), F G and F H are thus equal multiples of AE and AB (respectively) [Prop. 5.1]. And F G and K are equal multiples of AE and C (respectively). Thus, F H and K are equal multiples of AB and C (respectively). Thus, F H, K are equal multiples of AB, C. Again, since GH and K are equal multiples of EB and C, and EB (is) equal to C, GH (is) thus also equal to K. And K is not less than M . Thus, GH not less than M either. And F G (is) greater than D. Thus, the whole of F H is greater than D and M (added) together. But, D and M (added) together is equal to N , inasmuch as M is three times D, and M and D (added) together is four times D, and N is also four times D. Thus, M and D (added) together is equal to N . But, F H is greater than M and D. Thus, F H exceeds N . And K does not exceed N . And F H, K are equal multiples of AB, C, and N another random multiple of D. Thus, AB has a greater ratio to D than C (has) to D [Def. 5.7]. So, I say that D also has a greater ratio to C than D (has) to AB. For, similarly, by the same construction, we can show that N exceeds K, and N does not exceed F H. And N is a multiple of D, and F H, K other random equal multiples of AB, C (respectively). Thus, D has a greater

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τÕ ΑΒ. 'Αλλ¦ δ¾ τÕ ΑΕ τοà ΕΒ µε‹ζον œστω. τÕ δ¾ œλαττον τÕ ΕΒ πολλαπλασιαζόµενον œσται ποτ τοà ∆ µε‹ζον. πεπολλαπλασιάσθω, κሠœστω τÕ ΗΘ πολλαπλάσιον µν τοà ΕΒ, µε‹ζον δ τοà ∆· κሠÐσαπλασιόν ™στι τÕ ΗΘ τοà ΕΒ, τοσαυταπλάσιον γεγονέτω κሠτÕ µν ΖΗ τοà ΑΕ, τÕ δ Κ τοà Γ. еοίως δ¾ δείξοµεν, Óτι τ¦ ΖΘ, Κ τîν ΑΒ, Γ „σάκις ™στˆ πολλαπλάσια· κሠε„λήφθω еοίως τÕ Ν πολλαπλάσιον µν τοà ∆, πρώτως δ µε‹ζον τοà ΖΗ· éστε πάλιν τÕ ΖΗ τοà Μ οÜκ ™στιν œλασσον. µε‹ζον δ τÕ ΗΘ τοà ∆· Óλον ¥ρα τÕ ΖΘ τîν ∆, Μ, τουτέστι τοà Ν, Øπερέχει. τÕ δ Κ τοà Ν οÙχ Øπερέχει, ™πειδήπερ κሠτÕ ΖΗ µε‹ζον ×ν τοà ΗΘ, τουτέστι τοà Κ, τοà Ν οÙχ Øπερέχει. κሠæσαύτως κατακολουθοàντες το‹ς ™πάνω περαίνοµεν τ¾ν ¢πόδειξιν. Τîν ¥ρα ¢νίσων µεγεθîν τÕ µε‹ζον πρÕς τÕ αÙτÕ µείζονα λόγον œχει ½περ τÕ œλαττον· κሠτÕ αÙτÕ πρÕς τÕ œλαττον µείζονα λόγον œχει ½περ πρÕς τÕ µε‹ζον· Óπερ œδει δε‹ξαι.

ratio to C than D (has) to AB [Def. 5.5]. And so let AE be greater than EB. So, the lesser, EB, being multiplied, will sometimes be greater than D. Let it have been multiplied, and let GH be a multiple of EB (which is) greater than D. And as many times as GH is (divisible) by EB, so many times let F G also have become (divisible) by AE, and K by C. So, similarly (to the above), we can show that F H and K are equal multiples of AB and C (respectively). And, similarly (to the above), let the multiple N of D, (which is) the first (multiple) greater than F G, have been taken. So, F G is again not less than M . And GH (is) greater than D. Thus, the whole of F H exceeds D and M , that is to say N . And K does not exceed N , inasmuch as F G, which (is) greater than GH—that is to say, K—also does not exceed N . And, following the above (arguments), we (can) complete the proof in the same manner. Thus, for unequal magnitudes, the greater (magnitude) has a greater ratio than the lesser to the same (magnitude). And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater. (Which is) the very thing it was required to show.

θ΄.

Proposition 9

Τ¦ πρÕς τÕ αÙτÕ τÕν αÙτÕν œχοντα λÕγον ‡σα (Magnitudes) having the same ratio to the same ¢λλήλοις ™στίν· κሠπρÕς § τÕ αÙτÕ τÕν αÙτÕν ›χει (magnitude) are equal to one another. And those (magλόγον, ™κε‹να ‡σα ™στίν. nitudes) to which the same (magnitude) has the same ratio are equal.

Α

Β

A

Γ

B C

'Εχέτω γ¦ρ ˜κάτερον τîν Α, Β πρÕς τÕ Γ τÕν αÙτÕν λόγον· λέγω, Óτι ‡σον ™στˆ τÕ Α τù Β. Ε„ γ¦ρ µή, οÙκ ¨ν ˜κάτερον τîν Α, Β πρÕς τÕ Γ τÕν αÙτÕν εχε λόγον· œχει δέ· ‡σον ¥ρα ™στˆ τÕ Α τù Β. 'Εχέτω δ¾ πάλιν τÕ Γ πρÕς ˜κάτερον τîν Α, Β τÕν αÙτÕν λόγον· λέγω, Óτι ‡σον ™στˆ τÕ Α τù Β. Ε„ γ¦ρ µή, οÙκ ¨ν τÕ Γ πρÕς ˜κάτερον τîν Α, Β τÕν αÙτÕν εχε λόγον· œχει δέ· ‡σον ¥ρα ™στˆ τÕ Α τù Β. Τ¦ ¥ρα πρÕς τÕ αÙτÕ τÕν αÙτÕν œχοντα λόγον ‡σα ¢λλήλοις ™στίν· κሠπρÕς § τÕ αÙτÕ τÕν αÙτÕν œχει λόγον, ™κε‹να ‡σα ™στίν· Óπερ œδει δε‹ξαι.

For let A and B each have the same ratio to C. I say that A is equal to B. For if not, A and B would not each have the same ratio to C [Prop. 5.8]. But they do. Thus, A is equal to B. So, again, let C have the same ratio to each of A and B. I say that A is equal to B. For if not, C would not have the same ratio to each of A and B [Prop. 5.8]. But it does. Thus, A is equal to B. Thus, (magnitudes) having the same ratio to the same (magnitude) are equal to one another. And those (magnitudes) to which the same (magnitude) has the same ratio are equal. (Which is) the very thing it was required to show.

ι΄.

Proposition 10

Τîν πρÕς τÕ αÙτÕ λόγον ™χόντων τÕ µείζονα λόγον œχον ™κε‹νο µε‹ζόν ™στιν· πρÕς Ö δ τÕ αÙτÕ µείζονα

For (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is

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ELEMENTS BOOK 5

λόγον œχει, ™κε‹νο œλαττόν ™στιν.

Α

(the) greater. And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser.

Β

A

Γ

B C

'Εχέτω γ¦ρ τÕ Α πρÕς τÕ Γ µείζονα λόγον ½περ τÕ Β πρÕς τÕ Γ· λέγω, Óτι µε‹ζόν ™στι τÕ Α τοà Β. Ε„ γ¦ρ µή, ½τοι ‡σον ™στˆ τÕ Α τù Β À œλασσον. ‡σον µν οâν οÜκ ™στˆ τÕ Α τù Β· ˜κάτερον γ¦ρ ¨ν τîν Α, Β πρÕς τÕ Γ τÕν αÙτÕν εχε λόγον. οÙκ œχει δέ· οÙκ ¥ρα ‡σον ™στˆ τÕ Α τù Β. οÙδ µ¾ν œλασσόν ™στι τÕ Α τοà Β· τÕ Α γ¦ρ ¨ν πρÕς τÕ Γ ™λάσσονα λόγον εχεν ½περ τÕ Β πρÕς τÕ Γ. οÙκ œχει δέ· οÙκ ¥ρα œλασσόν ™στι τÕ Α τοà Β. ™δείχθη δ οÙδ ‡σον· µε‹ζον ¥ρα ™στˆ τÕ Α τοà Β. 'Εχέτω δ¾ πάλιν τÕ Γ πρÕς τÕ Β µείζονα λόγον ½περ τÕ Γ πρÕς τÕ Α· λέγω, Óτι œλασσόν ™στι τÕ Β τοà Α. Ε„ γ¦ρ µή, ½τοι ‡σον ™στˆν À µε‹ζον. ‡σον µν οâν οÜκ ™στι τÕ Β τù Α· τÕ Γ γ¦ρ ¨ν πρÕς ˜κάτερον τîν Α, Β τÕν αÙτÕν εχε λόγον. οÙκ œχει δέ· οÙκ ¥ρα ‡σον ™στˆ τÕ Α τù Β. οÙδ µ¾ν µε‹ζόν ™στι τÕ Β τοà Α· τÕ Γ γ¦ρ ¨ν πρÕς τÕ Β ™λάσσονα λόγον εχεν ½περ πρÕς τÕ Α. οÙκ œχει δέ· οÙκ ¥ρα µε‹ζον ™στι τÕ Β τοà Α. ™δείχθη δέ, Óτι οÙδ ‡σον· œλαττον ¥ρα ™στˆ τÕ Β τοà Α. Τîν ¥ρα πρÕς τÕ αÙτÕ λόγον ™χόντων τÕ µείζονα λόγον œχον µε‹ζόν ™στιν· κሠπρÕς Ö τÕ αÙτÕ µείζονα λόγον œχει, ™κε‹νο œλαττόν ™στιν· Óπερ œδει δε‹ξαι.

For let A have a greater ratio to C than B (has) to C. I say that A is greater than B. For if not, A is surely either equal to or less than B. In fact, A is not equal to B. For (then) A and B would each have the same ratio to C [Prop. 5.7]. But they do not. Thus, A is not equal to B. Neither, indeed, is A less than B. For (then) A would have a lesser ratio to C than B (has) to C [Prop. 5.8]. But it does not. Thus, A is not less than B. And it was shown not (to be) equal either. Thus, A is greater than B. So, again, let C have a greater ratio to B than C (has) to A. I say that B is less than A. For if not, (it is) surely either equal or greater. In fact, B is not equal to A. For (then) C would have the same ratio to each of A and B [Prop. 5.7]. But it does not. Thus, A is not equal to B. Neither, indeed, is B greater than A. For (then) C would have a lesser ratio to B than (it has) to A [Prop. 5.8]. But it does not. Thus, B is not greater than A. And it was shown that (it is) not equal (to A) either. Thus, B is less than A. Thus, for (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is (the) greater. And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser. (Which is) the very thing it was required to show.

ια΄.

Proposition 11†

Οƒ τù αÙτù λόγJ οƒ αÙτοˆ κሠ¢λλήλοις ε„σˆν οƒ αÙτοί.

Γ Α Ε ∆ Ζ Β Θ Κ Η Λ Μ Ν ”Εστωσαν γ¦ρ æς µν τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆, æς δ τÕ Γ πρÕς τÕ ∆, οÛτως τÕ Ε πρÕς τÕ Ζ· λέγω, Óτι ™στˆν æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Ε πρÕς τÕ Ζ. Ε„λήφθω γ¦ρ τîν Α, Γ, Ε „σάκις πολλαπλάσια τ¦ Η, Θ, Κ, τîν δ Β, ∆, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Λ, Μ, Ν. Κሠ™πεί ™στιν æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆, κሠε‡ληπται τîν µν Α, Γ „σάκις πολλαπλάσια τ¦ Η, Θ, τîν δ Β, ∆ ¢λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Λ, Μ, ε„ ¥ρα Øπερέχει τÕ Η τοà Λ, Øπερέχει κሠτÕ

(Ratios which are) the same with the same ratio are also the same with one another.

A B G L

C D H M

E F K N

For let it be that as A (is) to B, so C (is) to D, and as C (is) to D, so E (is) to F . I say that as A is to B, so E (is) to F . For let the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M , N of B, D, F (respectively). And since as A is to B, so C (is) to D, and the equal multiples G and H have been taken of A and C (respectively), and the other random equal multiples L and M of B and D (respectively), thus if G exceeds L then H also exceeds M , and if (G is) equal (to L then H is also)

141

ΣΤΟΙΧΕΙΩΝ ε΄.

ELEMENTS BOOK 5

Θ τοà Μ, καˆ ε„ ‡σον ™στίν, ‡σον, καˆ ε„ ™λλείπει, ™λλείπει. πάλιν, ™πεί ™στιν æς τÕ Γ πρÕς τÕ ∆, οÛτως τÕ Ε πρÕς τÕ Ζ, κሠε‡ληπται τîν Γ, Ε „σάκις πολλαπλάσια τ¦ Θ, Κ, τîν δ ∆, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Μ, Ν, ε„ ¥ρα Øπερέχει τÕ Θ τοà Μ, Øπερέχει κሠτÕ Κ τοà Ν, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλλατον, œλαττον. ¢λλ¦ ε„ Øπερε‹χε τÕ Θ τοà Μ, Øπερε‹χε κሠτÕ Η τοà Λ, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον· éστε καˆ ε„ Øπερέχει τÕ Η τοà Λ, Øπερέχει κሠτÕ Κ τοà Ν, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον. καί ™στι τ¦ µν Η, Κ τîν Α, Ε „σάκις πολλαπλάσια, τ¦ δ Λ, Ν τîν Β, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια· œστιν ¥ρα æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Ε πρÕς τÕ Ζ. Οƒ ¥ρα τù αÙτù λόγJ οƒ αÙτοˆ κሠ¢λλήλοις ε„σˆν οƒ αÙτοί· Óπερ œδει δε‹ξαι.

†

equal (to M ), and if (G is) less (than L then H is also) less (than M ) [Def. 5.5]. Again, since as C is to D, so E (is) to F , and the equal multiples H and K have been taken of C and E (respectively), and the other random equal multiples M and N of D and F (respectively), thus if H exceeds M then K also exceeds N , and if (H is) equal (to M then K is also) equal (to N ), and if (H is) less (than M then K is also) less (than N ) [Def. 5.5]. But if H was exceeding M then G was also exceeding L, and if (H was) equal (to M then G was also) equal (to L), and if (H was) less (than M then G was also) less (than L). And, hence, if G exceeds L then K also exceeds N , and if (G is) equal (to L then K is also) equal (to N ), and if (G is) less (than L then K is also) less (than N ). And G and K are equal multiples of A and E (respectively), and L and N other random equal multiples of B and F (respectively). Thus, as A is to B, so E (is) to F [Def. 5.5]. Thus, (ratios which are) the same with the same ratio are also the same with one another. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: γ : δ and γ : δ :: ǫ : ζ then α : β :: ǫ : ζ.

ιβ΄.

Proposition 12†

'Ε¦ν Ï Ðποσαοàν µεγέθη ¢νάλογον, œσται æς žν τîν If there are any number of magnitudes whatsoever ¹γουµένων πρÕς žν τîν ˜ποµένων, οÛτως ¤παντα τ¦ (which are) proportional then as one of the leading (mag¹γούµενα πρÕς ¤παντα τ¦ ˜πόµενα. nitudes is) to one of the following, so will all of the leading (magnitudes) be to all of the following.

Α Β Η Θ Κ

Γ ∆

Ε Ζ

A B

Λ Μ Ν

G H K

”Εστωσαν Ðποσαοàν µεγέθη ¢νάλογον τ¦ Α, Β, Γ, ∆, Ε, Ζ, æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆, κሠτÕ Ε πρÕς το Ζ· λέγω, Óτι ™στˆν æς τÕ Α πρÕς τÕ Β, οÛτως τ¦ Α, Γ, Ε πρÕς τ¦ Β, ∆, Ζ. Ε„λήφθω γ¦ρ τîν µν Α, Γ, Ε „σάκις πολλαπλάσια τ¦ Η, Θ, Κ, τîν δ Β, ∆, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Λ, Μ, Ν. Κሠ™πεί ™στιν æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆, κሠτÕ Ε πρÕς τÕ Ζ, κሠε‡ληπται τîν µν Α, Γ, Ε „σάκις πολλαπλάσια τ¦ Η, Θ, Κ τîν δ Β, ∆, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Λ, Μ, Ν, ε„ ¥ρα Øπερέχει τÕ Η τοà Λ, Øπερέχει κሠτÕ Θ τοà Μ, κሠτÕ Κ τοà Ν, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον. éστε καˆ

C D

E F L M N

Let there be any number of magnitudes whatsoever, A, B, C, D, E, F , (which are) proportional, (so that) as A (is) to B, so C (is) to D, and E to F . I say that as A is to B, so A, C, E (are) to B, D, F . For let the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M , N of B, D, F (respectively). And since as A is to B, so C (is) to D, and E to F , and the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M , N of B, D, F (respectively), thus if G exceeds L then H also exceeds M , and K (exceeds) N , and if (G is) equal (to L then H is also) equal (to M , and K to N ),

142

ΣΤΟΙΧΕΙΩΝ ε΄.

ELEMENTS BOOK 5

ε„ Øπερέχει τÕ Η τοà Λ, Øπερέχει κሠτ¦ Η, Θ, Κ τîν Λ, Μ, Ν, καˆ ε„ ‡σον, ‡σα, καˆ ε„ œλαττον, œλαττονα. καί ™στι τÕ µν Η κሠτ¦ Η, Θ, Κ τοà Α κሠτîν Α, Γ, Ε „σάκις πολλαπλάσια, ™πειδήπερ ™¦ν Ï Ðποσαοàν µεγέθη Ðποσωνοàν µεγεθîν ‡σων τÕ πλÁθος ›καστον ˜κάστου „σάκις πολλαπλάσιον, Ðσαπλάσιόν ™στιν žν τîν µεγεθîν ˜νός, τοσαυταπλάσια œσται κሠτ¦ πάντα τîν πάντων. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ Λ κሠτ¦ Λ, Μ, Ν τοà Β κሠτîν Β, ∆, Ζ „σάκις ™στˆ πολλαπλάσια· œστιν ¥ρα æς τÕ Α πρÕς τÕ Β, οÛτως τ¦ Α, Γ, Ε πρÕς τ¦ Β, ∆, Ζ. 'Ε¦ν ¥ρα Ï Ðποσαοàν µεγέθη ¢νάλογον, œσται æς žν τîν ¹γουµένων πρÕς žν τîν ˜ποµένων, οÛτως ¤παντα τ¦ ¹γούµενα πρÕς ¤παντα τ¦ ˜πόµενα· Óπερ œδει δε‹ξαι.

†

and if (G is) less (than L then H is also) less (than M , and K than N ) [Def. 5.5]. And, hence, if G exceeds L then G, H, K also exceed L, M , N , and if (G is) equal (to L then G, H, K are also) equal (to L, M , N ) and if (G is) less (than L then G, H, K are also) less (than L, M , N ). And G and G, H, K are equal multiples of A and A, C, E (respectively), inasmuch as if there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second) [Prop. 5.1]. So, for the same (reasons), L and L, M , N are also equal multiples of B and B, D, F (respectively). Thus, as A is to B, so A, C, E (are) to B, D, F (respectively). Thus, if there are any number of magnitudes whatsoever (which are) proportional then as one of the leading (magnitudes is) to one of the following, so will all of the leading (magnitudes) be to all of the following. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : α′ :: β : β ′ :: γ : γ ′ etc. then α : α′ :: (α + β + γ + · · · ) : (α′ + β ′ + γ ′ + · · · ).

ιγ΄.

Proposition 13†

'Ε¦ν πρîτον πρÕς δεύτερον τÕν αÙτÕν ŸχV λόγον κሠIf a first (magnitude) has the same ratio to a second τρίτον πρÕς τέταρτον, τρίτον δ πρÕς τέταρτον µείζονα that a third (has) to a fourth, and the third (magnitude) λόγον œχV À πέµπτον πρÕς ›κτον, κሠπρîτον πρÕς has a greater ratio to the fourth than a fifth (has) to a δεύτερον µείζονα λόγον ›ξει À πέµπτον πρÕς ›κτον. sixth, then the first (magnitude) will also have a greater ratio to the second than the fifth (has) to the sixth. Α Β Μ Ν

Γ ∆ Η Κ

Ε Ζ Θ Λ

A B M N

Πρîτον γ¦ρ τÕ Α πρÕς δεύτερον τÕ Β τÕν αÙτÕν ™χέτω λόγον κሠτρίτον τÕ Γ πρÕς τέταρτον τÕ ∆, τρίτον δ τÕ Γ πρÕς τέταρτον τÕ ∆ µείζονα λόγον ™χέτω À πέµπτον τÕ Ε πρÕς ›κτον τÕ Ζ. λέγω, Óτι κሠπρîτον τÕ Α πρÕς δεύτερον τÕ Β µείζονα λόγον ›ξει ½περ πέµπτον τÕ Ε πρÕς ›κτον τÕ Ζ. 'Επεˆ γ¦ρ œστι τιν¦ µν Γ, Ε „σάκις πολλαπλάσια, τîν δ ∆, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια, κሠτÕ µν τοà Γ πολλαπλάσιον τοà τοà ∆ πολλαπλασίου Øπερέχει, τÕ δ τοà Ε πολλαπλάσιον τοà τοà Ζ πολλαπλασίου οÙχ Øπερέχει, ε„λήφθω, κሠœστω τîν µν Γ, Ε „σάκις πολλαπλάσια τ¦ Η, Θ, τîν δ ∆, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Κ, Λ, éστε τÕ µν Η τοà Κ Øπερέχειν, τÕ δ Θ τοà Λ µ¾ Øπερέχειν· κሠÐσαπλάσιον µέν ™στι τÕ Η τοà Γ, τοσαυταπλάσιον œστω κሠτÕ Μ τοà Α, Ðσαπλάσιον δ τÕ Κ τοà ∆, τοσαυταπλάσιον œστω καˆ

C D G K

E F H L

For let a first (magnitude) A have the same ratio to a second B that a third C (has) to a fourth D, and let the third (magnitude) C have a greater ratio to the fourth D than a fifth E (has) to a sixth F . I say that the first (magnitude) A will also have a greater ratio to the second B than the fifth E (has) to the sixth F . For since there are some equal multiples of C and E, and other random equal multiples of D and F , (for which) the multiple of C exceeds the (multiple) of D, and the multiple of E does not exceed the multiple of F [Def. 5.7], let them have been taken. And let G and H be equal multiples of C and E (respectively), and K and L other random equal multiples of D and F (respectively), such that G exceeds K, but H does not exceed L. And as many times as G is (divisible) by C, so many times let M be (divisible) by A. And as many times as K (is divisible)

143

ΣΤΟΙΧΕΙΩΝ ε΄.

ELEMENTS BOOK 5

τÕ Ν τοà Β. Κሠ™πεί ™στιν æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆, κሠε‡ληπται τîν µν Α, Γ „σάκις πολλαπλάσια τ¦ Μ, Η, τîν δ Β, ∆ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Ν, Κ, ε„ ¥ρα Øπερέχει τÕ Μ τοà Ν, Øπερέχει κሠτÕ Η τοà Κ, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλλατον. Øπερέχει δ τÕ Η τοà Κ· Øπερέχει ¥ρα κሠτÕ Μ τοà Ν. τÕ δ Θ τοà Λ οÙχ Øπερέχει· καί ™στι τ¦ µν Μ, Θ τîν Α, Ε „σάκις πολλαπλάσια, τ¦ δ Ν, Λ τîν Β, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια· τÕ ¥ρα Α πρÕς τÕ Β µείζονα λόγον œχει ½περ τÕ Ε πρÕς τÕ Ζ. 'Ε¦ν ¥ρα πρîτον πρÕς δεύτερον τÕν αÙτÕν ŸχV λόγον κሠτρίτον πρÕς τέταρτον, τρίτον δ πρÕς τέταρτον µείζονα λόγον œχV À πέµπτον πρÕς ›κτον, κሠπρîτον πρÕς δεύτερον µείζονα λόγον ›ξει À πέµπτον πρÕς ›κτον· Óπερ œδει δε‹ξαι.

†

by D, so many times let N be (divisible) by B. And since as A is to B, so C (is) to D, and the equal multiples M and G have been taken of A and C (respectively), and the other random equal multiples N and K of B and D (respectively), thus if M exceeds N then G exceeds K, and if (M is) equal (to N then G is also) equal (to K), and if (M is) less (than N then G is also) less (than K) [Def. 5.5]. And G exceeds K. Thus, M also exceeds N . And H does not exceeds L. And M and H are equal multiples of A and E (respectively), and N and L other random equal multiples of B and F (respectively). Thus, A has a greater ratio to B than E (has) to F [Def. 5.7]. Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and a third (magnitude) has a greater ratio to a fourth than a fifth (has) to a sixth, then the first (magnitude) will also have a greater ratio to the second than the fifth (has) to the sixth. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: γ : δ and γ : δ > ǫ : ζ then α : β > ǫ : ζ.

ιδ΄.

Proposition 14†

'Ε¦ν πρîτον πρÕς δεύτερον τÕν αÙτÕν œχV λόγον κሠτρίτον πρÕς τέταρτον, τÕ δ πρîτον τοà τρίτου µε‹ζον Ï, κሠτÕ δεύτερον τοà τετάρτου µε‹ζον œσται, κ¨ν ‡σον, ‡σον, κ¨ν œλαττον, œλαττον.

If a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and the first (magnitude) is greater than the third, then the second will also be greater than the fourth. And if (the first magnitude is) equal (to the third then the second will also be) equal (to the fourth). And if (the first magnitude is) less (than the third then the second will also be) less (than the fourth).

Α Β

Γ ∆

A B

Πρîτον γ¦ρ τÕ Α πρÕς δεύτερον τÕ Β αÙτÕν ™χέτω λόγον κሠτρίτον τÕ Γ προς τέταρτον τÕ ∆, µε‹ζον δ œστω τÕ Α τοà Γ· λέγω, Óτι κሠτÕ Β τοà ∆ µε‹ζόν ™στιν. 'Επεˆ γ¦ρ τÕ Α τοà Γ µε‹ζόν ™στιν, ¥λλο δέ, Ö œτυχεν, [µέγεθος] τÕ Β, τÕ Α ¥ρα πρÕς τÕ Β µείζονα λόγον œχει ½περ τÕ Γ πρÕς τÕ Β. æς δ τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆· κሠτÕ Γ ¥ρα πρÕς τÕ ∆ µείζονα λόγον œχει ½περ τÕ Γ πρÕς τÕ Β. πρÕς Ö δ τÕ αÙτÕ µείζονα λόγον œχει, ™κε‹νο œλασσόν ™στιν· œλασσον ¥ρα τÕ ∆ τοà Β· éστε µε‹ζόν ™στι τÕ Β τοà ∆. `Οµοίως δ¾ δε‹ξοµεν, Óτι κ¨ν ‡σον Ï τÕ Α τù Γ, ‡σον œσται κሠτÕ Β τù ∆, κ¥ν œλασσον Ï τÕ Α τοà Γ, œλασσον œσται κሠτÕ Β τοà ∆. 'Ε¦ν ¥ρα πρîτον πρÕς δεύτερον τÕν αÙτÕν œχV λόγον κሠτρίτον πρÕς τέταρτον, τÕ δ πρîτον τοà τρίτου µε‹ζον Ï, κሠτÕ δεύτερον τοà τετάρτου µε‹ζον œσται, κ¨ν ‡σον, ‡σον, κ¨ν œλαττον, œλαττον· Óπερ œδει

C D

For let a first (magnitude) A have the same ratio to a second B that a third C (has) to a fourth D. And let A be greater than C. I say that B is also greater than D. For since A is greater than C, and B (is) another random [magnitude], A thus has a greater ratio to B than C (has) to B [Prop. 5.8]. And as A (is) to B, so C (is) to D. Thus, C also has a greater ratio to D than C (has) to B. And that (magnitude) to which the same (magnitude) has a greater ratio is the lesser [Prop. 5.10]. Thus, D (is) less than B. Hence, B is greater than D. So, similarly, we can show that even if A is equal to C then B will also be equal to D, and even if A is less than C then B will also be less than D. Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and the first (magnitude) is greater than the third, then the second will also be greater than the fourth. And if (the first magnitude is)

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δε‹ξαι.

†

equal (to the third then the second will also be) equal (to the fourth). And if (the first magnitude is) less (than the third then the second will also be) less (than the fourth). (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: γ : δ then α T γ as β T δ.

ιε΄.

Proposition 15†

Τ¦ µέρη το‹ς æσαύτως πολλαπλασίοις τÕν αÙτÕν œχει Parts have the same ratio as similar multiples, taken λόγον ληφθέντα κατάλληλα. in corresponding order.

Α ∆

Η

Θ

Κ Λ

Β Ε

A

Γ

H

B C

D

Ζ

K

L

E F

”Εστω γ¦ρ „σάκις πολλαπλάσιον τÕ ΑΒ τοà Γ κሠτο ∆Ε τοà Ζ· λέγω, Óτι ™στˆν æς τÕ Γ πρÕς τÕ Ζ, οÛτως τÕ ΑΒ πρÕς τÕ ∆Ε. 'Επεˆ γ¦ρ „σάκις ™στˆ πολλαπλάσιον τÕ ΑΒ τοà Γ κሠτÕ ∆Ε τοà Ζ, Óσα ¥ρα ™στˆν ™ν τù ΑΒ µεγέθη ‡σα τù Γ, τοσαàτα κሠ™ν τù ∆Ε ‡σα τù Ζ. διVρήσθω τÕ µν ΑΒ ε„ς τ¦ τù Γ ‡σα τ¦ ΑΗ, ΗΘ, ΘΒ, τÕ δ ∆Ε ε„ς τ¦ τù Ζ ‡σα τ¦ ∆Κ, ΚΛ, ΛΕ· œσται δ¾ ‡σον τÕ πλÁθος τîν ΑΗ, ΗΘ, ΘΒ, τù πλήθει τîν ∆Κ, ΚΛ, ΛΕ. κሠ™πεˆ ‡σα ™στˆ τ¦ ΑΗ, ΗΘ, ΘΒ ¢λλήλοις, œστι δ κሠτ¦ ∆Κ, ΚΛ, ΛΕ ‡σα ¢λλήλοις, œστιν ¥ρα æς τÕ ΑΗ πρÕς τÕ ∆Κ, οÛτως τÕ ΗΘ πρÕς τÕ ΚΛ, κሠτÕ ΘΒ πρÕς τÕ ΛΕ. œσται ¥ρα κሠæς žν τîν ¹γουµένων πρÕς žν τîν ˜ποµένων, οÛτως ¤παντα τ¦ ¹γουµένα πρÕς ¤παντα τ¦ ˜πόµενα· œστιν ¥ρα æς τÕ ΑΗ πρÕς τÕ ∆Κ, οÛτως τÕ ΑΒ πρÕς τÕ ∆Ε. ‡σον δ τÕ µν ΑΗ τù Γ, τÕ δ ∆Κ τù Ζ· œστιν ¥ρα æς τÕ Γ πρÕς τÕ Ζ οÛτως τÕ ΑΒ πρÕς τÕ ∆Ε. Τ¦ ¥ρα µέρη το‹ς æσαύτως πολλαπλασίοις τÕν αÙτÕν œχει λόγον ληφθέντα κατάλληλα· Óπερ œδει δε‹ξαι.

†

G

For let AB and DE be equal multiples of C and F (respectively). I say that as C is to F , so AB (is) to DE. For since AB and DE are equal multiples of C and F (respectively), thus as many magnitudes as there are in AB equal to C, so many (are there) also in DE equal to F . Let AB have been divided into (magnitudes) AG, GH, HB, equal to C, and DE into (magnitudes) DK, KL, LE, equal to F . So, the number of (magnitudes) AG, GH, HB will equal the number of (magnitudes) DK, KL, LE. And since AG, GH, HB are equal to one another, and DK, KL, LE are also equal to one another, thus as AG is to DK, so GH (is) to KL, and HB to LE [Prop. 5.7]. And, thus (for proportional magnitudes), as one of the leading (magnitudes) will be to one of the following, so all of the leading (magnitudes will be) to all of the following [Prop. 5.12]. Thus, as AG is to DK, so AB (is) to DE. And AG is equal to C, and DK to F . Thus, as C is to F , so AB (is) to DE. Thus, parts have the same ratio as similar multiples, taken in corresponding order. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that α : β :: m α : m β.

ι$΄.

Proposition 16†

'Ε¦ν τέσσαρα µεγέθη ¢νάλογον Ï, κሠ™ναλλ¦ξ ¢νάλογον œσται. ”Εστω τέσσαρα µεγέθη ¢νάλογον τ¦ Α, Β, Γ, ∆, æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆· λέγω, Óτι κሠ™ναλλ¦ξ [¢νάλογον] œσται, æς τÕ Α πρÕς τÕ Γ, οÛτως τÕ Β πρÕς τÕ ∆. Ε„λήφθω γ¦ρ τîν µν Α, Β „σάκις πολλαπλάσια τ¦ Ε, Ζ, τîν δ Γ, ∆ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Η, Θ.

If four magnitudes are proportional then they will also be proportional alternately. Let A, B, C and D be four proportional magnitudes, (such that) as A (is) to B, so C (is) to D. I say that they will also be [proportional] alternately, (so that) as A (is) to C, so B (is) to D. For let the equal multiples E and F have been taken of A and B (respectively), and the other random equal multiples G and H of C and D (respectively).

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Α Β Ε Ζ

Γ ∆ Η Θ

A B E F

Κሠ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ Ε τοà Α κሠτÕ Ζ τοà Β, τ¦ δ µέρη το‹ς æσαύτως πολλαπλασίοις τÕν αÙτÕν œχει λόγον, œστιν ¥ρα æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Ε πρÕς τÕ Ζ. æς δ τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆· κሠæς ¥ρα τÕ Γ πρÕς τÕ ∆, οÛτως τÕ Ε πρÕς τÕ Ζ. πάλιν, ™πεˆ τ¦ Η, Θ τîν Γ, ∆ „σάκις ™στˆ πολλαπλάσια, œστιν ¥ρα æς τÕ Γ πρÕς τÕ ∆, οÛτως τÕ Η πρÕς τÕ Θ. æς δ τÕ Γ πρÕς τÕ ∆, [οÛτως] τÕ Ε πρÕς τÕ Ζ· κሠæς ¥ρα τÕ Ε πρÕς τÕ Ζ, οÛτως τÕ Η πρÕς τÕ Θ. ™¦ν δ τέσσαρα µεγέθη ¢νάλογον Ï, τÕ δ πρîτον τοà τρίτου µε‹ζον Ï, κሠτÕ δεύτερον τοà τετάρτου µε‹ζον œσται, κ¨ν ‡σον, ‡σον, κ¥ν œλαττον, œλαττον. ε„ ¥ρα Øπερέχει τÕ Ε τοà Η, Øπερέχει κሠτÕ Ζ τοà Θ, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον. καί ™στι τ¦ µν Ε, Ζ τîν Α, Β „σάκις πολλαπλάσια, τ¦ δ Η, Θ τîν Γ, ∆ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια· œστιν ¥ρα æς τÕ Α πρÕς τÕ Γ, οÛτως τÕ Β πρÕς τÕ ∆. 'Ε¦ν ¥ρα τέσσαρα µεγέθη ¢νάλογον Ï, κሠ™ναλλ¦ξ ¢νάλογον œσται· Óπερ œδει δε‹ξαι.

†

C D G H

And since E and F are equal multiples of A and B (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as A is to B, so E (is) to F . But as A (is) to B, so C (is) to D. And, thus, as C (is) to D, so E (is) to F [Prop. 5.11]. Again, since G and H are equal multiples of C and D (respectively), thus as C is to D, so G (is) to H [Prop. 5.15]. But as C (is) to D, [so] E (is) to F . And, thus, as E (is) to F , so G (is) to H [Prop. 5.11]. And if four magnitudes are proportional, and the first is greater than the third then the second will also be greater than the fourth, and if (the first is) equal (to the third then the second will also be) equal (to the fourth), and if (the first is) less (than the third then the second will also be) less (than the fourth) [Prop. 5.14]. Thus, if E exceeds G then F also exceeds H, and if (E is) equal (to G then F is also) equal (to H), and if (E is) less (than G then F is also) less (than H). And E and F are equal multiples of A and B (respectively), and G and H other random equal multiples of C and D (respectively). Thus, as A is to C, so B (is) to D [Def. 5.5]. Thus, if four magnitudes are proportional then they will also be proportional alternately. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: γ : δ then α : γ :: β : δ.

ιζ΄.

Proposition 17†

'Ε¦ν συγκείµενα µεγέθη ¢νάλογον Ï, κሠδιαιρεθέντα ¢νάλογον œσται.

If composed magnitudes are proportional then they will also be proportional (when) separarted.

Α

Ε

Η Λ

Μ

Β

Γ

Θ

Κ

Ν

Ζ ∆

A Ξ

E

G

Π

L

”Εστω συγκείµενα µεγέθη ¢νάλογον τ¦ ΑΒ, ΒΕ, Γ∆, ∆Ζ, æς τÕ ΑΒ πρÕς τÕ ΒΕ, οÛτως τÕ Γ∆ πρÕς τÕ ∆Ζ· λέγω, Óτι κሠδιαιρεθέντα ¢νάλογον œσται, æς τÕ ΑΕ πρÕς τÕ ΕΒ, οÛτως τÕ ΓΖ πρÕς τÕ ∆Ζ. Ε„λήφθω γ¦ρ τîν µν ΑΕ, ΕΒ, ΓΖ, Ζ∆ „σάκις πολλαπλάσια τ¦ ΗΘ, ΘΚ, ΛΜ, ΜΝ, τîν δ ΕΒ, Ζ∆ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ ΚΞ, ΝΠ. Κሠ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ ΗΘ τοà ΑΕ κሠτÕ ΘΚ τοà ΕΒ, „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ

M

B

C

H

K

N

F D O P

Let AB, BE, CD, and DF be composed magnitudes (which are) proportional, (so that) as AB (is) to BE, so CD (is) to DF . I say that they will also be proportional (when) separated, (so that) as AE (is) to EB, so CF (is) to DF . For let the equal multiples GH, HK, LM , and M N have been taken of AE, EB, CF , and F D (respectively), and the other random equal multiples KO and N P of EB and F D (respectively).

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ΗΘ τοà ΑΕ κሠτÕ ΗΚ τοà ΑΒ. „σάκις δέ ™στι πολλαπλάσιον τÕ ΗΘ τοà ΑΕ κሠτÕ ΛΜ τοà ΓΖ· „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΗΚ τοà ΑΒ κሠτÕ ΛΜ τοà ΓΖ. πάλιν, ™πεˆ „σάκις ™στˆ πολλαπλάσιον τÕ ΛΜ τοà ΓΖ κሠτÕ ΜΝ τοà Ζ∆, „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΛΜ τοà ΓΖ κሠτÕ ΛΝ τοà Γ∆. „σάκις δ Ãν πολλαπλάσιον τÕ ΛΜ τοà ΓΖ κሠτÕ ΗΚ τοà ΑΒ· „σάκις ¥ρα ™στˆ πολλαπλάσιον τÕ ΗΚ τοà ΑΒ κሠτÕ ΛΝ τοà Γ∆. τ¦ ΗΚ, ΛΝ ¥ρα τîν ΑΒ, Γ∆ „σάκις ™στˆ πολλαπλάσια. πάλιν, ™πεˆ „σάκις ™στˆ πολλαπλασίον τÕ ΘΚ τοà ΕΒ κሠτÕ ΜΝ τοà Ζ∆, œστι δ κሠτÕ ΚΞ τοà ΕΒ „σάκις πολλαπλάσιον κሠτÕ ΝΠ τοà Ζ∆, κሠσυντεθν τÕ ΘΞ τοà ΕΒ „σάκις ™στˆ πολλαπλάσιον κሠτÕ ΜΠ τοà Ζ∆. κሠ™πεί ™στιν æς τÕ ΑΒ πρÕς τÕ ΒΕ, οÛτως τÕ Γ∆ πρÕς τÕ ∆Ζ, κሠε‡ληπται τîν µν ΑΒ, Γ∆ „σάκις πολλαπλάσια τ¦ ΗΚ, ΛΝ, τîν δ ΕΒ, Ζ∆ „σάκις πολλαπλάσια τ¦ ΘΞ, ΜΠ, ε„ ¥ρα Øπερέχει τÕ ΗΚ τοà ΘΞ, Øπερέχει κሠτÕ ΛΝ τοà ΜΠ, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον. Øπερεχέτω δ¾ τÕ ΗΚ τοà ΘΞ, κሠκοινοà ¢φαιρεθέντος τοà ΘΚ Øπερέχει ¥ρα κሠτÕ ΗΘ τοà ΚΞ. ¢λλα ε„ Øπερε‹χε τÕ ΗΚ τοà ΘΞ Øπερε‹χε κሠτÕ ΛΝ τοà ΜΠ· Øπερέχει ¥ρα κሠτÕ ΛΝ τοà ΜΠ, κሠκοινοà ¢φαιρεθέντος τοà ΜΝ Øπερέχει κሠτÕ ΛΜ τοà ΝΠ· éστε ε„ Øπερέχει τÕ ΗΘ τοà ΚΞ, Øπερέχει κሠτÕ ΛΜ τοà ΝΠ. еοίως δ¾ δε‹ξοµεν, Óτι κ¨ν ‡σον Ï τÕ ΗΘ τù ΚΞ, ‡σον œσται κሠτÕ ΛΜ τù ΝΠ, κ¨ν œλαττον, œλαττον. καί ™στι τ¦ µν ΗΘ, ΛΜ τîν ΑΕ, ΓΖ „σάκις πολλαπλάσια, τ¦ δ ΚΞ, ΝΠ τîν ΕΒ, Ζ∆ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια· œστιν ¥ρα æς τÕ ΑΕ πρÕς τÕ ΕΒ, οÛτως τÕ ΓΖ πρÕς τÕ Ζ∆. 'Ε¦ν ¥ρα συγκείµενα µεγέθη ¢νάλογον Ï, κሠδιαιρεθέντα ¢νάλογον œσται· Óπερ œδει δε‹ξαι.

†

And since GH and HK are equal multiples of AE and EB (respectively), GH and GK are thus equal multiples of AE and AB (respectively) [Prop. 5.1]. But GH and LM are equal multiples of AE and CF (respectively). Thus, GK and LM are equal multiples of AB and CF (respectively). Again, since LM and M N are equal multiples of CF and F D (respectively), LM and LN are thus equal multiples of CF and CD (respectively) [Prop. 5.1]. And LM and GK were equal multiples of CF and AB (respectively). Thus, GK and LN are equal multiples of AB and CD (respectively). Thus, GK, LN are equal multiples of AB, CD. Again, since HK and M N are equal multiples of EB and F D (respectively), and KO and N P are also equal multiples of EB and F D (respectively), then, added together, HO and M P are also equal multiples of EB and F D (respectively) [Prop. 5.2]. And since as AB (is) to BE, so CD (is) to DF , and the equal multiples GK, LN have been taken of AB, CD, and the equal multiples HO, M P of EB, F D, thus if GK exceeds HO then LN also exceeds M P , and if (GK is) equal (to HO then LN is also) equal (to M P ), and if (GK is) less (than HO then LN is also) less (than M P ) [Def. 5.5]. So let GK exceed HO, and thus, HK being taken away from both, GH exceeds KO. But if GK was exceeding HO then LN was also exceeding M P . Thus, LN also exceeds M P , and, M N being taken away from both, LM also exceeds N P . Hence, if GH exceeds KO then LM also exceeds N P . So, similarly, we can show that even if GH is equal to KO then LM will also be equal to N P , and even if (GH is) less (than KO then LM will also be) less (than N P ). And GH, LM are equal multiples of AE, CF , and KO, N P other random equal multiples of EB, F D. Thus, as AE is to EB, so CF (is) to F D [Def. 5.5]. Thus, if composed magnitudes are proportional then they will also be proportional (when) separarted. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α + β : β :: γ + δ : δ then α : β :: γ : δ.

ιη΄.

Proposition 18†

'Ε¦ν διVρηµένα µεγέθη ¢νάλογον Ï, κሠσυντεθέντα ¢νάλογον œσται.

If separated magnitudes are proportional then they will also be proportional (when) composed.

Α Γ

Ε

Β Ζ Η

A ∆

C

E

B F G

D

”Εστω διVρηµένα µεγέθη ¢νάλογον τ¦ ΑΕ, ΕΒ, ΓΖ, Let AE, EB, CF , and F D be separated magnitudes Ζ∆, æς τÕ ΑΕ πρÕς τÕ ΕΒ, οÛτως τÕ ΓΖ πρÕς τÕ Ζ∆· (which are) proportional, (so that) as AE (is) to EB, so λέγω, Óτι κሠσυντεθέντα ¢νάλογον œσται, æς τÕ ΑΒ CF (is) to F D. I say that they will also be proportional πρÕς τÕ ΒΕ, οÛτως τÕ Γ∆ πρÕς τÕ Ζ∆. (when) composed, (so that) as AB (is) to BE, so CD (is) 147

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Ε„ γ¦ρ µή ™στˆν æς τÕ ΑΒ πρÕς τÕ ΒΕ, οÛτως τÕ Γ∆ πρÕς τÕ ∆Ζ, œσται æς τÕ ΑΒ πρÕς τÕ ΒΕ, οÛτως τÕ Γ∆ ½τοι πρÕς œλασσόν τι τοà ∆Ζ À πρÕς µε‹ζον. ”Εστω πρότερον πρÕς œλασσον τÕ ∆Η. κሠ™πεί ™στιν æς τÕ ΑΒ πρÕς τÕ ΒΕ, οÛτως τÕ Γ∆ πρÕς τÕ ∆Η, συγκείµενα µεγέθη ¢νάλογόν ™στιν· éστε κሠδιαρεθέντα ¢νάλογον œσται. œστιν ¥ρα æς τÕ ΑΕ πρÕς τÕ ΕΒ, οÛτως τÕ ΓΗ πρÕς τÕ Η∆. Øπόκειται δ κሠæς τÕ ΑΕ πρÕς τÕ ΕΒ, οÛτως τÕ ΓΖ πρÕς τÕ Ζ∆. κሠæς ¥ρα τÕ ΓΗ πρÕς τÕ Η∆, οÛτως τÕ ΓΖ πρÕς τÕ Ζ∆· µε‹ζον δ τÕ πρîτον τÕ ΓΗ τοà τρίτου τοà ΓΖ· µε‹ζον ¥ρα κሠτÕ δεύτερον τÕ Η∆ τοà τετάρτου τοà Ζ∆. ¢λλ¦ κሠœλαττον· Óπερ ™στˆν ¢δύνατον· οÙκ ¥ρα ™στˆν æς τÕ ΑΒ πρÕς τÕ ΒΕ, οÛτως τÕ Γ∆ πρÕς œλασσον τοà Ζ∆. еοίως δ¾ δείξοµεν, Óτι οÙδ πρÕς µε‹ζον· πρÕς αÙτÕ ¥ρα. 'Ε¦ν ¥ρα διVρηµένα µεγέθη ¢νάλογον Ï, κሠσυντεθέντα ¢νάλογον œσται· Óπερ œδει δε‹ξαι.

†

to F D. For if (it is) not (the case that) as AB is to BE, so CD (is) to F D, then it will surely be (the case that) as AB (is) to BE, so CD is either to some (magnitude) less than F D, or (some magnitude) greater (than F D). Let it, first of all, be to (some magnitude) less (than F D), (namely) DG. And since composed magnitudes are proportional, (so that) as AB is to BE, so CD (is) to DG, they will thus also be proportional (when) separated [Prop. 5.17]. Thus, as AE is to EB, so CG (is) to GD. But it was also assumed that as AE (is) to EB, so CF (is) to F D. Thus, (it is) also (the case that) as CG (is) to GD, so CF (is) to F D [Prop. 5.11]. And the first (magnitude) CG (is) greater than the third CF . Thus, the second (magnitude) GD (is) also greater than the fourth F D [Prop. 5.14]. But (it is) also less. The very thing is impossible. Thus, (it is) not (the case that) as AB is to BE, so CD (is) to less than F D. Similarly, we can show that neither (is it the case) to greater (than F D). Thus, (it is the case) to the same (as F D). Thus, if separated magnitudes are proportional then they will also be proportional (when) composed. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: γ : δ then α + β : β :: γ + δ : δ.

ιθ΄.

Proposition 19†

'Ε¦ν Ï æς Óλον πρÕς Óλον, οÛτως ¢φαιρεθν πρÕς If as the whole is to the whole so the (part) taken ¢φαιρεθέν, κሠτÕ λοιπÕν πρÕς τÕ λοιπÕν œσται æς Óλον away is to the (part) taken away then the remainder to πρÕς Óλον. the remainder will also be as the whole (is) to the whole.

Α Γ

Ε

Β Ζ

A ∆

C

”Εστω γ¦ρ æς Óλον τÕ ΑΒ πρÕς Óλον τÕ Γ∆, οÛτως ¢φαιρεθν τÕ ΑΕ πρÕς ¢φειρεθν τÕ ΓΖ· λέγω, Óτι κሠλοιπÕν τÕ ΕΒ πρÕς λοιπÕν τÕ Ζ∆ œσται æς Óλον τÕ ΑΒ πρÕς Óλον τÕ Γ∆. 'Επεˆ γάρ ™στιν æς τÕ ΑΒ πρÕς τÕ Γ∆, οÛτως τÕ ΑΕ πρÕς τÕ ΓΖ, κሠ™ναλλ¦ξ æς τÕ ΒΑ πρÕς τÕ ΑΕ, οÛτως τÕ ∆Γ πρÕς τÕ ΓΖ. κሠ™πεˆ συγκείµενα µεγέθη ¢νάλογόν ™στιν, κሠδιαρεθέντα ¢νάλογον œσται, æς τÕ ΒΕ πρÕς τÕ ΕΑ, οÛτως τÕ ∆Ζ πρÕς τÕ ΓΖ· κሠ™ναλλάξ, æς τÕ ΒΕ πρÕς τÕ ∆Ζ, οÛτως τÕ ΕΑ πρÕς τÕ ΖΓ. æς δ τÕ ΑΕ πρÕς τÕ ΓΖ, οÛτως Øπόκειται Óλον τÕ ΑΒ πρÕς Óλον τÕ Γ∆. κሠλοιπÕν ¥ρα τÕ ΕΒ πρÕς λοιπÕν τÕ Ζ∆ œσται æς Óλον τÕ ΑΒ πρÕς Óλον τÕ Γ∆. 'Ε¦ν ¥ρα Ï æς Óλον πρÕς Óλον, οÛτως ¢φαιρεθν πρÕς ¢φαιρεθέν, κሠτÕ λοιπÕν πρÕς τÕ λοιπÕν œσται æς Óλον πρÕς Óλον [Óπερ œδει δε‹ξαι].

E

B F

D

For let the whole AB be to the whole CD as the (part) taken away AE (is) to the (part) taken away CF . I say that the remainder EB to the remainder F D will also be as the whole AB (is) to the whole CD. For since as AB is to CD, so AE (is) to CF , (it is) also (the case), alternately, (that) as BA (is) to AE, so DC (is) to CF [Prop. 5.16]. And since composed magnitudes are proportional then they will also be proportional (when) separated, (so that) as BE (is) to EA, so DF (is) to CF [Prop. 5.17]. Also, alternately, as BE (is) to DF , so EA (is) to F C [Prop. 5.16]. And it was assumed that as AE (is) to CF , so the whole AB (is) to the whole CD. And, thus, as the remainder EB (is) to the remainder F D, so the whole AB will be to the whole CD. Thus, if as the whole is to the whole so the (part) taken away is to the (part) taken away then the remain-

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[Κሠ™πεˆ ™δείχθη æς τÕ ΑΒ πρÕς τÕ Γ∆, οÛτως τÕ ΕΒ πρÕς τÕ Ζ∆, κሠ™ναλλ¦ξ æς τÕ ΑΒ πρÕς τÕ ΒΕ οÛτως τÕ Γ∆ πρÕς τÕ Ζ∆, συγκείµενα ¥ρα µεγέθη ¢νάλογόν ™στιν· ™δείχθη δ æς τÕ ΒΑ πρÕς τÕ ΑΕ, οÛτως τÕ ∆Γ πρÕς τÕ ΓΖ· καί ™στιν ¢ναστρέψαντι].

der to the remainder will also be as the whole (is) to the whole. [(Which is) the very thing it was required to show.] [And since it was shown (that) as AB (is) to CD, so EB (is) to F D, (it is) also (the case), alternately, (that) as AB (is) to BE, so CD (is) to F D. Thus, composed magnitudes are proportional. And it was shown (that) as BA (is) to AE, so DC (is) to CF . And (the latter) is converted (from the former).]

Πόρισµα.

Corollary‡

'Εκ δ¾ τούτου φανερόν, Óτι ™¦ν συγθείµενα µεγέθη ¢νάλογον Ï, κሠ¢ναστρέψαντι ¢νάλογον œσται· Óπερ œδει δε‹ξαι.

So (it is) clear, from this, that if composed magnitudes are proportional then they will also be proportional (when) converted. (Which is) the very thing it was required to show.

† ‡

In modern notation, this proposition reads that if α : β :: γ : δ then α : β :: α − γ : β − δ. In modern notation, this corollary reads that if α : β :: γ : δ then α : α − β :: γ : γ − δ.

κ΄.

Proposition 20†

'Ε¦ν Ï τρία µεγέθη κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος, If there are three magnitudes, and others of equal σύνδυο λαµβανόµενα κሠ™ν τù αÙτù λόγω, δι' ‡σου δ number to them, (being) also in the same ratio taken two τÕ πρîτον τοà τρίτου µε‹ζον Ï, κሠτÕ τέταρτον τοà by two, and (if), via equality, the first is greater than the ›κτου µε‹ζον œσται, κ¨ν ‡σον, ‡σον, κ¥ν œλαττον, œλαττον. third then the fourth will also be greater than the sixth. And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth). And if (the first is) less (than the third then the fourth will also be) less (than the sixth).

Α Β Γ

∆ Ε Ζ

A B C

”Εστω τρία µεγέθη τ¦ Α, Β, Γ, κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος τ¦ ∆, Ε, Ζ, σύνδυο λαµβανόµενα ™ν τù αÙτù λόγJ, æς µν τÕ Α πρÕς τÕ Β, οÛτως τÕ ∆ πρÕς τÕ Ε, æς δ τÕ Β πρÕς τÕ Γ, οÛτως τÕ Ε πρÕς τÕ Ζ, δι' ‡σου δ µε‹ζον œστω τÕ Α τοà Γ· λέγω, Óτι κሠτÕ ∆ τοà Ζ µε‹ζον œσται, κ¨ν ‡σον, ‡σον, κ¥ν œλαττον, œλαττον. 'Επεˆ γ¦ρ µε‹ζόν ™στι τÕ Α τοà Γ, ¥λλο δέ τι τÕ Β, τÕ δ µε‹ζον πρÕς τÕ αÙτÕ µείζονα λόγον œχει ½περ τÕ œλαττον, τÕ Α ¥ρα πρÕς τÕ Β µείζονα λόγον œχει ½περ τÕ Γ πρÕς τÕ Β. ¢λλ' æς µν τÕ Α πρÕς τÕ Β [οÛτως] τÕ ∆ πρÕς τÕ Ε, æς δ τÕ Γ πρÕς τÕ Β, ¢νάπαλιν οÛτως τÕ Ζ πρÕς τÕ Ε· κሠτÕ ∆ ¥ρα πρÕς τÕ Ε µείζονα λόγον œχει ½περ τÕ Ζ πρÕς τÕ Ε. τîν δ πρÕς τÕ αÙτÕ λόγον ™χόντων τÕ µείζονα λόγον œχον µε‹ζόν ™στιν. µε‹ζον ¥ρα τÕ ∆ τοà Ζ. еοίως δ¾ δείξοµεν, Óτι κ¨ν ‡σον Ï τÕ Α τù Γ, ‡σον œσται κሠτÕ ∆ τù Ζ, κ¨ν œλαττον, œλαττον. 'Ε¦ν ¥ρα Ï τρία µεγέθη κሠ¥λλα αÙτο‹ς ‡σα τÕ

D E F

Let A, B, and C be three magnitudes, and D, E, F other (magnitudes) of equal number to them, (being) in the same ratio taken two by two, (so that) as A (is) to B, so D (is) to E, and as B (is) to C, so E (is) to F . And let A be greater than C, via equality. I say that D will also be greater than F . And if (A is) equal (to C then D will also be) equal (to F ). And if (A is) less (than C then D will also be) less (than F ). For since A is greater than C, and B some other (magnitude), and the greater (magnitude) has a greater ratio than the lesser to the same (magnitude) [Prop. 5.8], A thus has a greater ratio to B than C (has) to B. But as A (is) to B, [so] D (is) to E. And, inversely, as C (is) to B, so F (is) to E [Prop. 5.7 corr.]. Thus, D also has a greater ratio to E than F (has) to E. And for (magnitudes) having a ratio to the same (magnitude), that having the greater ratio is greater [Prop. 5.10]. Thus, D (is)

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πλÁθος, σύνδυο λαµβανόµενα κሠ™ν τù αÙτù λόγω, δι' ‡σου δ τÕ πρîτον τοà τρίτου µε‹ζον Ï, κሠτÕ τέταρτον τοà ›κτου µε‹ζον œσται, κ¨ν ‡σον, ‡σον, κ¥ν œλαττον, œλαττον· Óπερ œδει δε‹ξαι.

†

greater than F . Similarly, we can show, that even if A is equal to C then D will also be equal to F , and even if (A is) less (than C then D will also be) less (than F ). Thus, if there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if), via equality, the first is greater than the third, then the fourth will also be greater than the sixth. And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth). And (if the first is) less (than the third then the fourth will also be) less (than the sixth). (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: δ : ǫ and β : γ :: ǫ : ζ then α T γ as δ T ζ.

κα΄.

Proposition 21†

'Ε¦ν Ï τρία µεγέθη κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος σύνδυο λαµβανόµενα κሠ™ν τù αÙτù λόγJ, Ï δ τεταραγµένη αÙτîν ¹ ¢ναλογία, δι' ‡σου δ τÕ πρîτον τοà τρίτου µε‹ζον Ï, κሠτÕ τέταρτον τοà ›κτου µε‹ζον œσται, κ¨ν ‡σον, ‡σον, κ¨ν œλαττον, œλαττον.

If there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if) their proportion (is) perturbed, and (if), via equality, the first is greater than the third then the fourth will also be greater than the sixth. And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth). And if (the first is) less (than the third then the fourth will also be) less (than the sixth).

Α Β Γ

∆ Ε Ζ

A B C

”Εστω τρία µεγέθη τ¦ Α, Β, Γ κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος τ¦ ∆, Ε, Ζ, σύνδυο λαµβανόµενα κሠ™ν τù αÙτù λόγJ, œστω δ τεταραγµένη αÙτîν ¹ ¢ναλογία, æς µν τÕ Α πρÕς τÕ Β, οÛτως τÕ Ε πρÕς τÕ Ζ, æς δ τÕ Β πρÕς τÕ Γ, οÛτως τÕ ∆ πρÕς τÕ Ε, δι' ‡σου δ τÕ Α τοà Γ µε‹ζον œστω· λέγω, Óτι κሠτÕ ∆ τοà Ζ µε‹ζον œσται, κ¨ν ‡σον, ‡σον, κ¨ν Ÿλαττον, Ÿλαττον. 'Επεˆ γ¦ρ µε‹ζόν ™στι τÕ Α τοà Γ, ¥λλο δέ τι τÕ Β, τÕ Α ¥ρα πρÕς τÕ Β µείζονα λόγον œχει ½περ τÕ Γ πρÕς τÕ Β. ¢λλ' æς µν τÕ Α πρÕς τÕ Β, οÛτως τÕ Ε πρÕς τÕ Ζ, æς δ τÕ Γ πρÕς τÕ Β, ¢νάπαλιν οÛτως τÕ Ε πρÕς τÕ ∆. κሠτÕ Ε ¥ρα πρÕς τÕ Ζ µείζονα λόγον œχει ½περ τÕ Ε πρÕς τÕ ∆. πρÕς Ö δ τÕ αÙτÕ µείζονα λόγον œχει, ™κε‹νο œλασσόν ™στιν· œλασσον ¥ρα ™στˆ τÕ Ζ τοà ∆· µε‹ζον ¥ρα ™στˆ τÕ ∆ τοà Ζ. еοίως δ¾ δείξοµεν, Óτι κ¨ν ‡σον Ï τÕ Α τù Γ, ‡σον œσται κሠτÕ ∆ τù Ζ, κ¨ν œλαττον, œλαττον. 'Ε¦ν ¥ρα Ï τρία µεγέθη κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος, σύνδυο λαµβανόµενα κሠ™ν τù αÙτù λόγJ, Ï δ τεταραγµένη αÙτîν ¹ ¢ναλογία, δι' ‡σου δ τÕ πρîτον τοà τρίτου µε‹ζον Ï, κሠτÕ τέταρτον τοà ›κτου µε‹ζον œσται, κ¨ν ‡σον, ‡σον, κ¨ν œλαττον, œλαττον· Óπερ œδει

D E F

Let A, B, and C be three magnitudes, and D, E, F other (magnitudes) of equal number to them, (being) in the same ratio taken two by two. And let their proportion be perturbed, (so that) as A (is) to B, so E (is) to F , and as B (is) to C, so D (is) to E. And let A be greater than C, via equality. I say that D will also be greater than F . And if (A is) equal (to C then D will also be) equal (to F ). And if (A is) less (than C then D will also be) less (than F ). For since A is greater than C, and B some other (magnitude), A thus has a greater ratio to B than C (has) to B [Prop. 5.8]. But as A (is) to B, so E (is) to F . And, inversely, as C (is) to B, so E (is) to D [Prop. 5.7 corr.]. Thus, E also has a greater ratio to F than E (has) to D. And that (magnitude) to which the same (magnitude) has a greater ratio is (the) lesser (magnitude) [Prop. 5.10]. Thus, F is less than D. Thus, D is greater than F . Similarly, we can show even if A is equal to C then D will also be equal to F , and even if (A is) less (than C then D will also be) less (than F ). Thus, if there are three magnitudes, and others of equal number to them, (being) also in the same ratio

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δε‹ξαι.

†

taken two by two, and (if) their proportion (is) perturbed, and (if), via equality, the first is greater than the third then the fourth will also be greater than the sixth. And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth). And if (the first is) less (than the third then the fourth will also be) less (than the sixth). (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: ǫ : ζ and β : γ :: δ : ǫ then α T γ as δ T ζ.

κβ΄.

Proposition 22†

'Ε¦ν Ï Ðποσαοàν µεγέθη κሠ¥λλα αÙτο‹ς ‡σα τÕ If there are any number of magnitudes whatsoever, πλÁθος, σύνδυο λαµβανόµενα κሠ™ν τù αÙτù λόγJ, κሠand (some) other (magnitudes) of equal number to them, δι' ‡σου ™ν τù αÙτù λόγJ œσται. (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality. Α ∆ Η Θ

Β Ε Κ Λ

Γ Ζ Μ Ν

A D G H

”Εστω Ðποσαοàν µεγέθη τ¦ Α, Β, Γ κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος τ¦ ∆, Ε, Ζ, σύνδυο λαµβανόµενα ™ν τù αÙτù λόγJ, æς µν τÕ Α πρÕς τÕ Β, οÛτως τÕ ∆ πρÕς τÕ Ε, æς δ τÕ Β πρÕς τÕ Γ, οÛτως τÕ Ε πρÕς τÕ Ζ· λέγω, Óτι κሠδι' ‡σου ™ν τù αÙτJ λόγJ œσται. Ε„λήφθω γ¦ρ τîν µν Α, ∆ „σάκις πολλαπλάσια τ¦ Η, Θ, τîν δ Β, Ε ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Κ, Λ, κሠœτι τîν Γ, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Μ, Ν. Κሠ™πεί ™στιν æς το Α πρÕς τÕ Β, οÛτως τÕ ∆ πρÕς το Ε, κሠε‡ληπται τîν µν Α, ∆ „σάκις πολλαπλάσια τ¦ Η, Θ, τîν δ Β, Ε ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Κ, Λ, œστιν ¥ρα æς τÕ Η πρÕς τÕ Κ, οÛτως τÕ Θ πρÕς τÕ Λ. δˆα τ¦ αÙτ¦ δ¾ κሠæς τÕ Κ πρÕς τÕ Μ, οÛτως τÕ Λ πρÕς τÕ Ν. ™πεˆ οâν τρία µεγέθη ™στˆ τ¦ Η, Κ, Μ, κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος τ¦ Θ, Λ, Ν, σύνδυο λαµβανόµενα κሠ™ν τù αÙτù λόγJ, δι' ‡σου ¥ρα, ε„ Øπερέχει τÕ Η τοà Μ, Øπερέχει κሠτÕ Θ τοà Ν, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον. καί ™στι τ¦ µν Η, Θ τîν Α, ∆ „σάκις πολλαπλάσια, τ¦ δ Μ, Ν τîν Γ, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια. œστιν ¥ρα æς τÕ Α πρÕς τÕ Γ, οÛτως τÕ ∆ πρÕς τÕ Ζ. 'Ε¦ν ¥ρα Ï Ðποσαοàν µεγέθη κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος, σύνδυο λαµβανόµενα ™ν τù αÙτù λόγJ, κሠδι' ‡σου ™ν τù αÙτù λόγJ œσται· Óπερ œδει δε‹ξαι.

B E K L

C F M N

Let there be any number of magnitudes whatsoever, A, B, C, and (some) other (magnitudes), D, E, F , of equal number to them, (which are) in the same ratio taken two by two, (so that) as A (is) to B, so D (is) to E, and as B (is) to C, so E (is) to F . I say that they will also be in the same ratio via equality. For let the equal multiples G and H have been taken of A and D (respectively), and the other random equal multiples K and L of B and E (respectively), and the yet other random equal multiples M and N of C and F (respectively). And since as A is to B, so D (is) to E, and the equal multiples G and H have been taken of A and D (respectively), and the other random equal multiples K and L of B and E (respectively), thus as G is to K, so H (is) to L [Prop. 5.4]. And, so, for the same (reasons), as K (is) to M , so L (is) to N . Therefore, since G, K, and M are three magnitudes, and H, L, and N other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, thus, via equality, if G exceeds M then H also exceeds N , and if (G is) equal (to M then H is also) equal (to N ), and if (G is) less (than M then H is also) less (than N ) [Prop. 5.20]. And G and H are equal multiples of A and D (respectively), and M and N other random equal multiples of C and F (respectively). Thus, as A is to C, so D (is) to F [Def. 5.5]. Thus, if there are any number of magnitudes whatsoever, and (some) other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality. (Which is) the very thing it was required to show.

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In modern notation, this proposition reads that if α : β :: ǫ : ζ and β : γ :: ζ : η and γ : δ :: η : θ then α : δ :: ǫ : θ.

κγ΄.

Proposition 23†

'Ε¦ν Ï τρία µεγέθη κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος σύνδυο λαµβανόµενα ™ν τù αÙτù λόγJ, Ï δ τεταραγµένη αÙτîν ¹ ¢ναλογία, κሠδι' ‡σου ™ν τù αÙτù λόγJ œσται. Γ Β Α

If there are three magnitudes, and others of equal number to them, (being) in the same ratio taken two by two, and (if) their proportion is perturbed, then they will also be in the same ratio via equality.

Ζ Ε ∆ Λ Θ Η Μ Κ Ν ”Εστω τρία µεγέθη τ¦ Α, Β, Γ κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος σύνδυο λαµβανόµενα ™ν τù αÙτù λόγJ τ¦ ∆, Ε, Ζ, œστω δ τεταραγµένη αÙτîν ¹ ¢ναλογία, æς µν τÕ Α πρÕς τÕ Β, οÛτως τÕ Ε πρÕς τÕ Ζ, æς δ τÕ Β πρÕς τÕ Γ, οÛτως τÕ ∆ πρÕς τÕ Ε· λέγω, Óτι ™στˆν æς τÕ Α πρÕς τÕ Γ, οÛτως τÕ ∆ πρÕς τÕ Ζ. Ε„λήφθω τîν µν Α, Β, ∆ „σάκις πολλαπλάσια τ¦ Η, Θ, Κ, τîν δ Γ, Ε, Ζ ¥λλα, § œτυχεν, „σάκις πολλαπλάσια τ¦ Λ, Μ, Ν. Κሠ™πεˆ „σάκις ™στˆ πολλαπλάσια τ¦ Η, Θ τîν Α, Β, τ¦ δ µέρη τοˆς æσαύτως πολλαπλασίοις τÕν αÙτÕν œχει λόγον, œστιν ¥ρα æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Η πρÕς τÕ Θ. δι¦ τ¦ αÙτ¦ δ¾ κሠæς τÕ Ε πρÕς τÕ Ζ, οÛτως τÕ Μ πρÕς τÕ Ν· καί ™στιν æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Ε πρÕς τÕ Ζ· κሠæς ¥ρα τÕ Η πρÕς τÕ Θ, οÛτως τÕ Μ πρÕς τÕ Ν. κሠ™πεί ™στιν æς τÕ Β πρÕς τÕ Γ, οÛτως τÕ ∆ πρÕς τÕ Ε, κሠ™ναλλ¦ξ æς τÕ Β πρÕς τÕ ∆, οÛτως τÕ Γ πρÕς τÕ Ε. κሠ™πεˆ τ¦ Θ, Κ τîν Β, ∆ „σάκις ™στˆ πολλαπλάσια, τ¦ δ µέρη το‹ς „σάκις πολλαπλασίοις τÕν αÙτÕν œχει λόγον, œστιν ¥ρα æς τÕ Β πρÕς τÕ ∆, οÛτως τÕ Θ πρÕς τÕ Κ. ¢λλ' æς τÕ Β πρÕς τÕ ∆, οÛτως τÕ Γ πρÕς τÕ Ε· κሠæς ¥ρα τÕ Θ πρÕς τÕ Κ, οÛτως τÕ Γ πρÕς τÕ Ε. πάλιν, ™πεˆ τ¦ Λ, Μ τîν Γ, Ε „σάκις ™στι πολλαπλάσια, œστιν ¥ρα æς τÕ Γ πρÕς τÕ Ε, οÛτως τÕ Λ πρÕς τÕ Μ. ¢λλ' æς τÕ Γ πρÕς τÕ Ε, οÛτως τÕ Θ πρÕς τÕ Κ· κሠæς ¥ρα τÕ Θ πρÕς τÕ Κ, οÛτως τÕ Λ πρÕς τÕ Μ, κሠ™ναλλ¦ξ æς τÕ Θ πρÕς τÕ Λ, τÕ Κ πρÕς τÕ Μ. ™δείχθη δ κሠæς τÕ Η πρÕς τÕ Θ, οÛτως τÕ Μ πρÕς τÕ Ν. ™πεˆ οâν τρία µεγέθη ™στˆ τ¦ Η, Θ, Λ, κሠ¥λλα αÙτοις ‡σα τÕ πλÁθος τ¦ Κ, Μ, Ν σύνδυο λαµβανόµενα ™ν τù αÙτù λόγJ, καί ™στιν αÙτîν τεταραγµένη ¹ ¢ναλογία, δι' ‡σου ¥ρα, ε„ Øπερέχει τÕ Η τοà Λ, Øπερέχει κሠτÕ Κ τοà Ν, καˆ ε„ ‡σον, ‡σον, καˆ ε„ œλαττον, œλαττον. καί ™στι τ¦ µν Η, Κ τîν Α, ∆ „σάκις πολλαπλάσια, τ¦ δ Λ, Ν τîν Γ, Ζ. œστιν ¥ρα æς τÕ Α πρÕς τÕ Γ, οÛτως τÕ ∆ πρÕς τÕ Ζ. 'Ε¦ν ¥ρα Ï τρία µεγέθη κሠ¥λλα αÙτο‹ς ‡σα τÕ πλÁθος σύνδυο λαµβανόµενα ™ν τù αÙτù λόγJ, Ï δ

A D G K

B E H M

C F L N

Let A, B, and C be three magnitudes, and D, E and F other (magnitudes) of equal number to them, (being) in the same ratio taken two by two. And let their proportion be perturbed, (so that) as A (is) to B, so E (is) to F , and as B (is) to C, so D (is) to E. I say that as A is to C, so D (is) to F . Let the equal multiples G, H, and K have been taken of A, B, and D (respectively), and the other random equal multiples L, M , and N of C, E, and F (respectively). And since G and H are equal multiples of A and B (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as A (is) to B, so G (is) to H. And, so, for the same (reasons), as E (is) to F , so M (is) to N . And as A is to B, so E (is) to F . And, thus, as G (is) to H, so M (is) to N [Prop. 5.11]. And since as B is to C, so D (is) to E, also, alternately, as B (is) to D, so C (is) to E [Prop. 5.16]. And since H and K are equal multiples of B and D (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as B is to D, so H (is) to K. But, as B (is) to D, so C (is) to E. And, thus, as H (is) to K, so C (is) to E [Prop. 5.11]. Again, since L and M are equal multiples of C and E (respectively), thus as C is to E, so L (is) to M [Prop. 5.15]. But, as C (is) to E, so H (is) to K. And, thus, as H (is) to K, so L (is) to M [Prop. 5.11]. Also, alternately, as H (is) to L, so K (is) to M [Prop. 5.16]. And it was also shown (that) as G (is) to H, so M (is) to N . Therefore, since G, H, and L are three magnitudes, and K, M , and N other (magnitudes) of equal number to them, (being) in the same ratio taken two by two, and their proportion is perturbed, thus, via equality, if G exceeds L then K also exceeds N , and if (G is) equal (to L then K is also) equal (to N ), and if (G is) less (than L then K is also) less (than N ) [Prop. 5.21]. And G and K are equal multiples of A and D (respectively), and L and N of C and F (respectively). Thus, as A (is) to C, so D (is) to F [Def. 5.5].

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τεταραγµένη αÙτîν ¹ ¢ναλογία, κሠδι' ‡σου ™ν τù αÙτù λόγJ œσται· Óπερ œδει δε‹ξαι.

†

Thus, if there are three magnitudes, and others of equal number to them, (being) in the same ratio taken two by two, and (if) their proportion is perturbed, then they will also be in the same ratio via equality. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: ǫ : ζ and β : γ :: δ : ǫ then α : γ :: δ : ζ.

κδ΄.

Proposition 24†

'Ε¦ν πρîτον πρÕς δεύτερον τÕν αÙτÕν œχV λόγον κሠτρίτον πρÕς τέταρτον, œχV δ κሠπέµπτον πρÕς δεύτερον τÕν αÙτÕν λόγον κሠ›κτον πρÕς τέταρτον, κሠσυντεθν πρîτον κሠπέµπτον πρÕς δεύτερον τÕν αÙτÕν ›ξει λόγον κሠτρίτον κሠ›κτον πρÕς τέταρτον.

If a first (magnitude) has to a second the same ratio that third (has) to a fourth, and a fifth (magnitude) also has to the second the same ratio that a sixth (has) to the fourth, then the first (magnitude) and the fifth, added together, will also have the same ratio to the second that the third (magnitude) and sixth (added together, have) to the fourth.

Β

Α Γ ∆ Ζ

Ε

B

Η

A C

G E

Θ

D F

Πρîτον γ¦ρ τÕ ΑΒ πρÕς δεύρερον τÕ Γ τÕν αÙτÕν ™χέτω λόγον κሠτρίτον τÕ ∆Ε πρÕς τέταρτον τÕ Ζ, ™χέτω δ κሠπέµπτον τÕ ΒΗ πρÕς δεύτερον τÕ Γ τÕν αÙτÕν λόγον κሠ›κτον τÕ ΕΘ πρÕς τέταρτον τÕ Ζ· λέγω, Óτι κሠσυντεθν πρîτον κሠπέµπτον τÕ ΑΗ πρÕς δεύτερον τÕ Γ τÕν αÙτÕν ›ξει λόγον, κሠτρίτον κሠ›κτον τÕ ∆Θ πρÕς τέταρτον τÕ Ζ. 'Επεˆ γάρ ™στιν æς τÕ ΒΗ πρÕς τÕ Γ, οÛτως τÕ ΕΘ πρÕς τÕ Ζ, ¢νάπαλιν ¥ρα æς τÕ Γ πρÕς τÕ ΒΗ, οÛτως τÕ Ζ πρÕς τÕ ΕΘ. ™πεˆ οâν ™στιν æς τÕ ΑΒ πρÕς τÕ Γ, οÛτως τÕ ∆Ε πρÕς τÕ Ζ, æς δ τÕ Γ πρÕς τÕ ΒΗ, οÛτως τÕ Ζ πρÕς τÕ ΕΘ, δι' ‡σου ¥ρα ™στˆν æς τÕ ΑΒ πρÕς τÕ ΒΗ, οÛτως τÕ ∆Ε πρÕς τÕ ΕΘ. κሠ™πεˆ διVρηµένα µεγέθη ¢νάλογόν ™στιν, κሠσυντεθέντα ¢νάλογον œσται· œστιν ¥ρα æς τÕ ΑΗ πρÕς τÕ ΗΒ, οÛτως τÕ ∆Θ πρÕς τÕ ΘΕ. œστι δ κሠæς τÕ ΒΗ πρÕς τÕ Γ, οÛτως τÕ ΕΘ πρÕς τÕ Ζ· δι' ‡σου ¥ρα ™στˆν æς τÕ ΑΗ πρÕς τÕ Γ, οÛτως τÕ ∆Θ πρÕς τÕ Ζ. 'Ε¦ν ¥ρα πρîτον πρÕς δεύτερον τÕν αÙτÕν œχV λόγον κሠτρίτον πρÕς τέταρτον, œχV δ κሠπέµπτον πρÕς δεύτερον τÕν αÙτÕν λόγον κሠ›κτον πρÕς τέτραρτον, κሠσυντεθν πρîτον κሠπέµπτον πρÕς δεύτερον τÕν αÙτÕν ›ξει λόγον κሠτρίτον κሠ›κτον πρÕς τέταρτον· Óπερ œδει δεˆξαι.

H

For let a first (magnitude) AB have the same ratio to a second C that a third DE (has) to a fourth F . And let a fifth (magnitude) BG also have the same ratio to the second C that a sixth EH (has) to the fourth F . I say that the first (magnitude) and the fifth, added together, AG, will also have the same ratio to the second C that the third (magnitude) and the sixth, (added together), DH, (has) to the fourth F . For since as BG is to C, so EH (is) to F , thus, inversely, as C (is) to BG, so F (is) to EH [Prop. 5.7 corr.]. Therefore, since as AB is to C, so DE (is) to F , and as C (is) to BG, so F (is) to EH, thus, via equality, as AB is to BG, so DE (is) to EH [Prop. 5.22]. And since separated magnitudes are proportional then they will also be proportional (when) composed [Prop. 5.18]. Thus, as AG is to GB, so DH (is) to HE. And, also, as BG is to C, so EH (is) to F . Thus, via equality, as AG is to C, so DH (is) to F [Prop. 5.22]. Thus, if a first (magnitude) has to a second the same ratio that a third (has) to a fourth, and a fifth (magnitude) also has to the second the same ratio that a sixth (has) to the fourth, then the first (magnitude) and the fifth, added together, will also have the same ratio to the second that the third (magnitude) and the sixth (added together, have) to the fourth. (Which is) the very thing it

153

ΣΤΟΙΧΕΙΩΝ ε΄.

ELEMENTS BOOK 5 was required to show.

†

In modern notation, this proposition reads that if α : β :: γ : δ and ǫ : β :: ζ : δ then α + ǫ : β :: γ + ζ : δ.

κε΄.

Proposition 25†

'Ε¦ν τέσσαρα µεγέθη ¢νάλογον Ï, τÕ µέγιστον [αÙτîν] κሠτÕ ™λάχιστον δύο τîν λοιπîν µείζονά ™στιν.

If four magnitudes are proportional then the (sum of the) largest and the smallest [of them] is greater than the (sum of the) remaining two (magnitudes).

Α Ε Γ Ζ

Η Θ

G

Β

A E H

∆

C F

”Εστω τέσσαρα µεγέθη ¢νάλογον τ¦ ΑΒ, Γ∆, Ε, Ζ, æς τÕ ΑΒ πρÕς τÕ Γ∆, οÛτως τÕ Ε πρÕς τÕ Ζ, œστω δ µέγιστον µν αÙτîν τÕ ΑΒ, ™λάχιστον δ τÕ Ζ· λέγω, Óτι τ¦ ΑΒ, Ζ τîν Γ∆, Ε µείζονά ™στιν. Κείσθω γ¦ρ τù µν Ε ‡σον τÕ ΑΗ, τù δ Ζ ‡σον τÕ ΓΘ. 'Επεˆ [οâν] ™στιν æς τÕ ΑΒ πρÕς τÕ Γ∆, οÛτως τÕ Ε πρÕς τÕ Ζ, ‡σον δ τÕ µν Ε τù ΑΗ, τÕ δ Ζ τù ΓΘ, œστιν ¥ρα æς τÕ ΑΒ πρÕς τÕ Γ∆, οÛτως τÕ ΑΗ πρÕς τÕ ΓΘ. κሠ™πεί ™στιν æς Óλον τÕ ΑΒ πρÕς Óλον τÕ Γ∆, οÛτως ¢φαιρεθν τÕ ΑΗ πρÕς ¢φαιρεθν τÕ ΓΘ, κሠλοιπÕν ¥ρα τÕ ΗΒ πρÕς λοιπÕν τÕ Θ∆ œσται æς Óλον τÕ ΑΒ πρÕς Óλον τÕ Γ∆. µε‹ζον δ τÕ ΑΒ τοà Γ∆· µε‹ζον ¥ρα κሠτÕ ΗΒ τοà Θ∆. κሠ™πεˆ ‡σον ™στˆ τÕ µν ΑΗ τù Ε, τÕ δ ΓΘ τù Ζ, τ¦ ¥ρα ΑΗ, Ζ ‡σα ™στˆ το‹ς ΓΘ, Ε. κሠ[™πεˆ] ™¦ν [¢νίσοις ‡σα προστεθÍ, τ¦ Óλα ¥νισά ™στιν, ™¦ν ¥ρα] τîν ΗΒ, Θ∆ ¢νίσων Ôντων κሠµείζονος τοà ΗΒ τù µν ΗΒ προστεθÍ τ¦ ΑΗ, Ζ, τù δ Θ∆ προστεθÍ τ¦ ΓΘ, Ε, συνάγεται τ¦ ΑΒ, Ζ µείζονα τîν Γ∆, Ε. 'Ε¦ν ¥ρα τέσσαρα µεγέθη ¢νάλογον Ï, τÕ µέγιστον αÙτîν κሠτÕ ™λάχιστον δύο τîν λοιπîν µείζονά ™στιν. Óπερ œδει δε‹ξαι. †

B

D

Let AB, CD, E, and F be four proportional magnitudes, (such that) as AB (is) to CD, so E (is) to F . And let AB be the greatest of them, and F the least. I say that AB and F is greater than CD and E. For let AG be made equal to E, and CH equal to F . [In fact,] since as AB is to CD, so E (is) to F , and E (is) equal to AG, and F to CH, thus as AB is to CD, so AG (is) to CH. And since the whole AB is to the whole CD as the (part) taken away AG (is) to the (part) taken away CH, thus the remainder GB will also be to the remainder HD as the whole AB (is) to the whole CD [Prop. 5.19]. And AB (is) greater than CD. Thus, GB (is) also greater than HD. And since AG is equal to E, and CH to F , thus AG and F is equal to CH and E. And [since] if [equal (magnitudes) are added to unequal (magnitudes) then the wholes are unequal, thus if] AG and F are added to GB, and CH and E to HD—GB and HD being unequal, and GB greater—it is inferred that AB and F (is) greater than CD and E. Thus, if four magnitudes are proportional then the (sum of the) largest and the smallest of them is greater than the (sum of the) remaining two (magnitudes). (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if α : β :: γ : δ, and α is the greatest and δ the least, then α + δ > β + γ.

154

ELEMENTS BOOK 6 Similar figures

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ELEMENTS BOOK 6

“Οροι.

Definitions

α΄. “Οµοια σχήµατα εÙθύγραµµά ™στιν, Óσα τάς τε γωνίας ‡σας œχει κατ¦ µίαν κሠτ¦ς περˆ τ¦ς ‡σας γωνίας πλευρ¦ς ¢νάλογον. β΄. ”Ακρον κሠµέσον λόγον εÙθε‹α τετµÁσθαι λέγεται, Óταν Ï æς ¹ Óλη πρÕς τÕ µε‹ζον τµÁµα, οÛτως τÕ µε‹ζον πρÕς τÕ œλαττÕν. γ΄. “Υψος ™στˆ πάντος σχήµατος ¹ ¢πÕ τÁς κορυφÁς ™πˆ τ¾ν βάσιν κάθετος ¢γοµένη.

1. Similar rectilinear figures are those (which) have (their) angles separately equal and the (corresponding) sides about the equal angles proportional. 2. A straight-line is said to have been cut in extreme and mean ratio when as the whole is to the greater segment so the greater (segment is) to the lesser. 3. The height of any figure is the (straight-line) drawn from the vertex perpendicular to the base.

α΄.

Proposition 1†

Τ¦ τρίγωνα κሠτ¦ παραλληλόγραµµα τ¦ ØπÕ τÕ Triangles and parallelograms which are of the same αÙτÕ Ûψος Ôντα πρÕς ¥λληλά ™στιν æς αƒ βάσεις. height are to one another as their bases.

Θ

Η

Ε

Α

Ζ

Β

Γ

∆

Κ

Λ

H

”Εστω τρίγωνα µν τ¦ ΑΒΓ, ΑΓ∆, παραλληλόγραµµα δ τ¦ ΕΓ, ΓΖ ØπÕ τÕ αÙτÕ Ûψος τÕ ΑΓ· λέγω, Óτι ™στˆν æς ¹ ΒΓ βάσις πρÕς τ¾ν Γ∆ βάσις, οÛτως τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΑΓ∆ τρίγωνον, κሠτÕ ΕΓ παραλληλόγραµµον πρÕς τÕ ΓΖ παραλληλόγραµµον. 'Εκβεβλήσθω γ¦ρ ¹ Β∆ ™φ' ˜κάτερα τ¦ µέρη ™πˆ τ¦ Θ, Λ σηµε‹α, κሠκείσθωσαν τÍ µν ΒΓ βάσει ‡σαι [Ðσαιδηποτοàν] αƒ ΒΗ, ΗΘ, τÍ δ Γ∆ βάσει ‡σαι Ðσαιδηποτοàν αƒ ∆Κ, ΚΛ, κሠ™πεζεύχθωσαν αƒ ΑΗ, ΑΘ, ΑΚ, ΑΛ. Κሠ™πεˆ ‡σαι ε„σˆν αƒ ΓΒ, ΒΗ, ΗΘ ¢λλήλαις, ‡σα ™στˆ κሠτ¦ ΑΘΗ, ΑΗΒ, ΑΒΓ τρίγωνα ¢λλήλοις. Ðσαπλασίων ¥ρα ™στˆν ¹ ΘΓ βάσις τÁς ΒΓ βάσεως, τοσαυταπλάσιόν ™στι κሠτÕ ΑΘΓ τρίγωνον τοà ΑΒΓ τριγώνου. δι¦ τ¦ αÙτ¦ δ¾ Ðσαπλασίων ™στˆν ¹ ΛΓ βάσις τÁς Γ∆ βάσεως, τοσαυταπλάσιόν ™στι κሠτÕ ΑΛΓ τρίγωνον τοà ΑΓ∆ τριγώνου· καˆ ε„ ‡ση ™στˆν ¹ ΘΓ βάσις τÍ ΓΛ βάσει, ‡σον ™στˆ κሠτÕ ΑΘΓ τρίγωνον τJ ΑΓΛ τριγώνJ, καˆ ε„ Øπερέχει ¹ ΘΓ βάσις τÁς ΓΛ βάσεως, Øπερέχει κሠτÕ ΑΘΓ τρίγωνον τοà ΑΓΛ τριγώνου, καˆ ε„ ™λάσσων, œλασσον. τεσσάρων δ¾ Ôντων µεγεθîν δύο µν βάσεων τîν ΒΓ, Γ∆, δύο δ τριγώνων τîν ΑΒΓ, ΑΓ∆ ε‡ληπται „σάκις πολλαπλάσια τÁς µν ΒΓ βάσεως

G

E

A

F

B

C

D

K

L

Let ABC and ACD be triangles, and EC and CF parallelograms, of the same height AC. I say that as base BC is to base CD, so triangle ABC (is) to triangle ACD, and parallelogram EC to parallelogram CF . For let the (straight-line) BD have been produced in each direction to points H and L, and let [any number] (of straight-lines) BG and GH be made equal to base BC, and any number (of straight-lines) DK and KL equal to base CD. And let AG, AH, AK, and AL have been joined. And since CB, BG, and GH are equal to one another, triangles AHG, AGB, and ABC are also equal to one another [Prop. 1.38]. Thus, as many times as base HC is (divisible by) base BC, so many times is triangle AHC also (divisible) by triangle ABC. So, for the same (reasons), as many times as base LC is (divisible) by base CD, so many times is triangle ALC also (divisible) by triangle ACD. And if base HC is equal to base CL then triangle AHC is also equal to triangle ACL [Prop. 1.38]. And if base HC exceeds base CL then triangle AHC also exceeds triangle ACL.‡ And if (HC is) less (than CL then AHC is also) less (than ACL). So, their being four magnitudes, two bases, BC and CD, and two trian-

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ΣΤΟΙΧΕΙΩΝ $΄.

ELEMENTS BOOK 6

κሠτοà ΑΒΓ τριγώνον ¼ τε ΘΓ βάσις κሠτÕ ΑΘΓ τρίγωνον, τÁς δ Γ∆ βάσεως κሠτοà Α∆Γ τριγώνου ¥λλα, § œτυχεν, „σάκις πολλαπλάσια ¼ τε ΛΓ βάσις κሠτÕ ΑΛΓ τρίγωνον· κሠδέδεικται, Óτι, ε„ Øπερέχει ¹ ΘΓ βάσις τÁς ΓΛ βάσεως, Øπερέχει κሠτÕ ΑΘΓ τρίγωνον τοà ΑΛΓ τριγώνου, καί ε„ ‡ση, ‡σον, καˆ ε„ œλασσων, œλασσον· œστιν ¥ρα æς ¹ ΒΓ βάσις πρÕς τ¾ν Γ∆ βάσιν, οÛτως τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΑΓ∆ τρίγωνον. Κሠ™πεˆ τοà µν ΑΒΓ τριγώνου διπλάσιόν ™στι τÕ ΕΓ παραλληλόγραµµον, τοà δ ΑΓ∆ τριγώνου διπλάσιόν ™στι τÕ ΖΓ παραλληλόγραµµον, τ¦ δ µέρη το‹ς æσαύτως πολλαπλασίοις τÕν αÙτÕν œχει λόγον, œστιν ¥ρα æς τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΑΓ∆ τρίγωνον, οÛτως τÕ ΕΓ παραλληλόγραµµον πρÕς τÕ ΖΓ παραλληλόγραµµον. ™πεˆ οâν ™δείχθη, æς µν ¹ ΒΓ βάσις πρÕς τ¾ν Γ∆, οÛτως τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΑΓ∆ τρίγωνον, æς δ τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΑΓ∆ τρίγωνον, οÛτως τÕ ΕΓ παραλληλόγραµµον πρÕς τÕ ΓΖ παραλληλόγραµµον, κሠæς ¥ρα ¹ ΒΓ βάσις πρÕς τ¾ν Γ∆ βάσιν, οÛτως τÕ ΕΓ παραλληλόγραµµον πρÕς τÕ ΖΓ παραλληλόγραµµον. Τ¦ ¥ρα τρίγωνα κሠτ¦ παραλληλόγραµµα τ¦ υπÕ τÕ αÙτÕ Ûψος Ôντα πρÕς ¥λληλά ™στιν æς αƒ βάσεις· Óπερ œδει δεˆξαι. †

gles, ABC and ACD, equal multiples have been taken of base BC and triangle ABC—(namely), base HC and triangle AHC—and other random equal multiples of base CD and triangle ADC—(namely), base LC and triangle ALC. And it has been shown that if base HC exceeds base CL then triangle AHC also exceeds triangle ALC, and if (HC is) equal (to CL then AHC is also) equal (to ALC), and if (HC is) less (than CL then AHC is also) less (than ALC). Thus, as base BC is to base CD, so triangle ABC (is) to triangle ACD [Def. 5.5]. And since parallelogram EC is double triangle ABC, and parallelogram F C is double triangle ACD [Prop. 1.34], and parts have the same ratio as similar multiples [Prop. 5.15], thus as triangle ABC is to triangle ACD, so parallelogram EC (is) to parallelogram F C. In fact, since it was shown that as base BC (is) to CD, so triangle ABC (is) to triangle ACD, and as triangle ABC (is) to triangle ACD, so parallelogram EC (is) to parallelogram CF , thus, also, as base BC (is) to base CD, so parallelogram EC (is) to parallelogram F C [Prop. 5.11]. Thus, triangles and parallelograms which are of the same height are to one another as their bases. (Which is) the very thing it was required to show.

As is easily demonstrated, this proposition holds even when the triangles, or parallelograms, do not share a common side, and/or are not

right-angled. ‡

This is a straight-forward generalization of Prop. 1.38.

β΄.

Proposition 2

'Ε¦ν τριγώνου παρ¦ µίαν τîν πλευρîν ¢χθÍ τις εÙθε‹α, ¢νάλογον τεµε‹ τ¦ς τοà τριγώνου πλευράς· κሠ™¦ν αƒ τοà τριγώνου πλευρሠ¢νάλογον τµηθîσιν, ¹ ™πˆ τ¦ς τﵦς ™πιζευγνυµένη εÙθε‹α παρ¦ τ¾ν λοιπ¾ν œσται τοà τριγώνου πλευράν.

If some straight-line is drawn parallel to one of the sides of a triangle, then it will cut the (other) sides of the triangle proportionally. And if (two of) the sides of a triangle are cut proportionally, then the straight-line joining the cutting (points) will be parallel to the remaining side of the triangle.

Α

A

∆

Ε Β

D

Γ

E B

C

Τριγώνου γ¦ρ τοà ΑΒΓ παράλληλος µι´ τîν For let DE have been drawn parallel to one of the πλευρîν τÍ ΒΓ ½χθω ¹ ∆Ε· λέγω, Óτι ™στˆν æς ¹ Β∆ sides BC of triangle ABC. I say that as BD is to DA, so 157

ΣΤΟΙΧΕΙΩΝ $΄.

ELEMENTS BOOK 6

πρÕς τ¾ν ∆Α, οÛτως ¹ ΓΕ πρÕς τ¾ν ΕΑ. 'Επεζεύχθωσαν γ¦ρ αƒ ΒΕ, Γ∆. ”Ισον ¥ρα ™στˆ τÕ Β∆Ε τρίγωνον τù Γ∆Ε τριγώνJ· ™πˆ γ¦ρ τÁς αÙτÁς βάσεώς ™στι τÁς ∆Ε κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ∆Ε, ΒΓ· ¥λλο δέ τι τÕ Α∆Ε τρίγωνον. τ¦ δ ‡σα πρÕς τÕ αÙτÕ τÕν αÙτÕν œχει λόγον· œστιν ¥ρα æς τÕ Β∆Ε τρίγωνον πρÕς τÕ Α∆Ε [τρίγωνον], οÛτως τÕ Γ∆Ε τρίγωνον πρÕς τÕ Α∆Ε τρίγωνον. αλλ' æς µν τÕ Β∆Ε τρίγωνον πρÕς τÕ Α∆Ε, οÛτως ¹ Β∆ πρÕς τ¾ν ∆Α· ØπÕ γ¦ρ τÕ αÙτÕ Ûψος Ôντα τ¾ν ¢πÕ τοà Ε ™πˆ τ¾ν ΑΒ κάθετον ¢γοµένην πρÕς ¥λληλά ε„σιν æς αƒ βάσεις. δι¦ τ¦ αÙτ¦ δ¾ æς τÕ Γ∆Ε τρίγωνον πρÕς τÕ Α∆Ε, οÛτως ¹ ΓΕ πρÕς τ¾ν ΕΑ· κሠæς ¥ρα ¹ Β∆ πρÕς τ¾ν ∆Α, οÛτως ¹ ΓΕ πρÕς τ¾ν ΕΑ. 'Αλλ¦ δ¾ αƒ τοà ΑΒΓ τριγώνου πλευραˆ αƒ ΑΒ, ΑΓ ¢νάλογον τετµήσθωσαν, æς ¹ Β∆ πρÕς τ¾ν ∆Α, οÛτως ¹ ΓΕ πρÕς τ¾ν ΕΑ, κሠ™πεζεύχθω ¹ ∆Ε· λέγω, Óτι παράλληλός ™στιν ¹ ∆Ε τÍ ΒΓ. Τîν γ¦ρ αÙτîν κατασκευασθέντων, ™πεί ™στιν æς ¹ Β∆ πρÕς τ¾ν ∆Α, οÛτως ¹ ΓΕ πρÕς τ¾ν ΕΑ, ¢λλ' æς µν ¹ Β∆ πρÕς τ¾ν ∆Α, οÛτως τÕ Β∆Ε τρίγωνον πρÕς τÕ Α∆Ε τρίγωνον, æς δ ¹ ΓΕ πρÕς τ¾ν ΕΑ, οÛτως τÕ Γ∆Ε τρίγωνον πρÕς τÕ Α∆Ε τρίγωνον, κሠæς ¥ρα τÕ Β∆Ε τρίγωνον πρÕς τÕ Α∆Ε τρίγωνον, οÛτως τÕ Γ∆Ε τρίγωνον πρÕς τÕ Α∆Ε τρίγωνον. ˜κάτερον ¥ρα τîν Β∆Ε, Γ∆Ε τριγώνων πρÕς τÕ Α∆Ε τÕν αÙτÕν œχει λόγον. ‡σον ¥ρα ™στˆ τÕ Β∆Ε τρίγωνον τù Γ∆Ε τριγώνJ· καί ε„σιν ™πˆ τ¾ς αÙτÁς βάσεως τÁς ∆Ε. τ¦ δ ‡σα τρίγωνα κሠ™πˆ τÁς αÙτÁς βάσεως Ôντα κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις ™στίν. παράλληλος ¥ρα ™στˆν ¹ ∆Ε τÍ ΒΓ. 'Ε¦ν ¥ρα τριγώνου παρ¦ µίαν τîν πλευρîν ¢χθÍ τις εÙθε‹α, ¢νάλογον τεµε‹ τ¦ς τοà τριγώνου πλευράς· κሠ™¦ν αƒ τοà τριγώνου πλευρሠ¢νάλογον τµηθîσιν, ¹ ™πˆ τ¦ς τﵦς ™πιζευγνυµένη εÙθε‹α παρ¦ τ¾ν λοιπ¾ν œσται τοà τριγώνου πλευράν· Óπερ œδει δε‹ξαι.

CE (is) to EA. For let BE and CD have been joined. Thus, triangle BDE is equal to triangle CDE. For they are on the same base DE and between the same parallels DE and BC [Prop. 1.38]. And ADE is some other triangle. And equal (magnitudes) have the same ratio to the same (magnitude) [Prop. 5.7]. Thus, as triangle BDE is to [triangle] ADE, so triangle CDE (is) to triangle ADE. But, as triangle BDE (is) to triangle ADE, so (is) BD to DA. For, having the same height—(namely), the (straight-line) drawn from E perpendicular to AB— they are to one another as their bases [Prop. 6.1]. So, for the same (reasons), as triangle CDE (is) to ADE, so CE (is) to EA. And, thus, as BD (is) to DA, so CE (is) to EA [Prop. 5.11]. And so, let the sides AB and AC of triangle ABC have been cut, (so that) as BD (is) to DA, so CE (is) to EA. And let DE have been joined. I say that DE is parallel to BC. For, by the same construction, since as BD is to DA, so CE (is) to EA, but as BD (is) to DA, so triangle BDE (is) to triangle ADE, and as CE (is) to EA, so triangle CDE (is) to triangle ADE [Prop. 6.1], thus, also, as triangle BDE (is) to triangle ADE, so triangle CDE (is) to triangle ADE [Prop. 5.11]. Thus, triangles BDE and CDE each have the same ratio to ADE. Thus, triangle BDE is equal to triangle CDE [Prop. 5.9]. And they are on the same base DE. And equal triangles, which are also on the same base, are also between the same parallels [Prop. 1.39]. Thus, DE is parallel to BC. Thus, if some straight-line is drawn parallel to one of the sides of a triangle, then it will cut the (other) sides of the triangle proportionally. And if (two of) the sides of a triangle are cut proportionally, then the straight-line joining the cutting (points) will be parallel to the remaining side of the triangle. (Which is) the very thing it was required to show.

γ΄.

Proposition 3

'Ε¦ν τριγώνου ¹ γωνία δίχα τµηθÍ, ¹ δ τέµνουσα τ¾ν γωνίαν εÙθε‹α τέµνV κሠτ¾ν βάσιν, τ¦ τÁς βάσεως τµήµατα τÕν αÙτÕν ›ξει λόγον τα‹ς λοιπα‹ς τοà τριγώνου πλευρα‹ς· κሠ™¦ν τ¦ τÁς βάσεως τµήµατα τÕν αÙτÕν œχV λόγον τα‹ς λοιπα‹ς τοà τριγώνου πλευρα‹ς, ¹ ¢πÕ τÁς κορυφÁς ™πˆ τ¾ν τοµ¾ν ™πιζευγνυµένη εÙθε‹α δίχα τεµε‹ τ¾ν τοà τριγώνου γωνίαν. ”Εστω τρίγωνον τÕ ΑΒΓ, κሠτετµήσθω ¹ ØπÕ ΒΑΓ γωνία δίχα ØπÕ τÁς Α∆ εÙθείας· λέγω, Óτι ™στˆν æς ¹ Β∆ πρÕς τ¾ν Γ∆, οÛτως ¹ ΒΑ πρÕς τ¾ν ΑΓ. ”Ηχθω γ¦ρ δι¦ τοà Γ τÍ ∆Α παράλληλος ¹ ΓΕ, καˆ

If an angle of a triangle is cut in half, and the straightline cutting the angle also cuts the base, then the segments of the base will have the same ratio as the remaining sides of the triangle. And if the segments of the base have the same ratio as the remaining sides of the triangle, then the straight-line joining the vertex to the cutting (point) will cut the angle of the triangle in half. Let ABC be a triangle. And let the angle BAC have been cut in half by the straight-line AD. I say that as BD is to CD, so BA (is) to AC. For let CE have been drawn through (point) C par-

158

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ELEMENTS BOOK 6

διαχθε‹σα ¹ ΒΑ συµπιπτέτω αÙτÍ κατ¦ τÕ Ε.

allel to DA. And, BA being drawn through, let it meet (CE) at (point) E.†

Ε

E

Α

Β

∆

A

Γ

B

Κሠ™πεˆ ε„ς παραλλήλους τ¦ς Α∆, ΕΓ εÙθε‹α ™νέπεσεν ¹ ΑΓ, ¹ ¥ρα ØπÕ ΑΓΕ γωνία ‡ση ™στˆ τÍ ØπÕ ΓΑ∆. ¢λλ' ¹ ØπÕ ΓΑ∆ τÍ ØπÕ ΒΑ∆ Øπόκειται ‡ση· κሠ¹ ØπÕ ΒΑ∆ ¥ρα τÍ ØπÕ ΑΓΕ ™στιν ‡ση. πάλιν, ™πεˆ ε„ς παραλλήλους τ¦ς Α∆, ΕΓ εÙθε‹α ™νέπεσεν ¹ ΒΑΕ, ¹ ™κτÕς γωνία ¹ ØπÕ ΒΑ∆ ‡ση ™στˆ τÍ ™ντÕς τÍ ØπÕ ΑΕΓ. ™δείχθη δ κሠ¹ ØπÕ ΑΓΕ τÍ ØπÕ ΒΑ∆ ‡ση· κሠ¹ ØπÕ ΑΓΕ ¥ρα γωνία τÊ ØπÕ ΑΕΓ ™στιν ‡ση· éστε κሠπλευρ¦ ¹ ΑΕ πλευρ´ τÍ ΑΓ ™στιν ‡ση. κሠ™πεˆ τριγώνου τοà ΒΓΕ παρ¦ µίαν τîν πλευρîν τ¾ν ΕΓ Ãκται ¹ Α∆, ¢νάλογον ¥ρα ™στˆν æς ¹ Β∆ πρÕς τ¾ν ∆Γ, οÛτως ¹ ΒΑ πρÕς τ¾ν ΑΕ. ‡ση δ ¹ ΑΕ τÍ ΑΓ· æς ¥ρα ¹ Β∆ πρÕς τ¾ν ∆Γ, οÛτως ¹ ΒΑ πρÕς τ¾ν ΑΓ. 'Αλλ¦ δ¾ œστω æς ¹ Β∆ πρÕς τ¾ν ∆Γ, οÛτως ¹ ΒΑ πρÕς τ¾ν ΑΓ, κሠ™πεζεύχθω ¹ Α∆· λέγω, Óτι δίχα τέτµηται ¹ ØπÕ ΒΑΓ γωνία ØπÕ τÁς Α∆ εÙθείας. Τîν γ¦ρ αÙτîν κατασκευασθέντων, ™πεί ™στιν æς ¹ Β∆ πρÕς τ¾ν ∆Γ, οÛτως ¹ ΒΑ πρÕς τ¾ν ΑΓ, ¢λλ¦ κሠæς ¹ Β∆ πρÕς τ¾ν ∆Γ, οÛτως ™στˆν ¹ ΒΑ πρÕς τ¾ν ΑΕ· τριγώνου γ¦ρ τοà ΒΓΕ παρ¦ µίαν τ¾ν ΕΓ Ãκται ¹ Α∆· κሠæς ¥ρα ¹ ΒΑ πρÕς τ¾ν ΑΓ, οÛτως ¹ ΒΑ πρÕς τ¾ν ΑΕ. ‡ση ¥ρα ¹ ΑΓ τÍ ΑΕ· éστε κሠγωνία ¹ ØπÕ ΑΕΓ τÍ ØπÕ ΑΓΕ ™στιν ‡ση. ¢λλ' ¹ µν ØπÕ ΑΕΓ τÍ ™κτÕς τÍ ØπÕ ΒΑ∆ [™στιν] ‡ση, ¹ δ ØπÕ ΑΓΕ τÍ ™ναλλ¦ξ τÍ ØπÕ ΓΑ∆ ™στιν ‡ση· κሠ¹ ØπÕ ΒΑ∆ ¥ρα τÍ ØπÕ ΓΑ∆ ™στιν ‡ση. ¹ ¥ρα ØπÕ ΒΑΓ γωνία δίχα τέτµηται ØπÕ τÁς Α∆ εÙθείας. 'Ε¦ν ¥ρα τριγώνου ¹ γωνία δίχα τµηθÍ, ¹ δ τέµνουσα τ¾ν γωνίαν εÙθε‹α τέµνV κሠτ¾ν βάσιν, τ¦ τÁς βάσεως τµήµατα τÕν αÙτÕν ›ξει λόγον τα‹ς λοιπα‹ς τοà τριγώνου πλευρα‹ς· κሠ™¦ν τ¦ τÁς βάσεως τµήµατα τÕν αÙτÕν œχV λόγον τα‹ς λοιπα‹ς τοà τριγώνου πλευρα‹ς, ¹ ¢πÕ τÁς κορυφÁς ™πˆ τ¾ν τοµ¾ν ™πιζευγνυµένη εÙθε‹α δίχα τέµνει τ¾ν τοà τριγώνου γωνίαν· Óπερ œδει δε‹ξαι.

D

C

And since the straight-line AC falls across the parallel (straight-lines) AD and EC, angle ACE is thus equal to CAD [Prop. 1.29]. But, (angle) CAD is assumed (to be) equal to BAD. Thus, (angle) BAD is also equal to ACE. Again, since the straight-line BAE falls across the parallel (straight-lines) AD and EC, the external angle BAD is equal to the internal (angle) AEC [Prop. 1.29]. And (angle) ACE was also shown (to be) equal to BAD. Thus, angle ACE is also equal to AEC. And, hence, side AE is equal to side AC [Prop. 1.6]. And since AD has been drawn parallel to one of the sides EC of triangle BCE, thus, proportionally, as BD is to DC, so BA (is) to AE [Prop. 6.2]. And AE (is) equal to AC. Thus, as BD (is) to DC, so BA (is) to AC. And so, let BD be to DC, as BA (is) to AC. And let AD have been joined. I say that angle BAC has been cut in half by the straight-line AD. For, by the same construction, since as BD is to DC, so BA (is) to AC, then also as BD (is) to DC, so BA is to AE. For AD has been drawn parallel to one (of the sides) EC of triangle BCE [Prop. 6.2]. Thus, also, as BA (is) to AC, so BA (is) to AE [Prop. 5.11]. Thus, AC (is) equal to AE [Prop. 5.9]. And, hence, angle AEC is equal to ACE [Prop. 1.5]. But, AEC [is] equal to the external (angle) BAD, and ACE is equal to the alternate (angle) CAD [Prop. 1.29]. Thus, (angle) BAD is also equal to CAD. Thus, angle BAC has been cut in half by the straight-line AD. Thus, if an angle of a triangle is cut in half, and the straight-line cutting the angle also cuts the base, then the segments of the base will have the same ratio as the remaining sides of the triangle. And if the segments of the base have the same ratio as the remaining sides of the triangle, then the straight-line joining the vertex to the cutting (point) will cut the angle of the triangle in half. (Which is) the very thing it was required to show.

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ELEMENTS BOOK 6

The fact that the two straight-lines meet follows because the sum of ACE and CAE is less than two right-angles, as can easily be demonstrated.

See Post. 5.

δ΄.

Proposition 4

Τîν „σογωνίων τριγώνων ¢νάλογόν ε„σιν αƒ πλευρሠFor equiangular triangles, the sides about the equal αƒ περˆ τ¦ς ‡σας γωνίας καˆ Ðµόλογοι αƒ ØπÕ τ¦ς ‡σας angles are proportional, and those (sides) subtending γωνίας Øποτείνουσαι. equal angles correspond.

Ζ

F

Α

A

∆ Β

Γ

D

Ε

B

”Εστω „σογώνια τρίγωνα τ¦ ΑΒΓ, ∆ΓΕ ‡σην œχοντα τ¾ν µν ØπÕ ΑΒΓ γωνίαν τÍ ØπÕ ∆ΓΕ, τ¾ν δ ØπÕ ΒΑΓ τÍ ØπÕ Γ∆Ε κሠœτι τ¾ν ØπÕ ΑΓΒ τÍ ØπÕ ΓΕ∆· λέγω, Óτι τîν ΑΒΓ, ∆ΓΕ τριγώνων ¢νάλογόν ε„σιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας καˆ Ðµόλογοι αƒ ØπÕ τ¦ς ‡σας γωνίας Øποτείνουσαι. Κείσθω γ¦ρ ™π' εÙθείας ¹ ΒΓ τÍ ΓΕ. κሠ™πεˆ αƒ ØπÕ ΑΒΓ, ΑΓΒ γωνίαι δύο Ñρθîν ™λάττονές ε„σιν, ‡ση δ ¹ ØπÕ ΑΓΒ τÍ ØπÕ ∆ΕΓ, αƒ ¥ρα ØπÕ ΑΒΓ, ∆ΕΓ δύο Ñρθîν ™λάττονές ε„σιν· αƒ ΒΑ, Ε∆ ¥ρα ™κβαλλόµεναι συµπεσοàνται. ™κβεβλήσθωσαν κሠσυµπιπτέτωσαν κατ¦ τÕ Ζ. Κሠ™πεˆ ‡ση ™στˆν ¹ ØπÕ ∆ΓΕ γωνία τÍ ØπÕ ΑΒΓ, παράλληλός ™στιν ¹ ΒΖ τÍ Γ∆. πάλιν, ™πεˆ ‡ση ™στˆν ¹ ØπÕ ΑΓΒ τÍ ØπÕ ∆ΕΓ, παράλληλός ™στιν ¹ ΑΓ τÍ ΖΕ. παραλληλόγραµµον ¥ρα ™στˆ τÕ ΖΑΓ∆· ‡ση ¥ρα ¹ µν ΖΑ τÍ ∆Γ, ¹ δ ΑΓ τÍ Ζ∆. κሠ™πεˆ τριγώνου τοà ΖΒΕ παρ¦ µίαν τ¾ν ΖΕ Ãκται ¹ ΑΓ, ™στιν ¥ρα æς ¹ ΒΑ πρÕς τ¾ν ΑΖ, οÛτως ¹ ΒΓ πρÕς τ¾ν ΓΕ. ‡ση δ ¹ ΑΖ τÍ Γ∆· æς ¥ρα ¹ ΒΑ πρÕς τ¾ν Γ∆, οÛτως ¹ ΒΓ πρÕς τ¾ν ΓΕ, κሠ™ναλλ¦ξ æς ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως ¹ ∆Γ πρÕς τ¾ν ΓΕ. πάλιν, ™πεˆ παράλληλός ™στιν ¹ Γ∆ τÍ ΒΖ, œστιν ¥ρα æς ¹ ΒΓ πρÕς τ¾ν ΓΕ, οÛτως ¹ Ζ∆ πρÕς τ¾ν ∆Ε. ‡ση δ ¹ Ζ∆ τÍ ΑΓ· æς ¥ρα ¹ ΒΓ πρÕς τ¾ν ΓΕ, οÛτως ¹ ΑΓ πρÕς τ¾ν ∆Ε, κሠ™ναλλ¦ξ æς ¹ ΒΓ πρÕς τ¾ν ΓΑ, οÛτως ¹ ΓΕ πρÕς τ¾ν Ε∆. ™πεˆ οâν ™δείχθη æς µν ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως ¹ ∆Γ πρÕς τ¾ν ΓΕ, æς δ ¹ ΒΓ πρÕς τ¾ν ΓΑ, οÛτως ¹ ΓΕ πρÕς τ¾ν Ε∆, δι' ‡σου ¥ρα æς ¹ ΒΑ πρÕς τ¾ν ΑΓ, οÛτως ¹ Γ∆ πρÕς τ¾ν ∆Ε.

C

E

Let ABC and DCE be equiangular triangles, having angle ABC equal to DCE, and (angle) BAC to CDE, and, further, (angle) ACB to CED. I say that, for triangles ABC and DCE, the sides about the equal angles are proportional, and those (sides) subtending equal angles correspond. Let BC be placed straight-on to CE. And since angles ABC and ACB are less than two right-angles [Prop 1.17], and ACB (is) equal to DEC, thus ABC and DEC are less than two right-angles. Thus, BA and ED, being produced, will meet [C.N. 5]. Let them have been produced, and let them meet at (point) F . And since angle DCE is equal to ABC, BF is parallel to CD [Prop. 1.28]. Again, since (angle) ACB is equal to DEC, AC is parallel to F E [Prop. 1.28]. Thus, F ACD is a parallelogram. Thus, F A is equal to DC, and AC to F D [Prop. 1.34]. And since AC has been drawn parallel to one (of the sides) F E of triangle F BE, thus as BA is to AF , so BC (is) to CE [Prop. 6.2]. And AF (is) equal to CD. Thus, as BA (is) to CD, so BC (is) to CE, and, alternately, as AB (is) to BC, so DC (is) to CE [Prop. 5.16]. Again, since CD is parallel to BF , thus as BC (is) to CE, so F D (is) to DE [Prop. 6.2]. And F D (is) equal to AC. Thus, as BC is to CE, so AC (is) to DE, and, alternately, as BC (is) to CA, so CE (is) to ED [Prop. 6.2]. Therefore, since it was shown that as AB (is) to BC, so DC (is) to CE, and as BC (is) to CA, so CE (is) to ED, thus, via equality, as BA (is) to AC, so CD (is) to DE [Prop. 5.22].

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ELEMENTS BOOK 6

Τîν ¥ρα „σογωνίων τριγώνων ¢νάλογόν ε„σιν αƒ Thus, for equiangular triangles, the sides about the πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας καˆ Ðµόλογοι αƒ ØπÕ equal angles are proportional, and those (sides) subtendτ¦ς ‡σας γωνίας Øποτείνουσαι· Óπερ œδει δε‹ξαι. ing equal angles correspond. (Which is) the very thing it was required to show.

ε΄.

Proposition 5

'Ε¦ν δύο τρίγωνα τ¦ς πλευρ¦ς ¢νάλογον œχV, „σογώνια œσται τ¦ τρίγωνα κሠ‡σας ›ξει τ¦ς γωνίας, Øφ' §ς αƒ Ðµόλογοι πλευρሠØποτείνουσιν.

If two triangles have proportional sides then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal.

∆

Α

Ε

Β

Γ

D

A

Ζ

E

Η

C

F

G

B

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ τ¦ς πλευρ¦ς ¢νάλογον œχοντα, æς µν τ¾ν ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τ¾ν ∆Ε πρÕς τ¾ν ΕΖ, æς δ τ¾ν ΒΓ πρÕς τ¾ν ΓΑ, οÛτως τ¾ν ΕΖ πρÕς τ¾ν Ζ∆, κሠœτι æς τ¾ν ΒΑ πρÕς τ¾ν ΑΓ, οÛτως τ¾ν Ε∆ πρÕς τ¾ν ∆Ζ. λέγω, Óτι „σογώνιόν ™στι τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ κሠ‡σας ›ξουσι τ¦ς γωνίας, Øφ' §ς αƒ Ðµόλογοι πλευρሠØποτείνουσιν, τ¾ν µν ØπÕ ΑΒΓ τÍ ØπÕ ∆ΕΖ, τ¾ν δ ØπÕ ΒΓΑ τÍ ØπÕ ΕΖ∆ κሠœτι τ¾ν ØπÕ ΒΑΓ τÍ ØπÕ Ε∆Ζ. Συνεστάτω γ¦ρ πρÕς τÍ ΕΖ εÙθείv κሠτο‹ς πρÕς αÙτÍ σηµείοις το‹ς Ε, Ζ τÍ µν ØπÕ ΑΒΓ γωνίv ‡ση ¹ ØπÕ ΖΕΗ, τÍ δ Øπο ΑΓΒ ‡ση ¹ ØπÕ ΕΖΗ· λοιπ¾ ¥ρα ¹ πρÕς τù Α λοιπÍ τÍ πρÕς τù Η ™στιν ‡ση. ”Ισογώνιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ΕΗΖ [τριγώνJ]. τîν ¥ρα ΑΒΓ, ΕΗΖ τριγώνων ¢νάλογόν ε„σιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας καˆ Ðµόλογοι αƒ ØπÕ τ¦ς ‡σας γωνίας Øποτείνουσαι· œστιν ¥ρα æς ¹ ΑΒ πρÕς τ¾ν ΒΓ, [οÛτως] ¹ ΗΕ πρÕς τ¾ν ΕΖ. ¢λλ' æς ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως Øπόκειται ¹ ∆Ε πρÕς τ¾ν ΕΖ· æς ¥ρα ¹ ∆Ε πρÕς τ¾ν ΕΖ, οÛτως ¹ ΗΕ πρÕς τ¾ν ΕΖ. ˜κατέρα ¥ρα τîν ∆Ε, ΗΕ πρÕς τ¾ν ΕΖ τÕν αÙτÕν œχει λόγον· ‡ση ¥ρα ™στˆν ¹ ∆Ε τÍ ΗΕ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ∆Ζ τÍ ΗΖ ™στιν ‡ση. ™πεˆ οâν ‡ση ™στˆν ¹ ∆Ε τÍ ΕΗ, κοιν¾ δ ¹ ΕΖ, δύο δ¾ αƒ ∆Ε, ΕΖ δυσˆ τα‹ς ΗΕ, ΕΖ ‡σαι ε„σίν· κሠβάσις ¹ ∆Ζ βάσει τÍ ΖΗ [™στιν] ‡ση· γωνία ¥ρα ¹ ØπÕ ∆ΕΖ γωνίv τÍ ØπÕ ΗΕΖ ™στιν ‡ση, κሠτÕ ∆ΕΖ τρίγωνον τù ΗΕΖ τριγώνJ ‡σον, καˆ αƒ λοιπαˆ

Let ABC and DEF be two triangles having proportional sides, (so that) as AB (is) to BC, so DE (is) to EF , and as BC (is) to CA, so EF (is) to F D, and, further, as BA (is) to AC, so ED (is) to DF . I say that triangle ABC is equiangular to triangle DEF , and (that the triangles) will have the angles which corresponding sides subtend equal. (That is), (angle) ABC (equal) to DEF , BCA to EF D, and, further, BAC to EDF . For let (angle) F EG, equal to angle ABC, and (angle) EF G, equal to ACB, have been constructed at points E and F (respectively) on the straight-line EF [Prop. 1.23]. Thus, the remaining (angle) at A is equal to the remaining (angle) at G [Prop. 1.32]. Thus, triangle ABC is equiangular to [triangle] EGF . Thus, for triangles ABC and EGF , the sides about the equal angles are proportional, and (those) sides subtending equal angles correspond [Prop. 6.4]. Thus, as AB is to BC, [so] GE (is) to EF . But, as AB (is) to BC, so, it was assumed, (is) DE to EF . Thus, as DE (is) to EF , so GE (is) to EF [Prop. 5.11]. Thus, DE and GE each have the same ratio to EF . Thus, DE is equal to GE [Prop. 5.9]. So, for the same (reasons), DF is also equal to GF . Therefore, since DE is equal to EG, and EF (is) common, the two (sides) DE, EF are equal to the two (sides) GE, EF (respectively). And base DF [is] equal to base F G. Thus, angle DEF is equal to angle GEF [Prop. 1.8], and triangle DEF (is) equal

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ELEMENTS BOOK 6

γωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν. ‡ση ¥ρα ™στˆ κሠ¹ µν ØπÕ ∆ΖΕ γωνία τÍ ØπÕ ΗΖΕ, ¹ δ ØπÕ Ε∆Ζ τÍ ØπÕ ΕΗΖ. κሠ™πεˆ ¹ µν ØπÕ ΖΕ∆ τÍ ØπÕ ΗΕΖ ™στιν ‡ση, ¢λλ' ¹ ØπÕ ΗΕΖ τÍ ØπÕ ΑΒΓ, κሠ¹ ØπÕ ΑΒΓ ¥ρα γωνία τÍ ØπÕ ∆ΕΖ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ ΑΓΒ τÍ ØπÕ ∆ΖΕ ™στιν ‡ση, κሠœτι ¹ πρÕς τù Α τÍ πρÕς τù ∆· „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ. 'Ε¦ν ¥ρα δύο τρίγωνα τ¦ς πλευρ¦ς ¢νάλογον œχV, „σογώνια œσται τ¦ τρίγωνα κሠ‡σας ›ξει τ¦ς γωνίας, Øφ' §ς αƒ Ðµόλογοι πλευρሠØποτείνουσιν· Óπερ œδει δε‹ξαι.

to triangle GEF , and the remaining angles (are) equal to the remaining angles which the equal sides subtend [Prop. 1.4]. Thus, angle DF E is also equal to GF E, and (angle) EDF to EGF . And since (angle) F ED is equal to GEF , and (angle) GEF to ABC, angle ABC is thus also equal to DEF . So, for the same (reasons), (angle) ACB is also equal to DF E, and, further, the (angle) at A to the (angle) at D. Thus, triangle ABC is equiangular to triangle DEF . Thus, if two triangles have proportional sides then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal. (Which is) the very thing it was required to show.

$΄.

Proposition 6

'Ε¦ν δύο τρίγωνα µίαν γωνίαν µι´ γωνίv ‡σην œχV, περˆ δ τ¦ς ‡σας γωνίας τ¦ς πλευρ¦ς ¢νάλογον, „σογώνια œσται τ¦ τρίγωνα κሠ‡σας ›ξει τ¦ς γωνίας, Øφ' §ς αƒ Ðµόλογοι πλευρሠØποτείνουσιν.

If two triangles have one angle equal to one angle, and the sides about the equal angles proportional, then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal.

∆

Α

D

A Η

Ε

Β

G

Ζ

E

Γ

B

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ µίαν γωνίαν τ¾ν ØπÕ ΒΑΓ µι´ γωνίv τÍ ØπÕ Ε∆Ζ ‡σην œχοντα, περˆ δ τ¦ς ‡σας γωνίας τ¦ς πλευρ¦ς ¢νάλογον, æς τ¾ν ΒΑ πρÕς τ¾ν ΑΓ, οÛτως τ¾ν Ε∆ πρÕς τ¾ν ∆Ζ· λέγω, Óτι „σογώνιόν ™στι τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ κሠ‡σην ›ξει τ¾ν ØπÕ ΑΒΓ γωνίαν τÍ ØπÕ ∆ΕΖ, τ¾ν δ ØπÕ ΑΓΒ τÍ ØπÕ ∆ΖΕ. Συνεστάτω γ¦ρ πρÕς τÍ ∆Ζ εÙθείv κሠτο‹ς πρÕς αÙτÍ σηµείοις το‹ς ∆, Ζ Ðποτέρv µν τîν ØπÕ ΒΑΓ, Ε∆Ζ ‡ση ¹ ØπÕ Ζ∆Η, τÍ δ ØπÕ ΑΓΒ ‡ση ¹ ØπÕ ∆ΖΗ· λοιπ¾ ¥ρα ¹ πρÕς τù Β γωνία λοιπÍ τÍ πρÕς τù Η ‡ση ™στίν. 'Ισογώνιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΗΖ τριγώνJ. ¢νάλογον ¥ρα ™στˆν æς ¹ ΒΑ πρÕς τ¾ν ΑΓ, οÛτως ¹ Η∆ πρÕς τ¾ν ∆Ζ. Øπόκειται δ κሠæς ¹ ΒΑ πρÕς τ¾ν ΑΓ, οÛτως ¹ Ε∆ πρÕς τ¾ν ∆Ζ· κሠæς ¥ρα ¹ Ε∆ πρÕς τ¾ν ∆Ζ, οÛτως ¹ Η∆ πρÕς τ¾ν ∆Ζ. ‡ση ¥ρα ¹ Ε∆ τÍ ∆Η· κሠκοιν¾ ¹ ∆Ζ· δύο δ¾ αƒ Ε∆, ∆Ζ δυσˆ τα‹ς

F

C

Let ABC and DEF be two triangles having one angle, BAC, equal to one angle, EDF (respectively), and the sides about the equal angles proportional, (so that) as BA (is) to AC, so ED (is) to DF . I say that triangle ABC is equiangular to triangle DEF , and will have angle ABC equal to DEF , and (angle) ACB to DF E. For let (angle) F DG, equal to each of BAC and EDF , and (angle) DF G, equal to ACB, have been constructed at the points D and F (respectively) on the straight-line AF [Prop. 1.23]. Thus, the remaining angle at B is equal to the remaining angle at G [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle DGF . Thus, proportionally, as BA (is) to AC, so GD (is) to DF [Prop. 6.4]. And it was also assumed that as BA is) to AC, so ED (is) to DF . And, thus, as ED (is) to DF , so GD (is) to DF [Prop. 5.11]. Thus, ED (is) equal to DG [Prop. 5.9]. And DF (is) common. So, the two (sides) ED, DF are equal to the two (sides) GD,

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ELEMENTS BOOK 6

Η∆, ∆Ζ ‡σας ε„σίν· κሠγωνία ¹ ØπÕ Ε∆Ζ γωνίv τÍ ØπÕ Η∆Ζ [™στιν] ‡ση· βάσις ¥ρα ¹ ΕΖ βάσει τÍ ΗΖ ™στιν ‡ση, κሠτÕ ∆ΕΖ τρίγωνον τù Η∆Ζ τριγώνJ ‡σον ™στίν, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σας œσονται, Ùφ' §ς ‡σας πλευρሠØποτείνουσιν. ‡ση ¥ρα ™στˆν ¹ µν ØπÕ ∆ΖΗ τÍ Øπο ∆ΖΕ, ¹ δ Øπο ∆ΗΖ τÍ ØπÕ ∆ΕΖ. ¢λλ' ¹ Øπο ∆ΖΗ τÍ ØπÕ ΑΓΒ ™στιν ‡ση· κሠ¹ ØπÕ ΑΓΒ ¥ρα τÍ ØπÕ ∆ΖΕ ™στιν ‡ση. Øπόκειται δ κሠ¹ ØπÕ ΒΑΓ τÍ ØπÕ Ε∆Ζ ‡ση· κሠλοιπη ¥ρα ¹ πρÕς τù Β λοιπÍ τÍ πρÕς τù Ε ‡ση ™στίν· „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ. 'Ε¦ν ¥ρα δύο τρίγωνα µίαν γωνίαν µι´ γωνίv ‡σην œχV, περˆ δ τ¦ς ‡σας γωνίας τ¦ς πλευρ¦ς ¢νάλογον, „σογώνια œσται τ¦ τρίγωνα κሠ‡σας ›ξει τ¦ς γωνίας, Øφ' §ς αƒ Ðµόλογοι πλευρሠØποτείνουσιν· Óπερ œδει δε‹ξαι.

DF (respectively). And angle EDF [is] equal to angle GDF . Thus, base EF is equal to base GF , and triangle DEF is equal to triangle GDF , and the remaining angles will be equal to the remaining angles which the equal sides subtend [Prop. 1.4]. Thus, (angle) DF G is equal to DF E, and (angle) DGF to DEF . But, (angle) DF G is equal to ACB. Thus, (angle) ACB is also equal to DF E. And (angle) BAC was also assumed (to be) equal to EDF . Thus, the remaining (angle) at B is equal to the remaining (angle) at E [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle DEF . Thus, if two triangles have one angle equal to one angle, and the sides about the equal angles proportional, then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal. (Which is) the very thing it was required to show.

ζ΄.

Proposition 7

'Ε¦ν δύο τρίγωνα µίαν γωνίαν µι´ γωνίv ‡σην œχV, περˆ δ ¥λλας γωνίας τ¦ς πλευρ¦ς ¢νάλογον, τîν δ λοιπîν ˜κατέραν ¤µα ½τοι ™λάσσονα À µ¾ ™λάσσονα ÑρθÁς, „σογώνια œσται τ¦ τρίγωνα κሠ‡σας ›ξει τ¦ς γωνίας, περˆ §ς ¢νάλογόν ε„σιν αƒ πλευραί.

If two triangles have one angle equal to one angle, and the sides about other angles proportional, and the remaining angles either both less than, or both not less than, right-angles, then the triangles will be equiangular, and will have the angles about which the sides are proportional equal.

Α

A ∆

Β

D

B

Ε Η

E Ζ

G

F

Γ

C

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ µίαν γωνίαν µι´ γωνίv ‡σην œχοντα τ¾ν ØπÕ ΒΑΓ τÍ ØπÕ Ε∆Ζ, περˆ δ ¥λλας γωνίας τ¦ς ØπÕ ΑΒΓ, ∆ΕΖ τ¦ς πλευρ¦ς ¢νάλογον, æς τ¾ν ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τ¾ν ∆Ε πρÕς τ¾ν ΕΖ, τîν δ λοιπîν τîν πρÕς το‹ς Γ, Ζ πρότερον ˜κατέραν ¤µα ™λάσσονα ÑρθÁς· λέγω, Óτι „σογώνιόν ™στι τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ, κሠ‡ση œσται ¹ ØπÕ ΑΒΓ γωνία τÍ ØπÕ ∆ΕΖ, κሠλοιπ¾ δηλονότι ¹ πρÕς τù Γ λοιπÍ τÍ πρÕς τù Ζ ‡ση. Ε„ γ¦ρ ¥νισός ™στιν ¹ ØπÕ ΑΒΓ γωνία τÍ ØπÕ ∆ΕΖ, µία αÙτîν µείζων ™στίν. œστω µείζων ¹ ØπÕ ΑΒΓ. κሠσυνεστάτω πρÕς τÍ ΑΒ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Β τÍ ØπÕ ∆ΕΖ γωνίv ‡ση ¹ ØπÕ ΑΒΗ.

Let ABC and DEF be two triangles having one angle, BAC, equal to one angle, EDF (respectively), and the sides about (some) other angles, ABC and DEF (respectively), proportional, (so that) as AB (is) to BC, so DE (is) to EF , and the remaining (angles) at C and F , first of all, both less than right-angles. I say that triangle ABC is equiangular to triangle DEF , and (that) angle ABC will be equal to DEF , and (that) the remaining (angle) at C (will be) manifestly equal to the remaining (angle) at F . For if angle ABC is not equal to (angle) DEF then one of them is greater. Let ABC be greater. And let (angle) ABG, equal to (angle) DEF , have been constructed

163

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ELEMENTS BOOK 6

Κሠ™πεˆ ‡ση ™στˆν ¹ µν Α γωνία τÍ ∆, ¹ δ ØπÕ ΑΒΗ τÍ ØπÕ ∆ΕΖ, λοιπ¾ ¥ρα ¹ ØπÕ ΑΗΒ λοιπÍ τÍ ØπÕ ∆ΖΕ ™στιν ‡ση. „σογώνιον ¥ρα ™στˆ τÕ ΑΒΗ τρίγωνον τù ∆ΕΖ τριγώνJ. œστιν ¥ρα æς ¹ ΑΒ πρÕς τ¾ν ΒΗ, οÛτως ¹ ∆Ε πρÕς τ¾ν ΕΖ. æς δ ¹ ∆Ε πρÕς τ¾ν ΕΖ, [οÛτως] Øπόκειται ¹ ΑΒ πρÕς τ¾ν ΒΓ· ¹ ΑΒ ¥ρα πρÕς ˜κατέραν τîν ΒΓ, ΒΗ τÕν αÙτÕν œχει λόγον· ‡ση ¥ρα ¹ ΒΓ τÍ ΒΗ. éστε κሠγωνία ¹ πρÕς τù Γ γωνίv τÍ ØπÕ ΒΗΓ ™στιν ‡ση. ™λάττων δ ÑρθÁς Øπόκειται ¹ πρÕς τù Γ· ™λάττων ¥ρα ™στˆν ÑρθÁς κሠØπÕ ΒΗΓ· éστε ¹ ™φεξÁς αÙτÍ γωνία ¹ ØπÕ ΑΗΒ µείζων ™στˆν ÑρθÁς. κሠ™δείχθη ‡ση οâσα τÍ πρÕς τù Ζ· κሠ¹ πρÕς τù Ζ ¥ρα µείζων ™στˆν ÑρθÁς. Øπόκειται δ ™λάσσων ÑρθÁς· Óπερ ™στˆν ¥τοπον. οÙκ ¥ρα ¥νισός ™στιν ¹ ØπÕ ΑΒΓ γωνία τÍ ØπÕ ∆ΕΖ· ‡ση ¥ρα. œστι δ κሠ¹ πρÕς τù Α ‡ση τÍ πρÕς τù ∆· κሠλοιπ¾ ¥ρα ¹ πρÕς τù Γ λοιπÍ τÍ πρÕς τù Ζ ‡ση ™στίν. „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ. 'Αλλ¦ δ¾ πάλιν Øποκείσθω ˜κατέρα τîν πρÕς το‹ς Γ, Ζ µ¾ ™λάσσων ÑρθÁς· λέγω πάλιν, Óτι κሠοÛτως ™στˆν „σογώνιον τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ. Τîν γ¦ρ αÙτîν κατασκευασθέντων еοίως δείξοµεν, Óτι ‡ση ™στˆν ¹ ΒΓ τÍ ΒΗ· éστε κሠγωνία ¹ πρÕς τù Γ τÍ ØπÕ ΒΗΓ ‡ση ™στίν. οÙκ ™λάττων δ ÑρθÁς ¹ πρÕς τù Γ· οÙκ ™λάττων ¥ρα ÑρθÁς οÙδ ¹ ØπÕ ΒΗΓ. τριγώνου δ¾ τοà ΒΗΓ αƒ δύο γωνίαι δύο Ñρθîν οÜκ ε„σιν ™λάττονες· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα πάλιν ¥νισός ™στιν ¹ ØπÕ ΑΒΓ γωνία τÍ ØπÕ ∆ΕΖ· ‡ση ¥ρα. œστι δ κሠ¹ πρÕς τù Α τÍ πρÕς τù ∆ ‡ση· λοιπ¾ ¥ρα ¹ πρÕς τù Γ λοιπÍ τÍ πρÕς τù Ζ ‡ση ™στίν. „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΕΖ τριγώνJ. 'Ε¦ν ¥ρα δύο τρίγωνα µίαν γωνίαν µι´ γωνίv ‡σην œχV, περˆ δ ¥λλας γωνίας τ¦ς πλευρ¦ς ¢νάλογον, τîν δ λοιπîν ˜κατέραν ¤µα ™λάττονα À µ¾ ™λάττονα ÑρθÁς, „σογώνια œσται τ¦ τρίγωνα κሠ‡σας ›ξει τ¦ς γωνίας, περˆ §ς ¢νάλογόν ε„σιν αƒ πλευραί· Óπερ œδει δε‹ξαι.

at the point B on the straight-line AB [Prop. 1.23]. And since angle A is equal to (angle) D, and (angle) ABG to DEF , the remaining (angle) AGB is thus equal to the remaining (angle) DF E [Prop. 1.32]. Thus, triangle ABG is equiangular to triangle DEF . Thus, as AB is to BG, so DE (is) to EF [Prop. 6.4]. And as DE (is) to EF , [so] it was assumed (is) AB to BC. Thus, AB has the same ratio to each of BC and BG [Prop. 5.11]. Thus, BC (is) equal to BG [Prop. 5.9]. And, hence, the angle at C is equal to angle BGC [Prop. 1.5]. And the angle at C was assumed (to be) less than a right-angle. Thus, (angle) BGC is also less than a right-angle. Hence, the adjacent angle to it, AGB, is greater than a right-angle [Prop. 1.13]. And (AGB) was shown to be equal to the (angle) at F . Thus, the (angle) at F is also greater than a right-angle. But it was assumed (to be) less than a rightangle. The very thing is absurd. Thus, angle ABC is not unequal to (angle) DEF . Thus, (it is) equal. And the (angle) at A is also equal to the (angle) at D. And thus the remaining (angle) at C is equal to the remaining (angle) at F [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle DEF . But, again, let each of the (angles) at C and F be assumed (to be) not less than a right-angle. I say, again, that triangle ABC is equiangular to triangle DEF in this case also. For, with the same construction, we can similarly show that BC is equal to BG. Hence, also, the angle at C is equal to (angle) BGC. And the (angle) at C (is) not less than a right-angle. Thus, BGC (is) not less than a right-angle either. So, for triangle BGC, the (sum of) two angles is not less than two right-angles. The very thing is impossible [Prop. 1.17]. Thus, again, angle ABC is not unequal to DEF . Thus, (it is) equal. And the (angle) at A is also equal to the (angle) at D. Thus, the remaining (angle) at C is equal to the remaining (angle) at F [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle DEF . Thus, if two triangles have one angle equal to one angle, and the sides about other angles proportional, and the remaining angles both less than, or both not less than, right-angles, then the triangles will be equiangular, and will have the angles about which the sides (are) proportional equal. (Which is) the very thing it was required to show.

η΄.

Proposition 8

'Ε¦ν ™ν ÑρθογωνίJ τριγώνJ ¢πό τÁς ÑρθÁς γωνίας If, in a right-angled triangle, a (straight-line) is drawn ™πˆ τ¾ν βάσιν κάθετος ¢χθÍ, τ¦ πρÕς τÍ καθέτJ τρίγωνα from the right-angle perpendicular to the base then the Óµοιά ™στι τù τε ÓλJ κሠ¢λλήλοις. triangles around the perpendicular are similar to the

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”Εστω τρίγωνον Ñρθογώνιον τÕ ΑΒΓ Ñρθ¾ν œχον whole (triangle), and to one another. τ¾ν ØπÕ ΒΑΓ γωνίαν, κሠ½χθω ¢πÕ τοà Α ™πˆ τ¾ν ΒΓ Let ABC be a right-angled triangle having the angle κάθετος ¹ Α∆· λέγω, Óτι Óµοιόν ™στιν ˜κάτερον τîν BAC a right-angle, and let AD have been drawn from ΑΒ∆, Α∆Γ τριγώνων ÓλJ τù ΑΒΓ κሠœτι ¢λλήλοις. A, perpendicular to BC [Prop. 1.12]. I say that triangles ABD and ADC are each similar to the whole (triangle) ABC and, further, to one another.

Α

Β

∆

A

Γ

B

'Επεˆ γ¦ρ ‡ση ™στˆν ¹ ØπÕ ΒΑΓ τÍ ØπÕ Α∆Β· Ñρθ¾ γ¦ρ ˜κατέρα· κሠκοιν¾ τîν δύο τριγώνων τοà τε ΑΒΓ κሠτοà ΑΒ∆ ¹ πρÕς τù Β, λοιπ¾ ¥ρα ¹ ØπÕ ΑΓΒ λοιπÍ τÍ Øπο ΒΑ∆ ™στιν ‡ση· „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ΑΒ∆ τριγώνJ. œστιν ¥ρα æς ¹ ΒΓ Øποτείνουσα τ¾ν Ñρθ¾ν τοà ΑΒΓ τριγώνου πρÕς τ¾ν ΒΑ Øποτείνουσαν τ¾ν Ñρθ¾ν τοà ΑΒ∆ τριγώνου, οÛτως αÙτ¾ ¹ ΑΒ Øποτείνουσα τ¾ν πρÕς τù Γ γωνίαν τοà ΑΒΓ τριγώνου πρÕς τ¾ν Β∆ Øποτείνουσαν τ¾ν ‡σην τ¾ν ØπÕ ΒΑ∆ τοà ΑΒ∆ τριγώνου, κሠœτι ¹ ΑΓ πρÕς τ¾ν Α∆ Øποτείνουσαν τ¾ν πρÕς τù Β γωνίαν κοιν¾ν τîν δύο τριγώνων. τÕ ΑΒΓ ¥ρα τρίγωνον τù ΑΒ∆ τριγώνJ „σογώνιόν τέ ™στι κሠτ¦ς περˆ τ¦ς ‡σας γωνίας πλευρ¦ς ¢νάλογον œχει. Óµοιον ¥µα [™στˆ] τÕ ΑΒΓ τρίγωνον τù ΑΒ∆ τριγώνJ. еοίως δ¾ δείξοµεν, Óτι κሠτù Α∆Γ τριγώνJ Óµοιόν ™στι τÕ ΑΒΓ τρίγωνον· ˜κάτερον ¥ρα τîν ΑΒ∆, Α∆Γ [τριγώνων] Óµοιόν ™στιν ÓλJ τù ΑΒΓ. Λέγω δή, Óτι κሠ¢λλήλοις ™στˆν Óµοια τ¦ ΑΒ∆, Α∆Γ τρίγωνα. 'Επεˆ γ¦ρ Ñρθ¾ ¹ ØπÕ Β∆Α ÑρθÍ τÍ ØπÕ Α∆Γ ™στιν ‡ση, ¢λλ¦ µ¾ν κሠ¹ ØπÕ ΒΑ∆ τÍ πρÕς τù Γ ™δείχθη ‡ση, κሠλοιπ¾ ¥ρα ¹ πρÕς τù Β λοιπÍ τÍ ØπÕ ∆ΑΓ ™στιν ‡ση· „σογώνιον ¥ρα ™στˆ τÕ ΑΒ∆ τρίγωνον τù Α∆Γ τριγώνJ. œστιν ¥ρα æς ¹ Β∆ τοà ΑΒ∆ τριγώνου Øποτείνουσα τ¾ν ØπÕ ΒΑ∆ πρÕς τ¾ν ∆Α τοà Α∆Γ τριγώνου Øποτείνουσαν τ¾ν πρÕς τù Γ ‡σην τÍ ØπÕ ΒΑ∆, οÛτως αÙτ¾ ¹ Α∆ τοà ΑΒ∆ τριγώνου Øποτείνουσα τ¾ν πρÕς τù Β γωνίαν πρÕς τ¾ν ∆Γ Øποτείνουσαν τ¾ν ØπÕ ∆ΑΓ τοà Α∆Γ τριγώνου ‡σην τÍ πρÕς τù Β, κሠœτι ¹ ΒΑ πρÕς τ¾ν ΑΓ Øποτείνουσαι τ¦ς Ñρθάς· Óµοιον ¥ρα ™στˆ τÕ ΑΒ∆ τρίγωνον τù Α∆Γ τριγώνJ. 'Ε¦ν ¥ρα ™ν ÑρθογωνίJ τριγώνJ ¢πÕ τÁς ÑρθÁς

D

C

For since (angle) BAC is equal to ADB—for each (are) right-angles—and the (angle) at B (is) common to the two triangles ABC and ABD, the remaining (angle) ACB is thus equal to the remaining (angle) BAD [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle ABD. Thus, as BC, subtending the right-angle in triangle ABC, is to BA, subtending the right-angle in triangle ABD, so the same AB, subtending the angle at C in triangle ABC, (is) to BD, subtending the equal (angle) BAD in triangle ABD, and, further, (so is) AC to AD, (both) subtending the angle at B common to the two triangles [Prop. 6.4]. Thus, triangle ABC is equiangular to triangle ABD, and has the sides about the equal angles proportional. Thus, triangle ABC [is] similar to triangle ABD [Def. 6.1]. So, similarly, we can show that triangle ADC is also similar to triangle ABC. Thus, [triangles] ABD and ADC are each similar to the whole (triangle) ABC. So I say that triangles ABD and ADC are also similar to one another. For since the right-angle BDA is equal to the rightangle ADC, and, indeed, (angle) BAD was also shown (to be) equal to the (angle) at C, thus the remaining (angle) at B is also equal to the remaining (angle) DAC [Prop. 1.32]. Thus, triangle ABD is equiangular to triangle ADC. Thus, as BD, subtending (angle) BAD in triangle ABD, is to DA, subtending the (angle) at C in triangle ADB, (which is) equal to (angle) BAD, so (is) the same AD, subtending the angle at B in triangle ABD, to DC, subtending (angle) DAC in triangle ADC, (which is) equal to the (angle) at B, and, further, (so is) BA to AC, (each) subtending right-angles [Prop. 6.4]. Thus,

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ELEMENTS BOOK 6

γωνίας ™πˆ τ¾ν βάσιν κάθετος ¢χθÍ, τ¦ πρÕς τÍ καθέτJ triangle ABD is similar to triangle ADC [Def. 6.1]. τρίγωνα Óµοιά ™στι τù τε ÓλJ κሠ¢λλήλοις [Óπερ œδει Thus, if, in a right-angled triangle, a (straight-line) δε‹ξαι]. is drawn from the right-angle perpendicular to the base then the triangles around the perpendicular are similar to the whole (triangle), and to one another. [(Which is) the very thing it was required to show.]

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι ™¦ν ™ν ÑρθογωνίJ τριγώνJ ¢πÕ τÁς ÑρθÁς γωνάις ™πˆ τ¾ν βάσις κάθετος ¢χθÍ, ¹ ¢χθε‹σα τîν τÁς βάσεως τµηµάτων µέση ¢νάλογόν ™στιν· Óπερ œδει δε‹ξαι.

So (it is) clear, from this, that if, in a right-angled triangle, a (straight-line) is drawn from the right-angle perpendicular to the base then the (straight-line so) drawn is in mean proportion to the pieces of the base.† (Which is) the very thing it was required to show.

†

In other words, the perpendicular is the geometric mean of the pieces.

θ΄.

Proposition 9

ΤÁς δοθείσης εÙθείας τÕ προσταχθν µέρος ¢φελε‹ν.

To cut off a prescribed part from a given straight-line.

Γ

C

Ε

E

∆ Α

D Ζ

Β

A

F

B

”Εστω ¹ δοθε‹σα εÙθε‹α ¹ ΑΒ· δε‹ δ¾ τÁς ΑΒ τÕ προσταχθν µέρος ¢φελε‹ν. 'Επιτετάχθω δ¾ τÕ τρίτον. [καˆ] διήθχω τις ¢πÕ τοà Α εÙθε‹α ¹ ΑΓ γωνίαν περιέχουσα µετ¦ τÁς ΑΒ τυχοàσαν· κሠε„λήφθω τυχÕν σηµε‹ον ™πˆ τÁς ΑΓ τÕ ∆, κሠκείσθωσαν τÍ Α∆ ‡σαι αƒ ∆Ε, ΕΓ. κሠ™πεζεύχθω ¹ ΒΓ, κሠδι¦ τοà Α παράλληλος αÙτÍ ½χθω ¹ ∆Ζ. 'Επεˆ οâν τριγώνου τοà ΑΒΓ παρ¦ µίαν τîν πλευρîν τ¾ν ΒΓ Ãκται ¹ Ζ∆, ¢νάλογον ¥ρα ™στˆν æς ¹ Γ∆ πρÕς τ¾ν ∆Α, οÛτως ¹ ΒΖ πρÕς τ¾ν ΖΑ. διπλÁ δ ¹ Γ∆ τÁς ∆Α· διπλÁ ¥ρα κሠ¹ ΒΖ τÁς ΖΑ· τριπλÁ ¥ρα ¹ ΒΑ τÁς ΑΖ. ΤÁς ¥ρα δοθείσης εÙθείας τÁς ΑΒ τÕ ™πιταχθν τρίτον µέρος ¢φÇρηται τÕ ΑΖ· Óπερ œδει ποιÁσαι.

Let AB be the given straight-line. So it is required to cut off a prescribed part from AB. So let a third (part) have been prescribed. [And] let some straight-line AC have been drawn from (point) A, encompassing a random angle with AB. And let a random point D have been taken on AC. And let DE and EC be made equal to AD [Prop. 1.3]. And let BC have been joined. And let DF have been drawn through D parallel to it [Prop. 1.31]. Therefore, since F D has been drawn parallel to one of the sides, BC, of triangle ABC, then, proportionally, as CD is to DA, so BF (is) to F A [Prop. 6.2]. And CD (is) double DA. Thus, BF (is) also double F A. Thus, BA (is) triple AF . Thus, the prescribed third part, AF , has been cut off from the given straight-line, AB. (Which is) the very thing it was required to do.

ι΄.

Proposition 10

Τ¾ν δοθε‹σαν εÙθε‹αν ¥τµητον τÍ δοθείσV τε166

To cut a given uncut straight-line similarly to a given

ΣΤΟΙΧΕΙΩΝ $΄.

ELEMENTS BOOK 6

τµηµένV еοίως τεµε‹ν.

cut (straight-line).

Γ

C

Ε Θ

∆

Α

Ζ

E

Η

Κ

H

D

Β

A

F

G

K

B

”Εστω ¹ µν δοθε‹σα εÙθε‹α ¤τµητος ¹ ΑΒ, ¹ δ τετµηµένη ¹ ΑΓ κατ¦ τ¦ ∆, Ε σηµε‹α, κሠκείσθωσαν éστε γωνίαν τυχοàσαν περιέχειν, κሠ™πεζεύχθω ¹ ΓΒ, κሠδι¦ τîν ∆, Ε τÍ ΒΓ παράλληλοι ½χθωσαν αƒ ∆Ζ, ΕΗ, δι¦ δ τοà ∆ τÍ ΑΒ παράλληλος ½χθω ¹ ∆ΘΚ. Παραλληλόγραµον ¥ρα ™στˆν ˜κάτερον τîν ΖΘ, ΘΒ· ‡ση ¥ρα ¹ µν ∆Θ τÍ ΖΗ, ¹ δ ΘΚ τÍ ΗΒ. κሠ™πεˆ τριγώνου τοà ∆ΚΓ παρ¦ µίαν τîν πλευρîν τ¾ν ΚΓ εÙθε‹α Ãκται ¹ ΘΕ, ¢νάλογον ¥ρα ™στˆν æς ¹ ΓΕ πρÕς τ¾ν Ε∆, οÛτως ¹ ΚΘ πρÕς τ¾ν Θ∆. ‡ση δ ¹ µν ΚΘ τÍ ΒΗ, ¹ δ Θ∆ τÍ ΗΖ. œστιν ¥ρα æς ¹ ΓΕ πρÕς τ¾ν Ε∆, οÛτως ¹ ΒΗ πρÕς τ¾ν ΗΖ. πάλιν, ™πεˆ τριγώνου τοà ΑΗΕ παρ¦ µίαν τîν πλευρîν τ¾ν ΗΕ Ãκται ¹ Ζ∆, ¢νάλογον ¥ρα ™στˆν æς ¹ Ε∆ πρÕς τ¾ν ∆Α, οÛτως ¹ ΗΖ πρÕς τ¾ν ΖΑ. ™δείχθη δ κሠæς ¹ ΓΕ πρÕς τ¾ν Ε∆, οÛτως ¹ ΒΗ πρÕς τ¾ν ΗΖ· œστιν ¥ρα æς µν ¹ ΓΕ πρÕς τ¾ν Ε∆, οÛτως ¹ ΒΗ πρÕς τ¾ν ΗΖ, æς δ ¹ Ε∆ πρÕς τ¾ν ∆Α, οÛτως ¹ ΗΖ πρÕς τ¾ν ΖΑ. `Η ¥ρα δοθε‹σα εÙθε‹α ¥τµητος ¹ ΑΒ τÍ δοθείσV εÙθείv τετµηµένV τÍ ΑΓ Ðµοίως τέτµηται· Óπερ œδει ποιÁσαι·

Let AB be the given uncut straight-line, and AC a (straight-line) cut at points D and E, and let (AC) be laid down so as to encompass a random angle (with AB). And let CB have been joined. And let DF and EG have been drawn through (points) D and E (respectively), parallel to BC, and let DHK have been drawn through (point) D, parallel to AB [Prop. 1.31]. Thus, F H and HB are each parallelograms. Thus, DH (is) equal to F G, and HK to GB [Prop. 1.34]. And since the straight-line HE has been drawn parallel to one of the sides, KC, of triangle DKC, thus, proportionally, as CE is to ED, so KH (is) to HD [Prop. 6.2]. And KH (is) equal to BG, and HD to GF . Thus, as CE is to ED, so BG (is) to GF . Again, since F D has been drawn parallel to one of the sides, GE, of triangle AGE, thus, proportionally, as ED is to DA, so GF (is) to F A [Prop. 6.2]. And it was also shown that as CE (is) to ED, so BG (is) to GF . Thus, as CE is to ED, so BG (is) to GF , and as ED (is) to DA, so GF (is) to F A. Thus, the given uncut straight-line, AB, has been cut similarly to the given cut straight-line, AC. (Which is) the very thing it was required to do.

ια΄.

Proposition 11

∆ύο δοθεισîν εÙθειîν τρίτην ¢νάλογον προσευρε‹ν. ”Εστωσαν αƒ δοθε‹σαι [δύο εÙθε‹αι] αƒ ΒΑ, ΑΓ κሠκείσθωσαν γωνίαν περιέχουσαι τυχοàσαν. δε‹ δ¾ τîν ΒΑ, ΑΓ τρίτην ¢νάλογον προσευρε‹ν. ™κβεβλήσθωσαν γ¦ρ ™πˆ τ¦ ∆, Ε σηµε‹α, κሠκείσθω τÍ ΑΓ ‡ση ¹ Β∆, κሠ™πεζεύχθω ¹ ΒΓ, κሠδι¦ τοà ∆ παράλληλος αÙτÍ ½χθω ¹ ∆Ε. 'Επεˆ οâν τριγώνου τοà Α∆Ε παρ¦ µίαν τîν πλευρîν τ¾ν ∆Ε Ãκται ¹ ΒΓ, ¢νάλογόν ™στιν æς ¹ ΑΒ πρÕς τ¾ν Β∆, οÛτως ¹ ΑΓ πρÕς τ¾ν ΓΕ. ‡ση δ ¹ Β∆ τÍ ΑΓ. œστιν ¥ρα æς ¹ ΑΒ πρÕς τ¾ν ΑΓ, οÛτως ¹ ΑΓ πρÕς τ¾ν

To find a third (straight-line) proportional to two given straight-lines. Let BA and AC be the [two] given [straight-lines], and let them be laid down encompassing a random angle. So it is required to find a third (straight-line) proportional to BA and AC. For let (BA and AC) have been produced to points D and E (respectively), and let BD be made equal to AC [Prop. 1.3]. And let BC have been joined. And let DE have been drawn through (point) D parallel to it [Prop. 1.31]. Therefore, since BC has been drawn parallel to one

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ELEMENTS BOOK 6

ΓΕ.

of the sides DE of triangle ADE, proportionally, as AB is to BD, so AC (is) to CE [Prop. 6.2]. And BD (is) equal to AC. Thus, as AB is to AC, so AC (is) to CE.

Α

A

Β

B Γ

C

∆

D

Ε

E

∆ύο ¥ρα δοθεισîν εÙθειîν τîν ΑΒ, ΑΓ τρίτη Thus, a third (straight-line), CE, has been found ¢νάλογον αÙτα‹ς προσεύρηται ¹ ΓΕ· Óπερ œδει ποιÁσαι. (which is) proportional to the two given straight-lines, AB and AC. (Which is) the very thing it was required to do.

ιβ΄.

Proposition 12

Τριîν δοθεισîν εÙθειîν τετάρτην ¢νάλογον προTo find a fourth (straight-line) proportional to three σευρε‹ν. given straight-lines.

Α Β Γ

A B C

Ε

G

Η

∆

Θ

E

Ζ

D

”Εστωσαν αƒ δοθε‹σαι τρε‹ς εÙθε‹αι αƒ Α, Β, Γ· δε‹ δ¾ τîν Α, Β, Γ τετράτην ¢νάλογον προσευρε‹ν. 'Εκκείσθωσαν δύο εÙθε‹αι αƒ ∆Ε, ∆Ζ γωνίαν περιέχουσαι [τυχοàσαν] τ¾ν ØπÕ Ε∆Ζ· κሠκείσθω τÍ µν Α ‡ση ¹ ∆Η, τÍ δ Β ‡ση ¹ ΗΕ, κሠœτι τÍ Γ ‡ση ¹ ∆Θ· κሠ™πιζευχθείσης τÁς ΗΘ παράλληλος αÙτÍ ½χθω δι¦ τοà Ε ¹ ΕΖ. 'Επεˆ οâν τριγώνου τοà ∆ΕΖ παρ¦ µίαν τ¾ν ΕΖ Ãκται ¹ ΗΘ, œστιν ¥ρα æς ¹ ∆Η πρÕς τ¾ν ΗΕ, οÛτως ¹ ∆Θ πρÕς τ¾ν ΘΖ. ‡ση δ ¹ µν ∆Η τÍ Α, ¹ δ ΗΕ τÍ

H

F

Let A, B, and C be the three given straight-lines. So it is required to find a fourth (straight-line) proportional to A, B, and C. Let the two straight-lines DE and DF be set out encompassing the [random] angle EDF . And let DG be made equal to A, and GE to B, and, further, DH to C [Prop. 1.3]. And GH being joined, let EF have been drawn through (point) E parallel to it [Prop. 1.31]. Therefore, since GH has been drawn parallel to one of the sides EF of triangle DEF , thus as DG is to GE,

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Β, ¹ δ ∆Θ τÍ Γ· œστιν ¥ρα æς ¹ Α πρÕς τ¾ν Β, οÛτως so DH (is) to HF [Prop. 6.2]. And DG (is) equal to A, ¹ Γ πρÕς τ¾ν ΘΖ. and GE to B, and DH to C. Thus, as A is to B, so C (is) Τριîν ¥ρα δοθεισîν εÙθειîν τîν Α, Β, Γ τετάρτη to HF . ¢νάλογον προσεύρηται ¹ ΘΖ· Óπερ œδει ποιÁσαι. Thus, a fourth (straight-line), HF , has been found (which is) proportional to the three given straight-lines, A, B, and C. (Which is) the very thing it was required to do.

ιγ΄.

Proposition 13

∆ύο δοθεισîν εÙθειîν µέσην ¢νάλογον προσευρε‹ν.

To find the (straight-line) in mean proportion to two given straight-lines.†

∆

Α

Β

D

Γ

A

”Εστωσαν αƒ δοθε‹σαι δύο εÙθε‹αι αƒ ΑΒ, ΒΓ· δε‹ δ¾ τîν ΑΒ, ΒΓ µέσην ¢νάλογον προσευρε‹ν. Κείσθωσαν ™π' εÙθείας, κሠγεγράφθω ™πˆ τÁς ΑΓ ¹µικύκλιον τÕ Α∆Γ, κሠ½χθω ¢πÕ τοà Β σηµείου τÍ ΑΓ εÙθείv πρÕς Ñρθ¦ς ¹ ΒΑ, κሠ™πεζεύχθωσαν αƒ Α∆, ∆Γ. 'Επεˆ ™ν ¹µικυκλίJ γωνία ™στˆν ¹ ØπÕ Α∆Γ, Ñρθή ™στιν. κሠ™πεˆ ™ν ÑρθογωνίJ τριγώνJ τù Α∆Γ ¢πÕ τÁς ÑρθÁς γωνίας ™πˆ τ¾ν βάσιν κάθετος Ãκται ¹ ∆Β, ¹ ∆Β ¥ρα τîν τÁς βάσεως τµηµάτων τîν ΑΒ, ΒΓ µέση ¢νάλογόν ™στιν. ∆ύο ¥ρα δοθεισîν εÙθειîν τîν ΑΒ, ΒΓ µέση ¢νάλογον προσεύρηται ¹ ∆Β· Óπερ œδει ποιÁσαι.

†

B

C

Let AB and BC be the two given straight-lines. So it is required to find the (straight-line) in mean proportion to AB and BC. Let (AB and BC) be laid down straight-on (with respect to one another), and let the semi-circle ADC have been drawn on AC [Prop. 1.10]. And let BD have been drawn from (point) B, at right-angles to AC [Prop. 1.11]. And let AD and DC have been joined. And since ADC is an angle in a semi-circle, it is a right-angle [Prop. 3.31]. And since, in the right-angled triangle ADC, the (straight-line) DB has been drawn from the right-angle perpendicular to the base, DB is thus the mean proportional to the pieces of the base, AB and BC [Prop. 6.8 corr.]. Thus, DB has been found (which is) in mean proportion to the two given straight-lines, AB and BC. (Which is) the very thing it was required to do.

In other words, to find the geometric mean of two given straight-lines.

ιδ΄.

Proposition 14

Τîν ‡σων τε κሠ‡σογωνίων παραλληλογράµµων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας· κሠïν „σογωνίων παραλληλογράµµων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας, ‡σα ™στˆν ™κε‹να. ”Εστω ‡σα τε κሠ„σογώνια παραλληλόγραµµα τ¦ ΑΒ, ΒΓ ‡σας œχοντα τ¦ς πρÕς τù Β γωνίας, κሠκείσθωσαν ™π' εÙθείας αƒ ∆Β, ΒΕ· ™π' εÙθείας ¥ρα ε„σˆ καˆ αƒ ΖΒ, ΒΗ. λέγω, Óτι τîν ΑΒ, ΒΓ ¢ντιπεπόνθασιν

For equal and equiangular parallelograms, the sides about the equal angles are reciprocally proportional. And those equiangular parallelograms for which the sides about the equal angles are reciprocally proportional are equal. Let AB and BC be equal and equiangular parallelograms having the angles at B equal. And let DB and BE be laid down straight-on (with respect to one an-

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ELEMENTS BOOK 6

αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας, τουτέστιν, Óτι ™στˆν other) [Prop. 1.14]. Thus, F B and BG are also straightæς ¹ ∆Β πρÕς τ¾ν ΒΕ, οÛτως ¹ ΗΒ πρÕς τ¾ν ΒΖ. on (with respect to one another). I say that the sides of AB and BC about the equal angles are reciprocally proportional, that is to say, that as DB is to BE, so GB (is) to BF .

Ε

Ζ

Α

Β

Γ

E

Η

F

∆

A

B

C

G

D

Συµπεπληρώσθω γ¦ρ τÕ ΖΕ παραλληλόγραµµον. ™πεˆ οâν ‡σον ™στˆ τÕ ΑΒ παραλληλόγραµµον τù ΒΓ παραλληλογράµµJ, ¥λλο δέ τι τÕ ΖΕ, œστιν ¥ρα æς τÕ ΑΒ πρÕς τÕ ΖΕ, οÛτως τÕ ΒΓ πρÕς τÕ ΖΕ. ¢λλ' æς µν τÕ ΑΒ πρÕς τÕ ΖΕ, οÛτως ¹ ∆Β πρÕς τ¾ν ΒΕ, æς δ τÕ ΒΓ πρÕς τÕ ΖΕ, οÛτως ¹ ΗΒ πρÕς τ¾ν ΒΖ· κሠæς ¥ρα ¹ ∆Β πρÕς τ¾ν ΒΕ, οÛτως ¹ ΗΒ πρÕς τ¾ν ΒΖ. τîν ¥ρα ΑΒ, ΒΓ παραλληλογράµµων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας. 'Αλλ¦ δ¾ œστω æς ¹ ∆Β πρÕς τ¾ν ΒΕ, οÛτως ¹ ΗΒ πρÕς τ¾ν ΒΖ· λέγω, Óτι ‡σον ™στˆ τÕ ΑΒ παραλληλόγραµµον τù ΒΓ παραλληλογράµµJ. 'Επεˆ γάρ ™στιν æς ¹ ∆Β πρÕς τ¾ν ΒΕ, οÛτως ¹ ΗΒ πρÕς τ¾ν ΒΖ, ¢λλ' æς µν ¹ ∆Β πρÕς τ¾ν ΒΕ, οÛτως τÕ ΑΒ παραλληλόγραµµον πρÕς τÕ ΖΕ παραλληλόγραµµον, æς δ ¹ ΗΒ πρÕς τ¾ν ΒΖ, οÛτως τÕ ΒΓ παραλληλόγραµµον πρÕς τÕ ΖΕ παραλληλόγραµµον, κሠæς ¥ρα τÕ ΑΒ πρÕς τÕ ΖΕ, οÛτως τÕ ΒΓ πρÕς τÕ ΖΕ· ‡σον ¥ρα ™στˆ τÕ ΑΒ παραλληλόγραµµον τù ΒΓ παραλληλογράµµJ. Τîν ¥ρα ‡σων τε κሠ„σογωνίων παραλληλογράµµων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας· κሠïν „σογωνίων παραλληλογράµµων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας, ‡σα ™στˆν ™κε‹να· Óπερ œδει δε‹ξαι.

For let the parallelogram F E have been filled in. Therefore, since parallelogram AB is equal to parallelogram BC, and F E (is) some other (parallelogram), thus as (parallelogram) AB is to F E, so (parallelogram) BC (is) to F E [Prop. 5.7]. But, as (parallelogram) AB (is) to F E, so DB (is) to BE, and as (parallelogram) BC (is) to F E, so GB (is) to BF [Prop. 6.1]. Thus, also, as DB (is) to BE, so GB (is) to BF . Thus, for parallelograms AB and BC, the sides about the equal angles are reciprocally proportional. And so, let DB be to BE, as GB (is) to BF . I say that parallelogram AB is equal to parallelogram BC. For since as DB is to BE, so GB (is) to BF , but as DB (is) to BE, so parallelogram AB (is) to parallelogram F E, and as GB (is) to BF , so parallelogram BC (is) to parallelogram F E [Prop. 6.1], thus, also, as (parallelogram) AB (is) to F E, so (parallelogram) BC (is) to F E [Prop. 5.11]. Thus, parallelogram AB is equal to parallelogram BC [Prop. 5.9]. Thus, for equal and equiangular parallelograms, the sides about the equal angles are reciprocally proportional. And those equiangular parallelograms for which the sides about the equal angles are reciprocally proportional are equal. (Which is) the very thing it was required to show.

ιε΄.

Proposition 15

Τîν ‡σων κሠµίαν µι´ ‡σην ™χόντων γωνίαν τριγώνων For equal triangles also having one angle equal to one ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας· (angle), the sides about the equal angles are reciprocally κሠïν µίαν µι´ ‡σην ™χόντων γωνίαν τριγώνων ¢ντι- proportional. And those triangles having one angle equal πεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας, ‡σα ™στˆν to one angle for which the sides about the equal angles

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ELEMENTS BOOK 6

™κε‹να. ”Εστω ‡σα τρίγωνα τ¦ ΑΒΓ, Α∆Ε µίαν µι´ ‡σην œχοντα γωνίαν τ¾ν ØπÕ ΒΑΓ τÍ ØπÕ ∆ΑΕ· λέγω, Óτι τîν ΑΒΓ, Α∆Ε τριγώνων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας, τουτέστιν, Óτι ™στˆν æς ¹ ΓΑ πρÕς τ¾ν Α∆, οÛτως ¹ ΕΑ πρÕς τ¾ν ΑΒ.

(are) reciprocally proportional are equal. Let ABC and ADE be equal triangles having one angle equal to one (angle), (namely) BAC (equal) to DAE. I say that, for triangles ABC and ADE, the sides about the equal angles are reciprocally proportional, that is to say, that as CA is to AD, so EA (is) to AB.

Β

B

Γ

Α

∆

C

A

Ε

D

Κείσθω γ¦ρ éστε ™π' εÙθείας εναι τ¾ν ΓΑ τÍ Α∆· ™π' εÙθείας ¥ρα ™στˆ κሠ¹ ΕΑ τÍ ΑΒ. κሠ™πεζεύχθω ¹ Β∆. 'Επεˆ οâν ‡σον ™στˆ τÕ ΑΒΓ τρίγωνον τù Α∆Ε τριγώνJ, ¥λλο δέ τι τÕ ΒΑ∆, œστιν ¥ρα æς τÕ ΓΑΒ τρίγωνον πρÕς τÕ ΒΑ∆ τρίγωνον, οÛτως τÕ ΕΑ∆ τρίγωνον πρÕς τÕ ΒΑ∆ τρίγωνον. ¢λλ' æς µν τÕ ΓΑΒ πρÕς τÕ ΒΑ∆, οÛτως ¹ ΓΑ πρÕς τ¾ν Α∆, æς δ τÕ ΕΑ∆ πρÕς τÕ ΒΑ∆, οÛτως ¹ ΕΑ πρÕς τ¾ν ΑΒ. κሠæς ¥ρα ¹ ΓΑ πρÕς τ¾ν Α∆, οÛτως ¹ ΕΑ πρÕς τ¾ν ΑΒ. τîν ΑΒΓ, Α∆Ε ¥ρα τριγώνων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας. 'Αλλ¦ δ¾ ¢ντιπεπονθέτωσαν αƒ πλευρሠτîν ΑΒΓ, Α∆Ε τριγώνων, κሠœστω æς ¹ ΓΑ πρÕς τ¾ν Α∆, οÛτως ¹ ΕΑ πρÕς τ¾ν ΑΒ· λέγω, Óτι ‡σον ™στˆ τÕ ΑΒΓ τρίγωνον τù Α∆Ε τριγώνJ. 'Επιζευχθείσης γ¦ρ πάλιν τÁς Β∆, ™πεί ™στιν æς ¹ ΓΑ πρÕς τ¾ν Α∆, οÛτως ¹ ΕΑ πρÕς τ¾ν ΑΒ, ¢λλ' æς µν ¹ ΓΑ πρÕς τ¾ν Α∆, οÛτως τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΒΑ∆ τρίγωνον, æς δ ¹ ΕΑ πρÕς τ¾ν ΑΒ, οÛτως τÕ ΕΑ∆ τρίγωνον πρÕς τÕ ΒΑ∆ τρίγωνον, æς ¥ρα τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΒΑ∆ τρίγωνον, οÛτως τÕ ΕΑ∆ τρίγωνον πρÕς τÕ ΒΑ∆ τρίγωνον. ˜κάτερον ¥ρα τîν ΑΒΓ, ΕΑ∆ πρÕς τÕ ΒΑ∆ τÕν αÙτÕν œχει λόγον. ‡σων ¥ρα ™στˆ τÕ ΑΒΓ [τρίγωνον] τù ΕΑ∆ τριγώνJ. Τîν ¥ρα ‡σων κሠµίαν µι´ ‡σην ™χόντων γωνίαν τριγώνων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας· κሠïς µίαν µι´ ‡σην ™χόντων γωνίαν τριγώνων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας, ™κε‹να ‡σα ™στˆν· Óπερ œδει δε‹ξαι.

E

For let CA be laid down so as to be straight-on (with respect) to AD. Thus, EA is also straight-on (with respect) to AB [Prop. 1.14]. And let BD have been joined. Therefore, since triangle ABC is equal to triangle ADE, and BAD (is) some other (triangle), thus as triangle CAB is to triangle BAD, so triangle EAD (is) to triangle BAD [Prop. 5.7]. But, as (triangle) CAB (is) to BAD, so CA (is) to AD, and as (triangle) EAD (is) to BAD, so EA (is) to AB [Prop. 6.1]. And thus, as CA (is) to AD, so EA (is) to AB. Thus, for triangles ABC and ADE, the sides about the equal angles (are) reciprocally proportional. And so, let the sides of triangles ABC and ADE be reciprocally proportional, and (thus) let CA be to AD, as EA (is) to AB. I say that triangle ABC is equal to triangle ADE. For, BD again being joined, since as CA is to AD, so EA (is) to AB, but as CA (is) to AD, so triangle ABC (is) to triangle BAD, and as EA (is) to AB, so triangle EAD (is) to triangle BAD [Prop. 6.1], thus as triangle ABC (is) to triangle BAD, so triangle EAD (is) to triangle BAD. Thus, (triangles) ABC and EAD each have the same ratio to BAD. Thus, [triangle] ABC is equal to triangle EAD [Prop. 5.9]. Thus, for equal triangles also having one angle equal to one (angle), the sides about the equal angles (are) reciprocally proportional. And those triangles having one angle equal to one angle for which the sides about the equal angles (are) reciprocally proportional are equal. (Which is) the very thing it was required to show.

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ELEMENTS BOOK 6 ι$΄.

Proposition 16

'Ε¦ν τέσσαρες εÙθε‹αι ¢νάλογον ðσιν, τÕ ØπÕ τîν ¥κρων περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ØπÕ τîν µέσων περιεχοµένJ ÑρθογωνίJ· κ¨ν τÕ ØπÕ τîν ¥κρων περιεχόµενον Ñρθογώνιον ‡σον Ï τù ØπÕ τîν µέσων περιεχοµένJ ÑρθογωνίJ, αƒ τέσσαρες εÙθε‹αι ¢νάλογον œσονται.

If four straight-lines are proportional, then the rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two). And if the rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two), then the four straight-lines will be proportional.

Θ

H

Η

Α Ε

G

Β

Γ

∆

A

Ζ

E

”Εστωσαν τέσσαρες εÙθε‹αι ¢νάλογον αƒ ΑΒ, Γ∆, Ε, Ζ, æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ Ε πρÕς τ¾ν Ζ· λέγω, Óτι τÕ ØπÕ τîν ΑΒ, Ζ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ØπÕ τîν Γ∆, Ε περιεχοµένJ ÑρθογωνίJ. ”Ηχθωσαν [γ¦ρ] ¢πÕ τîν Α, Γ σηµείων τα‹ς ΑΒ, Γ∆ εÙθείαις πρÕς Ñρθ¦ς αƒ ΑΗ, ΓΘ, κሠκείσθω τÍ µν Ζ ‡ση ¹ ΑΗ, τÍ δ Ε ‡ση ¹ ΓΘ. κሠσυµπεπληρώσθω τ¦ ΒΗ, ∆Θ παραλληλόγραµµα. Κሠ™πεί ™στιν æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ Ε πρÕς τ¾ν Ζ, ‡ση δ ¹ µν Ε τÍ ΓΘ, ¹ δ Ζ τÍ ΑΗ, œστιν ¥ρα æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΓΘ πρÕς τ¾ν ΑΗ. τîν ΒΗ, ∆Θ ¥ρα παραλληλογράµµων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας. ïν δ „σογωνίων παραλληλογράµµων ¢ντιπεπόνθασιν αƒ πλευραί αƒ περˆ τ¦ς ‡σας γωνάις, ‡σα ™στˆν ™κε‹να· ‡σον ¥ρα ™στˆ τÕ ΒΗ παραλληλόγραµµον τù ∆Θ παραλληλογράµµJ. καί ™στι τÕ µν ΒΗ τÕ ØπÕ τîν ΑΒ, Ζ· ‡ση γ¦ρ ¹ ΑΗ τÍ Ζ· τÕ δ ∆Θ τÕ ØπÕ τîν Γ∆, Ε· ‡ση γ¦ρ ¹ Ε τÍ ΓΘ· τÕ ¥ρα ØπÕ τîν ΑΒ, Ζ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ØπÕ τîν Γ∆, Ε περιεχόµενJ ÑρθογώνιJ. 'Αλλ¦ δ¾ τÕ ØπÕ τîν ΑΒ, Ζ περιεχόµενον Ñρθογώνιον ‡σον œστω τù ØπÕ τîν Γ∆, Ε περιεχοµένJ ÑρθογωνίJ. λέγω, Óτι αƒ τέσσαρες εÙθε‹αι ¢νάλογον œσονται, æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ Ε πρÕς τ¾ν Ζ. Τîν γ¦ρ αÙτîν κατασκευασθέντων, ™πεˆ τÕ ØπÕ τîν ΑΒ, Ζ ‡σον ™στˆ τù ØπÕ τîν Γ∆, Ε, καί ™στι τÕ µν ØπÕ τîν ΑΒ, Ζ τÕ ΒΗ· ‡ση γάρ ™στιν ¹ ΑΗ τÍ Ζ· τÕ δ ØπÕ τîν Γ∆, Ε τÕ ∆Θ· ‡ση γ¦ρ ¹ ΓΘ τÍ Ε· τÕ ¥ρα ΒΗ ‡σον ™στˆ τù ∆Θ. καί ™στιν „σογώνια. τîν δ ‡σων κሠ„σογωνίων παραλληλογράµµων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας. œστιν ¥ρα æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΓΘ πρÕς τ¾ν ΑΗ. ‡ση δ ¹ µν ΓΘ τÍ Ε, ¹ δ ΑΗ τÍ Ζ· œστιν ¥ρα æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ Ε πρÕς τ¾ν Ζ. 'Ε¦ν ¥ρα τέσσαρες εÙθε‹αι ¢νάλογον ðσιν, τÕ ØπÕ τîν ¥κρων περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ØπÕ

B

C

D

F

Let AB, CD, E, and F be four proportional straightlines, (such that) as AB (is) to CD, so E (is) to F . I say that the rectangle contained by AB and F is equal to the rectangle contained by CD and E. [For] let AG and CH have been drawn from points A and C at right-angles to the straight-lines AB and CD (respectively) [Prop. 1.11]. And let AG be made equal to F , and CH to E [Prop. 1.3]. And let the parallelograms BG and DH have been completed. And since as AB is to CD, so E (is) to F , and E (is) equal CH, and F to AG, thus as AB is to CD, so CH (is) to AG. Thus, for the parallelograms BG and DH, the sides about the equal angles are reciprocally proportional. And those equiangular parallelograms for which the sides about the equal angles are reciprocally proportional are equal [Prop. 6.14]. Thus, parallelogram BG is equal to parallelogram DH. And BG is the (rectangle contained) by AB and F . For AG (is) equal to F . And DH (is) the (rectangle contained) by CD and E. For E (is) equal to CH. Thus, the rectangle contained by AB and F is equal to the rectangle contained by CD and E. And so, let the rectangle contained by AB and F be equal to the rectangle contained by CD and E. I say that the four straight-lines will be proportional, (so that) as AB (is) to CD, so E (is) to F . For, with the same construction, since the (rectangle contained) by AB and F is equal to the (rectangle contained) by CD and E, and BG is the (rectangle contained) by AB and F . For AG is equal to F . And DH (is) the (rectangle contained) by CD and E. For CH (is) equal to E. BG is thus equal to DH. And they are equiangular. And for equal and equiangular parallelograms, the sides about the equal angles are reciprocally proportional [Prop. 6.14]. Thus, as AB is to CD, so CH (is) to AG. And CH (is) equal to E, and AG to F . Thus, as AB is to CD, so E (is) to F .

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ELEMENTS BOOK 6

τîν µέσων περιεχοµένJ ÑρθογωνίJ· κ¨ν τÕ ØπÕ τîν Thus, if four straight-lines are proportional, then the ¥κρων περιεχόµενον Ñρθογώνιον ‡σον Ï τù ØπÕ τîν rectangle contained by the (two) outermost is equal to µέσων περιεχοµένJ ÑρθογωνίJ, αƒ τέσσαρες εÙθε‹αι the rectangle contained by the middle (two). And if the ¢νάλογον œσονται· Óπερ œδει δε‹ξαι. rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two), then the four straight-lines will be proportional. (Which is) the very thing it was required to show.

ιζ΄.

Proposition 17

'Ε¦ν τρε‹ς εÙθε‹αι ¢νάλογον ðσιν, τÕ ØπÕ τîν ¥κρων περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ¢πÕ τÁς µέσης τετραγώνJ· κ¨ν τÕ ØπÕ τîν ¥κρων περιεχόµενον Ñρθογώνιον ‡σον Ï τù ¢πÕ τÁς µέσης τετραγώνJ, αƒ τρε‹ς εÙθε‹αι ¢νάλογον œσονται.

If three straight-lines are proportional, then the rectangle contained by the (two) outermost is equal to the square on the middle (one). And if the rectangle contained by the (two) outermost is equal to the square on the middle (one), then the three straight-lines will be proportional.

Α Β Γ

A B C

∆

”Εστωσαν τρε‹ς εÙθε‹αι ¢νάλογον αƒ Α, Β, Γ, æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Β πρÕς τ¾ν Γ· λέγω, Óτι τÕ ØπÕ τîν Α, Γ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ¢πÕ τÁς Β τετραγώνJ. Κείσθω τÍ Β ‡ση ¹ ∆. Κሠ™πεί ™στιν æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Β πρÕς τ¾ν Γ, ‡ση δ ¹ Β τÍ ∆, œστιν ¥ρα æς ¹ Α πρÕς τ¾ν Β, ¹ ∆ πρÕς τ¾ν Γ. ™¦ν δ τέσσαρες εÙθε‹αι ¢νάλογον ðσιν, τÕ ØπÕ τîν ¥κρων περιεχόµενον [Ñρθογώνιον] ‡σον ™στˆ τù ØπÕ τîν µέσων περιεχοµένJ ÑρθογωνίJ. τÕ ¥ρα ØπÕ τîν Α, Γ ‡σον ™στˆ τù ØπÕ τîν Β, ∆. ¢λλ¦ τÕ ØπÕ τîν Β, ∆ τÕ ¢πÕ τÁς Β ™στιν· ‡ση γ¦ρ ¹ Β τÍ ∆· τÕ ¥ρα ØπÕ τîν Α, Γ περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ¢πÕ τÁς Β τετραγώνJ. 'Αλλ¦ δ¾ τÕ ØπÕ τîν Α, Γ ‡σον œστω τù ¢πÕ τÁς Β· λέγω, Óτι ™στˆν æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Β πρÕς τ¾ν Γ. Τîν γ¦ρ αÙτîν κατασκευασθέντων, ™πεˆ τÕ ØπÕ τîν Α, Γ ‡σον ™στˆ τù ¢πÕ τÁς Β, ¢λλ¦ τÕ ¢πÕ τÁς Β τÕ ØπÕ τîν Β, ∆ ™στιν· ‡ση γ¦ρ ¹ Β τÍ ∆· τÕ ¥ρα ØπÕ τîν Α, Γ ‡σον ™στˆ τù ØπÕ τîν Β, ∆. ™¦ν δ τÕ ØπÕ τîν ¥κρων ‡σον Ï τù ØπÕ τîν µέσων, αƒ τέσσαρες εÙθε‹αι ¢νάλογόν ε„σιν. œστιν ¥ρα æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ ∆ πρÕς τ¾ν Γ. ‡ση δ ¹ Β τÍ ∆· æς ¥ρα ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Β πρÕς τ¾ν Γ. 'Ε¦ν ¥ρα τρε‹ς εÙθε‹αι ¢νάλογον ðσιν, τÕ ØπÕ τîν ¥κρων περιεχόµενον Ñρθογώνιον ‡σον ™στˆ τù ¢πÕ τÁς µέσης τετραγώνJ· κ¨ν τÕ ØπÕ τîν ¥κρων περιεχόµενον Ñρθογώνιον ‡σον Ï τù ¢πÕ τÁς µέσης τετραγώνJ, αƒ

D

Let A, B and C be three proportional straight-lines, (such that) as A (is) to B, so B (is) to C. I say that the rectangle contained by A and C is equal to the square on B. Let D be made equal to B [Prop. 1.3]. And since as A is to B, so B (is) to D, and B (is) equal to D, thus as A is to B, (so) D (is) to C. And if four straight-lines are proportional, then the [rectangle] contained by the (two) outermost is equal to the rectangle contained by the middle (two) [Prop. 6.16]. Thus, the (rectangle contained) by A and C is equal to the (rectangle contained) by B and D. But, the (rectangle contained) by B and D is the (square) on B. For B (is) equal to D. Thus, the rectangle contained by A and C is equal to the square on B. And so, let the (rectangle contained) by A and C be equal to the (square) on B. I say that as A is to B, so B (is) to C. For, with the same construction, since the (rectangle contained) by A and C is equal to the (square) on B. But, the (square) on B is the (rectangle contained) by B and D. For B (is) equal to D. The (rectangle contained) by A and C is thus equal to the (rectangle contained) by B and D. And if the (rectangle contained) by the (two) outermost is equal to the (rectangle contained) by the middle (two), then the four straight-lines are proportional [Prop. 6.16]. Thus, as A is to B, so D (is) to C. And B (is) equal to D. Thus, as A (is) to B, so B (is) to C.

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ELEMENTS BOOK 6

τρε‹ς εÙθε‹αι ¢νάλογον œσονται· Óπερ œδει δε‹ξαι.

Thus, if three straight-lines are proportional, then the rectangle contained by the (two) outermost is equal to the square on the middle (one). And if the rectangle contained by the (two) outermost is equal to the square on the middle (one), then the three straight-lines will be proportional. (Which is) the very thing it was required to show.

ιη΄.

Proposition 18

'ΑπÕ τÁς δοθείσης εÙθείας τù δοθέντι εÙθυγράµµJ Óµοιόν τε καˆ Ðµοίως κείµενον εÙθύγραµµον ¢ναγράψαι.

To describe a rectilinear figure similar, and similarly laid down, to a given rectilinear figure on a given straight-line.

Ε

E

Ζ

F Θ

Η Γ

∆

Α

Β

H G

C

”Εστω ¹ µν δοθε‹σα εÙθε‹α ¹ ΑΒ, τÕ δ δοθν εÙθύγραµµον τÕ ΓΕ· δε‹ δ¾ ¢πÕ τ¾ς ΑΒ εÙθείας τù ΓΕ εÙθυγράµµJ Óµοιόν τε καˆ Ðµοίως κείµενον εÙθύγραµµον ¢ναγράψαι. 'Επεζεύχθω ¹ ∆Ζ, κሠσυνεστάτω πρÕς τÍ ΑΒ εÙθείv κሠτο‹ς πρÕς αÙτÍ σηµείοις το‹ς Α, Β τÍ µν πρÕς τù Γ γωνίv ‡ση ¹ ØπÕ ΗΑΒ, τÍ δ ØπÕ Γ∆Ζ ‡ση ¹ ØπÕ ΑΒΗ. λοιπ¾ ¥ρα ¹ ØπÕ ΓΖ∆ τÍ ØπÕ ΑΗΒ ™στιν ‡ση· „σογώνιον ¥ρα ™στˆ τÕ ΖΓ∆ τρίγωνον τù ΗΑΒ τριγώνJ. ¢νάλογον ¥ρα ™στˆν æς ¹ Ζ∆ πρÕς τ¾ν ΗΒ, οÛτως ¹ ΖΓ πρÕς τ¾ν ΗΑ, κሠ¹ Γ∆ πρÕς τ¾ν ΑΒ. πάλιν συνεστάτω πρÕς τÍ ΒΗ εÙθείv κሠτο‹ς πρÕς αÙτÍ σηµείοις το‹ς Β, Η τÍ µν ØπÕ ∆ΖΕ γωνίv ‡ση ¹ ØπÕ ΒΗΘ, τÍ δ ØπÕ Ζ∆Ε ‡ση ¹ ØπÕ ΗΒΘ. λοιπ¾ ¥ρα ¹ πρÕς τù Ε λοιπÍ τÍ πρÕς τù Θ ™στιν ‡ση· „σογώνιον ¥ρα ™στˆ τÕ Ζ∆Ε τρίγωνον τù ΗΘΒ τριγώνJ· ¢νάλογον ¥ρα ™στˆν æς ¹ Ζ∆ πρÕς τ¾ν ΗΒ, οÛτως ¹ ΖΕ πρÕς τ¾ν ΗΘ κሠ¹ Ε∆ πρÕς τ¾ν ΘΒ. ™δείχθη δ κሠæς ¹ Ζ∆ πρÕς τ¾ν ΗΒ, οÛτως ¹ ΖΓ πρÕς τ¾ν ΗΑ κሠ¹ Γ∆ πρÕς τ¾ν ΑΒ· κሠæς ¥ρα ¹ ΖΓ πρÕς τ¾ν ΑΗ, οÛτως ¼ τε Γ∆ πρÕς τ¾ν ΑΒ κሠ¹ ΖΕ πρÕς τ¾ν ΗΘ κሠœτι ¹ ΕΑ πρÕς τ¾ν ΘΒ. κሠ™πεˆ ‡ση ™στˆν ¹ µν ØπÕ ΓΖ∆ γωνία τÍ ØπÕ ΑΗΒ, ¹ δ ØπÕ ∆ΖΕ τÍ ØπÕ ΒΗΘ, Óλη ¥ρα ¹ ØπÕ ΓΖΕ ÓλV τÍ ØπÕ ΑΗΘ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ Γ∆Ε τÍ ØπÕ ΑΒΘ ™στιν ‡ση. œστι δ κሠ¹ µν πρÕς τù Γ τÍ πρÕς τù Α ‡ση, ¹ δ πρÕς τù Ε τÍ πρÕς τù Θ. „σογώνιον ¥ρα ™στˆ τÕ ΑΘ τù ΓΕ· κሠτ¦ς περˆ τ¦ς ‡σας γωνίας αÙτîν πλευρ¦ς ¢νάλογον œχει· Óµοιον ¥ρα ™στˆ τÕ ΑΘ

D

A

B

Let AB be the given straight-line, and CE the given rectilinear figure. So it is required to describe a rectilinear figure similar, and similarly laid down, to the rectilinear figure CE on the straight-line AB. Let DF have been joined, and let GAB, equal to the angle at C, and ABG, equal to (angle) CDF , have been constructed at the points A and B (respectively) on the straight-line AB [Prop. 1.23]. Thus, the remaining (angle) CF D is equal to AGB [Prop. 1.32]. Thus, triangle F CD is equiangular to triangle GAB. Thus, proportionally, as F D is to GB, so F C (is) to GA, and CD to AB [Prop. 6.4]. Again, let BGH, equal to angle DF E, and GBH equal to (angle) F DE, have been constructed at the points G and B (respectively) on the straight-line BG [Prop. 1.23]. Thus, the remaining (angle) at E is equal to the remaining (angle) at H [Prop. 1.32]. Thus, triangle F DE is equiangular to triangle GHB. Thus, proportionally, as F D is to GB, so F E (is) to GH, and ED to HB [Prop. 6.4]. And it was also shown (that) as F D (is) to GB, so F C (is) to GA, and CD to AB. Thus, also, as F C (is) to AG, so CD (is) to AB, and F E to GH, and, further, ED to HB. And since angle CF D is equal to AGB, and DF E to BGH, thus the whole (angle) CF E is equal to the whole (angle) AGH. So, for the same (reasons), (angle) CDE is also equal to ABH. And the (angle) at C is also equal to the (angle) at A, and the (angle) at E to the (angle) at H. Thus, (figure) AH is equiangular to CE. And (the two figures) have the sides about

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ELEMENTS BOOK 6

εÙθύγραµµον τù ΓΕ εÙθυγράµµJ. 'ΑπÕ τÁς δοθείσης ¥ρα εÙθείας τÁς ΑΒ τù δοθέντι εÙθυγράµµJ τù ΓΕ Óµοιόν τε καˆ Ðµοίως κείµενον εÙθύγραµµον ¢ναγέγραπται τÕ ΑΘ· Óπερ œδει ποιÁσαι.

their equal angles proportional. Thus, the rectilinear figure AH is similar to the rectilinear figure CE [Def. 6.1]. Thus, the rectilinear figure AH, similar, and similarly laid down, to the given rectilinear figure CE has been constructed on the given straight-line AB. (Which is) the very thing it was required to do.

ιθ΄.

Proposition 19

Τ¦ Óµοια τρίγωνα πρÕς ¥λληλα ™ν διπλασίονι λόγJ Similar triangles are to one another in the squared† ™στˆ τîν еολόγων πλευρîν. ratio of (their) corresponding sides.

Α

A ∆

Β

Η

Γ

Ε

D

Ζ

B

”Εστω Óµοια τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ ‡σην œχοντα τ¾ν πρÕς τù Β γωνίαν τÍ πρÕς τù Ε, æς δ τ¾ν ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τ¾ν ∆Ε πρÕς τ¾ν ΕΖ, éστε еόλογον εναι τ¾ν ΒΓ τÍ ΕΖ· λέγω, Óτι τÕ ΑΒΓ τρίγωνον πρÕς τÕ ∆ΕΖ τρίγωνον διπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ. Ε„λήφθω γ¦ρ τîν ΒΓ, ΕΖ τρίτη ¢νάλογον ¹ ΒΗ, éστε εναι æς τ¾ν ΒΓ πρÕς τ¾ν ΕΖ, οÛτως τ¾ν ΕΖ πρÕς τ¾ν ΒΗ· κሠ™πεζεύχθω ¹ ΑΗ. 'Επεˆ οâν ™στιν æς ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως ¹ ∆Ε πρÕς τ¾ν ΕΖ, ™ναλλ¦ξ ¥ρα ™στˆν æς ¹ ΑΒ πρÕς τ¾ν ∆Ε, οÛτως ¹ ΒΓ πρÕς τ¾ν ΕΖ. ¢λλ' æς ¹ ΒΓ πρÕς ΕΖ, οÛτως ™στιν ¹ ΕΖ πρÕς ΒΗ. κሠæς ¥ρα ¹ ΑΒ πρÕς ∆Ε, οÛτως ¹ ΕΖ πρÕς ΒΗ· τîν ΑΒΗ, ∆ΕΖ ¥ρα τριγώνων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνάις. ïν δ µίαν µι´ ‡σην ™χόντων γωνίαν τριγώνων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνάις, ‡σα ™στˆν ™κε‹να. ‡σον ¥ρα ™στˆ τÕ ΑΒΗ τρίγωνον τù ∆ΕΖ τριγώνJ. κሠ™πεί ™στιν æς ¹ ΒΓ πρÕς τ¾ν ΕΖ, οÛτως ¹ ΕΖ πρÕς τ¾ν ΒΗ, ™¦ν δ τρε‹ς εÙθε‹αι ¢νάλογον ðσιν, ¹ πρώτη πρÕς τ¾ν τρίτην διπλασίονα λόγον œχει ½περ πρÕς τ¾ν δευτέραν, ¹ ΒΓ ¥ρα πρÕς τ¾ν ΒΗ διπλασίονα λόγον œχει ½περ ¹ ΓΒ πρÕς τ¾ν ΕΖ. æς δ ¹ ΓΒ πρÕς τ¾ν ΒΗ, οÛτως τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΑΒΗ τρίγωνον· κሠτÕ ΑΒΓ ¥ρα τρίγωνον πρÕς τÕ ΑΒΗ διπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ. ‡σον δ τÕ ΑΒΗ τρίγωνον τù ∆ΕΖ τριγώνJ. κሠτÕ ΑΒΓ ¥ρα τρίγωνον πρÕς τÕ ∆ΕΖ τρίγωνον διπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ. Τ¦ ¥ρα Óµοια τρίγωνα πρÕς ¥λληλα ™ν διπλασίονι λόγJ ™στˆ τîν еολόγων πλευρîν. [Óπερ œδει δε‹ξαι.]

G

C

E

F

Let ABC and DEF be similar triangles having the angle at B equal to the (angle) at E, and AB to BC, as DE (is) to EF , such that BC corresponds to EF . I say that triangle ABC has a squared ratio to triangle DEF with respect to (that side) BC (has) to EF . For let a third (straight-line), BG, have been taken (which is) proportional to BC and EF , so that as BC (is) to EF , so EF (is) to BG [Prop. 6.11]. And let AG have been joined. Therefore, since as AB is to BC, so DE (is) to EF , thus, alternately, as AB is to DE, so BC (is) to EF [Prop. 5.16]. But, as BC (is) to EF , so EF is to BG. And, thus, as AB (is) to DE, so EF (is) to BG. Thus, for triangles ABG and DEF , the sides about the equal angles are reciprocally proportional. And those triangles having one (angle) equal to one (angle) for which the sides about the equal angles are reciprocally proportional are equal [Prop. 6.15]. Thus, triangle ABG is equal to triangle DEF . And since as BC (is) to EF , so EF (is) to BG, and if three straight-lines are proportional then the first has a squared ratio to the third with respect to the second [Def. 5.9], BC thus has a squared ratio to BG with respect to (that) CB (has) to EF . And as CB (is) to BG, so triangle ABC (is) to triangle ABG [Prop. 6.1]. Thus, triangle ABC also has a squared ratio to (triangle) ABG with respect to (that side) BC (has) to EF . And triangle ABG (is) equal to triangle DEF . Thus, triangle ABC also has a squared ratio to triangle DEF with respect to (that side) BC (has) to EF . Thus, similar triangles are to one another in the squared ratio of (their) corresponding sides. [(Which

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ELEMENTS BOOK 6 is) the very thing it was required to show].

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι, ™¦ν τρε‹ς εÙθε‹αι So it is clear, from this, that if three straight-lines are ¢νάλογον ðσιν, œστιν æς ¹ πρώτη πρÕς τ¾ν τρίτην, οÛτως proportional, then as the first is to the third, so the figure τÕ ¢πÕ τÁς πρώτης εδος πρÕς τÕ ¢πÕ τÁς δευτέρας τÕ (described) on the first (is) to the similar, and similarly Óµοιον καˆ Ðµοίως ¢ναγραφόµενον. Óπερ œδει δε‹ξαι. described, (figure) on the second. (Which is) the very thing it was required to show. †

Literally, “double”.

κ΄.

Proposition 20

Τ¦ Óµοια πολύγωνα ε‡ς τε Óµοια τρίγωνα διαιρε‹ται Similar polygons can be divided into equal numbers κሠε„ς ‡σα τÕ πλÁθος καˆ Ðµόλογα το‹ς Óλοις, κሠτÕ of similar triangles corresponding (in proportion) to the πολύγωνον πρÕς τÕ πολύγωνον διπλασίονα λόγον œχει wholes, and one polygon has to the (other) polygon a ½περ ¹ еόλογος πλευρ¦ πρÕς τ¾ν еόλογον πλευράν. squared ratio with respect to (that) a corresponding side (has) to a corresponding side.

Α Β

A Ζ Ε

Μ

Η

E

Λ

Ν

Θ Γ

F B

G

M

Κ

L N

H

∆

C

”Εστω Óµοια πολύγωνα τ¦ ΑΒΓ∆Ε, ΖΗΘΚΛ, еόλογος δ œστω ¹ ΑΒ τÍ ΖΗ· λέγω, Óτι τ¦ ΑΒΓ∆Ε, ΖΗΘΚΛ πολύγωνα ε‡ς τε Óµοια τρίγωνα διαιρε‹ται κሠε„ς ‡σα τÕ πλÁθος καˆ Ðµόλογα το‹ς Óλοις, κሠτÕ ΑΒΓ∆Ε πολύγωνον πρÕς τÕ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον œχει ½περ ¹ ΑΒ πρÕς τ¾ν ΖΗ. 'Επεζεύχθωσαν αƒ ΒΕ, ΕΓ, ΗΛ, ΛΘ. Κሠ™πεˆ Óµοιόν ™στι τÕ ΑΒΓ∆Ε πολύγωνον τù ΖΗΘΚΛ πολυγώνJ, ‡ση ™στˆν ¹ ØπÕ ΒΑΕ γωνία τÍ ØπÕ ΗΖΛ. καί ™στιν æς ¹ ΒΑ πρÕς ΑΕ, οÛτως ¹ ΗΖ πρÕς ΖΛ. ™πεˆ οâν δύο τρίγωνά ™στι τ¦ ΑΒΕ, ΖΗΛ µίαν γωνίαν µι´ γωνίv ‡σην œχοντα, περˆ δ τ¦ς ‡σας γωνίας τ¦ς πλευρ¦ς ¢νάλογον, „σογώνιον ¥ρα ™στˆ τÕ ΑΒΕ τρίγωνον τù ΖΗΛ τριγώνJ· éστε καˆ Óµοιον· ‡ση ¥ρα ™στˆν ¹ ØπÕ ΑΒΕ γωνία τÍ ØπÕ ΖΗΛ. œστι δ κሠÓλη ¹ ØπÕ ΑΒΓ ÓλV τÍ ØπÕ ΖΗΘ ‡ση δι¦ τ¾ν еοιότητα τîν πολυγώνων· λοιπ¾ ¥ρα ¹ ØπÕ ΕΒΓ γωνία τÍ ØπÕ ΛΗΘ ™στιν ‡ση. κሠ™πεˆ δι¦ τ¾ν еοιότητα τîν ΑΒΕ, ΖΗΛ τριγώνων ™στˆν æς ¹ ΕΒ πρÕς ΒΑ, οÛτως ¹ ΛΗ πρÕς ΗΖ, ¢λλ¦ µ¾ν κሠδι¦ τ¾ν еοιότητα τîν πολυγώνων ™στˆν æς ¹ ΑΒ πρÕς ΒΓ, οÛτως ¹ ΖΗ πρÕς ΗΘ, δι' ‡σου ¥ρα ™στˆν æς ¹ ΕΒ πρÕς ΒΓ, οÛτως ¹

K

D

Let ABCDE and F GHKL be similar polygons, and let AB correspond to F G. I say that polygons ABCDE and F GHKL can be divided into equal numbers of similar triangles corresponding (in proportion) to the wholes, and (that) polygon ABCDE has a squared ratio to polygon F GHKL with respect to that AB (has) to F G. Let BE, EC, GL, and LH have been joined. And since polygon ABCDE is similar to polygon F GHKL, angle BAE is equal to angle GF L, and as BA is to AE, so GF (is) to F L [Def. 6.1]. Therefore, since ABE and F GL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, triangle ABE is thus equiangular to triangle F GL [Prop. 6.6]. Hence, (they are) also similar [Prop. 6.4, Def. 6.1]. Thus, angle ABE is equal to (angle) F GL. And the whole (angle) ABC is equal to the whole (angle) F GH, on account of the similarity of the polygons. Thus, the remaining angle EBC is equal to LGH. And since, on account of the similarity of triangles ABE and F GL, as EB is to BA, so LG (is) to GF , but also, on account of the similarity of the polygons, as AB is to BC, so F G (is) to GH, thus, via equality, as EB is to BC, so LG (is) to

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ELEMENTS BOOK 6

ΛΗ πρÕς ΗΘ, κሠπερˆ τ¦ς ‡σας γωνάις τ¦ς ØπÕ ΕΒΓ, ΛΗΘ αƒ πλευρሠ¢νάλογόν ε„σιν· „σογώνιον ¥ρα ™στˆ τÕ ΕΒΓ τρίγωνον τù ΛΗΘ τριγώνJ· éστε καˆ Óµοιόν ™στι τÕ ΕΒΓ τρίγωνον τù ΛΗΘ τριγώνω. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ΕΓ∆ τρίγωνον Óµοιόν ™στι τù ΛΘΚ τριγώνJ. τ¦ ¥ρα Óµοια πολύγωνα τ¦ ΑΒΓ∆Ε, ΖΗΘΚΛ ε‡ς τε Óµοια τρίγωνα διÇρηται κሠε„ς ‡σα τÕ πλÁθος. Λέγω, Óτι καˆ Ðµόλογα το‹ς Óλοις, τουτέστιν éστε ¢νάλογον εναι τ¦ τρίγωνα, κሠ¹γούµενα µν εναι τ¦ ΑΒΕ, ΕΒΓ, ΕΓ∆, ˜πόµενα δ αÙτîν τ¦ ΖΗΛ, ΛΗΘ, ΛΘΚ, κሠÓτι τÕ ΑΒΓ∆Ε πολύγωνον πρÕς τÕ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον œχει ½περ ¹ еόλογος πλευρ¦ πρÕς τ¾ν еόλογον πλευράν, τουτέστιν ¹ ΑΒ πρÕς τ¾ν ΖΗ. 'Επεζεύχθωσαν γ¦ρ αƒ ΑΓ, ΖΘ. κሠ™πεˆ δι¦ τ¾ν еοιότητα τîν πολυγώνων ‡ση ™στˆν ¹ ØπÕ ΑΒΓ γωνία τÍ ØπÕ ΖΗΘ, καί ™στιν æς ¹ ΑΒ πρÕς ΒΓ, οÛτως ¹ ΖΗ πρÕς ΗΘ, „σογώνιόν ™στι τÕ ΑΒΓ τρίγωνον τù ΖΗΘ τριγώνJ· ‡ση ¥ρα ™στˆν ¹ µν ØπÕ ΒΑΓ γωνία τÍ ØπÕ ΗΖΘ, ¹ δ ØπÕ ΒΓΑ τÍ ØπÕ ΗΘΖ. κሠ™πεˆ ‡ση ™στˆν ¹ ØπÕ ΒΑΜ γωνία τÍ ØπÕ ΗΖΝ, œστι δ κሠ¹ ØπÕ ΑΒΜ τÍ ØπÕ ΖΗΝ ‡ση, κሠλοιπ¾ ¥ρα ¹ ØπÕ ΑΜΒ λοιπÍ τÍ ØπÕ ΖΝΗ ‡ση ™στίν· „σογώνιον ¥ρα ™στˆ τÕ ΑΒΜ τρίγωνον τù ΖΗΝ τριγώνJ. еοίως δ¾ δε‹ξοµεν, Óτι κሠτÕ ΒΜΓ τρίγωνον „σογώνιόν ™στι τù ΗΝΘ τριγώνJ. ¢νάλογον ¥ρα ™στίν, æς µν ¹ ΑΜ πρÕς ΜΒ, οÛτως ¹ ΖΝ πρÕς ΝΗ, æς δ ¹ ΒΜ πρÕς ΜΓ, οÛτως ¹ ΗΝ πρÕς ΝΘ· éστε κሠδι' ‡σου, æς ¹ ΑΜ πρÕς ΜΓ, οÛτως ¹ ΖΝ πρÕς ΝΘ. ¢λλ' æς ¹ ΑΜ πρÕς ΜΓ, οÛτως τÕ ΑΒΜ [τρίγωνον] πρÕς τÕ ΜΒΓ, κሠτÕ ΑΜΕ πρÕς τÕ ΕΜΓ· πρÕς ¥λληλα γάρ ε„σιν æς αƒ βάσεις. κሠæς ¥ρα žν τîν ¹γουµένων πρÕς žν τîν ˜πόµενων, οÛτως ¤παντα τ¦ ¹γούµενα πρÕς ¤παντα τ¦ ˜πόµενα· æς ¥ρα τÕ ΑΜΒ τρίγωνον πρÕς τÕ ΒΜΓ, οÛτως τÕ ΑΒΕ πρÕς τÕ ΓΒΕ. αλλ' æς τÕ ΑΜΒ πρÕς τÕ ΒΜΓ, οÛτως ¹ ΑΜ πρÕς ΜΓ· κሠæς ¥ρα ¹ ΑΜ πρÕς ΜΓ, οÛτως τÕ ΑΒΕ τρίγωνον πρÕς τÕ ΕΒΓ τρίγωνον. δι¦ τ¦ αÙτ¦ δ¾ κሠæς ¹ ΖΝ πρÕς ΝΘ, οÛτως τÕ ΖΗΛ τρίγωνον πρÕς τÕ ΗΛΘ τρίγωνον. καί ™στιν æς ¹ ΑΜ πρÕς ΜΓ, οÛτως ¹ ΖΝ πρÕς ΝΘ· κሠæς ¥ρα τÕ ΑΒΕ τρίγωνον πρÕς τÕ ΒΕΓ τρίγωνον, οÛτως τÕ ΖΗΛ τρίγωνον πρÕς τÕ ΗΛΘ τρίγωνον, κሠ™ναλλ¦ξ æς τÕ ΑΒΕ τρίγωνον πρÕς τÕ ΖΗΛ τρίγωνον, οÛτως τÕ ΒΕΓ τρίγωνον πρÕς τÕ ΗΛΘ τρίγωνον. еοίως δ¾ δε‹ξοµεν ™πιζευχθεισîν τîν Β∆, ΗΚ, Óτι κሠæς τÕ ΒΕΓ τρίγωνον πρÕς τÕ ΛΗΘ τρίγωνον, οÛτως τÕ ΕΓ∆ τρίγωνον πρÕς τÕ ΛΘΚ τρίγωνον. κሠ™πεί ™στιν æς τÕ ΑΒΕ τρίγωνον πρÕς τÕ ΖΗΛ τρίγωνον, οÛτως τÕ ΕΒΓ πρÕς τÕ ΛΗΘ, κሠœτι τÕ ΕΓ∆ πρÕς τÕ ΛΘΚ, κሠæς ¥ρα žν τîν ¹γουµένων πρÕς žν τîν ˜ποµένων, οÛτως ¤παντα τ¦ ¹γούµενα πρÕς ¤παντα τ¦ ˜πόµενα· œστιν ¥ρα

GH [Prop. 5.22], the sides about the equal angles, EBC and LGH, are also proportional. Thus, triangle EBC is equiangular to triangle LGH [Prop. 6.6]. Hence, triangle EBC is also similar to triangle LGH [Prop. 6.4, Def. 6.1]. So, for the same (reasons), triangle ECD is also similar to triangle LHK. Thus, the similar polygons ABCDE and F GHKL have been divided into equal numbers of similar triangles. I also say that (the triangles) correspond (in proportion) to the wholes. That is to say, the triangles are proportional, ABE, EBC, and ECD are the leading (magnitudes), and their (associated) following (magnitudes are) F GL, LGH, and LHK (respectively). (I) also (say) that polygon ABCDE has a squared ratio to polygon F GHKL with respect to (that) a corresponding side (has) to a corresponding side—that is to say, (side) AB to F G. For let AC and F H have been joined. And since angle ABC is equal to F GH, and as AB is to BC, so F G (is) to GH, on account of the similarity of the polygons, triangle ABC is equiangular to triangle F GH [Prop. 6.6]. Thus, angle BAC is equal to GF H, and (angle) BCA to GHF . And since angle BAM is equal to GF N , and (angle) ABM is also equal to F GN (see earlier), the remaining (angle) AM B is thus also equal to the remaining (angle) F N G [Prop. 1.32]. Thus, triangle ABM is equiangular to triangle F GN . So, similarly, we can show that triangle BM C is equiangular to triangle GN H. Thus, proportionally, as AM is to M B, so F N (is) to N G, and as BM (is) to M C, so GN (is) to N H [Prop. 6.4]. Hence, also, via equality, as AM (is) to M C, so F N (is) to N H [Prop. 5.22]. But, as AM (is) to M C, so [triangle] ABM is to M BC, and AM E to EM C. For they are to one another as their bases [Prop. 6.1]. And as one of the leading (magnitudes) is to one of the following (magnitudes), so (the sum of) all the leading (magnitudes) is to (the sum of) all the following (magnitudes) [Prop. 5.12]. Thus, as triangle AM B (is) to BM C, so (triangle) ABE (is) to CBE. But, as (triangle) AM B (is) to BM C, so AM (is) to M C. Thus, also, as AM (is) to M C, so triangle ABE (is) to triangle EBC. And so, for the same (reasons), as F N (is) to N H, so triangle F GL (is) to triangle GLH. And as AM is to M C, so F N (is) to N H. Thus, also, as triangle ABE (is) to triangle BEC, so triangle F GL (is) to triangle GLH, and, alternately, as triangle ABE (is) to triangle F GL, so triangle BEC (is) to triangle GLH [Prop. 5.16]. So, similarly, we can also show, by joining BD and GK, that as triangle BEC (is) to triangle LGH, so triangle ECD (is) to triangle LHK. And since as triangle ABE is to triangle F GL, so (triangle) EBC (is) to LGH, and, further, (triangle) ECD to LHK, and also

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ELEMENTS BOOK 6

æς τÕ ΑΒΕ τρίγωνον πρÕς τÕ ΖΗΛ τρίγωνον, οÛτως τÕ ΑΒΓ∆Ε πολύγωνον πρÕς τÕ ΖΗΘΚΛ πολύγωνον. ¢λλ¦ τÕ ΑΒΕ τρίγωνον πρÕς τÕ ΖΗΛ τρίγωνον διπλασίονα λόγον œχει ½περ ¹ ΑΒ Ðµόλογος πλευρ¦ πρÕς τ¾ν ΖΗ Ðµόλογον πλευράν· τ¦ γ¦ρ Óµοια τρίγωνα ™ν διπλασίονι λόγJ ™στˆ τîν еολόγων πλευρîν. κሠτÕ ΑΒΓ∆Ε ¥ρα πολύγωνον πρÕς τÕ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον œχει ½περ ¹ ΑΒ Ðµόλογος πλευρ¦ πρÕς τ¾ν ΖΗ Ðµόλογον πλευράν. Τ¦ ¥ρα Óµοια πολύγωνα ε‡ς τε Óµοια τρίγωνα διαιρε‹ται κሠε„ς ‡σα τÕ πλÁθος καˆ Ðµόλογα το‹ς Óλοις, κሠτÕ πολύγωνον πρÕς τÕ πολύγωνον διπλασίονα λόγον œχει ½περ ¹ еόλογος πλευρ¦ πρÕς τ¾ν еόλογον πλευράν. [Óπερ œδει δε‹ξαι].

as one of the leading (magnitudes is) to one of the following, so (the sum of) all the leading (magnitudes is) to (the sum of) all the following [Prop. 5.12], thus as triangle ABE is to triangle F GL, so polygon ABCDE (is) to polygon F GHKL. But, triangle ABE has a squared ratio to triangle F GL with respect to (that) the corresponding side AB (has) to the corresponding side F G. For, similar triangles are in the squared ratio of corresponding sides [Prop. 6.14]. Thus, polygon ABCDE also has a squared ratio to polygon DEF GH with respect to (that) the corresponding side AB (has) to the corresponding side F G. Thus, similar polygons can be divided into equal numbers of similar triangles corresponding (in proportion) to the wholes, and one polygon has to the (other) polygon a squared ratio with respect to (that) a corresponding side (has) to a corresponding side. [(Which is) the very thing it was required to show].

Πόρισµα.

Corollary

`Ωσαύτως δ κሠ™πˆ τîν [еοίων] τετραπλεύρων δειχθήσεται, Óτι ™ν διπλασίονι λόγJ ε„σˆ τîν еολόγων πλευρîν. ™δείχθη δ κሠ™πˆ τîν τρίγώνων· éστε κሠκαθόλου τ¦ Óµοια εÙθύγραµµα σχήµατα πρÕς ¥λληλα ™ν διπλασίονι λόγJ ε„σˆ τîν еολόγων πλευρîν. Óπερ œδει δε‹ξαι.

And, in the same manner, it can also be shown for [similar] quadrilaterals that they are in the squared ratio of (their) corresponding sides. And it was also shown for triangles. Hence, in general, similar rectilinear figures are to one another in the squared ratio of (their) corresponding sides. (Which is) the very thing it was required to show.

κα΄.

Proposition 21

Τ¦ τù αÙτù εÙθυγράµµJ Óµοια κሠ¢λλήλοις ™στˆν (Rectilinear figures) similar to the same rectilinear figÓµοια. ure are also similar to one another.

Α

A

Β

Γ

B

C

”Εστω γ¦ρ ˜κάτερον τîν Α, Β εÙθυγράµµων τù Γ Let each of the rectilinear figures A and B be similar Óµοιον· λέγω, Óτι κሠτÕ Α τù Β ™στιν Óµοιον. to (the rectilinear figure) C. I say that A is also similar to 'Επεˆ γ¦ρ Óµοιόν ™στι τÕ Α τù Γ, „σογώνιόν τέ ™στιν B. αÙτù κሠτ¦ς περˆ τ¦ς ‡σας γωνίας πλευρ¦ς ¢νάλογον For since A is similar to C, (A) is equiangular to (C),

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ELEMENTS BOOK 6

œχει. πάλιν, ™πεˆ Óµοιόν ™στι τÕ Β τù Γ, „σογώνιόν τέ ™στιν αÙτù κሠτ¦ς περˆ τ¦ς ‡σας γωνίας πλευρ¦ς ¢νάλογον œχει. ˜κάτερον ¥ρα τîν Α, Β τù Γ „σογώνιόν τέ ™στι κሠτ¦ς περˆ τ¦ς ‡σας γωνίας πλευρ¦ς ¢νάλογον œχει [éστε κሠτÕ Α τù Β „σογώνιόν τέ ™στι κሠτ¦ς περˆ τ¦ς ‡σας γωνίας πλευρ¦ς ¢νάλογον œχει]. Óµοιον ¥ρα ™στˆ τÕ Α τù Β· Óπερ œδει δε‹ξαι.

and has the sides about the equal angles proportional [Def. 6.1]. Again, since B is similar to C, (B) is equiangular to (C), and has the sides about the equal angles proportional [Def. 6.1]. Thus, A and B are each equiangular to C, and have the sides about the equal angles proportional [hence, A is also equiangular to B, and has the sides about the equal angles proportional]. Thus, A is similar to B [Def. 6.1]. (Which is) the very thing it was required to show.

κβ΄.

Proposition 22

'Ε¦ν τέσσαρες εÙθε‹αι ¢νάλογον ðσιν, κሠτ¦ ¢π' αÙτîν εÙθύγραµµα Óµοιά τε καˆ Ðµοίως ¢ναγεγραµµένα ¢νάλογον œσται· κ¨ν τ¦ ¢π' αÙτîν εÙθύγραµµα Óµοιά τε καˆ Ðµοίως ¢ναγεγραµµένα ¢νάλογον Ï, κሠαÙτ¦ι αƒ εÙθε‹αι ¢νάλογον œσονται.

If four straight-lines are proportional, then similar, and similarly described, rectilinear figures (drawn) on them will also be proportional. And if similar, and similarly described, rectilinear figures (drawn) on them are proportional, then the straight-lines themselves will also be proportional.

Κ

Α

Ξ

Β

Μ

Ε

K

Λ

Ζ

L

Γ Ν Η

∆

A

B M

Θ

E

Σ

O

Ο Π

C N G

F

H S

P Q

Ρ

”Εστωσαν τέσσαρες εÙθε‹αι ¢νάλογον αƒ ΑΒ, Γ∆, ΕΖ, ΗΘ, æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΗΘ, κሠ¢ναγεγράφθωσαν ¢πÕ µν τîν ΑΒ, Γ∆ Óµοιά τε καˆ Ðµοίως κείµενα εÙθύγραµµα τ¦ ΚΑΒ, ΛΓ∆, ¢πÕ δ τîν ΕΖ, ΗΘ Óµοιά τε καˆ Ðµοίως κείµενα εÙθύγραµµα τ¦ ΜΖ, ΝΘ· λέγω, Óτι ™στˆν æς τÕ ΚΑΒ πρÕς τÕ ΛΓ∆, οÛτως τÕ ΜΖ πρÕς τÕ ΝΘ. Ε„λήφθω γ¦ρ τîν µν ΑΒ, Γ∆ τρίτη ¢νάλογον ¹ Ξ, τîν δ ΕΖ, ΗΘ τρίτη ¢νάλογον ¹ Ο. κሠ™πεί ™στιν æς µν ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΗΘ, æς δ ¹ Γ∆ πρÕς τ¾ν Ξ, οÛτως ¹ ΗΘ πρÕς τ¾ν Ο, δι' ‡σου ¥ρα ™στˆν æς ¹ ΑΒ πρÕς τ¾ν Ξ, οÛτως ¹ ΕΖ πρÕς τ¾ν Ο. ¢λλ' æς µν ¹ ΑΒ πρÕς τ¾ν Ξ, οÛτως [καˆ] τÕ ΚΑΒ πρÕς τÕ ΛΓ∆, æς δ ¹ ΕΖ πρÕς τ¾ν Ο, οÛτως τÕ ΜΖ πρÕς τÕ ΝΘ· κሠæς ¥ρα τÕ ΚΑΒ πρÕς τÕ ΛΓ∆, οÛτως

D

R

Let AB, CD, EF , and GH be four proportional straight-lines, (such that) as AB (is) to CD, so EF (is) to GH. And let the similar, and similarly laid out, rectilinear figures KAB and LCD have been described on AB and CD (respectively), and the similar, and similarly laid out, rectilinear figures M F and N H on EF and GH (respectively). I say that as KAB is to LCD, so M F (is) to N H. For let a third (straight-line) O have been taken (which is) proportional to AB and CD, and a third (straight-line) P proportional to EF and GH [Prop. 6.11]. And since as AB is to CD, so EF (is) to GH, and as CD (is) to O, so GH (is) to P , thus, via equality, as AB is to O, so EF (is) to P [Prop. 5.22]. But, as AB (is) to O, so [also] KAB (is) to LCD, and as EF (is) to P , so M F

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ELEMENTS BOOK 6

τÕ ΜΖ πρÕς τÕ ΝΘ. 'Αλλ¦ δ¾ œστω æς τÕ ΚΑΒ πρÕς τÕ ΛΓ∆, οÛτως τÕ ΜΖ πρÕς τÕ ΝΘ· λέγω, Óτι ™στˆ κሠæς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΗΘ. ε„ γ¦ρ µή ™στιν, æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΗΘ, œστω æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΠΡ, κሠ¢ναγεγράφθω ¢πÕ τÁς ΠΡ ÐποτέρJ τîν ΜΖ, ΝΘ Óµοιόν τε καˆ Ðµοίως κείµενον εÙθύγραµµον τÕ ΣΡ. 'Επεˆ οâν ™στιν æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΠΡ, κሠ¢ναγέγραπται ¢πÕ µν τîν ΑΒ, Γ∆ Óµοιά τε καˆ Ðµοίως κείµενα τ¦ ΚΑΒ, ΛΓ∆, ¢πÕ δ τîν ΕΖ, ΠΡ Óµοιά τε καˆ Ðµοίως κείµενα τ¦ ΜΖ, ΣΡ, œστιν ¥ρα æς τÕ ΚΑΒ πρÕς τÕ ΛΓ∆, οÛτως τÕ ΜΖ πρÕς τÕ ΣΡ. Øπόκειται δ κሠæς τÕ ΚΑΒ πρÕς τÕ ΛΓ∆, οÛτως τÕ ΜΖ πρÕς τÕ ΝΘ· κሠæς ¥ρα τÕ ΜΖ πρÕς τÕ ΣΡ, οÛτως τÕ ΜΖ πρÕς τÕ ΝΘ. τÕ ΜΖ ¥ρα πρÕς ˜κάτερον τîν ΝΘ, ΣΡ τÕν αÙτÕν œχει λόγον· ‡σον ¥ρα ™στˆ τÕ ΝΘ τù ΣΡ. œστι δ αÙτù καˆ Óµοιον καˆ Ðµοίως κείµενον· ‡ση ¥ρα ¹ ΗΘ τÍ ΠΡ. κሠ™πεί ™στιν æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΠΡ, ‡ση δ ¹ ΠΡ τÍ ΗΘ, œστιν ¥ρα æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΗΘ. 'Ε¦ν ¥ρα τέσσαρες εÙθε‹αι ¢νάλογον ðσιν, κሠτ¦ ¢π' αÙτîν εÙθύγραµµα Óµοιά τε καˆ Ðµοίως ¢ναγεγραµµένα ¢νάλογον œσται· κ¨ν τ¦ ¢π' αÙτîν εÙθύγραµµα Óµοιά τε καˆ Ðµοίως ¢ναγεγραµµένα ¢νάλογον Ï, κሠαÙτ¦ι αƒ εÙθε‹αι ¢νάλογον œσονται· Óπερ œδει δε‹ξαι.

(is) to N H [Prop. 5.19 corr.]. And, thus, as KAB (is) to LCD, so M F (is) to N H. And so let KAB be to LCD, as M F (is) to N H. I say also that as AB is to CD, so EF (is) to GH. For if as AB is to CD, so EF (is) not to GH, let AB be to CD, as EF (is) to QR [Prop. 6.12]. And let the rectilinear figure SR, similar, and similarly laid down, to either of M F or N H, have been described on QR [Props. 6.18, 6.21]. Therefore, since as AB is to CD, so EF (is) to QR, and the similar, and similarly laid out, (rectilinear figures) KAB and LCD have been described on AB and CD (respectively), and the similar, and similarly laid out, (rectilinear figures) M F and SR on EF and QR (resespectively), thus as KAB is to LCD, so M F (is) to SR (see above). And it was also assumed that as KAB (is) to LCD, so M F (is) to N H. Thus, also, as M F (is) to SR, so M F (is) to N H. Thus, M F has the same ratio to each of N H and SR. Thus, N H is equal to SR [Prop. 5.9]. And it is also similar, and similarly laid out, to it. Thus, GH (is) equal to QR. And since AB is to CD, as EF (is) to QR, and QR (is) equal to GH, thus as AB is to CD, so EF (is) to GH. Thus, if four straight-lines are proportional, then similar, and similarly described, rectilinear figures (drawn) on them will also be proportional. And if similar, and similarly described, rectilinear figures (drawn) on them are proportional, then the straight-lines themselves will also be proportional. (Which is) the very thing it was required to show.

κγ΄.

Proposition 23

Τ¦ „σογώνια παραλληλόγραµµα πρÕς ¥λληλα λόγον œχει τÕν συγκείµενον ™κ τîν πλευρîν. ”Εστω „σογώνια παραλληλόγραµµα τ¦ ΑΓ, ΓΖ ‡σην œχοντα τ¾ν ØπÕ ΒΓ∆ γωνίαν τÍ ØπÕ ΕΓΗ· λέγω, Óτι τÕ ΑΓ παραλληλόγραµµον πρÕς τÕ ΓΖ παραλληλόγραµµον λόγον œχει τÕν συγκείµενον ™κ τîν πλευρîν. Κείσθω γ¦ρ éστε ™π' εÙθείας εναι τ¾ν ΒΓ τÍ ΓΗ· ™π' εÙθε‹ας ¥ρα ™στˆ κሠ¹ ∆Γ τÍ ΓΕ. κሠσυµπεπληρώσθω τÕ ∆Η παραλληλόγραµµον, κሠ™κκείσθω τις εÙθε‹α ¹ Κ, κሠγεγονέτω æς µν ¹ ΒΓ πρÕς τ¾ν ΓΗ, οÛτως ¹ Κ πρÕς τ¾ν Λ, æς δ ¹ ∆Γ πρÕς τ¾ν ΓΕ, οÛτως ¹ Λ πρÕς τ¾ν Μ. Οƒ ¥ρα λόγοι τÁς τε Κ πρÕς τ¾ν Λ κሠτÁς Λ πρÕς τ¾ν Μ οƒ αÙτοί ε„σι το‹ς λόγοις τîν πλευρîν, τÁς τε ΒΓ πρÕς τ¾ν ΓΗ κሠτÁς ∆Γ πρÕς τ¾ν ΓΕ. ¢λλ' Ð τÁς Κ πρÕς Μ λόγος σύγκειται œκ τε τοà τÁς Κ πρÕς Λ λόγου κሠτοà τÁς Λ πρÕς Μ· éστε κሠ¹ Κ πρÕς τ¾ν Μ λόγον œχει τÕν συγκείµενον ™κ τîν πλευρîν. κሠ™πεί ™στιν æς

Equiangular parallelograms have to one another the ratio compounded† out of (the ratios of) their sides. Let AC and CF be equiangular parallelograms having angle BCD equal to ECG. I say that parallelogram AC has to parallelogram CF the ratio compounded out of (the ratios of) their sides. Let BC be laid down so as to be straight-on to CG. Thus, DC is also straight-on to CE [Prop. 1.14]. And let the parallelogram DG have been completed. And let some straight-line K have been laid down. And let it be that as BC (is) to CG, so K (is) to L, and as DC (is) to CE, so L (is) to M [Prop. 6.12]. Thus, the ratios of K to L and of L to M are the same as the ratios of the sides, (namely), BC to CG and DC to CE (respectively). But, the ratio of K to M is compounded out of the ratio of K to L and (the ratio) of L to M . Hence, K also has to M the ratio compounded out of (the ratios of) the sides (of the parallelograms). And since as BC is to CG, so parallelogram AC (is) to

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ELEMENTS BOOK 6

¹ ΒΓ πρÕς τ¾ν ΓΗ, οÛτως τÕ ΑΓ παραλληλόγραµµον πρÕς τÕ ΓΘ, ¢λλ' æς ¹ ΒΓ πρÕς τ¾ν ΓΗ, οÛτως ¹ Κ πρÕς τ¾ν Λ, κሠæς ¥ρα ¹ Κ πρÕς τ¾ν Λ, οÛτως τÕ ΑΓ πρÕς τÕ ΓΘ. πάλιν, ™πεί ™στιν æς ¹ ∆Γ πρÕς τ¾ν ΓΕ, οÛτως τÕ ΓΘ παραλληλόγραµµον πρÕς τÕ ΓΖ, ¢λλ' æς ¹ ∆Γ πρÕς τ¾ν ΓΕ, οÛτως ¹ Λ πρÕς τ¾ν Μ, κሠæς ¥ρα ¹ Λ πρÕς τ¾ν Μ, οÛτως τÕ ΓΘ παραλληλόγραµµον πρÕς τÕ ΓΖ παραλληλόγραµµον. ™πεˆ οâν ™δείχθη, æς µν ¹ Κ πρÕς τ¾ν Λ, οÛτως τÕ ΑΓ παραλληλόγραµµον πρÕς τÕ ΓΘ παραλληλόγραµµον, æς δ ¹ Λ πρÕς τ¾ν Μ, οÛτως τÕ ΓΘ παραλληλόγραµµον πρÕς τÕ ΓΖ παραλληλόγραµµον, δι' ‡σου ¥ρα ™στˆν æς ¹ Κ πρÕς τ¾ν Μ, οÛτως τÕ ΑΓ πρÕς τÕ ΓΖ παραλληλόγραµµον. ¹ δ Κ πρÕς τ¾ν Μ λόγον œχει τÕν συγκείµενον ™κ τîν πλευρîν· κሠτÕ ΑΓ ¥ρα πρÕς τÕ ΓΖ λόγον œχει τÕν συγκείµενον ™κ τîν πλευρîν.

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CH [Prop. 6.1], but as BC (is) to CG, so K (is) to L, thus, also, as K (is) to L, so (parallelogram) AC (is) to CH. Again, since as DC (is) to CE, so parallelogram CH (is) to CF [Prop. 6.1], but as DC (is) to CE, so L (is) to M , thus, also, as L (is) to M , so parallelogram CH (is) to parallelogram CF . Therefore, since it was shown that as K (is) to L, so parallelogram AC (is) to parallelogram CH, and as L (is) to M , so parallelogram CH (is) to parallelogram CF , thus, via equality, as K is to M , so (parallelogram) AC (is) to parallelogram CF [Prop. 5.22]. And K has to M the ratio compounded out of (the ratios of) the sides (of the parallelograms). Thus, (parallelogram) AC also has to (parallelogram) CF the ratio compounded out of (the ratio of) their sides.

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Τ¦ ¥ρα „σογώνια παραλληλόγραµµα πρÕς ¥λληλα Thus, equiangular parallelograms have to one another λόγον œχει τÕν συγκείµενον ™κ τîν πλευρîν· Óπερ œδει the ratio compounded out of (the ratio of) their sides. δε‹ξαι. (Which is) the very thing it was required to show. †

In modern terminology, if two ratios are “compounded” then they are multiplied together.

κδ΄.

Proposition 24

ΠαντÕς παραλληλογράµµου τ¦ περˆ τ¾ν διάµετρον παραλληλόγραµµα Óµοιά ™στι τù τε ÓλJ κሠ¢λλήλοις. ”Εστω παραλληλόγραµµον τÕ ΑΒΓ∆, διάµετρος δ αÙτοà ¹ ΑΓ, περˆ δ τ¾ν ΑΓ παραλληλόγραµµα œστω τ¦ ΕΗ, ΘΚ· λέγω, Óτι ˜κάτερον τîν ΕΗ, ΘΚ παραλληλογράµµων Óµοιόν ™στι ÓλJ τù ΑΒΓ∆ κሠ¢λλήλοις. 'Επεˆ γ¦ρ τριγώνου τοà ΑΒΓ παρ¦ µίαν τîν πλευρîν τ¾ν ΒΓ Ãκται ¹ ΕΖ, ¢νάλογόν ™στιν æς ¹ ΒΕ πρÕς τ¾ν ΕΑ, οÛτως ¹ ΓΖ πρÕς τ¾ν ΖΑ. πάλιν, ™πεˆ τριγώνου τοà ΑΓ∆ παρ¦ µίαν τ¾ν Γ∆ Ãκται ¹ ΖΗ, ¢νάλογόν ™στιν æς ¹ ΓΖ πρÕς τ¾ν ΖΑ, οÛτως ¹ ∆Η πρÕς τ¾ν ΗΑ. ¢λλ' æς ¹ ΓΖ πρÕς τ¾ν ΖΑ, οÛτως ™δείχθη κሠ¹ ΒΕ πρÕς τ¾ν ΕΑ· κሠæς ¥ρα ¹ ΒΕ πρÕς τ¾ν ΕΑ, οÛτως ¹ ∆Η πρÕς τ¾ν ΗΑ, κሠσυνθέντι ¥ρα æς ¹ ΒΑ πρÕς ΑΕ, οÛτως ¹ ∆Α πρÕς ΑΗ, κሠ™ναλλ¦ξ æς ¹ ΒΑ πρÕς τ¾ν

For every parallelogram, the parallelograms about the diagonal are similar to the whole, and to one another. Let ABCD be a parallelogram, and AC its diagonal. And let EG and HK be parallelograms about AC. I say that the parallelograms EG and HK are each similar to the whole (parallelogram) ABCD, and to one another. For since EF has been drawn parallel to one of the sides BC of triangle ABC, proportionally, as BE is to EA, so CF (is) to F A [Prop. 6.2]. Again, since F G has been drawn parallel to one (of the sides) CD of triangle ACD, proportionally, as CF is to F A, so DG (is) to GA [Prop. 6.2]. But, as CF (is) to F A, so it was also shown (is) BE to EA. And thus as BE (is) to EA, so DG (is) to GA. And, thus, compounding, as BA (is) to AE, so DA (is) to AG [Prop. 5.18]. And, alternately, as

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Α∆, οÛτως ¹ ΕΑ πρÕς τ¾ν ΑΗ. τîν ¥ρα ΑΒΓ∆, ΕΗ παραλληλογράµµων ¢νάλογόν ε„σιν αƒ πλευραˆ αƒ περˆ τ¾ν κοιν¾ν γωνίαν τ¾ν ØπÕ ΒΑ∆ κሠ™πεˆ παράλληλός ™στιν ¹ ΗΖ τÍ ∆Γ, ‡ση ™στˆν ¹ µν ØπÕ ΑΖΗ γωνία τÍ ØπÕ ∆ΓΑ· κሠκοιν¾ τîν δύο τριγώνων τîν Α∆Γ, ΑΗΖ ¹ ØπÕ ∆ΑΓ γωνία· „σογώνιον ¥ρα ™στˆ τÕ Α∆Γ τρίγωνον τù ΑΗΖ τριγώνJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ΑΓΒ τρίγωνον „σογώνιόν ™στι τù ΑΖΕ τριγώνJ, κሠÓλον τÕ ΑΒΓ∆ παραλληλόγραµµον τù ΕΗ παραλληλογράµµJ „σογώνιόν ™στιν. ¢νάλογον ¥ρα ™στˆν æς ¹ Α∆ πρÕς τ¾ν ∆Γ, οÛτως ¹ ΑΗ πρÕς τ¾ν ΗΖ, æς δ ¹ ∆Γ πρÕς τ¾ν ΓΑ, οÛτως ¹ ΗΖ πρÕς τ¾ν ΖΑ, æς δ ¹ ΑΓ πρÕς τ¾ν ΓΒ, οÛτως ¹ ΑΖ πρÕς τ¾ν ΖΕ, κሠœτι æς ¹ ΓΒ πρÕς τ¾ν ΒΑ, οÛτως ¹ ΖΕ πρÕς τ¾ν ΕΑ. κሠ™πεˆ ™δείχθη æς µν ¹ ∆Γ πρÕς τ¾ν ΓΑ, οÛτως ¹ ΗΖ πρÕς τ¾ν ΖΑ, æς δ ¹ ΑΓ πρÕς τ¾ν ΓΒ, οÛτως ¹ ΑΖ πρÕς τ¾ν ΖΕ, δι' ‡σου ¥ρα ™στˆν æς ¹ ∆Γ πρÕς τ¾ν ΓΒ, οÛτως ¹ ΗΖ πρÕς τ¾ν ΖΕ. τîν ¥ρα ΑΒΓ∆, ΕΗ παραλληλογράµµων ¢νάλογόν ε„σιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας· Óµοιον ¥ρα ™στˆ τÕ ΑΒΓ∆ παραλληλογράµµον τù ΕΗ παραλληλογράµµJ. δι¦ τ¦ αÙτ¦ δ¾ τÕ ΑΒΓ∆ παραλληλόγραµµον κሠτù ΚΘ παραλληλογράµµJ Óµοιόν ™στιν· ˜κάτερον ¥ρα τîν ΕΗ, ΘΚ παραλληλογράµµων τù ΑΒΓ∆ [παραλληλογράµµJ] Óµοιόν ™στιν. τ¦ δ τù αÙτù εÙθυγράµµJ Óµοια κሠ¢λλήλοις ™στˆν Óµοια· κሠτÕ ΕΗ ¥ρα παραλληλόγραµµον τù ΘΚ παραλληλογράµµJ Óµοιόν ™στιν.

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BA (is) to AD, so EA (is) to AG [Prop. 5.16]. Thus, for parallelograms ABCD and EG, the sides about the common angle BAD are proportional. And since GF is parallel to DC, angle AF G is equal to DCA [Prop. 1.29]. And angle DAC (is) common to the two triangles ADC and AGF . Thus, triangle ADC is equiangular to triangle AGF [Prop. 1.32]. So, for the same (reasons), triangle ACB is equiangular to triangle AF E, and the whole parallelogram ABCD is equiangular to parallelogram EG. Thus, proportionally, as AD (is) to DC, so AG (is) to GF , and as DC (is) to CA, so GF (is) to F A, and as AC (is) to CB, so AF (is) to F E, and, further, as CB (is) to BA, so F E (is) to EA [Prop. 6.4]. And since it was shown that as DC is to CA, so GF (is) to F A, and as AC (is) to CB, so AF (is) to F E, thus, via equality, as DC is to CB, so GF (is) to F E [Prop. 5.22]. Thus, for parallelograms ABCD and EG, the sides about the equal angles are proportional. Thus, parallelogram ABCD is similar to parallelogram EG [Def. 6.1]. So, for the same (reasons), parallelogram ABCD is also similar to parallelogram KH. Thus, parallelograms EG and HK are each similar to [parallelogram] ABCD. And (rectilinear figures) similar to the same rectilinear figure are also similar to one another [Prop. 6.21]. Thus, parallelogram EG is also similar to parallelogram HK.

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ΠαντÕς ¥ρα παραλληλογράµµου τ¦ περˆ τ¾ν διάµετρον παραλληλόγραµµα Óµοιά ™στι τù τε ÓλJ κሠ¢λλήλοις· Óπερ œδει δε‹ξαι.

Thus, for every parallelogram, the parallelograms about the diagonal are similar to the whole, and to one another. (Which is) the very thing it was required to show.

κε΄.

Proposition 25

Τù δοθέντι εÙθυγράµµJ Óµοιον κሠ¥λλJ τù δοθέντι ‡σον τÕ αÙτÕ συστήσασθαι.

To construct a single (rectilinear figure) similar to a given rectilinear figure, and equal to a different given rectilinear figure.

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Λ Ε Μ ”Εστω τÕ µν δοθν εÙθύγραµµον, ú δε‹ Óµοιον συστήσασθαι, τÕ ΑΒΓ, ú δ δε‹ ‡σον, τÕ ∆· δε‹ δ¾ τù µν ΑΒΓ Óµοιον, τù δ ∆ ‡σον τÕ αÙτÕ συστήσασθαι. Παραβεβλήσθω γ¦ρ παρ¦ µν τ¾ν ΒΓ τù ΑΒΓ τριγώνJ ‡σον παραλληλόγραµµον τÕ ΒΕ, παρ¦ δ τ¾ν ΓΕ τù ∆ ‡σον παραλληλόγραµµον τÕ ΓΜ ™ν γωνίv τÍ ØπÕ ΖΓΕ, ¼ ™στιν ‡ση τÍ ØπÕ ΓΒΛ. ™π' εÙθείας ¥ρα ™στˆν ¹ µν ΒΓ τÍ ΓΖ, ¹ δ ΛΕ τÍ ΕΜ. κሠε„λήφθω τîν ΒΓ, ΓΖ µέση ¢νάλογον ¹ ΗΘ, κሠ¢ναγεγράφθω ¢πÕ τÁς ΗΘ τù ΑΒΓ Óµοιόν τε καˆ Ðµοίως κείµενον τÕ ΚΗΘ. Κሠ™πεί ™στιν æς ¹ ΒΓ πρÕς τ¾ν ΗΘ, οÛτως ¹ ΗΘ πρÕς τ¾ν ΓΖ, ™¦ν δ τρε‹ς εÙθε‹αι ¢νάλογον ðσιν, œστιν æς ¹ πρώτη πρÕς τ¾ν τρίτην, οÛτως τÕ ¢πÕ τÁς πρώτης εδος πρÕς τÕ ¢πÕ τÁς δευτέρας τÕ Óµοιον καˆ Ðµοίως ¢ναγραφόµενον, œστιν ¥ρα æς ¹ ΒΓ πρÕς τ¾ν ΓΖ, οÛτως τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΚΗΘ τρίγωνον. ¢λλ¦ κሠæς ¹ ΒΓ πρÕς τ¾ν ΓΖ, οÛτως τÕ ΒΕ παραλληλόγραµµον πρÕς τÕ ΕΖ παραλληλόγραµµον. κሠæς ¥ρα τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΚΗΘ τρίγωνον, οÛτως τÕ ΒΕ παραλληλόγραµµον πρÕς τÕ ΕΖ παραλληλόγραµµον· ™ναλλ¦ξ ¥ρα æς τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΒΕ παραλληλόγραµµον, οÛτως τÕ ΚΗΘ τρίγωνον πρÕς τÕ ΕΖ παραλληλόγραµµον. ‡σον δ τÕ ΑΒΓ τρίγωνον τù ΒΕ παραλληλογράµµJ· ‡σον ¥ρα κሠτÕ ΚΗΘ τρίγωνον τù ΕΖ παραλληλογράµµJ. ¢λλ¦ τÕ ΕΖ παραλληλόγραµµον τù ∆ ™στιν ‡σον· κሠτÕ ΚΗΘ ¥ρα τù ∆ ™στιν ‡σον. œστι δ τÕ ΚΗΘ κሠτù ΑΒΓ Óµοιον. Τù ¥ρα δοθέντι εÙθυγράµµJ τù ΑΒΓ Óµοιον κሠ¥λλJ τù δοθέντι τù ∆ ‡σον τÕ αÙτÕ συνέσταται τÕ ΚΗΘ· Óπερ œδει ποιÁσαι.

L E M Let ABC be the given rectilinear figure to which it is required to construct a similar (rectilinear figure), and D the (rectilinear figure) to which (the constructed figure) is required (to be) equal. So it is required to construct a single (rectilinear figure) similar to ABC, and equal to D. For let the parallelogram BE, equal to triangle ABC, have been applied to (the straight-line) BC [Prop. 1.44], and the parallelogram CM , equal to D, (have been applied) to (the straight-line) CE, in the angle F CE, which is equal to CBL [Prop. 1.45]. Thus, BC is straight-on to CF , and LE to EM [Prop. 1.14]. And let the mean proportion GH have been taken of BC and CF [Prop. 6.13]. And let KGH, similar, and similarly laid out, to ABC have been described on GH [Prop. 6.18]. And since as BC is to GH, so GH (is) to CF , and if three straight-lines are proportional then as the first is to the third, so the figure (described) on the first (is) to the similar, and similarly described, (figure) on the second [Prop. 6.19 corr.], thus as BC is to CF , so triangle ABC (is) to triangle KGH. But, also, as BC (is) to CF , so parallelogram BE (is) to parallelogram EF [Prop. 6.1]. And, thus, as triangle ABC (is) to triangle KGH, so parallelogram BE (is) to parallelogram EF . Thus, alternately, as triangle ABC (is) to parallelogram BE, so triangle KGH (is) to parallelogram EF [Prop. 5.16]. And triangle ABC (is) equal to parallelogram BE. Thus, triangle KGH (is) also equal to parallelogram EF . But, parallelogram EF is equal to D. Thus, KGH is also equal to D. And KGH is also similar to ABC. Thus, a single (rectilinear figure) KGH has been constructed (which is) similar to the given rectilinear figure ABC, and equal to a different given (rectilinear figure) D. (Which is) the very thing it was required to do.

κ$΄.

Proposition 26

'Ε¦ν ¢πÕ παραλληλογράµµου παραλληλόγραµµον ¢φαιρεθÍ Óµοιόν τε τù ÓλJ καˆ Ðµοίως κείµενον κοιν¾ν γωνίαν œχον αÙτù, περˆ τ¾ν αÙτ¾ν διάµετρόν ™στι τù ÓλJ. 'ΑπÕ γ¦ρ παραλληλογράµµου τοà ΑΒΓ∆ παραλληλόγραµµον ¢φVρήσθω τÕ ΑΖ Óµοιον τù ΑΒΓ∆ καˆ

If from a parallelogram a(nother) parallelogram is subtracted (which is) similar, and similarly laid out, to the whole, having a common angle with it, then (the subtracted parallelogram) is about the same diagonal as the whole. For, from parallelogram ABCD, let (parallelogram)

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еοίως κείµενον κοιν¾ν γωνίαν œχον αÙτù τ¾ν ØπÕ ∆ΑΒ· AF have been subtracted (which is) similar, and similarly λέγω, Óτι περˆ τ¾ν αÙτ¾ν διάµετρόν ™στι τÕ ΑΒΓ∆ τù laid out, to ABCD, having the common angle DAB with ΑΖ. it. I say that ABCD is about the same diagonal as AF .

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Μ¾ γάρ, ¢λλ' ε„ δυνατόν, œστω [αÙτîν] διάµετρος ¹ ΑΘΓ, κሠ™κβληθε‹σα ¹ ΗΖ διήχθω ™πˆ τÕ Θ, κሠ½χθω δι¦ τοà Θ Ðπορέρv τîν Α∆, ΒΓ παράλληλος ¹ ΘΚ. 'Επεˆ οâν περˆ τ¾ν αÙτ¾ν διάµετρόν ™στι τÕ ΑΒΓ∆ τù ΚΗ, œστιν ¥ρα æς ¹ ∆Α πρÕς τ¾ν ΑΒ, οÛτως ¹ ΗΑ πρÕς τ¾ν ΑΚ. œστι δ κሠδι¦ τ¾ν еοιότητα τîν ΑΒΓ∆, ΕΗ κሠæς ¹ ∆Α πρÕς τ¾ν ΑΒ, οÛτως ¹ ΗΑ πρÕς τ¾ν ΑΕ· κሠæς ¥ρα ¹ ΗΑ πρÕς τ¾ν ΑΚ, οÛτως ¹ ΗΑ πρÕς τ¾ν ΑΕ. ¹ ΗΑ ¥ρα πρÕς ˜κατέραν τîν ΑΚ, ΑΕ τÕν αÙτÕν œχει λόγον. ‡ση ¥ρα ™στˆν ¹ ΑΕ τÍ ΑΚ ¹ ™λάττων τÍ µείζονι· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα οÜκ ™στι περˆ τ¾ν αÙτ¾ν διάµετρον τÕ ΑΒΓ∆ τù ΑΖ· περˆ τ¾ν αÙτ¾ν ¥ρα ™στˆ διάµετρον τÕ ΑΒΓ∆ παραλληλόγραµµον τù ΑΖ παραλληλογράµµJ. 'Ε¦ν ¥ρα ¢πÕ παραλληλογράµµου παραλληλόγραµµον ¢φαιρεθÍ Óµοιόν τε τù ÓλJ καˆ Ðµοίως κείµενον κοιν¾ν γωνίαν œχον αÙτù, περˆ τ¾ν αÙτ¾ν διάµετρόν ™στι τù ÓλJ· Óπερ œδει δε‹ξαι.

For (if) not, then, if possible, let AHC be [ABCD’s] diagonal. And producing GF , let it have been drawn through to (point) H. And let HK have been drawn through (point) H, parallel to either of AD or BC [Prop. 1.31]. Therefore, since ABCD is about the same diagonal as KG, thus as DA is to AB, so GA (is) to AK [Prop. 6.24]. And, on account of the similarity of ABCD and EG, also, as DA (is) to AB, so GA (is) to AE. Thus, also, as GA (is) to AK, so GA (is) to AE. Thus, GA has the same ratio to each of AK and AE. Thus, AE is equal to AK [Prop. 5.9], the lesser to the greater. The very thing is impossible. Thus, ABCD is not not about the same diagonal as AF . Thus, parallelogram ABCD is about the same diagonal as parallelogram AF . Thus, if from a parallelogram a(nother) parallelogram is subtracted (which is) similar, and similarly laid out, to the whole, having a common angle with it, then (the subtracted parallelogram) is about the same diagonal as the whole. (Which is) the very thing it was required to show.

κζ΄.

Proposition 27

Πάντων τîν παρ¦ τ¾ν αÙτ¾ν εÙθε‹αν παραβαλλοµένων παραλληλογράµµων κሠ™λλειπόντων ε‡δεσι παραλληλογράµµοις еοίοις τε καˆ Ðµοίως κειµένοις τù ¢πÕ τÁς ¹µισείας ¢ναγραφοµένJ µέγιστόν ™στι τÕ ¢πÕ τÁς ¹µισείας παραβαλλόµενον [παραλληλόγραµµον] Óµοιον ×ν τù ™λλείµµαντι. ”Εστω εÙθε‹α ¹ ΑΒ κሠτετµήσθω δίχα κατ¦ τÕ Γ, κሠπαραβεβλήσθω παρ¦ τ¾ν ΑΒ εÙθε‹αν τÕ Α∆ παραλληλόγραµµον ™λλε‹πον ε‡δει παραλληλογράµµJ τù ∆Β ¢ναγραφέντι ¢πÕ τÁς ¹µισείας τÁς ΑΒ, τουτέστι τÁς ΓΒ· λέγω, Óτι πάντων τîν παρ¦ τ¾ν ΑΒ παραβαλλοµένων παραλληλογράµµων κሠ™λλειπόντων ε‡δεσι [παραλλη-

For all parallelograms applied to the same straightline, and falling short by a parallelogrammic figure similar, and similarly laid out, to the (parallelogram) described on half (the straight-line), the greatest is the [parallelogram] applied to half (the straight-line), which (is) similar to (that parallelogram) by which it falls short. Let AB be the straight-line, and let it have been cut in half at (point) C [Prop. 1.10]. And let the parallelogram AD have been applied to the straight-line AB, falling short by the parallelogrammic figure DB, (which is) applied to half of AB—that is to say, CB. I say that of all the parallelograms applied to AB, and falling short by a

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ELEMENTS BOOK 6

λογράµµοις] еοίοις τε καˆ Ðµοίως κειµένοις τù ∆Β µέγιστόν ™στι τÕ Α∆. παραβεβλήσθω γ¦ρ παρ¦ τ¾ν ΑΒ εÙθε‹αν τÕ ΑΖ παραλληλόγραµµον ™λλε‹πον ε‡δει παραλληλογράµµJ τù ΖΒ ÐµοίJ τε καˆ Ðµοίως κειµένJ τù ∆Β· λέγω, Óτι µε‹ζόν ™στι τÕ Α∆ τοà ΑΖ.

∆

Ε Ν

Η

[parallelogrammic] figure similar, and similarly laid out, to DB, the greatest is AD. For let the parallelogram AF have been applied to the straight-line AB, falling short by the parallelogrammic figure F B, (which is) similar, and similarly laid out, to DB. I say that AD is greater than AF .

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'Επεˆ γ¦ρ Óµοιόν ™στι τÕ ∆Β παραλληλόγραµµον τù ΖΒ παραλληλογράµµJ, περˆ τ¾ν αÙτήν ε„σι διάµετρον. ½χθω αÙτîν διάµετρος ¹ ∆Β, κሠκαταγεγράφθω τÕ σχÁµα. 'Επεˆ οâν ‡σον ™στˆ τÕ ΓΖ τù ΖΕ, κοινÕν δ τÕ ΖΒ, Óλον ¥ρα τÕ ΓΘ ÓλJ τù ΚΕ ™στιν ‡σον. ¢λλ¦ τÕ ΓΘ τù ΓΗ ™στιν ‡σον, ™πεˆ κሠ¹ ΑΓ τÍ ΓΒ. κሠτÕ ΗΓ ¥ρα τù ΕΚ ™στιν ‡σον. κοινÕν προσκείσθω τÕ ΓΖ· Óλον ¥ρα τÕ ΑΖ τù ΛΜΝ γνώµονί ™στιν ‡σον· éστε τÕ ∆Β παραλληλόγραµµον, τουτέστι τÕ Α∆, τοà ΑΖ παραλληλογράµµου µε‹ζόν ™στιν. Πάντων ¥ρα τîν παρ¦ τ¾ν αÙτ¾ν εÙθε‹αν παραβαλλοµένων παραλληλογράµµων κሠ™λλειπόντων ε‡δεσι παραλληλογράµµοις еοίοις τε καˆ Ðµοίως κειµένοις τù ¢πÕ τÁς ¹µισείας ¢ναγραφοµένJ µέγιστόν ™στι τÕ ¢πÕ τÁς ¹µισείας παραβληθέν· Óπερ œδει δε‹ξαι.

For since parallelogram DB is similar to parallelogram F B, they are about the same diagonal [Prop. 6.26]. Let their (common) diagonal DB have been drawn, and let the (rest of the) figure have been described. Therefore, since (complement) CF is equal to (complement) F E [Prop. 1.43], and (parallelogram) F B is common, the whole (parallelogram) CH is thus equal to the whole (parallelogram) KE. But, (parallelogram) CH is equal to CG, since AC (is) also (equal) to CB [Prop. 6.1]. Thus, (parallelogram) GC is also equal to EK. Let (parallelogram) CF have been added to both. Thus, the whole (parallelogram) AF is equal to the gnomon LM N . Hence, parallelogram DB—that is to say, AD—is greater than parallelogram AF . Thus, for all parallelograms applied to the same straight-line, and falling short by a parallelogrammic figure similar, and similarly laid out, to the (parallelogram) described on half (the straight-line), the greatest is the [parallelogram] applied to half (the straight-line). (Which is) the very thing it was required to show.

κη΄.

Proposition 28†

Παρ¦ τ¾ν δοθε‹σαν εÙθε‹αν τù δοθέντι εÙθυγράµµJ ‡σον παραλληλόγραµµον παραβαλε‹ν ™λλε‹πον ε‡δει παραλληλογράµµJ еοίJ τù δοθέντι· δε‹ δ τÕ διδόµενον εÙθύγραµµον [ú δε‹ ‡σον παραβαλε‹ν] µ¾ µε‹ζον εναι τοà ¢πÕ τÁς ¹µισείας ¢ναγραφοµένου еοίου τù ™λλείµµατι [τοà τε ¢πÕ τÁς ¹µισείας κሠú δε‹ Óµοιον ™λλείπειν]. ”Εστω ¹ µν δοθε‹σα εÙθε‹α ¹ ΑΒ, τÕ δ δοθν εÙθύγραµµον, ú δε‹ ‡σον παρ¦ τ¾ν ΑΒ παραβαλε‹ν, τÕ Γ µ¾ µε‹ζον [×ν] τοà ¢πÕ τÁς ¹µισείας τÁς ΑΒ ¢να-

To apply a parallelogram, equal to a given rectilinear figure, to a given straight-line, (the applied parallelogram) falling short by a parallelogrammic figure similar to a given (parallelogram). It is necessary for the given rectilinear figure [to which it is required to apply an equal (parallelogram)] not to be greater than the (parallelogram) described on half (of the straight-line, which is) similar to the deficit. Let AB be the given straight-line, and C the given rectilinear figure to which the (parallelogram) applied to

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ELEMENTS BOOK 6

γραφοµένου еοίου τù ™λλείµµατι, ú δ δε‹ Óµοιον ™λλείπειν, τÕ ∆· δε‹ δ¾ παρ¦ τ¾ν δοθε‹σαν εÙθε‹αν τ¾ν ΑΒ τù δοθέντι εÙθυγράµµJ τù Γ ‡σον παραλληλόγραµµον παραβαλε‹ν ™λλε‹πον ε‡δει παραλληλογράµµJ еοίJ Ôντι τù ∆.

Θ

Η

AB is required (to be) equal, [being] not greater than the (parallelogram) described on half of AB (which is) similar to the deficit, and D the (parallelogram) to which the deficit is required (to be) similar. So it is required to apply a parallelogram, equal to the given rectilinear figure C, to the straight-line AB, falling short by a parallelogrammic figure which is similar to D.

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Κ Ν Τετµήσθω ¹ ΑΒ δίχα κατ¦ τÕ Ε σηµε‹ον, κሠ¢ναγεγράφθω ¢πÕ τÁς ΕΒ τù ∆ Óµοιον καˆ Ðµοίως κείµενον τÕ ΕΒΖΗ, κሠσυµπεπληρώσθω τÕ ΑΗ παραλληλόγραµµον. Ε„ µν οâν ‡σον ™στˆ τÕ ΑΗ τù Γ, γεγονÕς ¨ν ε‡η τÕ ™πιταχθέν· παραβέβληται γ¦ρ παρ¦ τ¾ν δοθε‹σαν εÙθε‹αν τ¾ν ΑΒ τù δοθέντι εÙθυγράµµJ τù Γ ‡σον παραλληλόγραµµον τÕ ΑΗ ™λλε‹πον ε‡δει παραλληλογράµµJ τù ΗΒ ÐµοίJ Ôντι τù ∆. ε„ δ οÜ, µε‹ζόν œστω τÕ ΘΕ τοà Γ. ‡σον δ τÕ ΘΕ τù ΗΒ· µε‹ζον ¥ρα κሠτÕ ΗΒ τοà Γ. ú δ¾ µε‹ζόν ™στι τÕ ΗΒ τοà Γ, ταύτV τÍ ØπεροχÍ ‡σον, τù δ ∆ Óµοιον καˆ Ðµοίως κείµενον τÕ αÙτÕ συνεστάτω τÕ ΚΛΜΝ. ¢λλ¦ τÕ ∆ τù ΗΒ [™στιν] Óµοιον· κሠτÕ ΚΜ ¥ρα τù ΗΒ ™στιν Óµοιον. œστω οâν еόλογος ¹ µν ΚΛ τÊ ΗΕ, ¹ δ ΛΜ τÍ ΗΖ. κሠ™πεˆ ‡σον ™στˆ τÕ ΗΒ το‹ς Γ, ΚΜ, µε‹ζον ¥ρα ™στˆ τÕ ΗΒ τοà ΚΜ· µείζων ¥ρα ™στˆ κሠ¹ µν ΗΕ τÁς ΚΛ, ¹ δ ΗΖ τÁς ΛΜ. κείσθω τÍ µν ΚΛ ‡ση ¹ ΗΞ, τÍ δ ΛΜ ‡ση ¹ ΗΟ, κሠσυµπεπληρώσθω τÕ ΞΗΟΠ παραλληλόγραµµον· ‡σον ¥ρα καˆ Óµοιον ™στι [τÕ ΗΠ] τù ΚΜ [¢λλ¦ τÕ ΚΜ τù ΗΒ Óµοιόν ™στιν]. κሠτÕ ΗΠ ¥ρα τù ΗΒ Óµοιόν ™στιν· περˆ τ¾ν αÙτ¾ν ¥ρα διάµετρόν ™στι τÕ ΗΠ τù ΗΒ. œστω αÙτîν διάµετρος ¹ ΗΠΒ, κሠκαταγεγράφθω τÕ σχÁµα. 'Επεˆ οâν ‡σον ™στˆ τÕ ΒΗ το‹ς Γ, ΚΜ, ïν τÕ ΗΠ τù ΚΜ ™στιν ‡σον, λοιπÕς ¥ρα Ð ΥΧΦ γνόµων λοιπù τù Γ ‡σος ™στίν. κሠ™πεˆ ‡σον ™στˆ τÕ ΟΡ τù ΞΣ, κοινÕν προσκείσθω τÕ ΠΒ· Óλον ¥ρα τÕ ΟΒ ÓλJ τù ΞΒ ‡σον ™στίν. ¢λλ¦ τÕ ΞΒ τù ΤΕ ™στιν ‡σον, ™πεˆ κሠπλευρ¦ ¹ ΑΕ πλευρ´ τÍ ΕΒ ™στιν ‡ση· κሠτÕ ΤΕ ¥ρα τù ΟΒ ™στιν ‡σον. κοινÕν προσκείσθω τÕ ΞΣ· Óλον ¥ρα τÕ ΤΣ ÓλJ τù ΦΧΥ γνώµονί ™στιν ‡σον. ¢λλ' Ð ΦΧΥ γνώµων τù Γ ™δείχθη ‡σος· κሠτÕ ΤΣ ¥ρα τù Γ ™στιν ‡σον.

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Let AB have been cut in half at point E [Prop. 1.10], and let (parallelogram) EBF G, (which is) similar, and similarly laid out, to (parallelogram) D, have been applied to EB [Prop. 6.18]. And let parallelogram AG have been completed. Therefore, if AG is equal to C then the thing prescribed has happened. For a parallelogram AG, equal to the given rectilinear figure C, has been applied to the given straight-line AB, falling short by a parallelogrammic figure GB which is similar to D. And if not, let HE be greater than C. And HE (is) equal to GB [Prop. 6.1]. Thus, GB (is) also greater than C. So, let (parallelogram) KLM N have been constructed (so as to be) both similar, and similarly laid out, to D, and equal o the excess by which GB is greater than C [Prop. 6.25]. But, GB [is] similar to D. Thus, KM is also similar to GB [Prop. 6.21]. Therefore, let KL correspond to GE, and LM to GF . And since (parallelogram) GB is equal to (figure) C and (parallelogram) KM , GB is thus greater than KM . Thus, GE is also greater than KL, and GF than LM . Let GO be made equal to KL, and GP to LM [Prop. 1.3]. And let the parallelogram OGP Q have been completed. Thus, [GQ] is equal and similar to KM [but, KM is similar to GB]. Thus, GQ is also similar to GB [Prop. 6.21]. Thus, GQ and GB are about the same diagonal [Prop. 6.26]. Let GQB be their (common) diagonal, and let the (remainder of the) figure have been described. Therefore, since BG is equal to C and KM , of which GQ is equal to KM , the remaining gnomon U XV is thus equal to the remainder C. And since (the complement) P R is equal to (the complement) OS [Prop. 1.43], let (parallelogram) QB have been added to both. Thus, the whole (parallelogram) P B is equal to the whole (par-

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Παρ¦ τ¾ν δοθε‹σαν ¥ρα εÙθε‹αν τ¾ν ΑΒ τù δοθέντι εÙθυγράµµJ τù Γ ‡σον παραλληλόγραµµον παραβέβληται τÕ ΣΤ ™λλε‹πον ε‡δει παραλληλογράµµJ τù ΠΒ ÐµοίJ Ôντι τù ∆ [™πειδήπερ τÕ ΠΒ τù ΗΠ Óµοιόν ™στιν]· Óπερ œδει ποιÁσαι.

†

allelogram) OB. But, OB is equal to T E, since side AE is equal to side EB [Prop. 6.1]. Thus, T E is also equal to P B. Let (parallelogram) OS have been added to both. Thus, the whole (parallelogram) T S is equal to the gnomon U XV . But, gnomon U XV was shown (to be) equal to C. Therefore, (parallelogram) T S is also equal to (figure) C. Thus, the parallelogram ST , equal to the given rectilinear figure C, has been applied to the given straightline AB, falling short by the parallelogrammic figure QB, which is similar to D [inasmuch as QB is similar to GQ [Prop. 6.24] ]. (Which is) the very thing it was required to do.

This proposition is a geometric solution of the quadratic equation x2 −α x+β = 0. Here, x is the ratio of a side of the deficit to the corresponding

side of figure D, α is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the deficit running along AB, and β is the ratio of the areas of figures C and D. The constraint corresponds to the condition β < α2 /4 for the equation to have real roots. Only the smaller root of the equation is found. The larger root can be found by a similar method.

κθ΄.

Proposition 29†

Παρ¦ τ¾ν δοθε‹σαν εÙθε‹αν τù δοθέντι εÙθυγράµµJ To apply a parallelogram, equal to a given rectilin‡σον παραλληλόγραµµον παραβαλε‹ν Øπερβάλλον ε‡δει ear figure, to a given straight-line, (the applied paralleloπαραλληλογράµµJ еοίJ τù δοθέντι. gram) overshooting by a parallelogrammic figure similar to a given (parallelogram). Ζ

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Η Ν Π Ξ ”Εστω ¹ µν δοθε‹σα εÙθε‹α ¹ ΑΒ, τÕ δ δοθν εÙθύγραµµον, ú δε‹ ‡σον παρ¦ τ¾ν ΑΒ παραβαλε‹ν, τÕ Γ, ú δ δε‹ Óµοιον Øπερβάλλειν, τÕ ∆· δε‹ δ¾ παρ¦ τ¾ν ΑΒ εÙθε‹αν τù Γ εÙθυγράµµJ ‡σον παραλληλόγραµµον παραβαλε‹ν Øπερβάλλον ε‡δει παραλληλογράµµJ еοίJ τù ∆. Τετµήσθω ¹ ΑΒ δίχα κατ¦ τÕ Ε, κሠ¢ναγεγράθω ¢πÕ τ¾ς ΕΒ τù ∆ Óµοιον καˆ Ðµοίως κείµενον παραλληλόγραµµον τÕ ΒΖ, κሠσυναµφοτέροις µν το‹ς ΒΖ, Γ ‡σον, τù δ ∆ Óµοιον καˆ Ðµοίως κείµενον τÕ αÙτÕ συνεστάτω τÕ ΗΘ. еόλογος δ œστω ¹ µν ΚΘ τÍ ΖΛ, ¹ δ ΚΗ τÍ ΖΕ. κሠ™πεˆ µε‹ζόν ™στι τÕ ΗΘ τοà ΖΒ, µείζων ¥ρα ™στˆ κሠ¹ µν ΚΘ τÁς ΖΛ, ¹ δ ΚΗ τÍ ΖΕ. ™κβεβλήσθωσαν αƒ ΖΛ, ΖΕ, κሠτÍ µν ΚΘ ‡ση œστω ¹ ΖΛΜ, τÍ δ ΚΗ ‡ση ¹ ΖΕΝ, κሠσυµπε-

P

N Q O G Let AB be the given straight-line, and C the given rectilinear figure to which the (parallelogram) applied to AB is required (to be) equal, and D the (parallelogram) to which the excess is required (to be) similar. So it is required to apply a parallelogram, equal to the given rectilinear figure C, to the given straight-line AB, overshooting by a parallelogrammic figure similar to D. Let AB have been cut in half at (point) E [Prop. 1.10], and let the parallelogram BF , (which is) similar, and similarly laid out, to D, have been applied to EB [Prop. 6.18]. And let (parallelogram) GH have been constructed (so as to be) both similar, and similarly laid out, to D, and equal to the sum of BF and C [Prop. 6.25]. And let KH correspond to F L, and KG to F E. And since (parallelogram) GH is greater than (parallelogram) F B,

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πληρώσθω τÕ ΜΝ· τÕ ΜΝ ¥ρα τù ΗΘ ‡σον τέ ™στι καˆ Óµοιον. ¢λλ¦ τÕ ΗΘ τù ΕΛ ™στιν Óµοιον· κሠτÕ ΜΝ ¥ρα τù ΕΛ Óµοιόν ™στιν· περˆ τ¾ν αÙτ¾ν ¥ρα διάµετρόν ™στι τÕ ΕΛ τù ΜΝ. ½χθω αÙτîν διάµετρος ¹ ΖΞ, κሠκαταγεγράφθω τÕ σχÁµα. 'Επεˆ ‡σον ™στˆ τÕ ΗΘ το‹ς ΕΛ, Γ, ¢λλ¦ τÕ ΗΘ τù ΜΝ ‡σον ™στίν, κሠτÕ ΜΝ ¥ρα το‹ς ΕΛ, Γ ‡σον ™στίν. κοινÕν ¢φVρήσθω τÕ ΕΛ· λοιπÕς ¥ρα Ð ΨΧΦ γνώµων τù Γ ™στιν ‡σος. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΕ τÍ ΕΒ, ‡σον ™στˆ κሠτÕ ΑΝ τù ΝΒ, τουτέστι τù ΛΟ. κοινÕν προσκείσθω τÕ ΕΞ· Óλον ¥ρα τÕ ΑΞ ‡σον ™στˆ τù ΦΧΨ γνώµονι. ¢λλ¦ Ð ΦΧΨ γνώµων τù Γ ‡σος ™στίν· κሠτÕ ΑΞ ¥ρα τù Γ ‡σον ™στίν. Παρ¦ τ¾ν δοθε‹σαν ¥ρα εÙθε‹αν τ¾ν ΑΒ τù δοθέντι εÙθυγράµµJ τù Γ ‡σον παραλληλόγραµµον παραβέβληται τÕ ΑΞ Øπερβάλλον ε‡δει παραλληλογράµµJ τù ΠΟ ÐµοίJ Ôντι τù ∆, ™πεˆ κሠτù ΕΛ ™στιν Óµοιον τÕ ΟΠ· Óπερ œδει ποιÁσαι.

†

KH is thus also greater than F L, and KG than F E. Let F L and F E have been produced, and let F LM be (made) equal to KH, and F EN to KG [Prop. 1.3]. And let (parallelogram) M N have been completed. Thus, M N is equal and similar to GH. But, GH is similar to EL. Thus, M N is also similar to EL [Prop. 6.21]. EL is thus about the same diagonal as M N [Prop. 6.26]. Let their (common) diagonal F O have been drawn, and let the (remainder of the) figure have been described. And since (parallelogram) GH is equal to (parallelogram) EL and (figure) C, but GH is equal to (parallelogram) M N , M N is thus also equal to EL and C. Let EL have been subtracted from both. Thus, the remaining gnomon U XV is equal to (figure) C. And since AE is equal to EB, (parallelogram) AN is also equal to (parallelogram) N B [Prop. 6.1], that is to say, (parallelogram) LP [Prop. 1.43]. Let (parallelogram) EO have been added to both. Thus, the whole (parallelogram) AO is equal to the gnomon U XV . But, the gnomon U XV is equal to (figure) C. Thus, (parallelogram) AO is also equal to (figure) C. Thus, the parallelogram AO, equal to the given rectilinear figure C, has been applied to the given straightline AB, overshooting by the parallelogrammic figure QP which is similar to D, since EL is also similar to P Q [Prop. 6.24]. (Which is) the very thing it was required to do.

This proposition is a geometric solution of the quadratic equation x2 +α x−β = 0. Here, x is the ratio of a side of the excess to the corresponding

side of figure D, α is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the excess running along AB, and β is the ratio of the areas of figures C and D. Only the positive root of the equation is found.

λ΄.

Proposition 30†

Τ¾ν δοθε‹σαν εÙθε‹αν πεπερασµένην ¥κρον κሠµέσον λόγον τεµε‹ν.

To cut a given finite straight-line in extreme and mean ratio.

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”Εστω ¹ δοθε‹σα εÙθε‹α πεπερασµένη ¹ ΑΒ· δε‹ δ¾ τ¾ν ΑΒ εÙθε‹αν ¥κρον κሠµέσον λόγον τεµε‹ν.

Let AB be the given finite straight-line. So it is required to cut the straight-line AB in extreme and mean

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ELEMENTS BOOK 6

'Αναγεγράφθω ¢πÕ τÁς ΑΒ τετράγωνον τÕ ΒΓ, κሠπαραβεβλήσθω παρ¦ τ¾ν ΑΓ τù ΒΓ ‡σον παραλληλόγραµµον τÕ Γ∆ Øπερβάλλον ε‡δει τù Α∆ еοίJ τù ΒΓ. Τετράγωνον δέ ™στι τÕ ΒΓ· τετράγωνον ¥ρα ™στι κሠτÕ Α∆. κሠ™πεˆ ‡σον ™στˆ τÕ ΒΓ τù Γ∆, κοινÕν ¢φVρήσθω τÕ ΓΕ· λοιπÕν ¥ρα τÕ ΒΖ λοιπù τù Α∆ ™στιν ‡σον. œστι δ αÙτù κሠ„σογώνιον· τîν ΒΖ, Α∆ ¥ρα ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας· œστιν ¥ρα æς ¹ ΖΕ πρÕς τ¾ν Ε∆, οÛτως ¹ ΑΕ πρÕς τ¾ν ΕΒ. ‡ση δ ¹ µν ΖΕ τÍ ΑΒ, ¹ δ Ε∆ τÍ ΑΕ. œστιν ¥ρα æς ¹ ΒΑ πρÕς τ¾ν ΑΕ, οÛτως ¹ ΑΕ πρÕς τ¾ν ΕΒ. µείζων δ ¹ ΑΒ τÁς ΑΕ· µείζων ¥ρα κሠ¹ ΑΕ τÁς ΕΒ. `Η ¥ρα ΑΒ εÙθε‹α ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Ε, κሠτÕ µε‹ζον αÙτÁς τµÁµά ™στι τÕ ΑΕ· Óπερ œδει ποιÁσαι.

†

ratio. Let the square BC have been described on AB [Prop. 1.46], and let the parallelogram CD, equal to BC, have been applied to AC, overshooting by the figure AD (which is) similar to BC [Prop. 6.29]. And BC is a square. Thus, AD is also a square. And since BC is equal to CD, let (rectangle) CE have been subtracted from both. Thus, the remaining (rectangle) BF is equal to the remaining (square) AD. And it is also equiangular to it. Thus, the sides of BF and AD about the equal angles are reciprocally proportional [Prop. 6.14]. Thus, as F E is to ED, so AE (is) to EB. And F E (is) equal to AB, and ED to AE. Thus, as BA is to AE, so AE (is) to EB. And AB (is) greater than AE. Thus, AE (is) also greater than EB [Prop. 5.14]. Thus, the straight-line AB has been cut in extreme and mean ratio at E, and AE is its greater piece. (Which is) the very thing it was required to do.

This method of cutting a straight-line is sometimes called the “Golden Section”—see Prop. 2.11.

λα΄.

Proposition 31

'Εν το‹ς Ñρθογωνίοις τριγώνοις τÕ ¢πÕ τÁς τ¾ν In right-angled triangles, the figure (drawn) on the Ñρθ¾ν γωνίαν Øποτεινούσης πλευρ©ς εδος ‡σον ™στˆ το‹ς side subtending the right-angle is equal to the (sum of ¢πÕ τîν τ¾ν Ñρθ¾ν γωνίαν περιεχουσîν πλευρîν ε‡δεσι the) similar, and similarly described, figures on the sides το‹ς еοίοις τε καˆ Ðµοίως ¢ναγραφοµένοις. surrounding the right-angle.

Α

Β

∆

A

Γ

B

”Εστω τρίγωνον Ñρθογώνιον τÕ ΑΒΓ Ñρθ¾ν œχον τ¾ν ØπÕ ΒΑΓ γωνίαν· λέγω, Óτι τÕ ¢πÕ τÁς ΒΓ εδος ‡σον ™στˆ το‹ς ¢πÕ τîν ΒΑ, ΑΓ ε‡δεσι το‹ς еοίοις τε καˆ Ðµοίως ¢ναγραφοµένοις. ”Ηχθω κάθετος ¹ Α∆. 'Επεˆ οâν ™ν ÑρθογωνίJ τριγώνJ τù ΑΒΓ ¢πÕ τÁς πρÕς τù Α ÑρθÁς γωνίας ™πˆ τ¾ν ΒΓ βάσιν κάθετος Ãκται ¹ Α∆, τ¦ ΑΒ∆, Α∆Γ πρÕς τÍ καθέτJ τρίγωνα Óµοιά ™στι τù τε ÓλJ τù ΑΒΓ κሠ¢λλήλοις. κሠ™πεˆ Óµοιόν ™στι τÕ ΑΒΓ τù ΑΒ∆, œστιν ¥ρα æς ¹ ΓΒ πρÕς τ¾ν ΒΑ, οÛτως ¹ ΑΒ πρÕς τ¾ν Β∆. κሠ™πεˆ τρε‹ς εÙθε‹αι ¢νάλογόν ε„σιν, œστιν æς ¹ πρώτη πρÕς τ¾ν τρίτην, οÛτως

D

C

Let ABC be a right-angled triangle having the angle BAC a right-angle. I say that the figure (drawn) on BC is equal to the (sum of the) similar, and similarly described, figures on BA and AC. Let the perpendicular AD have been drawn [Prop. 1.12]. Therefore, since, in the right-angled triangle ABC, the (straight-line) AD has been drawn from the rightangle at A perpendicular to the base BC, the triangles ABD and ADC about the perpendicular are similar to the whole (triangle) ABC, and to one another [Prop. 6.8]. And since ABC is similar to ABD, thus as BC is to BA, so AB (is) to BD [Def. 6.1]. And

189

ΣΤΟΙΧΕΙΩΝ $΄.

ELEMENTS BOOK 6

τÕ ¢πÕ τÁς πρώτης εδος πρÕς τÕ ¢πÕ τÁς δευτέρας τÕ Óµοιον καˆ Ðµοίως ¢ναγραφόµενον. æς ¥ρα ¹ ΓΒ πρÕς τ¾ν Β∆, οÛτως τÕ ¢πÕ τÁς ΓΒ εδος πρÕς τÕ ¢πÕ τÁς ΒΑ τÕ Óµοιον καˆ Ðµοίως ¢ναγραφόµενον. δι¦ τ¦ αÙτ¦ δ¾ κሠæς ¹ ΒΓ πρÕς τ¾ν Γ∆, οÛτως τÕ ¢πÕ τÁς ΒΓ εδος πρÕς τÕ ¢πÕ τÁς ΓΑ. éστε κሠæς ¹ ΒΓ πρÕς τ¦ς Β∆, ∆Γ, οÛτως τÕ ¢πÕ τÁς ΒΓ εδος πρÕς τ¦ ¢πÕ τîν ΒΑ, ΑΓ τ¦ Óµοια καˆ Ðµοίως ¢ναγραφόµενα. ‡ση δ ¹ ΒΓ τα‹ς Β∆, ∆Γ· ‡σον ¥ρα κሠτÕ ¥πÕ τÁς ΒΓ εδος το‹ς ¢πÕ τîν ΒΑ, ΑΓ ε‡δεσι το‹ς еοίοις τε καˆ Ðµοίως ¢ναγραφοµένοις. 'Εν ¥ρα το‹ς Ñρθογωνίοις τριγώνοις τÕ ¢πÕ τÁς τ¾ν Ñρθ¾ν γωνίαν Øποτεινούσης πλευρ©ς εδος ‡σον ™στˆ το‹ς ¢πÕ τîν τ¾ν Ñρθ¾ν γωνίαν περιεχουσîν πλευρîν ε‡δεσι το‹ς еοίοις τε καˆ Ðµοίως ¢ναγραφοµένοις· Óπερ œδει δε‹ξαι.

since three straight-lines are proportional, as the first is to the third, so the figure (drawn) on the first is to the similar, and similarly described, (figure) on the second [Prop. 6.19 corr.]. Thus, as CB (is) to BD, so the figure (drawn) on CB (is) to the similar, and similarly described, (figure) on BA. And so, for the same (reasons), as BC (is) to CD, so the figure (drawn) on BC (is) to the (figure) on CA. Hence, also, as BC (is) to BD and DC, so the figure (drawn) on BC (is) to the (sum of the) similar, and similarly described, (figures) on BA and AC [Prop. 5.24]. And BC is equal to BD and DC. Thus, the figure (drawn) on BC (is) also equal to the (sum of the) similar, and similarly described, figures on BA and AC [Prop. 5.9]. Thus, in right-angled triangles, the figure (drawn) on the side subtending the right-angle is equal to the (sum of the) similar, and similarly described, figures on the sides surrounding the right-angle. (Which is) the very thing it was required to show.

λβ΄.

Proposition 32

'Ε¦ν δύο τρίγωνα συντεθÍ κατ¦ µίαν γωνίαν τ¦ς δύο πλευρ¦ς τα‹ς δυσˆ πλευρα‹ς ¢νάλογον œχοντα éστε τ¦ς еολόγους αÙτîν πλευρ¦ς κሠπαραλλήλους εναι, αƒ λοιπሠτîν τριγώνων πλευρሠ™π' εÙθείας œσονται.

If two triangles, having two sides proportional to two sides, are placed together at a single angle such that the corresponding sides are also parallel, then the remaining sides of the triangles will be straight-on (with respect to one another).

∆

D

Α

Β

Γ

A

Ε

B

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΓΕ τ¦ς δύο πλευρ¦ς τ¦ς ΒΑ, ΑΓ τα‹ς δυσˆ πλευρα‹ς τα‹ς ∆Γ, ∆Ε ¢νάλογον œχοντα, æς µν τ¾ν ΑΒ πρÕς τ¾ν ΑΓ, οÛτως τ¾ν ∆Γ πρÕς τ¾ν ∆Ε, παράλληλον δ τ¾ν µν ΑΒ τÍ ∆Γ, τ¾ν δ ΑΓ τÍ ∆Ε· λέγω, Óτι ™π' εÙθείας ™στˆν ¹ ΒΓ τÍ ΓΕ. 'Επεˆ γ¦ρ παράλληλός ™στιν ¹ ΑΒ τÍ ∆Γ, κሠε„ς αÙτ¦ς ™µπέπτωκεν εÙθε‹α ¹ ΑΓ, αƒ ™ναλλ¦ξ γωνίαι αƒ ØπÕ ΒΑΓ, ΑΓ∆ ‡σαι ¢λλήλαις ε„σίν. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ Γ∆Ε τÍ ØπÕ ΑΓ∆ ‡ση ™στίν. éστε κሠ¹ ØπÕ ΒΑΓ τÍ ØπÕ Γ∆Ε ™στιν ‡ση. κሠ™πεˆ δύο τρίγωνά ™στι τ¦ ΑΒΓ, ∆ΓΕ µίαν γωνίαν τ¾ν πρÕς τù Α µι´ γωνίv τÍ πρÕς τù ∆ ‡σην œχοντα, περˆ δ τ¦ς ‡σας γωνίας τ¦ς

C

E

Let ABC and DCE be two triangles having the two sides BA and AC proportional to the two sides DC and DE—so that as AB (is) to AC, so DC (is) to DE—and (having side) AB parallel to DC, and AC to DE. I say that (side) BC is straight-on to CE. For since AB is parallel to DC, and the straight-line AC has fallen across them, the alternate angles BAC and ACD are equal to one another [Prop. 1.29]. So, for the same (reasons), CDE is also equal to ACD. And, hence, BAC is equal to CDE. And since ABC and DCE are two triangles having the one angle at A equal to the one angle at D, and the sides about the equal angles pro-

190

ΣΤΟΙΧΕΙΩΝ $΄.

ELEMENTS BOOK 6

πλευρ¦ς ¢νάλογον, æς τ¾ν ΒΑ πρÕς τ¾ν ΑΓ, οÛτως τ¾ν Γ∆ πρÕς τ¾ν ∆Ε, „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ∆ΓΕ τριγώνJ· ‡ση ¥ρα ¹ ØπÕ ΑΒΓ γωνία τÍ ØπÕ ∆ΓΕ. ™δείχθη δ κሠ¹ ØπÕ ΑΓ∆ τÍ ØπÕ ΒΑΓ ‡ση· Óλη ¥ρα ¹ ØπÕ ΑΓΕ δυσˆ τα‹ς ØπÕ ΑΒΓ, ΒΑΓ ‡ση ™στίν. κοιν¾ προσκείσθω ¹ ØπÕ ΑΓΒ· αƒ ¥ρα ØπÕ ΑΓΕ, ΑΓΒ τα‹ς ØπÕ ΒΑΓ, ΑΓΒ, ΓΒΑ ‡σαι ε„σίν. ¢λλ' αƒ ØπÕ ΒΑΓ, ΑΒΓ, ΑΓΒ δυσˆν Ñρθα‹ς ‡σαι ε„σίν· καˆ αƒ ØπÕ ΑΓΕ, ΑΓΒ ¥ρα δυσˆν Ñρθα‹ς ‡σαι ε„σίν. πρÕς δή τινι εÙθείv τÍ ΑΓ κሠτù πρÕς αÙτÍ σηµείJ τù Γ δύο εÙθε‹αι αƒ ΒΓ, ΓΕ µ¾ ™πˆ τ¦ αÙτ¦ µέρη κείµεναι τ¦ς ™φεξÁς γωνάις τ¦ς ØπÕ ΑΓΕ, ΑΓΒ δυσˆν Ñρθα‹ς ‡σας ποιοàσιν· ™π' εÙθείας ¥ρα ™στˆν ¹ ΒΓ τÍ ΓΕ. 'Ε¦ν ¥ρα δύο τρίγωνα συντεθÍ κατ¦ µίαν γωνίαν τ¦ς δύο πλευρ¦ς τα‹ς δυσˆ πλευρα‹ς ¢νάλογον œχοντα éστε τ¦ς еολόγους αÙτîν πλευρ¦ς κሠπαραλλήλους εναι, αƒ λοιπሠτîν τριγώνων πλευρሠ™π' εÙθείας œσονται· Óπερ œδει δε‹ξαι.

portional, (so that) as BA (is) to AC, so CD (is) to DE, triangle ABC is thus equiangular to triangle DCE [Prop. 6.6]. Thus, angle ABC is equal to DCE. And (angle) ACD was also shown (to be) equal to BAC. Thus, the whole (angle) ACE is equal to the two (angles) ABC and BAC. Let ACB have been added to both. Thus, ACE and ACB are equal to BAC, ACB, and CBA. But, BAC, ABC, and ACB are equal to two right-angles [Prop. 1.32]. Thus, ACE and ACB are also equal to two right-angles. Thus, the two straight-lines BC and CE, not lying on the same side, make the adjacent angles ACE and ACB equal to two right-angles at the point C on some straight-line AC. Thus, BC is straight-on to CE [Prop. 1.14]. Thus, if two triangles, having two sides proportional to two sides, are placed together at a single angle such that the corresponding sides are also parallel, then the remaining sides of the triangles will be straight-on (with respect to one another). (Which is) the very thing it was required to show.

λγ΄.

Proposition 33

'Εν το‹ς ‡σοις κύκλοις αƒ γωνίαι τÕν αÙτÕν œχουσι In equal circles, angles have the same ratio as the (raλόγον τα‹ς περιφερείαις, ™φ' ïν βεβήκασιν, ™άν τε πρÕς tio of the) circumferences on which they stand, whether το‹ς κέντροις ™άν τε πρÕς τα‹ς περιφερείαις ðσι βε- they are standing at the centers (of the circles) or at the βηκυ‹αι. circumferences.

∆

Α Η

Β

Θ Λ

Γ

D A

Κ

G

B

H L

Ε Ζ

Μ

C

Ν

K

”Εστωσαν ‡σοι κύκλοι οƒ ΑΒΓ, ∆ΕΖ, κሠπρÕς µν το‹ς κέντροις αÙτîν το‹ς Η, Θ γωνίαι œστωσαν αƒ ØπÕ ΒΗΓ, ΕΘΖ, πρÕς δ τα‹ς περιφερείαις αƒ ØπÕ ΒΑΓ, Ε∆Ζ· λέγω, Óτι ™στˆν æς ¹ ΒΓ περιφέρεια πρÕς τ¾ν ΕΖ περιφέρειαν, οÛτως ¼ τε ØπÕ ΒΗΓ γωνία πρÕς τ¾ν ØπÕ ΕΘΖ κሠ¹ ØπÕ ΒΑΓ πρÕς τ¾ν ØπÕ Ε∆Ζ. Κείσθωσαν γ¦ρ τÍ µν ΒΓ περιφερείv ‡σαι κατ¦ τÕ ˜ξÁς Ðσαιδηποτοàν αƒ ΓΚ, ΚΛ, τÍ δ ΕΖ περιφερείv ‡σαι Ðσαιδηποτοàν αƒ ΖΜ, ΜΝ, κሠ™πεζεύχθωσαν αƒ ΗΚ, ΗΛ, ΘΜ, ΘΝ. 'Επεˆ οâν ‡σαι ε„σˆν αƒ ΒΓ, ΓΚ, ΚΛ περιφέρειαι ¢λλήλαις, ‡σαι ε„σˆ καˆ αƒ ØπÕ ΒΗΓ, ΓΗΚ, ΚΗΛ γωνίαι ¢λλήλαις· Ðσαπλασίων ¥ρα ™στˆν ¹ ΒΛ περιφέρεια τÁς ΒΓ, τοσαυταπλασίων ™στˆ κሠ¹ ØπÕ ΒΗΛ γωνία τÁς ØπÕ ΒΗΓ. δι¦ τ¦ αÙτ¦ δ¾ κሠÐσαπλασίων ™στˆν ¹ ΝΕ πε-

E F

N M

Let ABC and DEF be equal circles, and let BGC and EHF be angles at their centers, G and H (respectively), and BAC and EDF (angles) at their circumferences. I say that as circumference BC is to circumference EF , so angle BGC (is) to EHF , and (angle) BAC to EDF . For let any number whatsoever of consecutive (circumferences), CK and KL, be made equal to circumference BC, and any number whatsoever, F M and M N , to circumference EF . And let GK, GL, HM , and HN have been joined. Therefore, since circumferences BC, CK, and KL are equal to one another, angles BGC, CGK, and KGL are also equal to one another [Prop. 3.27]. Thus, as many times as circumference BL is (divisible) by BC, so many times is angle BGL also (divisible) by BGC. And so, for

191

ΣΤΟΙΧΕΙΩΝ $΄.

ELEMENTS BOOK 6

ριφέρεια τÁς ΕΖ, τοσαυταπλασίων ™στˆ κሠ¹ ØπÕ ΝΘΕ γωνία τÁς ØπÕ ΕΘΖ. ε„ ¥ρα ‡ση ™στˆν ¹ ΒΛ περιφέρεια τÍ ΕΝ περιφερείv, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ΒΗΛ τÍ ØπÕ ΕΘΝ, καˆ ε„ µείζων ™στˆν ¹ ΒΛ περιφέρεια τÁς ΕΝ περιφερείας, µείζων ™στˆ κሠ¹ ØπÕ ΒΗΛ γωνία τÁς ØπÕ ΕΘΝ, καˆ ε„ ™λάσσων, ™λάσσων. τεσσάρων δ¾ Ôντων µεγεθîν, δύο µν περιφερειîν τîν ΒΓ, ΕΖ, δύο δ γωνιîν τîν ØπÕ ΒΗΓ, ΕΘΖ, ε‡ληπται τÁς µν ΒΓ περιφερείας κሠτÁς ØπÕ ΒΗΓ γωνίας „σάκις πολλαπλασίων ¼ τε ΒΛ περιφέρεια κሠ¹ ØπÕ ΒΗΛ γωνία, τÁς δ ΕΖ περιφερείας κሠτÁς ØπÕ ΕΘΖ γωνίας ¼ τε ΕΝ περιφέρια κሠ¹ ØπÕ ΕΘΝ γωνία. κሠδέδεικται, Óτι ε„ Øπερέχει ¹ ΒΛ περιφέρεια τÁς ΕΝ περιφερείας, Øπερέχει κሠ¹ ØπÕ ΒΗΛ γωνία τÁς Øπο ΕΘΝ γωνίας, καˆ ε„ ‡ση, ‡ση, καˆ ε„ ™λάσσων, ™λάσσων. œστιν ¥ρα, æς ¹ ΒΓ περιφέρεια πρÕς τ¾ν ΕΖ, οÛτως ¹ ØπÕ ΒΗΓ γωνία πρÕς τ¾ν ØπÕ ΕΘΖ. ¢λλ' æς ¹ ØπÕ ΒΗΓ γωνία πρÕς τ¾ν ØπÕ ΕΘΖ, οÛτως ¹ ØπÕ ΒΑΓ πρÕς τ¾ν ØπÕ Ε∆Ζ. διπλασία γ¦ρ ˜κατέρα ˜κατέρας. κሠæς ¥ρα ¹ ΒΓ περιφέρεια πρÕς τ¾ν ΕΖ περιφέρειαν, οÛτως ¼ τε ØπÕ ΒΗΓ γωνία πρÕς τ¾ν ØπÕ ΕΘΖ κሠ¹ ØπÕ ΒΑΓ πρÕς τ¾ν ØπÕ Ε∆Ζ. 'Εν ¥ρα το‹ς ‡σοις κύκλοις αƒ γωνίαι τÕν αÙτÕν œχουσι λόγον τα‹ς περιφερείαις, ™φ' ïν βεβήκασιν, ™άν τε πρÕς το‹ς κέντροις ™άν τε πρÕς τα‹ς περιφερείαις ðσι βεβηκυ‹αι· Óπερ œδει δε‹ξαι.

†

the same (reasons), as many times as circumference N E is (divisible) by EF , so many times is angle N HE also (divisible) by EHF . Thus, if circumference BL is equal to circumference EN then angle BGL is also equal to EHN [Prop. 3.27], and if circumference BL is greater than circumference EN then angle BGL is also greater than EHN ,† and if (BL is) less (than EN then BGL is also) less (than EHN ). So there are four magnitudes, two circumferences BC and EF , and two angles BGC and EHF . And equal multiples have been taken of circumference BC and angle BGC, (namely) circumference BL and angle BGL, and of circumference EF and angle EHF , (namely) circumference EN and angle EHN . And it has been shown that if circumference BL exceeds circumference EN then angle BGL also exceeds angle EHN , and if (BL is) equal (to EN then BGL is also) equal (to EHN ), and if (BL is) less (than EN then BGL is also) less (than EHN ). Thus, as circumference BC (is) to EF , so angle BGC (is) to EHF [Def. 5.5]. But as angle BGC (is) to EHF , so (angle) BAC (is) to EDF [Prop. 5.15]. For the former (are) double the latter (respectively) [Prop. 3.20]. Thus, also, as circumference BC (is) to circumference EF , so angle BGC (is) to EHF , and BAC to EDF . Thus, in equal circles, angles have the same ratio as the (ratio of the) circumferences on which they stand, whether they are standing at the centers (of the circles) or at the circumferences. (Which is) the very thing it was required to show.

This is a straight-forward generalization of Prop. 3.27

192

ELEMENTS BOOK 7 Elementary number theory†

† The

propositions contained in Books 7–9 are generally attributed to the school of Pythagoras.

193

ΣΤΟΙΧΕΙΩΝ ζ΄.

ELEMENTS BOOK 7

“Οροι.

Definitions

α΄. Μονάς ™στιν, καθ' ¿ν ›καστον τîν Ôντων žν λέγεται. β΄. 'ΑριθµÕς δ τÕ ™κ µονάδων συγκείµενον πλÁθος. γ΄. Μέρος ™στˆν ¢ριθµÕς ¢ριθµοà Ð ™λάσσων τοà µείζονος, Óταν καταµετρÍ τÕν µείζονα. δ΄. Μέρη δέ, Óταν µ¾ καταµετρÍ. ε΄. Πολλαπλάσιος δ Ð µείζων τοà ™λάσσονος, Óταν καταµετρÁται ØπÕ τοà ™λάσσονος. $΄. ”Αρτιος ¢ριθµός ™στιν Ð δίχα διαιρούµενος. ζ΄. ΠερισσÕς δ Ð µ¾ διαιρούµενος δίχα À [Ð] µονάδι διαφέρων ¢ρτίου ¢ριθµοà. η΄.'Αρτιάκις ¥ρτιος ¢ριθµός ™στιν Ð ØπÕ ¢ρτίου ¢ριθµοà µετρούµενος κατ¦ ¥ρτιον ¢ριθµόν. θ΄. ”Αρτιάκις δ περισσός ™στιν Ð ØπÕ ¢ρτίου ¢ριθµοà µετρούµενος κατ¦ περισσÕν ¢ριθµόν. ι΄. Περισσάκις δ περισσÕς ¢ριθµός ™στιν Ð ØπÕ περισσοà ¢ριθµοà µετρούµενος κατ¦ περισσÕν ¢ριθµόν. ια΄. Πρîτος ¢ριθµός ™στιν Ð µονάδι µόνV µετρούµενος. ιβ΄. Πρîτοι πρÕς ¢λλήλους ¢ριθµοί ε„σιν οƒ µονάδι µόνV µετρούµενοι κοινù µέτρJ. ιγ΄. Σύνθετος ¢ριθµός ™στιν Ð ¢ριθµù τινι µετρούµενος. ιδ΄. Σύνθετοι δ πρÕς ¢λλήλους ¢ριθµοί ε„σιν οƒ ¢ριθµù τινι µετρούµενοι κοινù µέτρJ. ιε΄. 'ΑριθµÕς ¢ριθµÕν πολλαπλασιάζειν λέγεται, Óταν, Óσαι ε„σˆν ™ν αÙτù µονάδες, τοσαυτάκις συντεθÍ Ð πολλαπλασιαζόµενος, κሠγένηταί τις. ι$΄. “Οταν δ δύο ¢ριθµοˆ πολλαπλασιάσαντες ¢λλήλους ποιîσί τινα, Ð γενόµενος ™πίπεδος καλε‹ται, πλευραˆ δ αÙτοà οƒ πολλαπλασιάσαντες ¢λλήλους ¢ριθµοί. ιζ΄. “Οταν δ τρε‹ς ¢ριθµοˆ πολλαπλασιάσαντες ¢λλήλους ποιîσί τινα, Ð γενόµενος στερεός ™στιν, πλευραˆ δ αÙτοà οƒ πολλαπλασιάσαντες ¢λλήλους ¢ριθµοί. ιη΄. Τετράγωνος ¢ριθµός ™στιν Ð „σάκις ‡σος À [Ð] ØπÕ δύο ‡σων ¢ριθµîν περιεχόµενος. ιθ΄. Κύβος δ Ð „σάκις ‡σος „σάκις À [Ð] ØπÕ τριîν ‡σων ¢ριθµîν περιεχόµενος. κ΄. 'Αριθµοˆ ¢νάλογόν ε„σιν, Óταν Ð πρîτος τοà δευτέρου καˆ Ð τρίτος τοà τετάρτου „σάκις Ï πολλαπλάσιος À τÕ αÙτÕ µέρος À τ¦ αÙτ¦ µέρη ðσιν. κα΄. “Οµοιοι ™πίπεδοι κሠστερεοˆ ¢ριθµοί ε„σιν οƒ ανάλογον œχοντες τ¦ς πλευράς. κβ΄. Τέλειος ¢ριθµός ™στιν Ð το‹ς ˜αυτοà µέρεσιν ‡σος êν.

1. A unit is (that) according to which each existing (thing) is said (to be) one. 2. And a number (is) a multitude composed of units.† 3. A number is part of a(nother) number, the lesser of the greater, when it measures the greater.‡ 4. But (the lesser is) parts (of the greater) when it does not measure it.§ 5. And the greater (number is) a multiple of the lesser when it is measured by the lesser. 6. An even number is one (which can be) divided in half. 7. And an odd number is one (which can)not (be) divided in half, or which differs from an even number by a unit. 8. An even-times-even number is one (which is) measured by an even number according to an even number.¶ 9. And an even-times-odd number is one (which is) measured by an even number according to an odd number.∗ 10. And an odd-times-odd number is one (which is) measured by an odd number according to an odd number.$ 11. A primek number is one (which is) measured by a unit alone. 12. Numbers prime to one another are those (which are) measured by a unit alone as a common measure. 13. A composite number is one (which is) measured by some number. 14. And numbers composite to one another are those (which are) measured by some number as a common measure. 15. A number is said to multiply a(nother) number when the (number being) multiplied is added (to itself) as many times as there are units in the former (number), and (thereby) some (other number) is produced. 16. And when two numbers multiplying one another make some (other number) then the (number so) created is called plane, and its sides (are) the numbers which multiply one another. 17. And when three numbers multiplying one another make some (other number) then the (number so) created is (called) solid, and its sides (are) the numbers which multiply one another. 18. A square number is an equal times an equal, or (a plane number) contained by two equal numbers. 19. And a cube (number) is an equal times an equal times an equal, or (a solid number) contained by three equal numbers.

194

ΣΤΟΙΧΕΙΩΝ ζ΄.

ELEMENTS BOOK 7 20. Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third (is) of the fourth. 21. Similar plane and solid numbers are those having proportional sides. 22. A perfect number is that which is equal to its own parts.††

†

In other words, a “number” is a positive integer greater than unity.

‡

In other words, a number a is part of another number b if their exists some number n such that n a = b.

§

In other words, a number a is parts of another number b (where a < b) if their exist distinct numbers, m and n, such that n a = m b.

¶

In other words. an even-times-even number is the product of two even numbers.

∗

In other words, an even-times-odd number is the product of an even and an odd number.

$

In other words, an odd-times-odd number is the product of two odd numbers.

k

Literally, “first”.

††

In other words, a perfect number is equal to the sum of its own factors.

α΄.

Proposition 1

∆ύο ¢ριθµîν ¢νίσων ™κκειµένων, ¢νθυφαιρουµένου δ ¢εˆ τοà ™λάσσονος ¢πÕ τοà µείζονος, ™¦ν Ð λειπόµενος µηδέποτε καταµετρÍ τÕν πρÕ ˜αυτοà, ›ως οá λειφθÍ µονάς, οƒ ™ξ ¢ρχÁς ¢ριθµοˆ πρîτοι πρÕς ¢λλ¾λους œσονται.

Two unequal numbers (being) laid down, and the lesser being continually subtracted, in turn, from the greater, if the remainder never measures the (number) preceding it, until a unit remains, then the original numbers will be prime to one another.

Α Ζ

A

Θ

H F

Γ Η

C G

Ε Β

E

∆

B

∆ύο γ¦ρ [¢νίσων] ¢ριθµîν τîν ΑΒ, Γ∆ ¢νθυφαιρουµένου ¢εˆ τοà ™λάσσονος ¢πÕ τοà µείζονος Ð λειπόµενος µηδέποτε καταµετρείτω τÕν πρÕ ˜αυτοà, ›ως οá λειφθÍ µονάς· λέγω, Óτι οƒ ΑΒ, Γ∆ πρîτοι πρÕς ¢λλήλους ε„σίν, τουτέστιν Óτι τοÝς ΑΒ, Γ∆ µον¦ς µόνη µετρε‹. Ε„ γ¦ρ µή ε„σιν οƒ ΑΒ, Γ∆ πρîτοι πρÕς ¢λλήλους, µετρήσει τις αÙτοÝς ¢ριθµός. µετρείτω, κሠœστω Ð Ε· καˆ Ð µν Γ∆ τÕν ΒΖ µετρîν λειπέτω ˜αυτοà ™λάσσονα τÕν ΖΑ, Ð δ ΑΖ τÕν ∆Η µετρîν λειπέτω ˜αυτοà ™λάσσονα τÕν ΗΓ, Ð δ ΗΓ τÕν ΖΘ µετρîν λεˆπέτω µονάδα τ¾ν ΘΑ. 'Επεˆ οâν Ð Ε τÕν Γ∆ µετρε‹, Ð δ Γ∆ τÕν ΒΖ µετρε‹, καˆ Ð Ε ¥ρα τÕν ΒΖ µετρε‹· µετρε‹ δ κሠÓλον τÕν ΒΑ· κሠλοιπÕν ¥ρα τÕν ΑΖ µετρήσει. Ð δ ΑΖ τÕν

D

For two [unequal] numbers, AB and CD, the lesser being continually subtracted, in turn, from the greater, let the remainder never measure the (number) preceding it, until a unit remains. I say that AB and CD are prime to one another—that is to say, that a unit alone measures (both) AB and CD. For if AB and CD are not prime to one another then some number will measure them. Let (some number) measure them, and let it be E. And let CD measuring BF leave F A less than itself, and let AF measuring DG leave GC less than itself, and let GC measuring F H leave a unit, HA. In fact, since E measures CD, and CD measures BF , E thus also measures BF .† And (E) also measures the whole of BA. Thus, (E) will also measure the remainder

195

ΣΤΟΙΧΕΙΩΝ ζ΄.

ELEMENTS BOOK 7

∆Η µετρε‹· καˆ Ð Ε ¥ρα τÕν ∆Η µετρε‹· µετρε‹ δ κሠÓλον τÕν ∆Γ· κሠλοιπÕν ¥ρα τÕν ΓΗ µετρήσει. Ð δ ΓΗ τÕν ΖΘ µετρε‹· καˆ Ð Ε ¥ρα τÕν ΖΘ µετρε‹· µετρε‹ δ κሠÓλον τÕν ΖΑ· κሠλοιπ¾ν ¥ρα τ¾ν ΑΘ µονάδα µετρήσει ¢ριθµÕς êν· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τοÝς ΑΒ, Γ∆ ¢ριθµοÝς µετρήσει τις ¢ριθµός· οƒ ΑΒ, Γ∆ ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι.

AF .‡ And AF measures DG. Thus, E also measures DG. And (E) also measures the whole of DC. Thus, (E) will also measure the remainder CG. And CG measures F H. Thus, E also measures F H. And (E) also measures the whole of F A. Thus, (E) will also measure the remaining unit AH, (despite) being a number. The very thing is impossible. Thus, some number does not measure (both) the numbers AB and CD. Thus, AB and CD are prime to one another. (Which is) the very thing it was required to show.

†

Here, use is made of the unstated common notion that if a measures b, and b measures c, then a also measures c, where all symbols denote numbers. ‡ Here, use is made of the unstated common notion that if a measures b, and a measures part of b, then a also measures the remainder of b, where all symbols denote numbers.

β΄.

Proposition 2

∆ύο ¢ριθµîν δοθέντων µ¾ πρώτων πρÕς ¢λλήλους τÕ µέγιστον αÙτîν κοινÕν µέτρον εØρε‹ν.

To find the greatest common measure of two given numbers (which are) not prime to one another.

Α Ε

A E

Γ Ζ

C F

Η Β

∆

G B

”Εστωσαν οƒ δοθέντες δύο ¢ριθµοˆ µ¾ πρîτοι πρÕς ¢λλήλους οƒ ΑΒ, Γ∆. δε‹ δ¾ τîν ΑΒ, Γ∆ τÕ µέγιστον κοινÕν µέτρον εØρε‹ν. Ε„ µν οâν Ð Γ∆ τÕν ΑΒ µετρε‹, µετρε‹ δ κሠ˜αυτόν, Ð Γ∆ ¥ρα τîν Γ∆, ΑΒ κοινÕν µέτρον ™στίν. κሠφανερόν, Óτι κሠµέγιστον· οÙδεˆς γ¦ρ µείζων τοà Γ∆ τÕν Γ∆ µετρήσει. Ε„ δ οÙ µετρε‹ Ð Γ∆ τÕν ΑΒ, τîν ΑΒ, Γ∆ ¢νθυφαιρουµένου ¢εˆ τοà ™λάσσονος ¢πÕ τοà µείζονος λειφθήσεταί τις ¢ριθµός, Öς µετρήσει τÕν πρÕ ˜αυτοà. µον¦ς µν γ¦ρ οÙ λειφθήσεται· ε„ δ µή, œσονται οƒ ΑΒ, Γ∆ πρîτοι πρÕς ¢λλήλους· Óπερ οÙχ Øπόκειται. λειφήσεταί τις ¥ρα ¢ριθµÕς, Öς µετρήσει τÕν πρÕ ˜αυτοà. καˆ Ð µν Γ∆ τÕν ΒΕ µετρîν λειπέτω ˜αυτοà ™λάσσονα τÕν ΕΑ, Ð δ ΕΑ τÕν ∆Ζ µετρîν λειπέτω ˜αυτοà ™λάσσονα τÕν ΖΓ, Ð δ ΓΖ τÕν ΑΕ µετρείτω. ™πεˆ οâν Ð ΓΖ τÕν ΑΕ µετρε‹, Ð δ ΑΕ τÕν ∆Ζ µετρε‹, καˆ Ð ΓΖ ¥ρα τÕν ∆Ζ µετρήσει. µετρε‹ δ κሠ˜αυτόν· κሠÓλον ¥ρα τÕν Γ∆ µετρήσει. Ð δ Γ∆ τÕν ΒΕ µετρε‹· καˆ Ð ΓΖ ¥ρα τÕν ΒΕ µετρε‹· µετρε‹ δ κሠτÕν ΕΑ·

D

Let AB and CD be the two given numbers (which are) not prime to one another. So it is required to find the greatest common measure of AB and CD. In fact, if CD measures AB, CD is thus a common measure of CD and AB, (since CD) also measures itself. And (it is) manifest that (it is) also the greatest (common measure). For nothing greater than CD can measure CD. But if CD does not measure AB then some number will remain from AB and CD, the lesser being continually subtracted, in turn, from the greater, which will measure the (number) preceding it. For a unit will not be left. But if not, AB and CD will be prime to one another [Prop. 7.1]. The very opposite thing was assumed. Thus, some number will remain which will measure the (number) preceding it. And let CD measuring BE leave EA less than itself, and let EA measuring DF leave F C less than itself, and let CF measure AE. Therefore, since CF measures AE, and AE measures DF , CF will thus also measure DF . And it also measures itself. Thus, it will

196

ΣΤΟΙΧΕΙΩΝ ζ΄.

ELEMENTS BOOK 7

κሠÓλον ¥ρα τÕν ΒΑ µετρήσει· µετρε‹ δ κሠτÕν Γ∆· Ð ΓΖ ¥ρα τοÝς ΑΒ, Γ∆ µετρε‹. Ð ΓΖ ¥ρα τîν ΑΒ, Γ∆ κοινÕν µέτρον ™στίν. λέγω δή, Óτι κሠµέγιστον. ε„ γ¦ρ µή ™στιν Ð ΓΖ τîν ΑΒ, Γ∆ µέγιστον κοινÕν µέτρον, µετρήσει τις τοÝς ΑΒ, Γ∆ ¢ριθµοÝς ¢ριθµÕς µείζων íν τοà ΓΖ. µετρείτω, κሠœστω Ð Η. κሠ™πεˆ Ð Η τÕν Γ∆ µετρε‹, Ð δ Γ∆ τÕν ΒΕ µετρε‹, καˆ Ð Η ¥ρα τÕν ΒΕ µετρε‹· µετρε‹ δ κሠÓλον τÕν ΒΑ· κሠλοιπÕν ¥ρα τÕν ΑΕ µετρήσει. Ð δ ΑΕ τÕν ∆Ζ µετρε‹· καˆ Ð Η ¥ρα τÕν ∆Ζ µετρήσει· µετρε‹ δ κሠÓλον τÕν ∆Γ· κሠλοιπÕν ¥ρα τÕν ΓΖ µετρήσει Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύνατον· οÙκ ¥ρα τοÝς ΑΒ, Γ∆ ¢ριθµοÝς ¢ριθµός τις µετρήσει µείζων íν τοà ΓΖ· Ð ΓΖ ¥ρα τîν ΑΒ, Γ∆ µέγιστόν ™στι κοινÕν µέτρον. [Óπερ œδει δε‹ξαι].

also measure the whole of CD. And CD measures BE. Thus, CF also measures BE. And it also measures EA. Thus, it will also measure the whole of BA. And it also measures CD. Thus, CF measures (both) AB and CD. Thus, CF is a common measure of AB and CD. So I say that (it is) also the greatest (common measure). For if CF is not the greatest common measure of AB and CD then some number which is greater than CF will measure the numbers AB and CD. Let it (so) measure (AB and CD), and let it be G. And since G measures CD, and CD measures BE, G thus also measures BE. And it also measures the whole of BA. Thus, it will also measure the remainder AE. And AE measures DF . Thus, G will also measure DF . And it also measures the whole of DC. Thus, it will also measure the remainder CF , the greater (measuring) the lesser. The very thing is impossible. Thus, some number which is greater than CF cannot measure the numbers AB and CD. Thus, CF is the greatest common measure of AB and CD. [(Which is) the very thing it was required to show].

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι ™¦ν ¢ριθµÕς δύο ¢ριθµοÝς µετρÍ, κሠτÕ µέγιστον αÙτîν κοινÕν µέτρον µετρήσει· Óπερ œδει δε‹ξαι.

So it is manifest, from this, that if a number measures two numbers then it will also measure their greatest common measure. (Which is) the very thing it was required to show.

γ΄.

Proposition 3

Τριîν ¢ριθµîν δοθέντων µ¾ πρώτων πρÕς ¢λλήλους τÕ µέγιστον αÙτîν κοινÕν µέτρον εØρε‹ν.

To find the greatest common measure of three given numbers (which are) not prime to one another.

Α

Β

Γ

∆

Ε

Ζ

A

”Εστωσαν οƒ δοθέντες τρε‹ς ¢ριθµοˆ µ¾ πρîτοι πρÕς ¢λλήλους οƒ Α, Β, Γ· δε‹ δ¾ τîν Α, Β, Γ τÕ µέγιστον κοινÕν µέτρον εØρε‹ν. Ε„λήφθω γ¦ρ δύο τîν Α, Β τÕ µέγιστον κοινÕν µέτρον Ð ∆· Ð δ¾ ∆ τÕν Γ ½τοι µετρε‹ À οÙ µετρε‹. µετρείτω πρότερον· µετρε‹ δέ κሠτοÝς Α, Β· Ð ∆ ¥ρα τοÝς Α, Β, Γ µετρε‹· Ð ∆ ¥ρα τîν Α, Β, Γ κοινÕν µέτρον

B

C

D

E

F

Let A, B, and C be the three given numbers (which are) not prime to one another. So it is required to find the greatest common measure of A, B, and C. For let the greatest common measure, D, of the two (numbers) A and B have been taken [Prop. 7.2]. So D either measures, or does not measure, C. First of all, let it measure (C). And it also measures A and B. Thus, D

197

ΣΤΟΙΧΕΙΩΝ ζ΄.

ELEMENTS BOOK 7

™στίν. λέγω δή, Óτι κሠµέγιστον. ε„ γ¦ρ µή ™στιν Ð ∆ τîν Α, Β, Γ µέγιστον κοινÕν µέτρον, µετρήσει τις τοÝς Α, Β, Γ ¢ριθµοÝς ¢ριθµÕς µείζων íν τοà ∆. µετρείτω, κሠœστω Ð Ε. ™πεˆ οâν Ð Ε τοÝς Α, Β, Γ µετρε‹, κሠτοÝς Α, Β ¥ρα µετρήσει· κሠτÕ τîν Α, Β ¥ρα µέγιστον κοινÕν µέτρον µετρήσει. τÕ δ τîν Α, Β µέγιστον κοινÕν µέτρον ™στˆν Ð ∆· Ð Ε ¥ρα τÕν ∆ µετρε‹ Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τοÝς Α, Β, Γ ¢ριθµοÝς ¢ριθµός τις µετρήσει µείζων íν τοà ∆· Ð ∆ ¥ρα τîν Α, Β, Γ µέγιστόν ™στι κοινÕν µέτρον. Μ¾ µετρείτω δ¾ Ð ∆ τÕν Γ· λέγω πρîτον, Óτι οƒ Γ, ∆ οÜκ ε„σι πρîτοι πρÕς ¢λλήλους. ™πεˆ γ¦ρ οƒ Α, Β, Γ οÜκ ε„σι πρîτοι πρÕς ¢λλήλους, µετρήσει τις αÙτοÝς ¢ριθµός. Ð δ¾ τοÝς Α, Β, Γ µετρîν κሠτοÝς Α, Β µετρήσει, κሠτÕ τîν Α, Β µέγιστον κοινÕν µέτρον τÕν ∆ µετρήσει· µετρε‹ δ κሠτÕν Γ· τοÝς ∆, Γ ¥ρα ¢ριθµοÝς ¢ριθµός τις µετρήσει· οƒ ∆, Γ ¥ρα οÜκ ε„σι πρîτοι πρÕς ¢λλήλους. ε„λήφθω οâν αÙτîν τÕ µέγιστον κοινÕν µέτρον Ð Ε. κሠ™πεˆ Ð Ε τÕν ∆ µετρε‹, Ð δ ∆ τοÝς Α, Β µετρε‹, καˆ Ð Ε ¥ρα τοÝς Α, Β µετρε‹· µετρε‹ δ κሠτÕν Γ· Ð Ε ¥ρα τοÝς Α, Β, Γ µετρε‹. Ð Ε ¥ρα τîν Α, Β, Γ κοινόν ™στι µέτρον. λέγω δή, Óτι κሠµέγιστον. ε„ γ¦ρ µή ™στιν Ð Ε τîν Α, Β, Γ τÕ µέγιστον κοινÕν µέτρον, µετρήσει τις τοÝς Α, Β, Γ ¢ριθµοÝς ¢ριθµÕς µείζων íν τοà Ε. µετρείτω, κሠœστω Ð Ζ. κሠ™πεˆ Ð Ζ τοÝς Α, Β, Γ µετρε‹, κሠτοÝς Α, Β µετρε‹· κሠτÕ τîν Α, Β ¥ρα µέγιστον κοινÕν µέτρον µετρήσει. τÕ δ τîν Α, Β µέγιστον κοινÕν µέτρον ™στˆν Ð ∆· Ð Ζ ¥ρα τÕν ∆ µετρε‹· µετρε‹ δ κሠτÕν Γ· Ð Ζ ¥ρα τοÝς ∆, Γ µετρε‹· κሠτÕ τîν ∆, Γ ¥ρα µέγιστον κοινÕν µέτρον µετρήσει. τÕ δ τîν ∆, Γ µέγιστον κοινÕν µέτρον ™στˆν Ð Ε· Ð Ζ ¥ρα τÕν Ε µετρε‹ Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τοÝς Α, Β, Γ ¢ριθµοÝς ¢ριθµός τις µετρήσει µείζων íν τοà Ε· Ð Ε ¥ρα τîν Α, Β, Γ µέγιστόν ™στι κοινÕν µέτρον· Óπερ œδει δε‹ξαι.

measures A, B, and C. Thus, D is a common measure of A, B, and C. So I say that (it is) also the greatest (common measure). For if D is not the greatest common measure of A, B, and C then some number greater than D will measure the numbers A, B, and C. Let it (so) measure (A, B, and C), and let it be E. Therefore, since E measures A, B, and C, it will thus also measure A and B. Thus, it will also measure the greatest common measure of A and B [Prop. 7.2 corr.]. And D is the greatest common measure of A and B. Thus, E measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, some number which is greater than D cannot measure the numbers A, B, and C. Thus, D is the greatest common measure of A, B, and C. So let D not measure C. I say, first of all, that C and D are not prime to one another. For since A, B, C are not prime to one another, some number will measure them. So the (number) measuring A, B, and C will also measure A and B, and it will also measure the greatest common measure, D, of A and B [Prop. 7.2 corr.]. And it also measures C. Thus, some number will measure the numbers D and C. Thus, D and C are not prime to one another. Therefore, let their greatest common measure, E, have been taken [Prop. 7.2]. And since E measures D, and D measures A and B, E thus also measures A and B. And it also measures C. Thus, E measures A, B, and C. Thus, E is a common measure of A, B, and C. So I say that (it is) also the greatest (common measure). For if E is not the greatest common measure of A, B, and C then some number greater than E will measure the numbers A, B, and C. Let it (so) measure (A, B, and C), and let it be F . And since F measures A, B, and C, it also measures A and B. Thus, it will also measure the greatest common measure of A and B [Prop. 7.2 corr.]. And D is the greatest common measure of A and B. Thus, F measures D. And it also measures C. Thus, F measures D and C. Thus, it will also measure the greatest common measure of D and C [Prop. 7.2 corr.]. And E is the greatest common measure of D and C. Thus, F measures E, the greater (measuring) the lesser. The very thing is impossible. Thus, some number which is greater than E does not measure the numbers A, B, and C. Thus, E is the greatest common measure of A, B, and C. (Which is) the very thing it was required to show.

δ΄.

Proposition 4

“Απας ¢ριθµÕς παντÕς ¢ριθµοà Ð ™λάσσων τοà Any number is either part or parts of any (other) numµείζονος ½τοι µέρος ™στˆν À µέρη. ber, the lesser of the greater. ”Εστωσαν δύο ¢ριθµοˆ οƒ Α, ΒΓ, κሠœστω ™λάσσων Let A and BC be two numbers, and let BC be the Ð ΒΓ· λέγω, Óτι Ð ΒΓ τοà Α ½τοι µέρος ™στˆν À µέρη. lesser. I say that BC is either part or parts of A.

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Οƒ Α, ΒΓ γ¦ρ ½τοι πρîτοι πρÕς ¢λλήλους ε„σˆν À For A and BC are either prime to one another, or not. οÜ. œστωσαν πρότερον οƒ Α, ΒΓ πρîτοι πρÕς ¢λλήλους. Let A and BC, first of all, be prime to one another. So διαρεθέντος δ¾ τοà ΒΓ ε„ς τ¦ς ™ν αÙτù µονάδας œσται separating BC into its constituent units, each of the units ˜κάστη µον¦ς τîν ™ν τù ΒΓ µέρος τι τοà Α· éστε µέρη in BC will be some part of A. Hence, BC is parts of A. ™στˆν Ð ΒΓ τοà Α.

Α

Β

B

Ε

E

Ζ

F

Γ

∆

A

C

D

Μ¾ œστωσαν δ¾ οƒ Α, ΒΓ πρîτοι πρÕς ¢λλήλους· Ð δ¾ ΒΓ τÕν Α ½τοι µετρε‹ À οÙ µετρε‹. ε„ µν οâν Ð ΒΓ τÕν Α µετρε‹, µέρος ™στˆν Ð ΒΓ τοà Α. ε„ δ οÜ, ε„λήφθω τîν Α, ΒΓ µέγιστον κοινÕν µέτρον Ð ∆, κሠδιVρήσθω Ð ΒΓ ε„ς τοÝς τù ∆ ‡σους τοÝς ΒΕ, ΕΖ, ΖΓ. κሠ™πεˆ Ð ∆ τÕν Α µετρε‹, µέρος ™στˆν Ð ∆ τοà Α· ‡σος δ Ð ∆ ˜κάστJ τîν ΒΕ, ΕΖ, ΖΓ· κሠ›καστος ¥ρα τîν ΒΕ, ΕΖ, ΖΓ τοà Α µέρος ™στίν· éστε µέρη ™στˆν Ð ΒΓ τοà Α. “Απας ¥ρα ¢ριθµÕς παντÕς ¢ριθµοà Ð ™λάσσων τοà µείζονος ½τοι µέρος ™στˆν À µέρη· Óπερ œδει δε‹ξαι.

So let A and BC be not prime to one another. So BC either measures, or does not measure, A. Therefore, if BC measures A then BC is part of A. And if not, let the greatest common measure, D, of A and BC have been taken [Prop. 7.2], and let BC have been divided into BE, EF , and F C, equal to D. And since D measures A, D is a part of A. And D is equal to each of BE, EF , and F C. Thus, BE, EF , and F C are also each part of A. Hence, BC is parts of A. Thus, any number is either part or parts of any (other) number, the lesser of the greater. (Which is) the very thing it was required to show.

ε΄.

Proposition 5†

'Ε¦ν ¢ριθµÕς ¢ριθµοà µέρος Ï, κሠ›τερος ˜τέρου If a number is part of a number, and another (numτÕ αÙτÕ µέρος Ï, κሠσυναµφότερος συναµφοτέρου τÕ ber) is the same part of another, then the sum (of the αÙτÕ µέρος œσται, Óπερ Ð εŒς τοà ˜νός. leading numbers) will also be the same part of the sum (of the following numbers) that one (number) is of another.

Β

B Ε

Η

Α

Γ

E G

Θ ∆

H

Ζ

A

'ΑριθµÕς γ¦ρ Ð Α [¢ριθµοà] τοà ΒΓ µέρος œστω, καˆ

199

C

D

F

For let a number A be part of a [number] BC, and

ΣΤΟΙΧΕΙΩΝ ζ΄.

ELEMENTS BOOK 7

›τερος Ð ∆ ˜τέρου τοà ΕΖ τÕ αÙτÕ µέρος, Óπερ Ð Α τοà ΒΓ· λέγω, Óτι κሠσυναµφότερος Ð Α, ∆ συναµφοτέρου τοà ΒΓ, ΕΖ τÕ αÙτÕ µέρος ™στίν, Óπερ Ð Α τοà ΒΓ. 'Επεˆ γάρ, Ö µέρος ™στˆν Ð Α τοà ΒΓ, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ∆ τοà ΕΖ, Óσοι ¥ρα ε„σˆν ™ν τù ΒΓ ¢ριθµοˆ ‡σοι τù Α, τοσοàτοί ε„σι κሠ™ν τù ΕΖ ¢ριθµοˆ ‡σοι τù ∆. διÍρήσθω Ð µν ΒΓ ε„ς τοÝς τù Α ‡σους τοÝς ΒΗ, ΗΓ, Ð δ ΕΖ ε„ς τοÝς τù ∆ ‡σους τοÝς ΕΘ, ΘΖ· œσται δ¾ ‡σον τÕ πλÁθος τîν ΒΗ, ΗΓ τù πλήθει τîν ΕΘ, ΘΖ. κሠ™πεˆ ‡σος ™στˆν Ð µν ΒΗ τù Α, Ð δ ΕΘ τù ∆, καˆ οƒ ΒΗ, ΕΘ ¥ρα το‹ς Α, ∆ ‡σοι. δι¦ τ¦ αÙτ¦ δ¾ καˆ οƒ ΗΓ, ΘΖ το‹ς Α, ∆. Óσοι ¥ρα [ε„σˆν] ™ν τù ΒΓ ¢ριθµοˆ ‡σοι τù Α, τοσοàτοί ε„σι κሠ™ν το‹ς ΒΓ, ΕΖ ‡σοι το‹ς Α, ∆. Ðσαπλασίων ¥ρα ™στˆν Ð ΒΓ τοà Α, τοσαυταπλασίων ™στˆ κሠσυναµφότερος Ð ΒΓ, ΕΖ συναµφοτέρου τοà Α, ∆. Ö ¥ρα µέρος ™στˆν Ð Α τοà ΒΓ, τÕ αÙτÕ µέρος ™στˆ κሠσυναµφότερος Ð Α, ∆ συναµφοτέρου τοà ΒΓ, ΕΖ· Óπερ œδει δε‹ξαι. †

another (number) D (be) the same part of another (number) EF that A (is) of BC. I say that the sum A, D is also the same part of the sum BC, EF that A (is) of BC. For since which(ever) part A is of BC, D is the same part of EF , thus as many numbers as are in BC equal to A, so many numbers are also in EF equal to D. Let BC have been divided into BG and GC, equal to A, and EF into EH and HF , equal to D. So the multitude of (divisions) BG, GC will be equal to the multitude of (divisions) EH, HF . And since BG is equal to A, and EH to D, thus BG, EH (is) also equal to A, D. So, for the same (reasons), GC, HF (is) also (equal) to A, D. Thus, as many numbers as [are] in BC equal to A, so many are also in BC, EF equal to A, D. Thus, as many times as BC is (divisible) by A, so many times is the sum BC, EF also (divisible) by the sum A, D. Thus, which(ever) part A is of BC, the sum A, D is also the same part of the sum BC, EF . (Which is) the very thing it was required to show.

In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then (a + c) = (1/n) (b + d), where all symbols denote numbers.

$΄.

Proposition 6†

'Ε¦ν ¢ριθµÕς ¢ριθµοà µέρη Ï, κሠ›τερος ˜τέρου τ¦ If a number is parts of a number, and another (numαÙτ¦ µέρη Ï, κሠσυναµφότερος συναµφοτέρου τ¦ αÙτ¦ ber) is the same parts of another, then the sum (of the µέρη œσται, Óπερ Ð εŒς τοà ˜νός. leading numbers) will also be the same parts of the sum (of the following numbers) that one (number) is of another.

Α

A ∆

Η Β

D G

Θ Γ

Ε

H B

Ζ

'ΑριθµÕς γ¦ρ Ð ΑΒ ¢ριθµοà τοà Γ µέρη œστω, κሠ›τερος Ð ∆Ε ˜τέρου τοà Ζ τ¦ αÙτ¦ µέρη, ¤περ Ð ΑΒ τοà Γ· λέγω, Óτι κሠσυναµφότερος Ð ΑΒ, ∆Ε συναµφοτέρου τοà Γ, Ζ τ¦ αÙτ¦ µέρη ™στίν, ¤περ Ð ΑΒ τοà Γ. 'Επεˆ γάρ, § µέρη ™στˆν Ð ΑΒ τοà Γ, τ¦ αÙτ¦ µέρη καˆ Ð ∆Ε τοà Ζ, Óσα ¥ρα ™στˆν ™ν τù ΑΒ µέρη τοà Γ, τοσαàτά ™στι κሠ™ν τù ∆Ε µέρη τοà Ζ. διVρήσθω Ð µν ΑΒ ε„ς τ¦ τοà Γ µέρη τ¦ ΑΗ, ΗΒ, Ð δ ∆Ε ε„ς τ¦ τοà Ζ µέρη τ¦ ∆Θ, ΘΕ· œσται δ¾ ‡σον τÕ πλÁθος τîν ΑΗ, ΗΒ τù πλήθει τîν ∆Θ, ΘΕ. κሠ™πεί, Ö µέρος

C

E

F

For let a number AB be parts of a number C, and another (number) DE (be) the same parts of another (number) F that AB (is) of C. I say that the sum AB, DE is also the same parts of the sum C, F that AB (is) of C. For since which(ever) parts AB is of C, DE (is) also the same parts of F , thus as many parts of C as are in AB, so many parts of F are also in DE. Let AB have been divided into the parts of C, AG and GB, and DE into the parts of F , DH and HE. So the multitude of (divisions) AG, GB will be equal to the multitude of (divisions) DH,

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ELEMENTS BOOK 7

™στˆν Ð ΑΗ τοà Γ, τÕ αØτÕ µέρος ™στˆ καˆ Ð ∆Θ τοà Ζ, Ö ¥ρα µέρος ™στˆν Ð ΑΗ τοà Γ, τÕ αÙτÕ µέρος ™στˆ κሠσυναµφότερος Ð ΑΗ, ∆Θ συναµφοτέρου τοà Γ, Ζ. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ö µέρος ™στˆν Ð ΗΒ τοà Γ, τÕ αÙτÕ µέρος ™στˆ κሠσυναµφότερος Ð ΗΒ, ΘΕ συναµφοτέρου τοà Γ, Ζ. § ¥ρα µέρη ™στˆν Ð ΑΒ τοà Γ, τ¦ αÙτ¦ µέρη ™στˆ κሠσυναµφότερος Ð ΑΒ, ∆Ε συναµφοτέρου τοà Γ, Ζ· Óπερ œδει δε‹ξαι. †

HE. And since which(ever) part AG is of C, DH is also the same part of F , thus which(ever) part AG is of C, the sum AG, DH is also the same part of the sum C, F [Prop. 7.5]. And so, for the same (reasons), which(ever) part GB is of C, the sum GB, HE is also the same part of the sum C, F . Thus, which(ever) parts AB is of C, the sum AB, DE is also the same parts of the sum C, F . (Which is) the very thing it was required to show.

In modern notation, this proposition states that if a = (m/n) b and c = (m/n) d then (a + c) = (m/n) (b + d), where all symbols denote

numbers.

ζ΄.

Proposition 7

'Ε¦ν ¢ριθµÕς ¢ριθµοà µέρος Ï, Óπερ ¢φαιρεθεˆς ¢φαιρεθέντος, καˆ Ð λοιπÕς τοà λοιποà τÕ αÙτÕ µέρος œσται, Óπερ Ð Óλος τοà Óλου.

If a number is that part of a number that a (part) taken away (is) of a (part) taken away, then the remainder will also be the same part of the remainder that the whole (is) of the whole.

Α Ε Η

Β

A E

Γ

Ζ

∆

G

'ΑριθµÕς γ¦ρ Ð ΑΒ ¢ριθµοà τοà Γ∆ µέρος œστω, Óπερ ¢φαιρεθεˆς Ð ΑΕ ¢φαιρεθέντος τοà ΓΖ· λέγω, Óτι κሠλοιπÕς Ð ΕΒ λοιποà τοà Ζ∆ τÕ αÙτÕ µέρος ™στίν, Óπερ Óλος Ð ΑΒ Óλου τοà Γ∆. •Ο γ¦ρ µέρος ™στˆν Ð ΑΕ τοà ΓΖ, τÕ αÙτÕ µέρος œστω καˆ Ð ΕΒ τοà ΓΗ. κሠ™πεί, Ö µέρος ™στˆν Ð ΑΕ τοà ΓΖ, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΕΒ τοà ΓΗ, Ö ¥ρα µέρος ™στˆν Ð ΑΕ τοà ΓΖ, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΑΒ τοà ΗΖ. Ö δ µέρος ™στˆν Ð ΑΕ τοà ΓΖ, τÕ αÙτÕ µέρος Øπόκειται καˆ Ð ΑΒ τοà Γ∆· Ö ¥ρα µέρος ™στˆ καˆ Ð ΑΒ τοà ΗΖ, τÕ αÙτÕ µέρος ™στˆ κሠτοà Γ∆· ‡σος ¥ρα ™στˆν Ð ΗΖ τù Γ∆. κοινÕς ¢φVρήσθω Ð ΓΖ· λοιπÕς ¥ρα Ð ΗΓ λοιπù τù Ζ∆ ™στιν ‡σος. κሠ™πεί, Ö µέρος ™στˆν Ð ΑΕ τοà ΓΖ, τÕ αÙτÕ µέρος [™στˆ] καˆ Ð ΕΒ τοà ΗΓ, ‡σος δ Ð ΗΓ τù Ζ∆, Ö ¥ρα µέρος ™στˆν Ð ΑΕ τοà ΓΖ, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΕΒ τοà Ζ∆. ¢λλ¦ Ö µέρος ™στˆν Ð ΑΕ τοà ΓΖ, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΑΒ τοà Γ∆· κሠλοιπÕς ¥ρα Ð ΕΒ λοιποà τοà Ζ∆ τÕ αÙτÕ µέρος ™στίν, Óπερ Óλος Ð ΑΒ Óλου τοà Γ∆· Óπερ œδει δε‹ξαι.

†

B C

F

D

For let a number AB be that part of a number CD that a (part) taken away AE (is) of a part taken away CF . I say that the remainder EB is also the same part of the remainder F D that the whole AB (is) of the whole CD. For which(ever) part AE is of CF , let EB also be the same part of CG. And since which(ever) part AE is of CF , EB is also the same part of CG, thus which(ever) part AE is of CF , AB is also the same part of GF [Prop. 7.5]. And which(ever) part AE is of CF , AB is also assumed (to be) the same part of CD. Thus, also, which(ever) part AB is of GF , (AB) is also the same part of CD. Thus, GF is equal to CD. Let CF have been subtracted from both. Thus, the remainder GC is equal to the remainder F D. And since which(ever) part AE is of CF , EB [is] also the same part of GC, and GC (is) equal to F D, thus which(ever) part AE is of CF , EB is also the same part of F D. But, which(ever) part AE is of CF , AB is also the same part of CD. Thus, the remainder EB is also the same part of the remainder F D that the whole AB (is) of the whole CD. (Which is) the very thing it was required to show.

In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then (a − c) = (1/n) (b − d), where all symbols denote numbers.

η΄.

Proposition 8†

'Ε¦ν ¢ριθµÕς ¢ριθµοà µέρη Ï, ¤περ ¢φαιρεθεˆς If a number is those parts of a number that a (part) ¢φαιρεθέντος, καˆ Ð λοιπÕς τοà λοιποà τ¦ αÙτ¦ µέρη taken away (is) of a (part) taken away, then the remain-

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ELEMENTS BOOK 7

œσται, ¤περ Ð Óλος τοà Óλου.

der will also be the same parts of the remainder that the whole (is) of the whole.

Γ

Ζ

Η

ΜΚ

Α

Λ

∆

C

F

ΝΘ

G

M K

Β

A

L

Ε

'ΑριθµÕς γ¦ρ Ð ΑΒ ¢ριθµοà τοà Γ∆ µέρη œστω, ¤περ ¢φαιρεθεˆς Ð ΑΕ ¢φαιρεθέντος τοà ΓΖ· λέγω, Óτι κሠλοιπÕς Ð ΕΒ λοιποà τοà Ζ∆ τ¦ αÙτ¦ µέρη ™στίν, ¤περ Óλος Ð ΑΒ Óλου τοà Γ∆. Κείσθω γ¦ρ τù ΑΒ ‡σος Ð ΗΘ, § ¥ρα µέρη ™στˆν Ð ΗΘ τοà Γ∆, τ¦ αÙτ¦ µέρη ™στˆ καˆ Ð ΑΕ τοà ΓΖ. διVρήσθω Ð µν ΗΘ ε„ς τ¦ τοà Γ∆ µέρη τ¦ ΗΚ, ΚΘ, Ð δ ΑΕ ε„ς τ¦ τοà ΓΖ µέρη τ¦ ΑΛ, ΛΕ· œσται δ¾ ‡σον τÕ πλÁθος τîν ΗΚ, ΚΘ τù πλήθει τîν ΑΛ, ΛΕ. κሠ™πεί, Ö µέρος ™στˆν Ð ΗΚ τοà Γ∆, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΑΛ τοà ΓΖ, µείζων δ Ð Γ∆ τοà ΓΖ, µείζων ¥ρα καˆ Ð ΗΚ τοà ΑΛ. κείσθω τù ΑΛ ‡σος Ð ΗΜ. Ö ¥ρα µέρος ™στˆν Ð ΗΚ τοà Γ∆, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΗΜ τοà ΓΖ· κሠλοιπÕς ¥ρα Ð ΜΚ λοιποà τοà Ζ∆ τÕ αÙτÕ µέρος ™στίν, Óπερ Óλος Ð ΗΚ Óλου τοà Γ∆. πάλιν ™πεί, Ö µέρος ™στˆν Ð ΚΘ τοà Γ∆, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΕΛ τοà ΓΖ, µείζων δ Ð Γ∆ τοà ΓΖ, µείζων ¥ρα καˆ Ð ΘΚ τοà ΕΛ. κείσθω τù ΕΛ ‡σος Ð ΚΝ. Ö ¥ρα µέρος ™στˆν Ð ΚΘ τοà Γ∆, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΚΝ τοà ΓΖ· κሠλοιπÕς ¥ρα Ð ΝΘ λοιποà τοà Ζ∆ τÕ αÙτÕ µέρος ™στίν, Óπερ Óλος Ð ΚΘ Óλου τοà Γ∆. ™δείχθη δ κሠλοιπÕς Ð ΜΚ λοιποà τοà Ζ∆ τÕ αÙτÕ µέρος êν, Óπερ Óλος Ð ΗΚ Óλου τοà Γ∆· κሠσυναµφότερος ¥ρα Ð ΜΚ, ΝΘ τοà ∆Ζ τ¦ αÙτ¦ µέρη ™στίν, ¤περ Óλος Ð ΘΗ Óλου τοà Γ∆. ‡σος δ συναµφότερος µν Ð ΜΚ, ΝΘ τù ΕΒ, Ð δ ΘΗ τù ΒΑ· κሠλοιπÕς ¥ρα Ð ΕΒ λοιποà τοà Ζ∆ τ¦ αÙτ¦ µέρη ™στίν, ¤περ Óλος Ð ΑΒ Óλου τοà Γ∆· Óπερ œδει δε‹ξαι.

D

NH E

B

For let a number AB be those parts of a number CD that a (part) taken away AE (is) of a (part) taken away CF . I say that the remainder EB is also the same parts of the remainder F D that the whole AB (is) of the whole CD. For let GH be laid down equal to AB. Thus, which(ever) parts GH is of CD, AE is also the same parts of CF . Let GH have been divided into the parts of CD, GK and KH, and AE into the part of CF , AL and LE. So the multitude of (divisions) GK, KH will be equal to the multitude of (divisions) AL, LE. And since which(ever) part GK is of CD, AL is also the same part of CF , and CD (is) greater than CF , GK (is) thus also greater than AL. Let GM be made equal to AL. Thus, which(ever) part GK is of CD, GM is also the same part of CF . Thus, the remainder M K is also the same part of the remainder F D that the whole GK (is) of the whole CD [Prop. 7.5]. Again, since which(ever) part KH is of CD, EL is also the same part of CF , and CD (is) greater than CF , HK (is) thus also greater than EL. Let KN be made equal to EL. Thus, which(ever) part KH (is) of CD, KN is also the same part of CF . Thus, the remainder N H is also the same part of the remainder F D that the whole KH (is) of the whole CD [Prop. 7.5]. And the remainder M K was also shown to be the same part of the remainder F D that the whole GK (is) of the whole CD. Thus, the sum M K, N H is the same parts of DF that the whole HG (is) of the whole CD. And the sum M K, N H (is) equal to EB, and HG to BA. Thus, the remainder EB is also the same parts of the remainder F D that the whole AB (is) of the whole CD. (Which is) the very thing it was required to show.

†

In modern notation, this proposition states that if a = (m/n) b and c = (m/n) d then (a − c) = (m/n) (b − d), where all symbols denote numbers.

θ΄.

Proposition 9†

'Ε¦ν ¢ριθµÕς ¢ριθµοà µέρος Ï, κሠ›τερος ˜τέρου If a number is part of a number, and another (numτÕ αÙτÕ µέρος Ï, κሠ™ναλλάξ, Ö µέρος ™στˆν À µέρη Ð ber) is the same part of another, also, alternately,

202

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ELEMENTS BOOK 7

πρîτος τοà τρίτου, τÕ αÙτÕ µέρος œσται À τ¦ αÙτ¦ µέρη which(ever) part, or parts, the first (number) is of the καˆ Ð δεύτερος τοà τετάρτου. third, the second (number) will also be the same part, or the same parts, of the fourth.

Ε

E

Β

B Θ

Η

Α

Γ

∆

H G

Ζ

A

'ΑριθµÕς γ¦ρ Ð Α ¢ριθµοà τοà ΒΓ µέρος œστω, κሠ›τερος Ð ∆ ˜τέρου τοà ΕΖ τÕ αÙτÕ µέρος, Óπερ Ð Α τοà ΒΓ· λέγω, Óτι κሠ™ναλλάξ, Ö µέρος ™στˆν Ð Α τοà ∆ À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΒΓ τοà ΕΖ À µέρη. 'Επεˆ γ¦ρ Ö µέρος ™στˆν Ð Α τοà ΒΓ, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ∆ τοà ΕΖ, Óσοι ¥ρα ε„σˆν ™ν τù ΒΓ ¢ριθµοˆ ‡σοι τù Α, τοσοàτοί ε„σι κሠ™ν τù ΕΖ ‡σοι τù ∆. διVρήσθω Ð µν ΒΓ ε„ς τοÝς τù Α ‡σους τοÝς ΒΗ, ΗΓ, Ð δ ΕΖ ε„ς τοÝς τù ∆ ‡σους τοÝς ΕΘ, ΘΖ· œσται δ¾ ‡σον τÕ πλÁθος τîν ΒΗ, ΗΓ τù πλήθει τîν ΕΘ, ΘΖ. Κሠ™πεˆ ‡σοι ε„σˆν οƒ ΒΗ, ΗΓ ¢ριθµοˆ ¢λλήλοις, ε„σˆ δ καˆ οƒ ΕΘ, ΘΖ ¢ριθµοˆ ‡σοι ¢λλήλοις, καί ™στιν ‡σον τÕ πλÁθος τîν ΒΗ, ΗΓ τù πλήθει τîν ΕΘ, ΘΖ, Ö ¥ρα µέρος ™στˆν Ð ΒΗ τοà ΕΘ À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΗΓ τοà ΘΖ À τ¦ αÙτ¦ µέρη· éστε καˆ Ö µέρος ™στˆν Ð ΒΗ τοà ΕΘ À µέρη, τÕ αÙτÕ µέρος ™στˆ κሠσυναµφότερος Ð ΒΓ συναµφοτέρου τοà ΕΖ À τ¦ αÙτ¦ µέρη. ‡σος δ Ð µν ΒΗ τù Α, Ð δ ΕΘ τù ∆· Ö ¥ρα µέρος ™στˆν Ð Α τοà ∆ À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΒΓ τοà ΕΖ À τ¦ αÙτ¦ µέρη· Óπερ œδει δε‹ξαι.

†

C

D

F

For let a number A be part of a number BC, and another (number) D (be) the same part of another EF that A (is) of BC. I say that, also, alternately, which(ever) part, or parts, A is of D, BC is also the same part, or parts, of EF . For since which(ever) part A is of BC, D is also the same part of EF , thus as many numbers as are in BC equal to A, so many are also in EF equal to D. Let BC have been divided into BG and GC, equal to A, and EF into EH and HF , equal to D. So the multitude of (divisions) BG, GC will be equal to the multitude of (divisions) EH, HF . And since the numbers BG and GC are equal to one another, and the numbers EH and HF are also equal to one another, and the multitude of (divisions) BG, GC is equal to the multitude of (divisions) EH, HC, thus which(ever) part, or parts, BG is of EH, GC is also the same part, or the same parts, of HF . And hence, which(ever) part, or parts, BG is of EH, the sum BC is also the same part, or the same parts, of the sum EF [Props. 7.5, 7.6]. And BG (is) equal to A, and EH to D. Thus, which(ever) part, or parts, A is of D, BC is also the same part, or the same parts, of EF . (Which is) the very thing it was required to show.

In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote

numbers.

ι΄.

Proposition 10†

'Ε¦ν ¢ριθµÕς ¢ριθµοà µέρη Ï, κሠ›τερος ˜τέρου τ¦ αÙτ¦ µέρη Ï, κሠ™ναλλάξ, § µέρη ™στˆν Ð πρîτος τοà τρίτου À µέρος, τ¦ αÙτ¦ µέρη œσται καˆ Ð δεύτερος τοà τετάρτου À τÕ αÙτÕ µέρος. 'ΑριθµÕς γ¦ρ Ð ΑΒ ¢ριθµοà τοà Γ µέρη œστω, κሠ›τερος Ð ∆Ε ˜τέρου τοà Ζ τ¦ αÙτ¦ µέρη· λέγω, Óτι καˆ

If a number is parts of a number, and another (number) is the same parts of another, also, alternately, which(ever) parts, or part, the first (number) is of the third, the second will also be the same parts, or the same part, of the fourth. For let a number AB be parts of a number C, and

203

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ELEMENTS BOOK 7

™ναλλάξ, § µέρη ™στˆν Ð ΑΒ τοà ∆Ε À µέρος, τ¦ αÙτ¦ µέρη ™στˆ καˆ Ð Γ τοà Ζ À τÕ αÙτÕ µέρος.

another (number) DE (be) the same parts of another F . I say that, also, alternately, which(ever) parts, or part, AB is of DE, C is also the same parts, or the same part, of F .

∆

D

Α

A Θ

Η Β

Ε

Γ

H G B

Ζ

'Επεˆ γάρ, § µέρη ™στˆν Ð ΑΒ τοà Γ, τ¦ αÙτ¦ µέρη ™στˆ καˆ Ð ∆Ε τοà Ζ, Óσα ¥ρα ™στˆν ™ν τù ΑΒ µέρη τοà Γ, τοσαàτα κሠ™ν τù ∆Ε µέρη τοà Ζ. διVρήσθω Ð µν ΑΒ ε„ς τ¦ τοà Γ µέρη τ¦ ΑΗ, ΗΒ, Ð δ ∆Ε ε„ς τ¦ τοà Ζ µέρη τ¦ ∆Θ, ΘΕ· œσται δ¾ ‡σον τÕ πλÁθος τîν ΑΗ, ΗΒ τù πλήθει τîν ∆Θ, ΘΕ. κሠ™πεί, Ö µέρος ™στˆν Ð ΑΗ τοà Γ, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ∆Θ τοà Ζ, κሠ™ναλλάξ, Ö µέρος ™στˆν Ð ΑΗ τοà ∆Θ À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð Γ τοà Ζ À τ¦ αÙτ¦ µέρη. δι¦ τ¦ αÙτ¦ δ¾ καί, Ö µέρος ™στˆν Ð ΗΒ τοà ΘΕ À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð Γ τοà Ζ À τ¦ αÙτ¦ µέρη· éστε καί [Ö µέρος ™στˆν Ð ΑΗ τοà ∆Θ À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΗΒ τοà ΘΕ À τ¦ αÙτ¦ µέρη· καˆ Ö ¥ρα µέρος ™στˆν Ð ΑΗ τοà ∆Θ À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΑΒ τοà ∆Ε À τ¦ αÙτ¦ µέρη· ¢λλ' Ö µέρος ™στˆν Ð ΑΗ τοà ∆Θ À µέρη, τÕ αÙτÕ µέρος ™δείχθη καˆ Ð Γ τοà Ζ À τ¦ αÙτ¦ µέρη, καˆ] § [¥ρα] µέρη ™στˆν Ð ΑΒ τοà ∆Ε À µέρος, τ¦ αÙτ¦ µέρη ™στˆ καˆ Ð Γ τοà Ζ À τÕ αÙτÕ µέρος· Óπερ œδει δε‹ξαι.

†

C

E

F

For since which(ever) parts AB is of C, DE is also the same parts of F , thus as many parts of C as are in AB, so many parts of F (are) also in DE. Let AB have been divided into the parts of C, AG and GB, and DE into the parts of F , DH and HE. So the multitude of (divisions) AG, GB will be equal to the multitude of (divisions) DH, HE. And since which(ever) part AG is of C, DH is also the same part of F , also, alternately, which(ever) part, or parts, AG is of DH, C is also the same part, or the same parts, of F [Prop. 7.9]. And so, for the same (reasons), which(ever) part, or parts, GB is of HE, C is also the same part, or the same parts, of F [Prop. 7.9]. And so [which(ever) part, or parts, AG is of DH, GB is also the same part, or the same parts, of HE. And thus, which(ever) part, or parts, AG is of DH, AB is also the same part, or the same parts, of DE [Props. 7.5, 7.6]. But, which(ever) part, or parts, AG is of DH, C was also shown (to be) the same part, or the same parts, of F . And, thus] which(ever) parts, or part, AB is of DE, C is also the same parts, or the same part, of F . (Which is) the very thing it was required to show.

In modern notation, this proposition states that if a = (m/n) b and c = (m/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote

numbers.

ια΄.

Proposition 11

'Εαν Ï æς Óλος πρÕς Óλον, οÛτως ¢φαιρεθεˆς πρÕς ¢φαιρεθέντα, καˆ Ð λοιπÕς πρÕς τÕν λοιπÕν œσται, æς Óλος πρÕς Óλον. ”Εστω æς Óλος Ð ΑΒ πρÕς Óλον τÕν Γ∆, οÛτως ¢φαιρεθεˆς Ð ΑΕ πρÕς ¢φαιρεθέντα τÕν ΓΖ· λέγω, Óτι κሠλοιπÕς Ð ΕΒ πρÕς λοιπÕν τÕν Ζ∆ ™στιν, æς Óλος Ð ΑΒ πρÕς Óλον τÕν Γ∆.

If as the whole (of a number) is to the whole (of another), so a (part) taken away (is) to a (part) taken away, then the remainder will also be to the remainder as the whole (is) to the whole. Let the whole AB be to the whole CD as the (part) taken away AE (is) to the (part) taken away CF . I say that the remainder EB is to the remainder F D as the whole AB (is) to the whole CD.

204

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Γ

Α

C

Ζ

F A

Ε Β

E

∆

B

'Επεί ™στιν æς Ð ΑΒ πρÕς τÕν Γ∆, οÛτως Ð ΑΕ πρÕς τÕν ΓΖ, Ö ¥ρα µέρος ™στˆν Ð ΑΒ τοà Γ∆ À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð ΑΕ τοà ΓΖ À τ¦ αÙτ¦ µέρη. κሠλοιπÕς ¥ρα Ð ΕΒ λοιποà τοà Ζ∆ τÕ αÙτÕ µέρος ™στˆν À µέρη, ¤περ Ð ΑΒ τοà Γ∆. œστιν ¥ρα æς Ð ΕΒ πρÕς τÕν Ζ∆, οÛτως Ð ΑΒ πρÕς τÕν Γ∆· Óπερ œδει δε‹ξαι. †

D

(For) since as AB is to CD, so AE (is) to CF , thus which(ever) part, or parts, AB is of CD, AE is also the same part, or the same parts, of CF [Def. 7.20]. Thus, the remainder EB is also the same part, or parts, of the remainder F D that AB (is) of CD [Props. 7.7, 7.8]. Thus, as EB is to F D, so AB (is) to CD [Def. 7.20]. (Which is) the very thing it was required to show.

In modern notation, this proposition states that if a : b :: c : d then a : b :: a − c : b − d, where all symbols denote numbers.

ιβ΄.

Proposition 12†

'Ε¦ν ðσιν Ðποσοιοàν ¢ριθµοˆ ¢νάλογον, œσται æς εŒς τîν ¹γουµένων πρÕς ›να τîν ˜ποµένων, οÛτως ¤παντες οƒ ¹γούµενοι πρÕς ¤παντας τοÝς ˜ποµένους.

If any multitude whatsoever of numbers are proportional then as one of the leading (numbers is) to one of the following so (the sum of) all of the leading (numbers) will be to (the sum of)all of the following.

Α

Β

Γ

∆

A

”Εστωσαν Ðποσοιοàν ¢ριθµοˆ ¢νάλογον οƒ Α, Β, Γ, ∆, æς Ð Α πρÕς τÕν Β, οÛτως Ð Γ πρÕς τÕν ∆· λέγω, Óτι ™στˆν æς Ð Α πρÕς τÕν Β, οÛτως οƒ Α, Γ πρÕς τοÝς Β, ∆. 'Επεˆ γάρ ™στιν æς Ð Α πρÕς τÕν Β, οÛτως Ð Γ πρÕς τÕν ∆, Ö ¥ρα µέρος ™στˆν Ð Α τοà Β À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð Γ τοà ∆ À µέρη. κሠσυναµφότερος ¥ρα Ð Α, Γ συναµφοτέρου τοà Β, ∆ τÕ αÙτÕ µέρος ™στˆν À τ¦ αÙτ¦ µέρη, ¤περ Ð Α τοà Β. œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως οƒ Α, Γ πρÕς τοÝς Β, ∆· Óπερ œδει δε‹ξαι.

B

C

D

Let any multitude whatsoever of numbers, A, B, C, D, be proportional, (such that) as A (is) to B, so C (is) to D. I say that as A is to B, so A, C (is) to B, D. For since as A is to B, so C (is) to D, thus which(ever) part, or parts, A is of B, C is also the same part, or parts, of D [Def. 7.20]. Thus, the sum A, C is also the same part, or the same parts, of the sum B, D that A (is) of B [Props. 7.5, 7.6]. Thus, as A is to B, so A, C (is) to B, D [Def. 7.20]. (Which is) the very thing it was required to show.

205

ΣΤΟΙΧΕΙΩΝ ζ΄. †

ELEMENTS BOOK 7

In modern notation, this proposition states that if a : b :: c : d then a : b :: a + c : b + d, where all symbols denote numbers.

ιγ΄.

Proposition 13†

'Ε¦ν τέσσαρες ¢ριθµοˆ ¢νάλογον ðσιν, κሠ™ναλλ¦ξ ¢νάλογον œσονται.

If four numbers are proportional then they will also be proportional alternately.

Α

Β

Γ

∆

A

”Εστωσαν τέσσαρες ¢ριθµοˆ ¢νάλογον οƒ Α, Β, Γ, ∆, æς Ð Α πρÕς τÕν Β, οÛτως Ð Γ πρÕς τÕν ∆· λέγω, Óτι κሠ™ναλλ¦ξ ¢νάλογον œσονται, æς Ð Α πρÕς τÕν Γ, οÛτως Ð Β πρÕς τÕν ∆. 'Επεˆ γάρ ™στιν æς Ð Α πρÕς τÕν Β, οÛτως Ð Γ πρÕς τÕν ∆, Ö ¥ρα µέρος ™στˆν Ð Α τοà Β À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð Γ τοà ∆ À τ¦ αÙτ¦ µέρη. ™ναλλ¦ξ ¥ρα, Ö µέρος ™στˆν Ð Α τοà Γ À µέρη, τÕ αÙτÕ µέρος ™στˆ καˆ Ð Β τοà ∆ À τ¦ αÙτ¦ µέρη. œστιν ¥ρα æς Ð Α πρÕς τÕν Γ, οÛτως Ð Β πρÕς τÕν ∆· Óπερ œδει δε‹ξαι. †

B

C

D

Let the four numbers A, B, C, and D be proportional, (such that) as A (is) to B, so C (is) to D. I say that they will also be proportional alternately, (such that) as A (is) to C, so B (is) to D. For since as A is to B, so C (is) to D, thus which(ever) part, or parts, A is of B, C is also the same part, or the same parts, of D [Def. 7.20]. Thus, alterately, which(ever) part, or parts, A is of C, B is also the same part, or the same parts, of D [Props. 7.9, 7.10]. Thus, as A is to C, so B (is) to D [Def. 7.20]. (Which is) the very thing it was required to show.

In modern notation, this proposition states that if a : b :: c : d then a : c :: b : d, where all symbols denote numbers.

ιδ΄.

Proposition 14†

'Ε¦ν ðσιν Ðποσοιοàν ¢ριθµοˆ κሠ¥λλοι αÙτο‹ς ‡σοι If there are any multitude of numbers whatsoever, τÕ πλÁθος σύνδυο λαµβανόµενοι κሠ™ν τù αÙτù λόγJ, and (some) other (numbers) of equal multitude to them, κሠδι' ‡σου ™ν τù αÙτù λόγù œσονται. (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality.

Α Β Γ

∆ Ε Ζ

A B C

”Εστωσαν Ðποσοιοàν ¢ριθµοˆ οƒ Α, Β, Γ κሠ¥λλοι αÙτο‹ς ‡σοι τÕ πλÁθος σύνδυο λαµβανόµενοι ™ν τù αÙτù λόγJ οƒ ∆, Ε, Ζ, æς µν Ð Α πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Ε, æς δ Ð Β πρÕς τÕν Γ, οÛτως Ð Ε πρÕς τÕν Ζ· λέγω, Óτι κሠδι' ‡σου ™στˆν æς Ð Α πρÕς τÕν Γ, οÛτως Ð ∆ πρÕς τÕν Ζ. 'Επεˆ γάρ ™στιν æς Ð Α πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Ε, ™ναλλ¦ξ ¥ρα ™στˆν æς Ð Α πρÕς τÕν ∆, οÛτως Ð Β πρÕς τÕν Ε. πάλιν, ™πεί ™στιν æς Ð Β πρÕς τÕν Γ, οÛτως

D E F

Let there be any multitude of numbers whatsoever, A, B, C, and (some) other (numbers), D, E, F , of equal multitude to them, (which are) in the same ratio taken two by two, (such that) as A (is) to B, so D (is) to E, and as B (is) to C, so E (is) to F . I say that also, via equality, as A is to C, so D (is) to F . For since as A is to B, so D (is) to E, thus, alternately, as A is to D, so B (is) to E [Prop. 7.13]. Again, since as B is to C, so E (is) to F , thus, alternately, as B is

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Ð Ε πρÕς τÕν Ζ, ™ναλλ¦ξ ¥ρα ™στˆν æς Ð Β πρÕς τÕν Ε, οÛτως Ð Γ πρÕς τÕν Ζ. æς δ Ð Β πρÕς τÕν Ε, οÛτως Ð Α πρÕς τÕν ∆· κሠæς ¥ρα Ð Α πρÕς τÕν ∆, οÛτως Ð Γ πρÕς τÕν Ζ· ™ναλλ¦ξ ¥ρα ™στˆν æς Ð Α πρÕς τÕν Γ, οÛτως Ð ∆ πρÕς τÕν Ζ· Óπερ œδει δε‹ξαι. †

to E, so C (is) to F [Prop. 7.13]. And as B (is) to E, so A (is) to D. Thus, also, as A (is) to D, so C (is) to F . Thus, alternately, as A is to C, so D (is) to F [Prop. 7.13]. (Which is) the very thing it was required to show.

In modern notation, this proposition states that if a : b :: d : e and b : c :: e : f then a : c :: d : f , where all symbols denote numbers.

ιε΄.

Proposition 15

'Ε¦ν µον¦ς ¢ριθµόν τινα µετρÍ, „σακις δ ›τερος ¢ριθµÕς ¥λλον τιν¦ ¢ριθµÕν µετρÍ, κሠ™ναλλ¦ξ „σάκις ¹ µον¦ς τÕν τρίτον ¢ριθµÕν µετρήσει καˆ Ð δεύτερος τÕν τέταρτον.

If a unit measures some number, and another number measures some other number as many times, then, also, alternately, the unit will measure the third number as many times as the second (number measures) the fourth.

Α ∆

Β Ε

Η

Θ Κ

Γ

A Λ

Ζ

E

H

G K

C L

F

D

Μον¦ς γ¦ρ ¹ Α ¢ριθµόν τινα τÕν ΒΓ µετρείτω, „σάκις δ ›τερος ¢ριθµÕς Ð ∆ ¥λλον τιν¦ ¢ριθµÕν τÕν ΕΖ µετρείτω· λέγω, Óτι κሠ™ναλλ¦ξ „σάκις ¹ Α µον¦ς τÕν ∆ ¢ριθµÕν µετρε‹ καˆ Ð ΒΓ τÕν ΕΖ. 'Επεˆ γ¦ρ „σάκις ¹ Α µον¦ς τÕν ΒΓ ¢ριθµÕν µετρε‹ καˆ Ð ∆ τÕν ΕΖ, Óσαι ¥ρα ε„σˆν ™ν τù ΒΓ µονάδες, τοσοàτοί ε„σι κሠ™ν τù ΕΖ ¢ριθµοˆ ‡σοι τù ∆. διVρήσθω Ð µν ΒΓ ε„ς τ¦ς ™ν ˜αυτù µονάδας τ¦ς ΒΗ, ΗΘ, ΘΓ, Ð δ ΕΖ ε„ς τοÝς τù ∆ ‡σους τοÝς ΕΚ, ΚΛ, ΛΖ. œσται δ¾ ‡σον τÕ πλÁθος τîν ΒΗ, ΗΘ, ΘΓ τù πλήθει τîν ΕΚ, ΚΛ, ΛΖ. κሠ™πεˆ ‡σαι ε„σˆν αƒ ΒΗ, ΗΘ, ΘΓ µονάδες ¢λλήλαις, ε„σˆ δ καˆ οƒ ΕΚ, ΚΛ, ΛΖ ¢ριθµοˆ ‡σοι ¢λλήλοις, καί ™στιν ‡σον τÕ πλÁθος τîν ΒΗ, ΗΘ, ΘΓ µονάδων τù πλήθει τîν ΕΚ, ΚΛ, ΛΖ ¢ριθµîν, œσται ¥ρα æς ¹ ΒΗ µον¦ς πρÕς τÕν ΕΚ ¢ριθµόν, οÛτως ¹ ΗΘ µον¦ς πρÕς τÕν ΚΛ ¢ριθµÕν κሠ¹ ΘΓ µον¦ς πρÕς τÕν ΛΖ ¢ριθµόν. œσται ¥ρα κሠæς εŒς τîν ¹γουµένων πρÕς ›να τîν ˜ποµένων, οÛτως ¤παντες οƒ ¹γούµενοι πρÕς ¤παντας τοÝς ˜ποµένους· œστιν ¥ρα æς ¹ ΒΗ µον¦ς πρÕς τÕν ΕΚ ¢ριθµόν, οÛτως Ð ΒΓ πρÕς τÕν ΕΖ. ‡ση δ ¹ ΒΗ µον¦ς τÍ Α µονάδι, Ð δ ΕΚ ¢ριθµÕς τù ∆ ¢ριθµù. œστιν ¥ρα æς ¹ Α µον¦ς πρÕς τÕν ∆ ¢ριθµόν, οÛτως Ð ΒΓ πρÕς τÕν ΕΖ. „σάκις ¥ρα ¹ Α µον¦ς τÕν ∆ ¢ριθµÕν µετρε‹ καˆ Ð ΒΓ τÕν ΕΖ· Óπερ œδει δε‹ξαι.

†

B

For let a unit A measure some number BC, and let another number D measure some other number EF as many times. I say that, also, alternately, the unit A also measures the number D as many times as BC (measures) EF . For since the unit A measures the number BC as many times as D (measures) EF , thus as many units as are in BC, so many numbers are also in EF equal to D. Let BC have been divided into its constituent units, BG, GH, and HC, and EF into the (divisions) EK, KL, and LF , equal to D. So the multitude of (units) BG, GH, HC will be equal to the multitude of (divisions) EK, KL, LF . And since the units BG, GH, and HC are equal to one another, and the numbers EK, KL, and LF are also equal to one another, and the multitude of the (units) BG, GH, HC is equal to the multitude of the numbers EK, KL, LF , thus as the unit BG (is) to the number EK, so the unit GH will be to the number KL, and the unit HC to the number LF . And thus, as one of the leading (numbers is) to one of the following, so (the sum of) all of the leading will be to (the sum of) all of the following [Prop. 7.12]. Thus, as the unit BG (is) to the number EK, so BC (is) to EF . And the unit BG (is) equal to the unit A, and the number EK to the number D. Thus, as the unit A is to the number D, so BC (is) to EF . Thus, the unit A measures the number D as many times as BC (measures) EF [Def. 7.20]. (Which is) the very thing it was required to show.

This proposition is a special case of Prop. 7.9.

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ELEMENTS BOOK 7 ι$΄.

Proposition 16†

Ε¦ν δύο ¢ριθµοˆ πολλαπλασιάσαντες ¢λλήλους ποιîσί τινας, οƒ γενόµενοι ™ξ αÙτîν ‡σοι ¢λλήλοις œσονται.

If two numbers multiplying one another make some (numbers) then the (numbers) generated from them will be equal to one another.

Α Β Γ ∆ Ε

A B C D E

”Εστωσαν δύο ¢ριθµοˆ οƒ Α, Β, καˆ Ð µν Α τÕν Β πολλαπλασιάσας τÕν Γ ποιείτω, Ð δ Β τÕν Α πολλαπλασιάσας τÕν ∆ ποιείτω· λέγω, Óτι ‡σος ™στˆν Ð Γ τù ∆. 'Επεˆ γ¦ρ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν, Ð Β ¥ρα τÕν Γ µετρε‹ κατ¦ τ¦ς ™ν τù Α µονάδας. µετρε‹ δ κሠ¹ Ε µον¦ς τÕν Α ¢ριθµÕν κατ¦ τ¦ς ™ν αÙτù µονάδας· „σάκις ¥ρα ¹ Ε µον¦ς τÕν Α ¢ριθµÕν µετρε‹ καˆ Ð Β τÕν Γ. ™ναλλ¦ξ ¥ρα „σάκις ¹ Ε µον¦ς τÕν Β ¢ριθµÕν µετρε‹ καˆ Ð Α τÕν Γ. πάλιν, ™πεˆ Ð Β τÕν Α πολλαπλασιάσας τÕν ∆ πεποίηκεν, Ð Α ¥ρα τÕν ∆ µετρε‹ κατ¦ τ¦ς ™ν τù Β µονάδας. µετρε‹ δ κሠ¹ Ε µον¦ς τÕν Β κατ¦ τ¦ς ™ν αÙτù µονάδας· „σάκις ¥ρα ¹ Ε µον¦ς τÕν Β ¢ριθµÕν µετρε‹ καˆ Ð Α τÕν ∆. „σάκις δ ¹ Ε µον¦ς τÕν Β ¢ριθµÕν ™µέτρει καˆ Ð Α τÕν Γ· „σάκις ¥ρα Ð Α ˜κάτερον τîν Γ, ∆ µετρε‹. ‡σος ¥ρα ™στˆν Ð Γ τù ∆· Óπερ œδει δε‹ξαι. †

Let A and B be two numbers. And let A make C (by) multiplying B, and let B make D (by) multiplying A. I say that C is equal to D. For since A has made C (by) multiplying B, B thus measures C according to the units in A [Def. 7.15]. And the unit E also measures the number A according to the units in it. Thus, the unit E measures the number A as many times as B (measures) C. Thus, alternately, the unit E measures the number B as many times as A (measures) C [Prop. 7.15]. Again, since B has made D (by) multiplying A, A thus measures D according to the units in B [Def. 7.15]. And the unit E also measures B according to the units in it. Thus, the unit E measures the number B as many times as A (measures) D. And the unit E was measuring the number B as many times as A (measures) C. Thus, A measures each of C and D an equal number of times. Thus, C is equal to D. (Which is) the very thing it was required to show.

In modern notation, this proposition states that a b = b a, where all symbols denote numbers.

ιζ΄.

Proposition 17†

'Ε¦ν ¢ριθµÕς δύο ¢ριθµοÝς πολλαπλασιάσας ποιÍ τινας, οƒ γενόµενοι ™ξ αÙτîν τÕν αÙτÕν ›ξουσι λόγον το‹ς πολλαπλασιασθε‹σιν.

If a number multiplying two numbers makes some (numbers) then the (numbers) generated from them will have the same ratio as the multiplied (numbers).

Α Β ∆ Ζ

A B D F

Γ Ε

C E

'ΑριθµÕς γ¦ρ Ð Α δύο ¢ριθµοÝς τοÝς Β, Γ πολλαFor let the number A make (the numbers) D and πλασιάσας τοÝς ∆, Ε ποιείτω· λέγω, Óτι ™στˆν æς Ð Β E (by) multiplying the two numbers B and C (respecπρÕς τÕν Γ, οÛτως Ð ∆ πρÕς τÕν Ε. tively). I say that as B is to C, so D (is) to E. 'Επεˆ γ¦ρ Ð Α τÕν Β πολλαπλασιάσας τÕν ∆ For since A has made D (by) multiplying B, B thus πεποίηκεν, Ð Β ¥ρα τÕν ∆ µετρε‹ κατ¦ τ¦ς ™ν τù Α measures D according to the units in A [Def. 7.15]. And

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µονάδας. µετρε‹ δ κሠ¹ Ζ µον¦ς τÕν Α ¢ριθµÕν κατ¦ τ¦ς ™ν αÙτù µονάδας· „σάκις ¥ρα ¹ Ζ µον¦ς τÕν Α ¢ριθµÕν µετρε‹ καˆ Ð Β τÕν ∆. œστιν ¥ρα æς ¹ Ζ µον¦ς πρÕς τÕν Α ¢ριθµόν, οÛτως Ð Β πρÕς τÕν ∆. δι¦ τ¦ αÙτ¦ δ¾ κሠæς ¹ Ζ µον¦ς πρÕς τÕν Α ¢ριθµόν, οÛτως Ð Γ πρÕς τÕν Ε· κሠæς ¥ρα Ð Β πρÕς τÕν ∆, οÛτως Ð Γ πρÕς τÕν Ε. ™ναλλ¦ξ ¥ρα ™στˆν æς Ð Β πρÕς τÕν Γ, οÛτως Ð ∆ πρÕς τÕν Ε· Óπερ œδει δε‹ξαι. †

the unit F also measures the number A according to the units in it. Thus, the unit F measures the number A as many times as B (measures) D. Thus, as the unit F is to the number A, so B (is) to D [Def. 7.20]. And so, for the same (reasons), as the unit F (is) to the number A, so C (is) to E. And thus, as B (is) to D, so C (is) to E. Thus, alternately, as B is to C, so D (is) to E [Prop. 7.13]. (Which is) the very thing it was required to show.

In modern notation, this proposition states that if d = a b and e = a c then d : e :: b : c, where all symbols denote numbers.

ιη΄.

Proposition 18†

'Ε¦ν δύο ¢ριθµοˆ ¢ριθµόν τινα πολλαπλασιάσαντες If two numbers multiplying some number make some ποιîσί τινας, οƒ γενόµενοι ™ξ αÙτîν τÕν αÙτÕν ›ξουσι (other numbers) then the (numbers) generated from λόγον το‹ς πολλαπλασιάσασιν. them will have the same ratio as the multiplying (numbers).

Α Β Γ ∆ Ε

A B C D E

∆ύο γ¦ρ ¢ριθµοˆ οƒ Α, Β ¢ριθµόν τινα τÕν Γ πολλαπλασιάσαντες τοÝς ∆, Ε ποιείτωσαν· λέγω, Óτι ™στˆν æς Ð Α πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Ε. 'Επεˆ γ¦ρ Ð Α τÕν Γ πολλαπλασιάσας τÕν ∆ πεποίηκεν, καˆ Ð Γ ¥ρα τÕν Α πολλαπλασιάσας τÕν ∆ πεποίηκεν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Γ τÕν Β πολλαλασιάσας τÕν Ε πεποίηκεν. ¢ριθµÕς δ¾ Ð Γ δύο ¢ριθµοÝς τοÝς Α, Β πολλαπλασιάσας τοÝς ∆, Ε πεποίηκεν. œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Ε· Óπερ œδει δε‹ξαι. †

For let the two numbers A and B make (the numbers) D and E (respectively, by) multiplying the number C. I say that as A is to B, so D (is) to E. For since A has made D (by) multiplying C, C has thus also made D (by) multiplying A [Prop. 7.16]. So, for the same (reasons), C has also made E (by) multiplying B. So the number C has made the two numbers D and E (by) multiplying A and B (respectively). Thus, as A is to B, so D (is) to E [Prop. 7.17]. (Which is) the very thing it was required to show.

In modern notation, this propositions states that if a c = d and b c = e then a : b :: d : e, where all symbols denote numbers.

ιθ΄.

Proposition 19†

'Ε¦ν τέσσαρες ¢ριθµοˆ ¢νάλογον ðσιν, Ð ™κ πρώτου κሠτετάρτου γενόµενος ¢ριθµÕς ‡σος œσται τù ™κ δευτέρου κሠτρίτου γενοµένJ ¢ριθµù· κሠ™¦ν Ð ™κ πρώτου κሠτετάρτου γενόµενος ¢ριθµÕς ‡σος Ï τù ™κ δευτέρου κሠτρίτου, οƒ τέσσασρες ¢ριθµοˆ ¢νάλογον œσονται. ”Εστωσαν τέσσαρες ¢ριθµοˆ ¢νάλογον οƒ Α, Β, Γ, ∆, æς Ð Α πρÕς τÕν Β, οÛτως Ð Γ πρÕς τÕν ∆, καˆ Ð µν Α τÕν ∆ πολλαπλασιάσας τÕν Ε ποιείτω, Ð δ Β τÕν Γ πολλαπλασιάσας τÕν Ζ ποιείτω· λέγω, Óτι ‡σος ™στˆν Ð Ε τù Ζ.

If four number are proportional then the number created from (multiplying) the first and fourth will be equal to the number created from (multiplying) the second and third. And if the number created from (multiplying) the first and fourth is equal to the (number created) from (multiplying) the second and third then the four numbers will be proportional. Let A, B, C, and D be four proportional numbers, (such that) as A (is) to B, so C (is) to D. And let A make E (by) multiplying D, and let B make F (by) multiplying C. I say that E is equal to F .

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Α

Β

Γ

ELEMENTS BOOK 7

∆

Ε

Ζ

Η

A

`Ο γ¦ρ Α τÕν Γ πολλαπλασιάσας τÕν Η ποιείτω. ™πεˆ οâν Ð Α τÕν Γ πολλαπλασιάσας τÕν Η πεποίηκεν, τÕν δ ∆ πολλαπλασιάσας τÕν Ε πεποίηκεν, ¢ριθµÕς δ¾ Ð Α δύο ¢ριθµοÝς τοÝς Γ, ∆ πολλαπλασιάσας τούς Η, Ε πεποίηκεν. œστιν ¥ρα æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Η πρÕς τÕν Ε. ¢λλ' æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Α πρÕς τÕν Β· κሠæς ¥ρα Ð Α πρÕς τÕν Β, οÛτως Ð Η πρÕς τÕν Ε. πάλιν, ™πεˆ Ð Α τÕν Γ πολλαπλασιάσας τÕν Η πεποίηκεν, ¢λλ¦ µ¾ν καˆ Ð Β τÕν Γ πολλαπλασιάσας τÕν Ζ πεποίηκεν, δύο δ¾ ¢ριθµοˆ οƒ Α, Β ¢ριθµόν τινα τÕν Γ πολλαπλασιάσαντες τοÝς Η, Ζ πεποιήκασιν. œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως Ð Η πρÕς τÕν Ζ. ¢λλ¦ µ¾ν κሠæς Ð Α πρÕς τÕν Β, οÛτως Ð Η πρÕς τÕν Ε· κሠæς ¥ρα Ð Η πρÕς τÕν Ε, οÛτως Ð Η πρÕς τÕν Ζ. Ð Η ¥ρα πρÕς ˜κάτερον τîν Ε, Ζ τÕν αÙτÕν œχει λόγον· ‡σος ¥ρα ™στˆν Ð Ε τù Ζ. ”Εστω δ¾ πάλιν ‡σος Ð Ε τù Ζ· λέγω, Óτι ™στˆν æς Ð Α πρÕς τÕν Β, οÛτως Ð Γ πρÕς τÕν ∆. Τîν γ¦ρ αÙτîν κατασκευασθέντων, ™πεˆ „σος ™στˆν Ð Ε τù Ζ, œστιν ¥ρα æς Ð Η πρÕς τÕν Ε, οÛτως Ð Η πρÕς τÕν Ζ. ¢λλ' æς µν Ð Η πρÕς τÕν Ε, οÛτως Ð Γ πρÕς τÕν ∆, æς δ Ð Η πρÕς τÕν Ζ, οÛτως Ð Α πρÕς τÕν Β. κሠæς ¥ρα Ð Α πρÕς τÕν Β, οÛτως Ð Γ πρÕς τÕν ∆· Óπερ œδει δε‹ξαι. †

B

C

D

E

F

G

For let A make G (by) multiplying C. Therefore, since A has made G (by) multiplying C, and has made E (by) multiplying D, the number A has made G and E by multiplying the two numbers C and D (respectively). Thus, as C is to D, so G (is) to E [Prop. 7.17]. But, as C (is) to D, so A (is) to B. Thus, also, as A (is) to B, so G (is) to E. Again, since A has made G (by) multiplying C, but, in fact, B has also made F (by) multiplying C, the two numbers A and B have made G and F (respectively, by) multiplying some number C. Thus, as A is to B, so G (is) to F [Prop. 7.18]. But, also, as A (is) to B, so G (is) to E. And thus, as G (is) to E, so G (is) to F . Thus, G has the same ratio to each of E and F . Thus, E is equal to F [Prop. 5.9]. So, again, let E be equal to F . I say that as A is to B, so C (is) to D. For, with the same construction, since E is equal to F , thus as G is to E, so G (is) to F [Prop. 5.7]. But, as G (is) to E, so C (is) to D [Prop. 7.17]. And as G (is) to F , so A (is) to B [Prop. 7.18]. And, thus, as A (is) to B, so C (is) to D. (Which is) the very thing it was required to show.

In modern notation, this proposition reads that if a : b :: c : d then a d = b c, and vice versa, where all symbols denote numbers.

κ΄.

Proposition 20

Οƒ ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάσσων τÕν ™λάσσονα. ”Εστωσαν γ¦ρ ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Β οƒ Γ∆, ΕΖ· λέγω, Óτι „σάκις Ð Γ∆ τÕν Α µετρε‹ καˆ Ð ΕΖ τÕν Β.

The least numbers of those (numbers) having the same ratio measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser. For let CD and EF be the least numbers having the same ratio as A and B (respectively). I say that CD measures A the same number of times as EF (measures) B.

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ELEMENTS BOOK 7

Β

Γ Η

A

Ε

B

C

E H

Θ

G

Ζ

F

∆

D

`Ο Γ∆ γ¦ρ τοà Α οÜκ ™στι µέρη. ε„ γ¦ρ δυνατόν, œστω· καˆ Ð ΕΖ ¥ρα τοà Β τ¦ αÙτ¦ µέρη ™στίν, ¤περ Ð Γ∆ τοà Α. Óσα ¥ρα ™στˆν ™ν τù Γ∆ µέρη τοà Α, τοσαàτά ™στι κሠ™ν τù ΕΖ µέρη τοà Β. διVρήσθω Ð µν Γ∆ ε„ς τ¦ τοà Α µέρη τ¦ ΓΗ, Η∆, Ð δ ΕΖ ε„ς τ¦ τοà Β µέρη τ¦ ΕΘ, ΘΖ· œσται δ¾ ‡σον τÕ πλÁθος τîν ΓΗ, Η∆ τù πλήθει τîν ΕΘ, ΘΖ. κሠ™πεˆ ‡σοι ε„σˆν οƒ ΓΗ, Η∆ ¢ριθµοˆ ¢λλήλοις, ε„σˆ δ καˆ οƒ ΕΘ, ΘΖ ¢ριθµοˆ ‡σοι ¢λλήλοις, καί ™στιν ‡σον τÕ πλÁθος τîν ΓΗ, Η∆ τù πλήθει τîν ΕΘ, ΘΖ, œστιν ¥ρα æς Ð ΓΗ πρÕς τÕν ΕΘ, οÛτως Ð Η∆ πρÕς τÕν ΘΖ. œσται ¥ρα κሠæς εŒς τîν ¹γουµένων πρÕς ›να τîν ˜ποµένων, οÛτως ¤παντες οƒ ¹γούµενοι πρÕς ¤παντας τοÝς ˜ποµένους. œστιν ¥ρα æς Ð ΓΗ πρÕς τÕν ΕΘ, οÛτως Ð Γ∆ πρÕς τÕν ΕΖ· οƒ ΓΗ, ΕΘ ¥ρα το‹ς Γ∆, ΕΖ ™ν τù αÙτù λόγJ ε„σˆν ™λάσσονες Ôντες αÙτîν· Óπερ ™στˆν ¢δύνατον· Øπόκεινται γ¦ρ οƒ Γ∆, ΕΖ ™λάχιστοι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς. οÙκ ¥ρα µέρη ™στˆν Ð Γ∆ τοà Α· µέρος ¥ρα. καˆ Ð ΕΖ τοà Β τÕ αÙτÕ µέρος ™στίν, Óπερ Ð Γ∆ τοà Α· „σάκις ¥ρα Ð Γ∆ τÕν Α µετρε‹ καˆ Ð ΕΖ τÕν Β· Óπερ œδει δε‹ξαι.

For CD is not parts of A. For, if possible, let it be (parts of A). Thus, EF is also the same parts of B that CD (is) of A [Def. 7.20, Prop. 7.13]. Thus, as many parts of A as are in CD, so many parts of B are also in EF . Let CD have been divided into the parts of A, CG and GD, and EF into the parts of B, EH and HF . So the multitude of (divisions) CG, GD will be equal to the multitude of (divisions) EH, HF . And since the numbers CG and GD are equal to one another, and the numbers EH and HF are also equal to one another, and the multitude of (divisions) CG, GD is equal to the multitude of (divisions) EH, HF , thus as CG is to EH, so GD (is) to HF . Thus, as one of the leading (numbers is) to one of the following, so will (the sum of) all of the leading (numbers) be to (the sum of) all of the following [Prop. 7.12]. Thus, as CG is to EH, so CD (is) to EF . Thus, CG and EH are in the same ratio as CD and EF , being less than them. The very thing is impossible. For CD and EF were assumed (to be) the least of those (numbers) having the same ratio as them. Thus, CD is not parts of A. Thus, (it is) a part (of A) [Prop. 7.4]. And EF is the same part of B that CD (is) of A [Def. 7.20, Prop 7.13]. Thus, CD measures A the same number of times that EF (measures) B. (Which is) the very thing it was required to show.

κα΄.

Proposition 21

Οƒ πρîτοι πρÕς ¢λλήλους ¢ριθµοˆ ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς. ”Εστωσαν πρîτοι πρÕς ¢λλήλους ¢ριθµοˆ οƒ Α, Β· λέγω, Óτι οƒ Α, Β ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς. Ε„ γ¦ρ µή, œσονταί τινες τîν Α, Β ™λάσσονες ¢ριθµοˆ ™ν τù αÙτù λόγJ Ôντες το‹ς Α, Β. œστωσαν οƒ Γ, ∆.

Numbers prime to one another are the least of those (numbers) having the same ratio as them. Let A and B be numbers prime to one another. I say that A and B are the least of those (numbers) having the same ratio as them. For if not, then there will be some numbers, less than A and B, which are in the same ratio as A and B. Let them be C and D.

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ELEMENTS BOOK 7

Γ

∆

Ε

A

B

C

D

E

'Επεˆ οâν οƒ ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάττων τÕν ™λάττονα, τουτέστιν Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον, „σάκις ¥ρα Ð Γ τÕν Α µετρε‹ καˆ Ð ∆ τÕν Β. Ðσάκις δ¾ Ð Γ τÕν Α µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ε. καˆ Ð ∆ ¥ρα τÕν Β µετρε‹ κατ¦ τ¦ς ™ν τù Ε µονάδας. κሠ™πεˆ Ð Γ τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù Ε µονάδας, καί Ð Ε ¥ρα τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù Γ µονάδας. δι¦ τ¦ αÙτ¦ δ¾ Ð Ε κሠτÕν Β µετρε‹ κατ¦ τ¦ς ™ν τù ∆ µονάδας. Ð Ε ¥ρα τοÝς Α, Β µετρε‹ πρώτους Ôντας πρÕς ¢λλήλους· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα œσονταί τινες τîν Α, Β ™λάσσονες ¢ριθµοˆ ™ν τù αÙτù λόγJ Ôντες το‹ς Α, Β. οƒ Α, Β ¥ρα ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς· Óπερ œδει δε‹ξαι.

Therefore, since the least numbers of those (numbers) having the same ratio measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following—C thus measures A the same number of times that D (measures) B [Prop. 7.20]. So as many times as C measures A, so many units let there be in E. Thus, D also measures B according to the units in E. And since C measures A according to the units in E, E thus also measures A according to the units in C [Prop. 7.16]. So, for the same (reasons), E also measures B according to the units in D [Prop. 7.16]. Thus, E measures A and B, which are prime to one another. The very thing is impossible. Thus, there cannot be any numbers, less than A and B, which are in the same ratio as A and B. Thus, A and B are the least of those (numbers) having the same ratio as them. (Which is) the very thing it was required to show.

κβ΄.

Proposition 22

Οƒ ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς πρîτοι πρÕς ¢λλήλους ε„σίν.

The least numbers of those (numbers) having the same ratio as them are prime to one another.

Α Β Γ ∆ Ε

A B C D E

”Εστωσαν ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς οƒ Α, Β· λέγω, Óτι οƒ Α, Β πρîτοι πρÕς ¢λλήλους ε„σίν. Ε„ γ¦ρ µή ε„σι πρîτοι πρÕς ¢λλήλους, µετρήσει τις αÙτοÝς ¢ριθµός. µετρείτω, κሠœστω Ð Γ. κሠÐσάκις µν Ð Γ τÕν Α µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù ∆,

Let A and B be the least numbers of those (numbers) having the same ratio as them. I say that A and B are prime to one another. For if they are not prime to one another then some number will measure them. Let it (so measure them), and let it be C. And as many times as C measures A, so

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ELEMENTS BOOK 7

Ðσάκις δ Ð Γ τÕν Β µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ε. 'Επεˆ Ð Γ τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù ∆ µονάδας, Ð Γ ¥ρα τÕν ∆ πολλαπλασιάσας τÕν Α πεποίηκεν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Γ τÕν Ε πολλαπλασιάσας τÕν Β πεποίηκεν. ¢ριθµÕς δ¾ Ð Γ δύο ¢ριθµοÝς τοàς ∆, Ε πολλαπλασιάσας τοÝς Α, Β πεποίηκεν· œστιν ¥ρα æς Ð ∆ πρÕς τÕν Ε, οÛτως Ð Α πρÕς τÕν Β· οƒ ∆, Ε ¥ρα το‹ς Α, Β ™ν τù αÙτù λόγJ ε„σˆν ™λάσσονες Ôντες αÙτîν· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τοÝς Α, Β ¢ριθµοÝς ¢ριθµός τις µετρήσει. οƒ Α, Β ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι.

many units let there be in D. And as many times as C measures B, so many units let there be in E. Since C measures A according to the units in D, C has thus made A (by) multiplying D [Def. 7.15]. So, for the same (reasons), C has also made B (by) multiplying E. So the number C has made A and B (by) multiplying the two numbers D and E (respectively). Thus, as D is to E, so A (is) to B [Prop. 7.17]. Thus, D and E are in the same ratio as A and B, being less than them. The very thing is impossible. Thus, some number does not measure the numbers A and B. Thus, A and B are prime to one another. (Which is) the very thing it was required to show.

κγ΄.

Proposition 23

'Ε¦ν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους ðσιν, Ð If two numbers are prime to one another then a numτÕν ›να αÙτîν µετρîν ¢ριθµÕς πρÕς τÕν λοιπÕν πρîτος ber measuring one of them will be prime to the remaining œσται. (one).

Α

Β

Γ

A

∆

B

C

D

.

”Εστωσαν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους οƒ Α, Β, τÕν δ Α µετρείτω τις ¢ριθµÕς Ð Γ· λέγω, Óτι καˆ οƒ Γ, Β πρîτοι πρÕς ¢λλήλους ε„σίν. Ε„ γ¦ρ µή ε„σιν οƒ Γ, Β πρîτοι πρÕς ¢λλήλους, µετρήσει [τις] τοÝς Γ, Β ¢ριθµός. µετείτω, κሠœστω Ð ∆. ™πεˆ Ð ∆ τÕν Γ µετρε‹, Ð δ Γ τÕν Α µετρε‹, καˆ Ð ∆ ¥ρα τÕν Α µετρε‹. µετρε‹ δ κሠτÕν Β· Ð ∆ ¥ρα τοÝς Α, Β µετρε‹ πρώτους Ôντας πρÕς ¢λλήλους· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τοÝς Γ, Β ¢ριθµοÝς ¢ριθµός τις µετρήσει. οƒ Γ, Β ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι.

Let A and B be two numbers (which are) prime to one another, and let some number C measure A. I say that C and B are also prime to one another. For if C and B are not prime to one another then [some] number will measure C and B. Let it (so) measure (them), and let it be D. Since D measures C, and C measures A, D thus also measures A. And (D) also measures B. Thus, D measures A and B, which are prime to one another. The very thing is impossible. Thus, some number does not measure the numbers C and B. Thus, C and B are prime to one another. (Which is) the very thing it was required to show.

κδ΄.

Proposition 24

'Ε¦ν δύο ¢ριθµοˆ πρός τινα ¢ριθµÕν πρîτοι ðσιν, καˆ Ð ™ξ αÙτîν γενόµενος πρÕς τÕν αÙτÕν πρîτος œσται.

If two numbers are prime to some number then the number created from (multiplying) the former (two numbers) will also be prime to the latter (number).

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ELEMENTS BOOK 7

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∆

Ε

Ζ

A

B

C

D

E

F

∆ύο γ¦ρ ¢ριθµοˆ οƒ Α, Β πρός τινα ¢ριθµÕν τÕν Γ πρîτοι œστωσαν, καˆ Ð Α τÕν Β πολλαπλασιάσας τÕν ∆ ποιείτω· λέγω, Óτι οƒ Γ, ∆ πρîτοι πρÕς ¢λλήλους ε„σίν. Ε„ γ¦ρ µή ε„σιν οƒ Γ, ∆ πρîτοι πρÕς ¢λλήλους, µετρήσει [τις] τοÝς Γ, ∆ ¢ριθµός. µετρείτω, κሠœστω Ð Ε. κሠ™πεˆ οƒ Γ, ∆ πρîτοι πρÕς ¢λλήλους ε„σίν, τÕν δ Γ µετρε‹ τις ¢ριθµÕς Ð Ε, οƒ Α, Ε ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. Ðσάκις δ¾ Ð Ε τÕν ∆ µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ζ· καˆ Ð Ζ ¥ρα τÕν ∆ µετρε‹ κατ¦ τ¦ς ™ν τù Ε µονάδας. Ð Ε ¥ρα τÕν Ζ πολλαπλασιάσας τÕν ∆ πεποίηκεν. ¢λλ¦ µ¾ν καˆ Ð Α τÕν Β πολλαπλασιάσας τÕν ∆ πεποίηκεν· ‡σος ¥ρα ™στˆν Ð ˜κ τîν Ε, Ζ τù ™κ τîν Α, Β. ™¦ν δ Ð ØπÕ τîν ¥κρων ‡σος Ï τù ØπÕ τîν µέσων, οƒ τέσσαρες ¢ριθµοˆ ¢νάλογόν ε„σιν· œστιν ¥ρα æς Ð Ε πρÕς τÕν Α, οÛτως Ð Β πρÕς τÕν Ζ. οƒ δ Α, Ε πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάσσων τÕν ™λάσσονα, τουτέστιν Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον· Ð Ε ¥ρα τÕν Β µετρε‹. µετρε‹ δ κሠτÕν Γ· Ð Ε ¥ρα τοÝς Β, Γ µετρε‹ πρώτους Ôντας πρÕς ¢λλήλους· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τοÝς Γ, ∆ ¢ριθµοÝς ¢ριθµός τις µετρήσει. οƒ Γ, ∆ ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι.

For let A and B be two numbers (which are both) prime to some number C. And let A make D (by) multiplying B. I say that C and D are prime to one another. For if C and D are not prime to one another then [some] number will measure C and D. Let it (so) measure them, and let it be E. And since C and A are prime to one another, and some number E measures C, A and E are thus prime to one another [Prop. 7.23]. So as many times as E measures D, so many units let there be in F . Thus, F also measures D according to the units in E [Prop. 7.16]. Thus, E has made D (by) multiplying F [Def. 7.15]. But, in fact, A has also made D (by) multiplying B. Thus, the (number created) from (multiplying) E and F is equal to the (number created) from (multiplying) A and B. And if the (rectangle contained) by the (two) outermost is equal to the (rectangle contained) by the middle (two) then the four numbers are proportional [Prop. 6.15]. Thus, as E is to A, so B (is) to F . And A and E (are) prime (to one another). And (numbers) prime (to one another) are also the least (of those numbers having the same ratio) [Prop. 7.21]. And the least numbers of those (numbers) having the same ratio measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures B. And it also measures C. Thus, E measures B and C, which are prime to one another. The very thing is impossible. Thus, some number cannot measure the numbers C and D. Thus, C and D are prime to one another. (Which is) the very thing it was required to show.

κε΄.

Proposition 25

'Ε¦ν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους ðσιν, Ð If two numbers are prime to one another then the ™κ τοà ˜νÕς αÙτîν γενόµενος πρÕς τÕν λοιπÕν πρîτος number created from (squaring) one of them will be œσται. prime to the remaining number. ”Εστωσαν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους οƒ Α, Let A and B be two numbers (which are) prime to 214

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ELEMENTS BOOK 7

Β, καˆ Ð Α ˜αυτÕν πολλαπλασιάσας τÕν Γ ποιείτω· λέγω, one another. And let A make C (by) multiplying itself. I Óτι οƒ Β, Γ πρîτοι πρÕς ¢λλ¾λους ε„σίν. say that B and C are prime to one another.

Α

Β

Γ

∆

A

B

C

D

Κείσθω γ¦ρ τù Α ‡σος Ð ∆. ™πεˆ οƒ Α, Β πρîτοι πρÕς ¢λλήλους ε„σίν, ‡σος δ Ð Α τù ∆, καί οƒ ∆, Β ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· ˜κάτερος ¥ρα τîν ∆, Α πρÕς τÕν Β πρîτός ™στιν· καˆ Ð ™κ τîν ∆, Α ¥ρα γενόµενος πρÕς τÕν Β πρîτος œσται. Ð δ ™κ τîν ∆, Α γενόµενος ¢ριθµός ™στιν Ð Γ. οƒ Γ, Β ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι.

For let D be made equal to A. Since A and B are prime to one another, and A (is) equal to D, D and B are thus also prime to one another. Thus, D and A are each prime to B. Thus, the (number) created from (multilying) D and A will also be prime to B [Prop. 7.24]. And C is the number created from (multiplying) D and A. Thus, C and B are prime to one another. (Which is) the very thing it was required to show.

κ$΄.

Proposition 26

'Ε¦ν δύο ¢ριθµοˆ πρÕς δύο ¢ριθµοÝς ¢µφότεροι πρÕς ˜κάτερον πρîτοι ðσιν, καˆ οƒ ™ξ αÙτîν γενόµενοι πρîτοι πρÕς ¢λλήλους œσονται.

If two numbers are both prime to each of two numbers then the (numbers) created from (multiplying) them will also be prime to one another.

Α Β Ε Ζ

Γ ∆

A B E F

∆ύο γ¦ρ ¢ριθµοˆ οƒ Α, Β πρÕς δύο ¢ριθµοÝς τοÝς Γ, ∆ ¢µφότεροι πρÕς ˜κάτερον πρîτοι œστωσαν, καˆ Ð µν Α τÕν Β πολλαπλασιάσας τÕν Ε ποιείτω, Ð δ Γ τÕν ∆ πολλαπλασιάσας τÕν Ζ ποιείτω· λέγω, Óτι οƒ Ε, Ζ πρîτοι πρÕς ¢λλ¾λους ε„σίν. 'Επεˆ γ¦ρ ˜κάτερος τîν Α, Β πρÕς τÕν Γ πρîτός ™στιν, καˆ Ð ™κ τîν Α, Β ¥ρα γενόµενος πρÕς τÕν Γ πρîτος œσται. Ð δ ™κ τîν Α, Β γενόµενός ™στιν Ð Ε· οƒ Ε, Γ ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. δι¦ τ¦ αÙτ¦ δ¾ καˆ οƒ Ε, ∆ πρîτοι πρÕς ¢λλήλους ε„σίν. ˜κάτερος ¥ρα τîν Γ, ∆ πρÕς τÕν Ε πρîτός ™στιν. καˆ Ð ™κ τîν Γ, ∆ ¥ρα γενόµενος πρÕς τÕν Ε πρîτος œσται. Ð δ ™κ τîν Γ, ∆ γενόµενός ™στιν Ð Ζ. οƒ Ε, Ζ ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι.

C D

For let two numbers, A and B, both be prime to each of two numbers, C and D. And let A make E (by) multiplying B, and let C make F (by) multiplying D. I say that E and F are prime to one another. For since A and B are each prime to C, the (number) created from (multiplying) A and B will thus also be prime to C [Prop. 7.24]. And E is the (number) created from (multiplying) A and B. Thus, E and C are prime to one another. So, for the same (reasons), E and D are also prime to one another. Thus, C and D are each prime to E. Thus, the (number) created from (multiplying) C and D will also be prime to E [Prop. 7.24]. And F is the (number) created from (multiplying) C and D. Thus, E and F are prime to one another. (Which is) the very thing it was required to show.

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ELEMENTS BOOK 7 κζ΄.

Proposition 27†

'Ε¦ν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους ðσιν, κሠπολλαπλασιάσας ˜κάτερος ˜αυτÕν ποιÍ τινα, οƒ γενόµενοι ™ξ αÙτîν πρîτοι πρÕς ¢λλήλους œσονται, κ¨ν οƒ ™ξ ¢ρχÁς τοÝς γενοµένους πολλαπλασιάσαντες ποιîσί τινας, κ¢κε‹νοι πρîτοι πρÕς ¢λλήλους œσονται [κሠ¢εˆ περˆ τοÝς ¥κρους τοàτο συµβαίνει].

If two numbers are prime to one another and each makes some (number by) multiplying itself then the numbers created from them will be prime to one another, and if the original (numbers) make some (more numbers by) multiplying the created (numbers) then these will also be prime to one another [and this always happens with the extremes].

Α

Β

Γ

∆

Ε

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A

”Εστωσαν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους οƒ Α, Β, καˆ Ð Α ˜αυτÕν µν πολλαπλασιάσας τÕν Γ ποιείτω, τÕν δ Γ πολλαπλασιάσας τÕν ∆ ποιείτω, Ð δ Β ˜αυτÕν µν πολλαπλασιάσας τÕν Ε ποιείτω, τÕν δ Ε πολλαπλασιάσας τÕν Ζ ποιείτω· λέγω, Óτι ο† τε Γ, Ε καˆ οƒ ∆, Ζ πρîτοι πρÕς ¢λλήλους ε„σίν. 'Επεˆ γ¦ρ οƒ Α, Β πρîτοι πρÕς ¢λλήλους ε„σίν, καˆ Ð Α ˜αυτÕν πολλαπλασιάσας τÕν Γ πεποίηκεν, οƒ Γ, Β ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. ™πεˆ οâν οƒ Γ, Β πρîτοι πρÕς ¢λλήλους ε„σίν, καˆ Ð Β ˜αυτÕν πολλαπλασιάσας τÕν Ε πεποίηκεν, οƒ Γ, Ε ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. πάλιν, ™πεˆ οƒ Α, Β πρîτοι πρÕς ¢λλήλους ε„σίν, καˆ Ð Β ˜αυτÕν πολλαπλασιάσας τÕν Ε πεποίηκεν, οƒ Α, Ε ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. ™πεˆ οâν δύο ¢ριθµοˆ οƒ Α, Γ πρÕς δύο ¢ριθµοÝς τοÝς Β, Ε ¢µφότεροι πρÕς ˜κάτερον πρîτοί ε„σιν, καˆ Ð ™κ τîν Α, Γ ¥ρα γενόµενος πρÕς τÕν ™κ τîν Β, Ε πρîτός ™στιν. καί ™στιν Ð µν ™κ τîν Α, Γ Ð ∆, Ð δ ™κ τîν Β, Ε Ð Ζ. οƒ ∆, Ζ ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι. †

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Let A and B be two numbers prime to one another, and let A make C (by) multiplying itself, and let it make D (by) multiplying C. And let B make E (by) multiplying itself, and let it make F by multiplying E. I say that C and E, and D and F , are prime to one another. For since A and B are prime to one another, and A has made C (by) multiplying itself, C and B are thus prime to one another [Prop. 7.25]. Therefore, since C and B are prime to one another, and B has made E (by) multiplying itself, C and E are thus prime to one another [Prop. 7.25]. Again, since A and B are prime to one another, and B has made E (by) multiplying itself, A and E are thus prime to one another [Prop. 7.25]. Therefore, since the two numbers A and C are both prime to each of the two numbers B and E, the (number) created from (multiplying) A and C is thus prime to the (number created) from (multiplying) B and E [Prop. 7.26]. And D is the (number created) from (multiplying) A and C, and F the (number created) from (multiplying) B and E. Thus, D and F are prime to one another. (Which is) the very thing it was required to show.

In modern notation, this proposition states that if a is prime to b, then a2 is also prime to b2 , as well as a3 to b3 , etc., where all symbols denote

numbers.

κη΄.

Proposition 28

'Ε¦ν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους ðσιν, κሠIf two numbers are prime to one another then their συναµφότερος πρÕς ˜κάτερον αÙτîν πρîτος œσται· κሠsum will also be prime to each of them. And if the sum ™¦ν συναµφότερος πρÕς ›να τιν¦ αÙτîν πρîτος Ï, καˆ οƒ (of two numbers) is prime to any one of them then the ™ξ ¢ρχÁς ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους œσονται. original numbers will also be prime to one another. 216

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Συγκείσθωσαν γ¦ρ δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους οƒ ΑΒ, ΒΓ· λέγω, Óτι κሠσυναµφότερος Ð ΑΓ πρÕς ˜κάτερον τîν ΑΒ, ΒΓ πρîτός ™στιν. Ε„ γ¦ρ µή ε„σιν οƒ ΓΑ, ΑΒ πρîτοι πρÕς ¢λλήλους, µετρήσει τις τοÝς ΓΑ, ΑΒ ¢ριθµός. µετρείτω, κሠœστω Ð ∆. ™πεˆ οâν Ð ∆ τοÝς ΓΑ, ΑΒ µετρε‹, κሠλοιπÕν ¥ρα τÕν ΒΓ µετρήσει. µετρε‹ δ κሠτÕν ΒΑ· Ð ∆ ¥ρα τοÝς ΑΒ, ΒΓ µετρε‹ πρώτους Ôντας πρÕς ¢λλήλους· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τοÝς ΓΑ, ΑΒ ¢ριθµοÝς ¢ριθµός τις µετρήσει· οƒ ΓΑ, ΑΒ ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. δι¦ τ¦ αÙτ¦ δ¾ καˆ οƒ ΑΓ, ΓΒ πρîτοι πρÕς ¢λλήλους ε„σίν. Ð ΓΑ ¥ρα πρÕς ˜κάτερον τîν ΑΒ, ΒΓ πρîτός ™στιν. ”Εστωσαν δ¾ πάλιν οƒ ΓΑ, ΑΒ πρîτοι πρÕς ¢λλήλους· λέγω, Óτι καˆ οƒ ΑΒ, ΒΓ πρîτοι πρÕς ¢λλ¾λους ε„σίν. Ε„ γ¦ρ µή ε„σιν οƒ ΑΒ, ΒΓ πρîτοι πρÕς ¢λλήλους, µετρήσει τις τοÝς ΑΒ, ΒΓ ¢ριθµός. µετρείτω, κሠœστω Ð ∆. κሠ™πεˆ Ð ∆ ˜κάτερον τîν ΑΒ, ΒΓ µετρε‹, κሠÓλον ¥ρα τÕν ΓΑ µετρήσει. µετρε‹ δ κሠτÕν ΑΒ· Ð ∆ ¥ρα τοÝς ΓΑ, ΑΒ µετρε‹ πρώτους Ôντας πρÕς ¢λλήλους· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τοÝς ΑΒ, ΒΓ ¢ριθµοÝς ¢ριθµός τις µετρήσει. οƒ ΑΒ, ΒΓ ¥ρα πρîτοι πρÕς ¢λλ¾λους ε„σίν· Óπερ œδει δε‹ξαι.

For let the two numbers, AB and BC, (which are) prime to one another, be laid down together. I say that their sum AC is also prime to each of AB and BC. For if CA and AB are not prime to one another then some number will measure CA and AB. Let it (so) measure (them), and let it be D. Therefore, since D measures CA and AB, it will thus also measure the remainder BC. And it also measures BA. Thus, D measures AB and BC, which are prime to one another. The very thing is impossible. Thus, some number cannot measure (both) the numbers CA and AB. Thus, CA and AB are prime to one another. So, for the same (reasons), AC and CB are also prime to one another. Thus, CA is prime to each of AB and BC. So, again, let CA and AB be prime to one another. I say that AB and BC are also prime to one another. For if AB and BC are not prime to one another then some number will measure AB and BC. Let it (so) measure (them), and let it be D. And since D measures each of AB and BC, it will thus also measure the whole of CA. And it also measures AB. Thus, D measures CA and AB, which are prime to one another. The very thing is impossible. Thus, some number cannot measure (both) the numbers AB and BC. Thus, AB and BC are prime to one another. (Which is) the very thing it was required to show.

κθ΄.

Proposition 29

“Απας πρîτος ¢ριθµÕς πρÕς ¤παντα ¢ριθµόν, Öν µ¾ µετρε‹, πρîτός ™στιν.

Every prime number is prime to every number which it does not measure.

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A B C

”Εστω πρîτος ¢ριθµÕς Ð Α κሠτÕν Β µ¾ µετρείτω· λέγω, Óτι οƒ Β, Α πρîτοι πρÕς ¢λλήλους ε„σίν. Ε„ γ¦ρ µή ε„σιν οƒ Β, Α πρîτοι πρÕς ¢λλήλους, µετρήσει τις αÙτοÝς ¢ριθµός. µετρείτω Ð Γ. ™πεˆ Ð Γ τÕν Β µετρε‹, Ð δ Α τÕν Β οÙ µετρε‹, Ð Γ ¥ρα τù Α οÜκ ™στιν Ð αÙτός. κሠ™πεˆ Ð Γ τοÝς Β, Α µετρε‹, κሠτÕν Α ¥ρα µετρε‹ πρîτον Ôντα µ¾ íν αÙτù Ð αÙτός· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τοÝς Β, Α µετρήσει τις ¢ριθµός. οƒ Α, Β ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι.

Let A be a prime number, and let it not measure B. I say that B and A are prime to one another. For if B and A are not prime to one another then some number will measure them. Let C measure (them). Since C measures B, and A does not measure B, C is thus not the same as A. And since C measures B and A, it thus also measures A, which is prime, (despite) not being the same as it. The very thing is impossible. Thus, some number cannot measure (both) B and A. Thus, A and B are prime to one another. (Which is) the very thing it was required to

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λ΄.

Proposition 30

'Ε¦ν δύο ¢ριθµοˆ πολλαπλασιάσαντες ¢λλήλους If two numbers make some (number by) multiplying ποιîσί τινα, τÕν δ γενόµενον ™ξ αÙτîν µετρÍ τις πρîτος one another, and some prime number measures the num¢ριθµός, κሠ›να τîν ™ξ ¢ρχÁς µετρήσει. ber (so) created from them, then it will also measure one of the original (numbers).

Α Β Γ ∆ Ε

A B C D E

∆ύο γ¦ρ ¢ριθµοˆ οƒ Α, Β πολλαπλασιάσαντες ¢λλήλους τÕν Γ ποιείτωσαν, τÕν δ Γ µετρείτω τις πρîτος ¢ριθµÕς Ð ∆· λέγω, Óτι Ð ∆ ›να τîν Α, Β µετρε‹. ΤÕν γ¦ρ Α µ¾ µετρείτω· καί ™στι πρîτος Ð ∆· οƒ Α, ∆ ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. κሠÐσάκις Ð ∆ τÕν Γ µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ε. ™πεˆ οâν Ð ∆ τÕν Γ µετρε‹ κατ¦ τ¦ς ™ν τù Ε µονάδας, Ð ∆ ¥ρα τÕν Ε πολλαπλασιάσας τÕν Γ πεποίηκεν. ¢λλ¦ µ¾ν καˆ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν· ‡σος ¥ρα ™στˆν Ð ™κ τîν ∆, Ε τù ™κ τîν Α, Β. œστιν ¥ρα æς Ð ∆ πρÕς τÕν Α, οÛτως Ð Β πρÕς τÕν Ε. οƒ δ ∆, Α πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάσσων τÕν ™λάσσονα, τουτέστιν Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ™πόµενος τÕν ˜πόµενον· Ð ∆ ¥ρα τÕν Β µετρε‹. еοίως δ¾ δείξοµεν, Óτι κሠ™¦ν τÕν Β µ¾ µετρÍ, τÕν Α µετρήσει. Ð ∆ ¥ρα ›να τîν Α, Β µετρε‹· Óπερ œδει δε‹ξαι.

For let two numbers A and B make C (by) multiplying one another, and let some prime number D measure C. I say that D measures one of A and B. For let it not measure A. And since D is prime, A and D are thus prime to one another [Prop. 7.29]. And as many times as D measures C, so many units let there be in E. Therefore, since D measures C according to the units E, D has thus made C (by) multiplying E [Def. 7.15]. But, in fact, A has also made C (by) multiplying B. Thus, the (number created) from (multiplying) D and E is equal to the (number created) from (multiplying) A and B. Thus, as D is to A, so B (is) to E [Prop. 7.19]. And A and D (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, D measures B. So, similarly, we can also show that if (D) does not measure B then it will measure A. Thus, D measures one of A and B. (Which is) the very thing it was required to show.

λα΄.

Proposition 31

“Απας σύνθεντος ¢ριθµÕς ØπÕ πρώτου τινÕς ¢ριθµοà µετρε‹ται.

Every composite number is measured by some prime number. Let A be a composite number. I say that A is measured ”Εστω σύνθεντος ¢ριθµÕς Ð Α· λέγω, Óτι Ð Α ØπÕ by some prime number. πρώτου τινÕς ¢ριθµοà µετρε‹ται. For since A is composite, some number will measure 'Επεˆ γ¦ρ σύνθετός ™στιν Ð Α, µετρήσει τις αÙτÕν it. Let it (so) measure (A), and let it be B. And if B

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¢ριθµός. µετρείτω, κሠœστω Ð Β. καˆ ε„ µν πρîτός ™στιν Ð Β, γεγονÕς ¨ν ε‡η τÕ ™πιταχθέν. ε„ δ σύνθετος, µετρήσει τις αÙτÕν ¢ριθµός. µετρείτω, κሠœστω Ð Γ. κሠ™πεˆ Ð Γ τÕν Β µετρε‹, Ð δ Β τÕν Α µετρε‹, καˆ Ð Γ ¥ρα τÕν Α µετρε‹. καˆ ε„ µν πρîτός ™στιν Ð Γ, γεγονÕς ¨ν ε‡η τÕ ™πιταχθέν. ε„ δ σύνθετος, µετρήσει τις αÙτÕν ¢ριθµός. τοιαύτης δ¾ γινοµένης ™πισκέψεως ληφθήσεταί τις πρîτος ¢ριθµός, Öς µετρήσει. ε„ γ¦ρ οÙ ληφθήσεται, µετρήσουσι τÕν Α ¢ριθµÕν ¥πειροι ¢ριθµοί, ïν ›τερος ˜τέρου ™λάσσων ™στίν· Óπερ ™στˆν ¢δύνατον ™ν ¢ριθµο‹ς. ληφθήσεταί τις ¥ρα πρîτος ¢ριθµός, Öς µετρήσει τÕν πρÕ ˜αυτοà, Öς κሠτÕν Α µετρήσει.

Α Β Γ

is prime then that which was prescribed has happened. And if (B is) composite then some number will measure it. Let it (so) measure (B), and let it be C. And since C measures B, and B measures A, C thus also measures A. And if C is prime then that which was prescribed has happened. And if (C is) composite then some number will measure it. So, in this manner of continued investigation, some prime number will be found which will measure (the number preceding it, which will also measure A). And if (such a number) cannot be found then the number A will be measured by an infinite (series of) numbers, each of which is less than the preceding. The very thing is impossible for numbers. Thus, some prime number will (eventually) be found which will measure the (number) preceding it, which will also measure A.

A B C

“Απας ¥ρα σύνθεντος ¢ριθµÕς ØπÕ πρώτου τινÕς ¢ριθµοà µετρε‹ται· Óπερ œδει δε‹ξαι.

Thus, every composite number is measured by some prime number. (Which is) the very thing it was required to show.

λβ΄.

Proposition 32

“Απας ¢ριθµÕς ½τοι πρîτός ™στιν À ØπÕ πρώτου Every number is either prime or is measured by some τινÕς ¢ριθµοà µετρε‹ται. prime number.

Α

A

”Εστω ¢ριθµÕς Ð Α· λέγω, Óτι Ð Α ½τοι πρîτός ™στιν À ØπÕ πρώτου τινÕς ¢ριθµοà µετρε‹ται. Ε„ µν οâν πρîτός ™στιν Ð Α, γεγονÕς ¨ν ε‡η τό ™πιταχθέν. ε„ δ σύνθεντος, µετρήσει τις αÙτÕν πρîτος ¢ριθµός. “Απας ¥ρα ¢ριθµÕς ½τοι πρîτός ™στιν À ØπÕ πρώτου τινÕς ¢ριθµοà µετρε‹ται· Óπερ œδει δε‹ξαι.

Let A be a number. I say that A is either prime or is measured by some prime number. In fact, if A is prime then that which was prescribed has happened. And if (it is) composite then some prime number will measure it [Prop. 7.31]. Thus, every number is either prime or is measured by some prime number. (Which is) the very thing it was required to show.

λγ΄.

Proposition 33

'Αριθµîν δοθέντων Ðποσωνοàν εØρε‹ν τοÝς ™λαχίστους τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς. ”Εστωσαν οƒ δοθέντες Ðποσοιοàν ¢ριθµοˆ οƒ Α, Β, Γ· δε‹ δ¾ εØρε‹ν τοÝς ™λαχίστους τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Β, Γ. Οƒ Α, Β, Γ γ¦ρ ½τοι πρîτοι πρÕς ¢λλήλους ε„σˆν À οÜ. ε„ µν οâν οƒ Α, Β, Γ πρîτοι πρÕς ¢λλήλους ε„σίν, ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς.

To find the least of those (numbers) having the same ratio as any given multitude of numbers. Let A, B, and C be any given multitude of numbers. So it is required to find the least of those (numbers) having the same ratio as A, B, and C. For A, B, and C are either prime to one another, or not. In fact, if A, B, and C are prime to one another then they are the least of those (numbers) having the same ratio as them [Prop. 7.22].

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M

Ε„ δ οÜ, ε„λήφθω τîν Α, Β, Γ τÕ µέγιστον κοινÕν µέτρον Ð ∆, κሠÐσάκις Ð ∆ ›καστον τîν Α, Β, Γ µετρε‹, τοσαàται µονάδες œστωσαν ™ν ˜κάστJ τîν Ε, Ζ, Η. κሠ›καστος ¥ρα τîν Ε, Ζ, Η ›καστον τîν Α, Β, Γ µετρε‹ κατ¦ τ¦ς ™ν τù ∆ µονάδας. οƒ Ε, Ζ, Η ¥ρα τοÝς Α, Β, Γ „σάκις µετροàσιν· οƒ Ε, Ζ, Η ¥ρα το‹ς Α, Β, Γ ™ν τù αÙτù λόγJ ε„σίν. λέγω δή, Óτι κሠ™λάχιστοι. ε„ γ¦ρ µή ε„σιν οƒ Ε, Ζ, Η ™λάχιστοι τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Β, Γ, œσονται [τινες] τîν Ε, Ζ, Η ™λάσσονες ¢ριθµοˆ ™ν τù αÙτù λόγJ Ôντες το‹ς Α, Β, Γ. œστωσαν οƒ Θ, Κ, Λ· „σάκις ¥ρα Ð Θ τÕν Α µετρε‹ κሠ˜κάτερος τîν Κ, Λ ˜κάτερον τîν Β, Γ. Ðσάκις δ Ð Θ τÕν Α µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Μ· κሠ˜κάτερος ¥ρα τîν Κ, Λ ˜κάτερον τîν Β, Γ µετρε‹ κατ¦ τ¦ς ™ν τù Μ µονάδας. κሠ™πεˆ Ð Θ τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù Μ µονάδας, καˆ Ð Μ ¥ρα τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù Θ µονάδας. δι¦ τ¦ αÙτ¦ δ¾ Ð Μ κሠ˜κάτερον τîν Β, Γ µετρε‹ κατ¦ τ¦ς ™ν ˜κατέρJ τîν Κ, Λ µονάδας· Ð Μ ¥ρα τοÝς Α, Β, Γ µετρε‹. κሠ™πεˆ Ð Θ τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù Μ µονάδας, Ð Θ ¥ρα τÕν Μ πολλαπλασιάσας τÕν Α πεποίηκεν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Ε τÕν ∆ πολλαπλασιάσας τÕν Α πεποίηκεν. ‡σος ¥ρα ™στˆν Ð ™κ τîν Ε, ∆ τù ™κ τîν Θ, Μ. œστιν ¥ρα æς Ð Ε πρÕς τÕν Θ, οÛτως Ð Μ πρÕς τÕν ∆. µε‹ζων δ Ð Ε τοà Θ· µείζων ¥ρα καˆ Ð Μ τοà ∆. κሠµετρε‹ τοÝς Α, Β, Γ· Óπερ ™στˆν ¢δύνατον· Øπόκειται γ¦ρ Ð ∆ τîν Α, Β, Γ τÕ µέγιστον κοινÕν µέτρον. οÙκ ¥ρα œσονταί τινες τîν Ε, Ζ, Η ™λάσσονες ¢ριθµοˆ ™ν τù αÙτù λόγJ Ôντες το‹ς Α, Β, Γ. οƒ Ε, Ζ, Η ¥ρα ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Β, Γ· Óπερ œδει δε‹ξαι.

And if not, let the greatest common measure, D, of A, B, and C have be taken [Prop. 7.3]. And as many times as D measures A, B, C, so many units let there be in E, F , G, respectively. And thus E, F , G measure A, B, C, respectively, according to the units in D [Prop. 7.15]. Thus, E, F , G measure A, B, C (respectively) an equal number of times. Thus, E, F , G are in the same ratio as A, B, C (respectively) [Def. 7.20]. So I say that (they are) also the least (of those numbers having the same ratio as A, B, C). For if E, F , G are not the least of those (numbers) having the same ratio as A, B, C (respectively), then there will be [some] numbers less than E, F , G which are in the same ratio as A, B, C (respectively). Let them be H, K, L. Thus, H measures A the same number of times that K, L also measure B, C, respectively. And as many times as H measures A, so many units let there be in M . Thus, K, L measure B, C, respectively, according to the units in M . And since H measures A according to the units in M , M thus also measures A according to the units in H [Prop. 7.15]. So, for the same (reasons), M also measures B, C according to the units in K, L, respectively. Thus, M measures A, B, and C. And since H measures A according to the units in M , H has thus made A (by) multiplying M . So, for the same (reasons), E has also made A (by) multiplying D. Thus, the (number created) from (multiplying) E and D is equal to the (number created) from (multiplying) H and M . Thus, as E (is) to H, so M (is) to D [Prop. 7.19]. And E (is) greater than H. Thus, M (is) also greater than D [Prop. 5.13]. And (M ) measures A, B, and C. The very thing is impossible. For D was assumed (to be) the greatest common measure of A, B, and C. Thus, there cannot be any numbers less than E, F , G which are in the same ratio as A, B, C (respectively). Thus, E, F , G are the least of (those numbers) having the same ratio as A, B, C (respectively). (Which is) the very thing it was required to show.

λδ΄.

Proposition 34

∆ύο ¢ριθµîν δοθέντων εØρε‹ν, Öν ™λάχιστον µετροàσιν ¢ριθµόν. ”Εστωσαν οƒ δοθέντες δύο ¢ριθµοˆ οƒ Α, Β· δε‹ δ¾

To find the least number which two given numbers (both) measure. Let A and B be the two given numbers. So it is re-

220

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ELEMENTS BOOK 7

εØρε‹ν, Öν ™λάχιστον µετροàσιν ¢ριθµόν.

Α Γ ∆ Ε

quired to find the least number which they (both) measure.

Β

A C D E

Ζ

Οƒ Α, Β γ¦ρ ½τοι πρîτοι πρÕς ¢λλήλους ε„σˆν À οÜ. œστωσαν πρότερον οƒ Α, Β πρîτοι πρÕς ¢λλήλους, καˆ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ ποιείτω· καˆ Ð Β ¥ρα τÕν Α πολλαπλασιάσας τÕν Γ πεποίηκεν. οƒ Α, Β ¥ρα τÕν Γ µετροàσιν. λέγω δή, Óτι κሠ™λάχιστον. ε„ γ¦ρ µή, µετρήσουσί τινα ¢ριθµÕν οƒ Α, Β ™λάσσονα Ôντα τοà Γ. µετρείτωσαν τÕν ∆. κሠÐσάκις Ð Α τÕν ∆ µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ε, Ðσάκις δ Ð Β τÕν ∆ µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ζ. Ð µν Α ¥ρα τÕν Ε πολλαπλασιάσας τÕν ∆ πεποίηκεν, Ð δ Β τÕν Ζ πολλαπλασιάσας τÕν ∆ πεποίηκεν· ‡σος ¥ρα ™στˆν Ð ™κ τîν Α, Ε τù ™κ τîν Β, Ζ. œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως Ð Ζ πρÕς τÕν Ε. οƒ δ Α, Β πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάσσων τÕν ™λάσσονα· Ð Β ¥ρα τÕν Ε µετρε‹, æς ˜πόµενος ˜πόµενον. κሠ™πεˆ Ð Α τοÝς Β, Ε πολλαπλασιάσας τοÝς Γ, ∆ πεποίηκεν, œστιν ¥ρα æς Ð Β πρÕς τÕν Ε, οÛτως Ð Γ πρÕς τÕν ∆. µετρε‹ δ Ð Β τÕν Ε· µετρε‹ ¥ρα καˆ Ð Γ τÕν ∆ Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα οƒ Α, Β µετροàσί τινα ¢ριθµÕν ™λάσσονα Ôντα τοà Γ. Ð Γ ¥ρα ™λάχιστος íν ØπÕ τîν Α, Β µετρε‹ται.

Α Ζ Γ ∆ Η

B

F

For A and B are either prime to one another, or not. Let them, first of all, be prime to one another. And let A make C (by) multiplying B. Thus, B has also made C (by) multiplying A [Prop. 7.16]. Thus, A and B (both) measure C. So I say that (C) is also the least (number which they both measure). For if not, A and B will (both) measure some (other) number which is less than C. Let them (both) measure D (which is less than C). And as many times as A measures D, so many units let there be in E. And as many times as B measures D, so many units let there be in F . Thus, A has made D (by) multiplying E, and B has made D (by) multiplying F . Thus, the (number created) from (multiplying) A and E is equal to the (number created) from (multiplying) B and F . Thus, as A (is) to B, so F (is) to E [Prop. 7.19]. And A and B are prime (to one another), and prime (numbers) are the least (of those numbers having the same ratio) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser [Prop. 7.20]. Thus, B measures E, as the following (number measuring) the following. And since A has made C and D (by) multiplying B and E (respectively), thus as B is to E, so C (is) to D [Prop. 7.17]. And B measures E. Thus, C also measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, A and B do not (both) measure some number which is less than C. Thus, C is the least (number) which is measured by (both) A and B.

Β Ε

A F C D G

Θ

B E

H

Μ¾ œστωσαν δ¾ οƒ Α, Β πρîτοι πρÕς ¢λλήλους, So let A and B be not prime to one another. And κሠε„λήφθωσαν ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον let the least numbers, F and E, have been taken having ™χόντων το‹ς Α, Β οƒ Ζ, Ε· ‡σος ¥ρα ™στˆν Ð ™κ τîν the same ratio as A and B (respectively) [Prop. 7.33]. 221

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Α, Ε τù ™κ τîν Β, Ζ. καˆ Ð Α τÕν Ε πολλαπλασιάσας τÕν Γ ποιείτω· καˆ Ð Β ¥ρα τÕν Ζ πολλαπλασιάσας τÕν Γ πεποίηκεν· οƒ Α, Β ¥ρα τÕν Γ µετροàσιν. λέγω δή, Óτι κሠ™λάχιστον. ε„ γ¦ρ µή, µετρήσουσί τινα ¢ριθµÕν οƒ Α, Β ™λάσσονα Ôντα τοà Γ. µετρείτωσαν τÕν ∆. κሠÐσάκις µν Ð Α τÕν ∆ µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Η, Ðσάκις δ Ð Β τÕν ∆ µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Θ. Ð µν Α ¥ρα τÕν Η πολλαπλασιάσας τÕν ∆ πεποίηκεν, Ð δ Β τÕν Θ πολλαπλασιάσας τÕν ∆ πεποίηκεν. ‡σος ¥ρα ™στˆν Ð ™κ τîν Α, Η τù ™κ τîν Β, Θ· œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως Ð Θ πρÕς τÕν Η. æς δ Ð Α πρÕς τÕν Β, οÛτως Ð Ζ πρÕς τÕν Ε· κሠæς ¥ρα Ð Ζ πρÕς τÕν Ε, οÛτως Ð Θ πρÕς τÕν Η. οƒ δ Ζ, Ε ™λάχιστοι, οƒ δ ™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάσσων τÕν ™λάσσονα· Ð Ε ¥ρα τÕν Η µετρε‹. κሠ™πεˆ Ð Α τοÝς Ε, Η πολλαπλασιάσας τοÝς Γ, ∆ πεποίηκεν, œστιν ¥ρα æς Ð Ε πρÕς τÕν Η, οÛτως Ð Γ πρÕς τÕν ∆. Ð δ Ε τÕν Η µετρε‹· καˆ Ð Γ ¥ρα τÕν ∆ µετρε‹ Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα οƒ Α, Β µετρήσουσί τινα ¢ριθµÕν ™λάσσονα Ôντα τοà Γ. Ð Γ ¥ρα ™λάχιστος íν ØπÕ τîν Α, Β µετρε‹ται· Óπερ œπει δε‹ξαι.

Thus, the (number created) from (multiplying) A and E is equal to the (number created) from (multiplying) B and F [Prop. 7.19]. And let A make C (by) multiplying E. Thus, B has also made C (by) multiplying F . Thus, A and B (both) measure C. So I say that (C) is also the least (number which they both measure). For if not, A and B will (both) measure some number which is less than C. Let them (both) measure D (which is less than C). And as many times as A measures D, so many units let there be in G. And as many times as B measures D, so many units let there be in H. Thus, A has made D (by) multiplying G, and B has made D (by) multiplying H. Thus, the (number created) from (multiplying) A and G is equal to the (number created) from (multiplying) B and H. Thus, as A is to B, so H (is) to G [Prop. 7.19]. And as A (is) to B, so F (is) to E. Thus, also, as F (is) to E, so H (is) to G. And F and E are the least (numbers having the same ratio as A and B), and the least (numbers) measure those (numbers) having the same ratio an equal number of times, the greater (measuring) the greater, and the lesser the lesser [Prop. 7.20]. Thus, E measures G. And since A has made C and D (by) multiplying E and G (respectively), thus as E is to G, so C (is) to D [Prop. 7.17]. And E measures G. Thus, C also measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, A and B do not (both) measure some (number) which is less than C. Thus, C (is) the least (number) which is measured by (both) A and B. (Which is) the very thing it was required to show.

λε΄.

Proposition 35

'Ε¦ν δύο ¢ριθµοˆ ¢ριθµόν τινα µετρîσιν, καˆ Ð If two numbers (both) measure some number then the ™λάχιστος Øπ' αÙτîν µετρούµενος τÕν αÙτÕν µετρήσει. least (number) measured by them will also measure the same (number).

Α Γ

Β Ζ

A C

∆

Ε

B F

D

E

∆ύο γ¦ρ ¢ριθµοˆ οƒ Α, Β ¢ριθµόν τινα τÕν Γ∆ µετρείτωσαν, ™λάχιστον δ τÕν Ε· λέγω, Óτι καˆ Ð Ε τÕν Γ∆ µετρε‹. Ε„ γ¦ρ οÙ µετρε‹ Ð Ε τÕν Γ∆, Ð Ε τÕν ∆Ζ µετρîν λειπέτω ˜αυτοà ™λάσσονα τÕν ΓΖ. κሠ™πεˆ οƒ Α, Β τÕν Ε µετροàσιν, Ð δ Ε τÕν ∆Ζ µετρε‹, καˆ οƒ Α, Β ¥ρα τÕν ∆Ζ µετρήσουσιν. µετροàσι δ κሠÓλον τÕν Γ∆· κሠλοιπÕν ¥ρα τÕν ΓΖ µετρήσουσιν ™λάσσονα Ôντα τοà Ε· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα οÙ µετρε‹ Ð Ε τÕν Γ∆· µετρε‹ ¥ρα· Óπερ œδει δε‹ξαι.

For let two numbers, A and B, (both) measure some number CD, and (let) E (be the) least (number measured by both A and B). I say that E also measures CD. For if E does not measure CD then let E leave CF less than itself (in) measuring CD. And since A and B (both) measure E, and E measures DF , A and B will thus also measure DF . And (A and B) also measure the whole of CD. Thus, they will also measure the remainder CF , which is less than E. The very thing is impossible. Thus, E cannot not measure CD. Thus, (E) measures

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ELEMENTS BOOK 7 (CD). (Which is) the very thing it was required to show.

λ$΄.

Proposition 36

Τριîν ¢ριθµîν δοθέντων εØρε‹ν, Öν ™λάχιστον µεTo find the least number which three given numbers τροàσιν ¢ριθµόν. (all) measure. ”Εστωσαν οƒ δοθέντες τρε‹ς ¢ριθµοˆ οƒ Α, Β, Γ· δε‹ Let A, B, and C be the three given numbers. So it is δ¾ εØρε‹ν, Öν ™λάχιστον µετροàσιν ¢ριθµόν. required to find the least number which they (all) measure.

Α Β Γ ∆ Ε Ζ

A B C D E F

Ε„λήφθω γ¦ρ ØπÕ δύο τîν Α, Β ™λάχιστος µετρούµενος Ð ∆. Ð δ¾ Γ τÕν ∆ ½τοι µετρε‹ À οÙ µετρε‹. µετρείτω πρότερον. µετροàσι δ καˆ οƒ Α, Β τÕν ∆. οƒ Α, Β, Γ ¥ρα τÕν ∆ µετροàσιν. λέγω δή, Óτι κሠ™λάχιστον. ε„ γ¦ρ µή, µετρήσουσιν [τινα] ¢ριθµÕν οƒ Α, Β, Γ ™λάσσονα Ôντα τοà ∆. µετρείτωσαν τÕν Ε. ™πεˆ οƒ Α, Β, Γ τÕν Ε µετροàσιν, καˆ οƒ Α, Β ¥ρα τÕν Ε µετροàσιν. καˆ Ð ™λάχιστος ¥ρα ØπÕ τîν Α, Β µετρούµενος [τÕν Ε] µετρήσει. ™λάχιστος δ ØπÕ τîν Α, Β µετρούµενός ™στιν Ð ∆· Ð ∆ ¥ρα τÕν Ε µετρήσει Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα οƒ Α, Β, Γ µετρήσουσί τινα ¢ριθµÕν ™λάσσονα Ôντα τοà ∆· οƒ Α, Β, Γ ¥ρα ™λάχιστον τÕν ∆ µετροàσιν. Μ¾ µετρείτω δ¾ πάλιν Ð Γ τÕν ∆, κሠε„λήφθω ØπÕ τîν Γ, ∆ ™λάχιστος µετρούµενος ¢ριθµÕς Ð Ε. ™πεˆ οƒ Α, Β τÕν ∆ µετροàσιν, Ð δ ∆ τÕν Ε µετρε‹, καˆ οƒ Α, Β ¥ρα τÕν Ε µετροàσιν. µετρε‹ δ καˆ Ð Γ [τÕν Ε· καˆ] οƒ Α, Β, Γ ¥ρα τÕν Ε µετροàσιν. λέγω δή, Óτι κሠ™λάχιστον. ε„ γ¦ρ µή, µετρήσουσί τινα οƒ Α, Β, Γ ™λάσσονα Ôντα τοà Ε. µετρείτωσαν τÕν Ζ. ™πεˆ οƒ Α, Β, Γ τÕν Ζ µετροàσιν, καˆ οƒ Α, Β ¥ρα τÕν Ζ µετροàσιν· καˆ Ð ™λάχιστος ¥ρα ØπÕ τîν Α, Β µετρούµενος τÕν Ζ µετρήσει. ™λάχιστος δ ØπÕ τîν Α, Β µετρούµενός ™στιν Ð ∆· Ð ∆ ¥ρα τÕν Ζ µετρε‹. µετρε‹ δ καˆ Ð Γ τÕν Ζ· οƒ ∆, Γ ¥ρα τÕν Ζ µετροàσιν· éστε καˆ Ð ™λάχιστος ØπÕ τîν ∆, Γ µετρούµενος τÕν Ζ µετρήσει. Ð δ ™λάχιστος ØπÕ τîν Γ, ∆ µετρούµενός ™στιν Ð Ε· Ð Ε ¥ρα τÕν Ζ µετρε‹ Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα οƒ Α, Β, Γ µετρήσουσί τινα ¢ριθµÕν ™λάσσονα Ôντα τοà Ε. Ð Ε ¥ρα ™λάχιστος íν ØπÕ τîν Α, Β, Γ µετρε‹ται· Óπερ œδει δε‹ξαι.

For let the least (number), D, measured by the two (numbers) A and B have been taken [Prop. 7.34]. So C either measures, or does not measure, D. Let it, first of all, measure (D). And A and B also measure D. Thus, A, B, and C (all) measure D. So I say that (D is) also the least (number measured by A, B, and C). For if not, A, B, and C will (all) measure [some] number which is less than D. Let them measure E (which is less than D). Since A, B, and C (all) measure E then A and B thus also measure E. Thus, the least (number) measured by A and B will also measure [E] [Prop. 7.35]. And D is the least (number) measured by A and B. Thus, D will measure E, the greater (measuring) the lesser. The very thing is impossible. Thus, A, B, and C cannot (all) measure some number which is less than D. Thus, A, B, and C (all) measure the least (number) D. So, again, let C not measure D. And let the least number, E, measured by C and D have been taken [Prop. 7.34]. Since A and B measure D, and D measures E, A and B thus also measure E. And C also measures [E]. Thus, A, B, and C [also] measure E. So I say that (E is) also the least (number measured by A, B, and C). For if not, A, B, and C will (all) measure some (number) which is less than E. Let them measure F (which is less than E). Since A, B, and C (all) measure F , A and B thus also measure F . Thus, the least (number) measured by A and B will also measure F [Prop. 7.35]. And D is the least (number) measured by A and B. Thus, D measures F . And C also measures F . Thus, D and C (both) measure F . Hence, the least (number) measured by D and C will also measure F [Prop. 7.35]. And E

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ELEMENTS BOOK 7 is the least (number) measured by C and D. Thus, E measures F , the greater (measuring) the lesser. The very thing is impossible. Thus, A, B, and C cannot measure some number which is less than E. Thus, E (is) the least (number) which is measured by A, B, and C. (Which is) the very thing it was required to show.

λζ΄.

Proposition 37

'Ε¦ν ¢ριθµÕς Øπό τινος ¢ριθµοà µετρÁται, Ð µετρούµενος еώνυµον µέρος ›ξει τù µετροàντι.

If a number is measured by some number then the (number) measured will have a part called the same as the measuring (number).

Α Β Γ ∆

A B C D

'ΑριθµÕς γάρ Ð Α Øπό τινος ¢ριθµοà τοà Β µετρείσθω· λέγω, Óτι Ð Α еώνυµον µέρος œχει τù Β. `Οσάκις γ¦ρ Ð Β τÕν Α µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Γ. ™πεˆ Ð Β τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù Γ µονάδας, µετρε‹ δ κሠ¹ ∆ µον¦ς τÕν Γ ¢ριθµÕν κατ¦ τ¦ς ™ν αÙτù µονάδας, „σάκις ¥ρα ¹ ∆ µον¦ς τÕν Γ ¢ριθµÕν µετρε‹ καˆ Ð Β τÕν Α. ™ναλλ¦ξ ¥ρα „σάκις ¹ ∆ µον¦ς τÕν Β ¢ριθµÕν µετρε‹ καˆ Ð Γ τÕν Α· Ö ¥ρα µέρος ™στˆν ¹ ∆ µον¦ς τοà Β ¢ριθµοà, τÕ αÙτÕ µέρος ™στˆ καˆ Ð Γ τοà Α. ¹ δ ∆ µον¦ς τοà Β ¢ριθµοà µέρος ™στˆν еώνυµον αÙτù· καˆ Ð Γ ¥ρα τοà Α µέρος ™στˆν еώνυµον τù Β. éστε Ð Α µέρος œχει τÕν Γ Ðµώνυµον Ôντα τù Β· Óπερ œδει δε‹ξαι.

For let the number A be measured by some number B. I say that A has a part called the same as B. For as many times as B measures A, so many units let there be in C. Since B measures A according to the units in C, and the unit D also measures C according to the units in it, thus the unit D measures the number C as many times as B (measures) A. Thus, alternately, the unit D measures the number B as many times as C (measures) A [Prop. 7.15]. Thus, which(ever) part the unit D is of the number B, C is also the same part of A. And the unit D is a part of the number B called the same as it (i.e., a Bth part). Thus, C is also a part of A called the same as B (i.e., C is the Bth part of A). Hence, A has a part C which is called the same as B (i.e., A has a Bth part). (Which is) the very thing it was required to show.

λη΄.

Proposition 38

'Ε¦ν ¢ριθµος µέρος œχV Ðτιοàν, ØπÕ Ðµωνύµου ¢ριθµοà µετρηθήσεται τù µέρει.

If a number has any part whatever then it will be measured by a number called the same as the part.

Α Β Γ ∆

A B C D

'ΑριθµÕς γ¦ρ Ð Α µέρος ™χέτω Ðτιοàν τÕν Β, κሠτù For let the number A have any part whatever, B. And Β µέρει еώνυµος œστω [¢ριθµÕς] Ð Γ· λέγω, Óτι Ð Γ let the [number] C be called the same as the part B (i.e., 224

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ELEMENTS BOOK 7

τÕν Α µετρε‹. 'Επεˆ γ¦ρ Ð Β τοà Α µέρος ™στˆν еώνυµον τù Γ, œστι δ κሠ¹ ∆ µον¦ς τοà Γ µέρος еώνυµον αÙτù, Ö ¥ρα µέρος ™στˆν ¹ ∆ µον¦ς τοà Γ ¢ριθµοà, τÕ αÙτÕ µέρος ™στˆ καˆ Ð Β τοà Α· „σάκις ¥ρα ¹ ∆ µον¦ς τÕν Γ ¢ριθµÕν µετρε‹ καˆ Ð Β τÕν Α. ™ναλλ¦ξ ¥ρα „σάκις ¹ ∆ µον¦ς τÕν Β ¢ριθµÕν µετρε‹ καˆ Ð Γ τÕν Α. Ð Γ ¥ρα τÕν Α µετρε‹· Óπερ œδει δεˆξαι.

B is the Cth part of A). I say that C measures A. For since B is a part of A called the same as C, and the unit D is also a part of C called the same as it (i.e., D is the Cth part of C), thus which(ever) part the unit D is of the number C, B is also the same part of A. Thus, the unit D measures the number C as many times as B (measures) A. Thus, alternately, the unit D measures the number B as many times as C (measures) A [Prop. 7.15]. Thus, C measures A. (Which is) the very thing it was required to show.

λθ΄.

Proposition 39

'ΑριθµÕν εÙρε‹ν, Öς ™λάχιστος íν ›ξει τ¦ δοθέντα µέρη.

To find the least number that will have given parts.

Α

Β ∆

Γ

A

Ε

B D

Ζ

C E

F Η

G

Θ

H

”Εστω τ¦ δοθέντα µέρη τ¦ Α, Β, Γ· δε‹ δ¾ ¢ριθµÕν εØρε‹ν, Öς ™λάχιστος íν ›ξει τ¦ Α, Β, Γ µέρη. ”Εστωσαν γ¦ρ το‹ς Α, Β, Γ µέρεσιν еώνυµοι ¢ριθµοˆ οƒ ∆, Ε, Ζ, κሠε„λήφθω ØπÕ τîν ∆, Ε, Ζ ™λάχιστος µετρούµενος ¢ριθµÕς Ð Η. `Ο Η ¥ρα еώνυµα µέρη œχει το‹ς ∆, Ε, Ζ. το‹ς δ ∆, Ε, Ζ Ðµώνυµα µέρη ™στˆ τ¦ Α, Β, Γ· Ð Η ¥ρα œχει τ¦ Α, Β, Γ µέρη. λέγω δή, Óτι κሠ™λάχιστος êν, ε„ γ¦ρ µή, œσται τις τοà Η ™λάσσων ¢ριθµός, Öς ›ξει τ¦ Α, Β, Γ µέρη. œστω Ð Θ. ™πεˆ Ð Θ œχει τ¦ Α, Β, Γ µέρη, Ð Θ ¥ρα ØπÕ Ðµωνύµων ¢ριθµîν µετρηθήσεται το‹ς Α, Β, Γ µέρεσιν. το‹ς δ Α, Β, Γ µέρεσιν еώνυµοι ¢ριθµοί ε„σιν οƒ ∆, Ε, Ζ· Ð Θ ¥ρα ØπÕ τîν ∆, Ε, Ζ µετρε‹ται. καί ™στιν ™λάσσων τοà Η· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα œσται τις τοà Η ™λάσσων ¢ριθµός, Öς ›ξει τ¦ Α, Β, Γ µέρη· Óπερ œδει δε‹ξαι.

Let A, B, and C be the given parts. So it is required to find the least number which will have the parts A, B, and C (i.e., an Ath part, a Bth part, and a Cth part). For let D, E, and F be numbers having the same names as the parts A, B, and C (respectively). And let the least number, G, measured by D, E, and F , have been taken [Prop. 7.36]. Thus, G has parts called the same as D, E, and F [Prop. 7.37]. And A, B, and C are parts called the same as D, E, and F (respectively). Thus, G has the parts A, B, and C. So I say that (G) is also the least (number having the parts A, B, and C). For if not, there will be some number less than G which will have the parts A, B, and C. Let it be H. Since H has the parts A, B, and C, H will thus be measured by numbers called the same as the parts A, B, and C [Prop. 7.38]. And D, E, and F are numbers called the same as the parts A, B, and C (respectively). Thus, H is measured by D, E, and F . And (H) is less than G. The very thing is impossible. Thus, there cannot be some number less than G which will have the parts A, B, and C. (Which is) the very thing it was required to show.

225

226

ELEMENTS BOOK 8 Continued proportion†

† The

propositions contained in Books 7–9 are generally attributed to the school of Pythagoras.

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ELEMENTS BOOK 8

α΄.

Proposition 1

'Ε¦ν ðσιν Ðσοιδηποτοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον, οƒ If there are any multitude whatsoever of continuously δ ¥κροι αÙτîν πρîτοι πρÕς ¢λλήλους ðσιν, ™λάχιστοί proportional numbers, and the outermost of them are ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς. prime to one another, then the (numbers) are the least of those (numbers) having the same ratio as them.

Ε Ζ Η Θ

Α Β Γ ∆

A B C D

E F G H

”Εστωσαν Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον οƒ Α, Β, Γ, ∆, οƒ δ ¥κροι αÙτîν οƒ Α, ∆, πρîτοι πρÕς ¢λλήλους œστωσαν· λέγω, Óτι οƒ Α, Β, Γ, ∆ ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς. Ε„ γ¦ρ µή, œστωσαν ™λάττονες τîν Α, Β, Γ, ∆ οƒ Ε, Ζ, Η, Θ ™ν τù αÙτù λόγJ Ôντες αÙτο‹ς. κሠ™πεˆ οƒ Α, Β, Γ, ∆ ™ν τù αÙτù λόγJ ε„σˆ το‹ς Ε, Ζ, Η, Θ, καί ™στιν ‡σον τÕ πλÁθος [τîν Α, Β, Γ, ∆] τù πλήθει [τîν Ε, Ζ, Η, Θ], δι' ‡σου ¥ρα ™στˆν æς Ð Α πρÕς τÕν ∆, Ð Ε πρÕς τÕν Θ. οƒ δ Α, ∆ πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι ¢ριθµοˆ µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάσσων τÕν ™λάσσονα, τουτέστιν Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον. µετρε‹ ¥ρα Ð Α τÕν Ε Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα οƒ Ε, Ζ, Η, Θ ™λάσσονες Ôντες τîν Α, Β, Γ, ∆ ™ν τù αÙτù λόγJ ε„σˆν αÙτο‹ς. οƒ Α, Β, Γ, ∆ ¥ρα ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς· Óπερ œδει δε‹ξαι.

Let A, B, C, D be any multitude whatsoever of continuously proportional numbers. And let the outermost of them, A and D, be prime to one another. I say that A, B, C, D are the least of those (numbers) having the same ratio as them. For if not, let E, F , G, H be less than A, B, C, D (respectively), being in the same ratio as them. And since A, B, C, D are in the same ratio as E, F , G, H, and the multitude [of A, B, C, D] is equal to the multitude [of E, F , G, H], thus, via equality, as A is to D, (so) E (is) to H [Prop. 7.14]. And A and D (are) prime (to one another). And prime (numbers are) also the least of those (numbers having the same ratio as them) [Prop. 7.21]. And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, A measures E, the greater (measuring) the lesser. The very thing is impossible. Thus, E, F , G, H, being less than A, B, C, D, are not in the same ratio as them. Thus, A, B, C, D are the least of those (numbers) having the same ratio as them. (Which is) the very thing it was required to show.

β΄.

Proposition 2

ΑριθµοÝς εØρε‹ν ˜ξÁς ¢νάλογον ™λαχίστους, Óσους ¨ν ™πιτάξV τις, ™ν τù δοθέντι λόγJ. ”Εστω Ð δοθεˆς λόγος ™ν ™λάχίστοις ¢ριθµο‹ς Ð τοà Α πρÕς τÕν Β· δε‹ δ¾ ¢ριθµοÝς εØρε‹ν ˜ξÁς ¢νάλογον ™λαχίστους, Óσους ¥ν τις ™πιτάξV, ™ν τù τοà Α πρÕς τÕν Β λόγJ. 'Επιτετάχθωσαν δ¾ τέσσαρες, καˆ Ð Α ˜αυτÕν πολλαπλασιάσας τÕν Γ ποιείτω, τÕν δ Β πολλαπλασιάσας τÕν ∆ ποιείτω, κሠœτι Ð Β ˜αυτÕν πολλαπλασιάσας τÕν Ε ποιείτω, κሠœτι Ð Α τοÝς Γ, ∆, Ε πολλαπλασιάσας τοÝς Ζ, Η, Θ ποιείτω, Ð δ Β τÕν Ε πολλαπλασιάσας τÕν Κ ποιείτω.

To find the least numbers, as many as may be prescribed, (which are) continuously proportional in a given ratio. Let the given ratio, (expressed) in the least numbers, be that of A to B. So it is required to find the least numbers, as many as may be prescribed, (which are) in the ratio of A to B. Let four (numbers) have been prescribed. And let A make C (by) multiplying itself, and let it make D (by) multiplying B. And, further, let B make E (by) multiplying itself. And, further, let A make F , G, H (by) multiplying C, D, E. And let B make K (by) multiplying E.

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ΣΤΟΙΧΕΙΩΝ η΄.

Α Β

ELEMENTS BOOK 8

Γ ∆ Ε

A B

Ζ Η Θ Κ

C D E

F G H K

Κሠ™πεˆ Ð Α ˜αυτÕν µν πολλαπλασιάσας τÕν Γ πεποίηκεν, τÕν δ Β πολλαπλασιάσας τÕν ∆ πεποίηκεν, œστιν ¥ρα æς Ð Α πρÕς τÕν Β, [οÛτως] Ð Γ πρÕς τÕν ∆. πάλιν, ™πεˆ Ð µν Α τÕν Β πολλαπλασιάσας τÕν ∆ πεποίηκεν, Ð δ Β ˜αυτÕν πολλαπλασιάσας τÕν Ε πεποίηκεν, ˜κάτερος ¥ρα τîν Α, Β τÕν Β πολλαπλασιάσας ˜κάτερον τîν ∆, Ε πεποίηκεν. œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Ε. ¢λλ' æς Ð Α πρÕς τÕν Β, Ð Γ πρÕς τÕν ∆· κሠæς ¥ρα Ð Γ πρÕς τÕν ∆, Ð ∆ πρÕς τÕν Ε. κሠ™πεˆ Ð Α τοÝς Γ, ∆ πολλαπλασιάσας τοÝς Ζ, Η πεποίηκεν, œστιν ¥ρα æς Ð Γ πρÕς τÕν ∆, [οÛτως] Ð Ζ πρÕς τÕν Η. æς δ Ð Γ πρÕς τÕν ∆, οÛτως Ãν Ð Α πρÕς τÕν Β· κሠæς ¥ρα Ð Α πρÕς τÕν Β, Ð Ζ πρÕς τÕν Η. πάλιν, ™πεˆ Ð Α τοÝς ∆, Ε πολλαπλασιάσας τοÝς Η, Θ πεποίηκεν, œστιν ¥ρα æς Ð ∆ πρÕς τÕν Ε, Ð Η πρÕς τÕν Θ. ¢λλ' æς Ð ∆ πρÕς τÕν Ε, Ð Α πρÕς τÕν Β. κሠæς ¥ρα Ð Α πρÕς τÕν Β, οÛτως Ð Η πρÕς τÕν Θ. κሠ™πεˆ οƒ Α, Β τÕν Ε πολλαπλασιάσαντες τοÝς Θ, Κ πεποιήκασιν, œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως Ð Θ πρÕς τÕν Κ. ¢λλ' æς Ð Α πρÕς τÕν Β, οÛτως Ó τε Ζ πρÕς τÕν Η καˆ Ð Η πρÕς τÕν Θ. κሠæς ¥ρα Ð Ζ πρÕς τÕν Η, οÛτως Ó τε Η πρÕς τÕν Θ καˆ Ð Θ πρÕς τÕν Κ· οƒ Γ, ∆, Ε ¥ρα καˆ οƒ Ζ, Η, Θ, Κ ¢νάλογόν ε„σιν ™ν τù τοà Α πρÕς τÕν Β λόγJ. λέγω δή, Óτι κሠ™λάχιστοι. ™πεˆ γ¦ρ οƒ Α, Β ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς, οƒ δ ™λάχιστοι τîν τÕν αÙτÕν λόγον ™χόντων πρîτοι πρÕς ¢λλήλους ε„σίν, οƒ Α, Β ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. κሠ˜κάτερος µν τîν Α, Β ˜αυτÕν πολλαπλασιάσας ˜κάτερον τîν Γ, Ε πεποίηκεν, ˜κάτερον δ τîν Γ, Ε πολλαπλασιάσας ˜κάτερον τîν Ζ, Κ πεποίηκεν· οƒ Γ, Ε ¥ρα καˆ οƒ Ζ, Κ πρîτοι πρÕς ¢λλήλους ε„σίν. ™¦ν δ ðσιν Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον, οƒ δ ¥κροι αÙτîν πρîτοι πρÕς ¢λλήλους ðσιν, ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς. οƒ Γ, ∆, Ε ¥ρα καˆ οƒ Ζ, Η, Θ, Κ ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Β· Óπερ œδει δε‹ξαι.

And since A has made C (by) multiplying itself, and has made D (by) multiplying B, thus as A is to B, [so] C (is) to D [Prop. 7.17]. Again, since A has made D (by) multiplying B, and B has made E (by) multiplying itself, A, B have thus made D, E, respectively, (by) multiplying B. Thus, as A is to B, so D (is) to E [Prop. 7.18]. But, as A (is) to B, (so) C (is) to D. And thus as C (is) to D, (so) D (is) to E. And since A has made F , G (by) multiplying C, D, thus as C is to D, [so] F (is) to G [Prop. 7.17]. And as C (is) to D, so A was to B. And thus as A (is) to B, (so) F (is) to G. Again, since A has made G, H (by) multiplying D, E, thus as D is to E, (so) G (is) to H [Prop. 7.17]. But, as D (is) to E, (so) A (is) to B. And thus as A (is) to B, so G (is) to H. And since A, B have made H, K (by) multiplying E, thus as A is to B, so H (is) to K. But, as A (is) to B, so F (is) to G, and G to H. And thus as F (is) to G, so G (is) to H, and H to K. Thus, C, D, E and F , G, H, K are (both continuously) proportional in the ratio of A to B. So I say that (they are) also the least (sets of numbers continuously proportional in that ratio). For since A and B are the least of those (numbers) having the same ratio as them, and the least of those (numbers) having the same ratio are prime to one another [Prop. 7.22], A and B are thus prime to one another. And A, B have made C, E, respectively, (by) multiplying themselves, and have made F , K by multiplying C, E, respectively. Thus, C, E and F , K are prime to one another [Prop. 7.27]. And if there are any multitude whatsoever of continuously proportional numbers, and the outermost of them are prime to one another, then the (numbers) are the least of those (numbers) having the same ratio as them [Prop. 8.1]. Thus, C, D, E and F , G, H, K are the least of those (continuously proportional sets of numbers) having the same ratio as A and B. (Which is) the very thing it was required to show.

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ELEMENTS BOOK 8 Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι ™¦ν τρε‹ς ¢ριθµοˆ ˜ξÁς ¢νάλογον ™λάχιστοι ðσι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς, οƒ ¥κρον αÙτîν τετράγωνοί ε„σιν, ™¦ν δ τέσσαρες, κύβοι.

So it is clear, from this, that if three continuously proportional numbers are the least of those (numbers) having the same ratio as them, then the outermost of them are square, and, if four, cube.

γ΄.

Proposition 3

'Ε¦ν ðσιν Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον ™λάχισIf there are any multitude whatsoever of continuously τοι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς, οƒ ¥κροι αÙτîν proportional numbers, (which are) the least of those πρîτοι πρÕς ¢λλήλους ε„σίν. (numbers) having the same ratio as them, then the outermost of them are prime to one another.

Α Β Γ ∆

Ε Ζ

Η Θ Κ

A B C D

Λ Μ Ν Ξ

E F

G H K

L M N O

”Εστωσαν Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον ™λάχιστοι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς οƒ Α, Β, Γ, ∆· λέγω, Óτι οƒ ¥κροι αÙτîν οƒ Α, ∆ πρîτοι πρÕς ¢λλήλους ε„σίν. Ε„λήφθωσαν γ¦ρ δύο µν ¢ριθµοˆ ™λάχιστοι ™ν τù τîν Α, Β, Γ, ∆ λόγJ οƒ Ε, Ζ, τρε‹ς δ οƒ Η, Θ, Κ, κሠ˜ξÁς ˜νˆ πλείους, ›ως τÕ λαµβανόµενον πλÁθος ‡σον γένηται τù πλήθει τîν Α, Β, Γ, ∆. ε„λήφθωσαν κሠœστωσαν οƒ Λ, Μ, Ν, Ξ. Κሠ™πεˆ οƒ Ε, Ζ ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς, πρîτοι πρÕς ¢λλήλους ε„σίν. κሠ™πεˆ ˜κάτερος τîν Ε, Ζ ˜αυτÕν µν πολλαπλασιάσας ˜κάτερον τîν Η, Κ πεποίηκεν, ˜κάτερον δ τîν Η, Κ πολλαπλασιάσας ˜κάτερον τîν Λ, Ξ πεποίηκεν, καˆ οƒ Η, Κ ¥ρα καˆ οƒ Λ, Ξ πρîτοι πρÕς ¢λλήλους ε„σίν. κሠ™πεˆ οƒ Α, Β, Γ, ∆ ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς, ε„σˆ δ καˆ οƒ Λ, Μ, Ν, Ξ ™λάχιστοι ™ν τù αÙτù λόγJ Ôντες το‹ς Α, Β, Γ, ∆, καί ™στιν ‡σον τÕ πλÁθος τîν Α, Β, Γ, ∆ τù πλήθει τîν Λ, Μ, Ν, Ξ, ›καστος ¥ρα τîν Α, Β, Γ, ∆ ˜κάστJ τîν Λ, Μ, Ν, Ξ ‡σος ™στίν· ‡σος ¥ρα ™στˆν Ð µν Α τù Λ, Ð δ ∆ τù Ξ. καί ε„σιν οƒ Λ, Ξ πρîτοι πρÕς ¢λλήλους. καˆ οƒ Α, ∆ ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι.

Let A, B, C, D be any multitude whatsoever of continuously proportional numbers, (which are) the least of those (numbers) having the same ratio as them. I say that the outermost of them, A and D, are prime to one another. For let the two least (numbers) E, F (which are) in the same ratio as A, B, C, D have been taken [Prop. 7.33]. And the three (least numbers) G, H, K [Prop. 8.2]. And (so on), successively increasing by one, until the multitude of (numbers) taken is made equal to the multitude of A, B, C, D. Let them have been taken, and let them be L, M , N , O. And since E and F are the least of those (numbers) having the same ratio as them, they are prime to one another [Prop. 7.22]. And since E, F have made G, K, respectively, (by) multiplying themselves [Prop. 8.2 corr.], and have made L, O (by) multiplying G, K, respectively, G, K and L, O are thus also prime to one another [Prop. 7.27]. And since A, B, C, D are the least of those (numbers) having the same ratio as them, and L, M , N , O are also the least (of those numbers having the same ratio as them), being in the same ratio as A, B, C, D, and the multitude of A, B, C, D is equal to the multitude of L, M , N , O, thus A, B, C, D are equal to L, M , N , O, respectively. Thus, A is equal to L, and D to O. And L and O are prime to one another. Thus, A and D are also

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ELEMENTS BOOK 8 prime to one another. (Which is) the very thing it was required to show.

δ΄.

Proposition 4

Λόγων δοθέντων Ðποσωνοàν ™ν ™λαχίστοις ¢ριθµο‹ς ¢ριθµοÝς εØρε‹ν ˜ξÁς ¢νάλογον ™λαχίστους ™ν το‹ς δοθε‹σι λόγοις.

For any multitude whatsoever of given ratios, (expressed) in the least numbers, to find the least numbers continuously proportional in these given ratios.

Α Γ Ε

Β ∆ Ζ

A C E

B D F

Ν Ξ Μ Ο

Θ Η Κ Λ

N O M P

H G K L

”Εστωσαν οƒ δοθέντες λόγοι ™ν ™λαχίστοις ¢ριθµο‹ς Ó τε τοà Α πρÕς τÕν Β καˆ Ð τοà Γ πρÕς τÕν ∆ κሠœτι Ð τοà Ε πρÕς τÕν Ζ· δε‹ δ¾ ¢ριθµοÝς εØρε‹ν ˜ξÁς ¢νάλογον ™λαχίστους œν τε τù τοà Α πρÕς τÕν Β λόγJ κሠ™ν τù τοà Γ πρÕς τÕν ∆ κሠœτι τù τοà Ε πρÕς τÕν Ζ. Ε„λήφθω γ¦ρ Ð ØπÕ τîν Β, Γ ™λάχιστος µετρούµενος ¢ριθµÕς Ð Η. κሠÐσάκις µν Ð Β τÕν Η µετρε‹, τοσαυτάκις καˆ Ð Α τÕν Θ µετρείτω, Ðσάκις δ Ð Γ τÕν Η µετρε‹, τοσαυτάκις καˆ Ð ∆ τÕν Κ µετρείτω. Ð δ Ε τÕν Κ ½τοι µετρε‹ À οÙ µετρε‹. µετρείτω πρότερον. κሠÐσάκις Ð Ε τÕν Κ µετρε‹, τοσαυτάκις καˆ Ð Ζ τÕν Λ µετρείτω. κሠ™πεˆ „σάκις Ð Α τÕν Θ µετρε‹ καˆ Ð Β τÕν Η, œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως Ð Θ πρÕς τÕν Η. δι¦ τ¦ αÙτ¦ δ¾ κሠæς Ð Γ πρÕς τÕν ∆, οÛτως Ð Η πρÕς τÕν Κ, κሠœτι æς Ð Ε πρÕς τÕν Ζ, οÛτως Ð Κ πρÕς τÕν Λ· οƒ Θ, Η, Κ, Λ ¥ρα ˜ξÁς ¢νάλογόν ε„σιν œν τε τù τοà Α πρÕς τÕν Β κሠ™ν τù τοà Γ πρÕς τÕν ∆ κሠœτι ™ν τù τοà Ε πρÕς τÕν Ζ λόγJ. λέγω δή, Óτι κሠ™λάχιστοι. ε„ γ¦ρ µή ε„σιν οƒ Θ, Η, Κ, Λ ˜ξÁς ¢νάλογον ™λάχιστοι œν τε το‹ς τοà Α πρÕς τÕν Β κሠτοà Γ πρÕς τÕν ∆ κሠ™ν τù τοà Ε πρÕς τÕν Ζ λόγοις, œστωσαν οƒ Ν, Ξ, Μ, Ο. κሠ™πεί ™στιν æς Ð Α πρÕς τÕν Β, οÛτως Ð Ν πρÕς τÕν Ξ, οƒ δ Α, Β ™λάχιστοι, οƒ δ ™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάσσων τÕν ™λάσσονα, τουτέστιν Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον, Ð Β ¥ρα τÕν Ξ µετρε‹. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Γ τÕν Ξ µετρε‹· οƒ Β, Γ ¥ρα τÕν Ξ µετροàσιν· καˆ Ð ™λάχιστος ¥ρα ØπÕ τîν Β, Γ µετρούµενος τÕν Ξ µετρήσει. ™λάχιστος δ ØπÕ τîν Β, Γ µετρε‹ται Ð Η· Ð Η ¥ρα τÕν Ξ µετρε‹ Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύντατον. οÙκ ¥ρα

Let the given ratios, (expressed) in the least numbers, be the (ratios) of A to B, and of C to D, and, further, of E to F . So it is required to find the least numbers continuously proportional in the ratio of A to B, and of C to B, and, further, of E to F . For let the least number, G, measured by (both) B and C have be taken [Prop. 7.34]. And as many times as B measures G, so many times let A also measure H. And as many times as C measures G, so many times let D also measure K. And E either measures, or does not measure, K. Let it, first of all, measure (K). And as many times as E measures K, so many times let F also measure L. And since A measures H the same number of times that B also (measures) G, thus as A is to B, so H (is) to G [Def. 7.20, Prop. 7.13]. And so, for the same (reasons), as C (is) to D, so G (is) to K, and, further, as E (is) to F , so K (is) to L. Thus, H, G, K, L are continuously proportional in the ratio of A to B, and of C to D, and, further, of E to F . So I say that (they are) also the least (numbers continuously proportional in these ratios). For if H, G, K, L are not the least numbers continuously proportional in the ratios of A to B, and of C to D, and of E to F , let N , O, M , P be (the least such numbers). And since as A is to B, so N (is) to O, and A and B are the least (numbers which have the same ratio as them), and the least (numbers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20], B thus measures O. So, for the same (reasons), C also measures O. Thus, B and C (both) measure O. Thus, the least number measured by

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ELEMENTS BOOK 8

œσονταί τινες τîν Θ, Η, Κ, Λ ™λάσσονες ¢ριθµοˆ ˜ξÁς œν τε τù τοà Α πρÕς τÕν Β κሠτù τοà Γ πρÕς τÕν ∆ κሠœτι τù τοà Ε πρÕς τÕν Ζ λόγù.

(both) B and C will also measure O [Prop. 7.35]. And G (is) the least number measured by (both) B and C. Thus, G measures O, the greater (measuring) the lesser. The very thing is impossible. Thus, there cannot be any numbers less than H, G, K, L (which are) continuously (proportional) in the ratio of A to B, and of C to D, and, further, of E to F .

Α Γ Ε

Β ∆ Ζ

A C E

B D F

Ν Ξ Μ Ο

Θ Η Κ

N O M P

H G K

Π Ρ Σ Τ

Q R S T

Μ¾ µετρείτω δ¾ Ð Ε τÕν Κ, κሠε„λήφθω ØπÕ τîν Ε, Κ ™λάχιστος µετρούµενος ¢ριθµÕς Ð Μ. κሠÐσάκις µν Ð Κ τÕν Μ µετρε‹, τοσαυτάκις κሠ˜κάτερος τîν Θ, Η ˜κάτερον τîν Ν, Ξ µετρείτω, Ðσάακις δ Ð Ε τÕν Μ µετρε‹, τοσαυτάκις καˆ Ð Ζ τÕν Ο µετρείτω. ™πεˆ „σάκις Ð Θ τÕν Ν µετρε‹ καˆ Ð Η τÕν Ξ, œστιν ¥ρα æς Ð Θ πρÕς τÕν Η, οÛτως Ð Ν πρÕς τÕν Ξ. æς δ Ð Θ πρÕς τÕν Η, οÛτως Ð Α πρÕς τÕν Β· κሠæς ¥ρα Ð Α πρÕς τÕν Β, οÛτως Ð Ν πρÕς τÕν Ξ. δι¦ τ¦ αÙτ¦ δ¾ κሠæς Ð Γ πρÕς τÕν ∆, οÛτως Ð Ξ πρÕς τÕν Μ. πάλιν, ™πεˆ „σάκις Ð Ε τÕν Μ µετρε‹ καˆ Ð Ζ τÕν Ο, œστιν ¥ρα æς Ð Ε πρÕς τÕν Ζ, οÛτως Ð Μ πρÕς τÕν Ο· οƒ Ν, Ξ, Μ, Ο ¥ρα ˜ξÁς ¢νάλογόν ε„σιν ™ν το‹ς τοà τε Α πρÕς τÕν Β κሠτοà Γ πρÕς τÕν ∆ κሠœτι τοà Ε πρÕς τÕν Ζ λόγοις. λέγω δή, Óτι κሠ™λάχιστοι ™ν το‹ς Α Β, Γ ∆, Ε Ζ λόγοις. ε„ γ¦ρ µή, œσονταί τινες τîν Ν, Ξ, Μ, Ο ™λάσσονες ¢ριθµοˆ ˜ξÁς ¢νάλογον ™ν το‹ς Α Β, Γ ∆, Ε Ζ λόγοις. œστωσαν οƒ Π, Ρ, Σ, Τ. κሠ™πεί ™στιν æς Ð Π πρÕς τÕν Ρ, οÛτως Ð Α πρÕς τÕν Β, οƒ δ Α, Β ™λάχιστοι, οƒ δ ™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας αÙτο‹ς „σάκις Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον, Ð Β ¥ρα τÕν Ρ µετρε‹. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Γ τÕν Ρ µετρε‹· οƒ Β, Γ ¥ρα τÕν Ρ µετροàσιν. καˆ Ð ™λάχιστος ¥ρα ØπÕ τîν Β, Γ µετούµενος τÕν Ρ µετρήσει. ™λάχιστος δ ØπÕ τîν Β, Γ µετρούµενος ™στιν Ð Η· Ð Η ¥ρα τÕν Ρ µετρε‹. καί ™στιν æς Ð Η πρÕς τÕν Ρ, οÛτως Ð Κ πρÕς τÕν Σ· καˆ Ð Κ ¥ρα τÕν Σ µετρε‹. µετρε‹ δ καˆ Ð Ε τÕν Σ· οƒ Ε, Κ

So let E not measure K. And let the least number, M , measured by (both) E and K have been taken [Prop. 7.34]. And as many times as K measures M , so many times let H, G also measure N , O, respectively. And as many times as E measures M , so many times let F also measure P . Since H measures N the same number of times as G (measures) O, thus as H is to G, so N (is) to O [Def. 7.20, Prop. 7.13]. And as H (is) to G, so A (is) to B. And thus as A (is) to B, so N (is) to O. And so, for the same (reasons), as C (is) to D, so O (is) to M . Again, since E measures M the same number of times as F (measures) P , thus as E is to F , so M (is) to P [Def. 7.20,Prop. 7.13]. Thus, N , O, M , P are continuously proportional in the ratios of A to B, and of C to D, and, further, of E to F . So I say that (they are) also the least (numbers) in the ratios of A B, C D, E F . For if not, then there will be some numbers less than N , O, M , P (which are) continuously proportional in the ratios of A B, C D, E F . Let them be Q, R, S, T . And since as Q is to R, so A (is) to B, and A and B (are) the least (numbers having the same ratio as them), and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20], B thus measures R. So, for the same (reasons), C also measures R. Thus, B and C (both) measure R. Thus, the least (number) measured by (both) B and C will also measure R [Prop. 7.35]. And G

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ELEMENTS BOOK 8

¥ρα τÕν Σ µετροàσιν. καˆ Ð ™λάχιστος ¥ρα ØπÕ τîν Ε, Κ µετρούµενος τÕν Σ µετρήσει. ™λάχιστος δ ØπÕ τîν Ε, Κ µετρούµενός ™στιν Ð Μ· Ð Μ ¥ρα τÕν Σ µετρε‹ Ð µείζων τÕν ™λάσσονα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα œσονταί τινες τîν Ν, Ξ, Μ, Ο ™λάσσονες ¢ριθµοˆ ˜ξÁς ¢νάλογον œν τε το‹ς τοà Α πρÕς τÕν Β κሠτοà Γ πρÕς τÕν ∆ κሠœτι τοà Ε πρÕς τÕν Ζ λόγοις· οƒ Ν, Ξ, Μ, Ο ¥ρα ˜ξÁς ¢νάλογον ™λάχιστοί ε„σιν ™ν το‹ς Α Β, Γ ∆, Ε Ζ λόγοις· Óπερ œδει δε‹ξαι.

is the least number measured by (both) B and C. Thus, G measures R. And as G is to R, so K (is) to S. Thus, K also measures S [Def. 7.20]. And E also measures S [Prop. 7.20]. Thus, E and K (both) measure S. Thus, the least (number) measured by (both) E and K will also measure S [Prop. 7.35]. And M is the least (number) measured by (both) E and K. Thus, M measures S, the greater (measuring) the lesser. The very thing is impossible. Thus there cannot be any numbers less than N , O, M , P (which are) continuously proportional in the ratios of A to B, and of C to D, and, further, of E to F . Thus, N , O, M , P are the least (numbers) continuously proportional in the ratios of A B, C D, E F . (Which is) the very thing it was required to show.

ε΄.

Proposition 5

Οƒ ™πίπεδοι ¢ριθµοˆ πρÕς ¢λλήλους λόγον œχουσι Plane numbers have to one another the ratio compounτÕν συγκείµενον ™κ τîν πλευρîν. ded† out of (the ratios of) their sides.

Α Β Γ Ε

A B ∆ Ζ

C E

Η Θ Κ

G H K

Λ

L

”Εστωσαν ™πίπεδοι ¢ριθµοˆ οƒ Α, Β, κሠτοà µν Α πλευρሠœστωσαν οƒ Γ, ∆ ¢ριθµοί, τοà δ Β οƒ Ε, Ζ· λέγω, Óτι Ð Α πρÕς τÕν Β λόγον œχει τÕν συγκείµενον ™κ τîν πλευρîν. Λόγων γ¦ρ δοθέντων τοà τε Öν œχει Ð Γ πρÕς τÕν Ε καˆ Ð ∆ πρÕς τÕν Ζ ε„λήφθωσαν ¢ριθµοˆ ˜ξÁς ™λάχιστοι ™ν το‹ς Γ Ε, ∆ Ζ λόγοις, οƒ Η, Θ, Κ, éστε εναι æς µν τÕν Γ πρÕς τÕν Ε, οÛτως τÕν Η πρÕς τÕν Θ, æς δ τÕν ∆ πρÕς τÕν Ζ, οÛτως τÕν Θ πρÕς τÕν Κ. καˆ Ð ∆ τÕν Ε πολλαπλασιάσας τÕν Λ ποιείτω. Κሠ™πεˆ Ð ∆ τÕν µν Γ πολλαπλασιάσας τÕν Α πεποίηκεν, τÕν δ Ε πολλαπλασιάσας τÕν Λ πεποίηκεν, œστιν ¥ρα æς Ð Γ πρÕς τÕν Ε, οÛτως Ð Α πρÕς τÕν Λ. æς δ Ð Γ πρÕς τÕν Ε, οÛτως Ð Η πρÕς τÕν Θ· κሠæς ¥ρα Ð Η πρÕς τÕν Θ, οÛτως Ð Α πρÕς τÕν Λ. πάλιν, ™πεˆ Ð Ε τÕν ∆ πολλαπλασιάσας τÕν Λ πεποίηκεν, ¢λλ¦ µ¾ν κሠτÕν Ζ πολλαπλασιάσας τÕν Β πεποίηκεν, œστιν

D F

Let A and B be plane numbers, and let C, D be the sides of A, and E, F (the sides) of B. I say that A has to B the ratio compounded out of (the ratios of) their sides. For given the ratios which C has to E, and D (has) to F , let the least numbers, G, H, K, continuously proportional in the ratios C E, D F have been taken [Prop. 8.4], so that as C is to E, so G (is) to H, and as D (is) to F , so H (is) to K. And let D make L (by) multiplying E. And since D has made A (by) multiplying C, and has made L (by) multiplying E, thus as C is to E, so A (is) to L [Prop. 7.17]. And as C (is) to E, so G (is) to H. And thus as G (is) to H, so A (is) to L. Again, since E has made L (by) multiplying D [Prop. 7.16], but, in fact, has also made B (by) multiplying F , thus as D is to F , so L (is) to B [Prop. 7.17]. But, as D (is) to F , so H (is) to K. And thus as H (is) to K, so L (is) to B. And it was also shown that as G (is) to H, so A (is) to L. Thus, via

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¥ρα æς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Λ πρÕς τÕν Β. ¢λλ' æς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Θ πρÕς τÕν Κ· κሠæς ¥ρα Ð Θ πρÕς τÕν Κ, οÛτως Ð Λ πρÕς τÕν Β. ™δείχθη δ κሠæς Ð Η πρÕς τÕν Θ, οÛτως Ð Α πρÕς τÕν Λ· δι' ‡σου ¥ρα ™στˆν æς Ð Η πρÕς τÕν Κ, [οÛτως] Ð Α πρÕς τÕν Β. Ð δ Η πρÕς τÕν Κ λόγον œχει τÕν συγκείµενον ™κ τîν πλευρîν· καˆ Ð Α ¥ρα πρÕς τÕν Β λόγον œχει τÕν συγκείµενον ™κ τîν πλευρîν· Óπερ œδει δε‹ξαι. †

equality, as G is to K, [so] A (is) to B [Prop. 7.14]. And G has to K the ratio compounded out of (the ratios of) the sides (of A and B). Thus, A also has to B the ratio compounded out of (the ratios of) the sides (of A and B). (Which is) the very thing it was required to show.

i.e., multiplied.

$΄.

Proposition 6

'Ε¦ν ðσιν Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον, Ð δ If there are any multitude whatsoever of continuously πρîτος τÕν δεύτερον µ¾ µετρÍ, οÙδ ¥λλος οÙδεˆς proportional numbers, and the first does not measure the οÙδένα µετρήσει. second, then no other (number) will measure any other (number) either.

Α Β Γ ∆ Ε

A B C D E

Ζ Η Θ

F G H

”Εστωσαν Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον οƒ Α, Β, Γ, ∆, Ε, Ð δ Α τÕν Β µ¾ µετρείτω· λέγω, Óτι οÙδ ¥λλος οÙδεˆς οÙδένα µετρήσει. “Οτι µν οâν οƒ Α, Β, Γ, ∆, Ε ˜ξÁς ¢λλήλους οÙ µετροàσιν, φανερόν· οÙδ γ¦ρ Ð Α τÕν Β µετρε‹. λέγω δή, Óτι οÙδ ¥λλος οÙδεˆς οÙδένα µετρήσει. ε„ γ¦ρ δυνατόν, µετρείτω Ð Α τÕν Γ. κሠÓσοι ε„σˆν οƒ Α, Β, Γ, τοσοàτοι ε„λήφθωσαν ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Β, Γ οƒ Ζ, Η, Θ. κሠ™πεˆ οƒ Ζ, Η, Θ ™ν τù αÙτù λόγJ ε„σˆ το‹ς Α, Β, Γ, καί ™στιν ‡σον τÕ πλÁθος τîν Α, Β, Γ τù πλήθει τîν Ζ, Η, Θ, δι' ‡σου ¥ρα ™στˆν æς Ð Α πρÕς τÕν Γ, οÛτως Ð Ζ πρÕς τÕν Θ. κሠ™πεί ™στιν æς Ð Α πρÕς τÕν Β, οÛτως Ð Ζ πρÕς τÕν Η, οÙ µετρε‹ δ Ð Α τÕν Β, οÙ µετρε‹ ¥ρα οÙδ Ð Ζ τÕν Η· οÙκ ¥ρα µονάς ™στιν Ð Ζ· ¹ γ¦ρ µον¦ς πάντα ¢ριθµÕν µετρε‹. καί ε„σιν οƒ Ζ, Θ πρîτοι πρÕς ¢λλήλους [οÙδ Ð Ζ ¥ρα τÕν Θ µετρε‹]. καί ™στιν æς Ð Ζ πρÕς τÕν Θ, οÛτως Ð Α πρÕς τÕν Γ· οÙδ Ð Α ¥ρα τÕν Γ µετρε‹. еοίως δ¾ δείξοµεν, Óτι οÙδ ¥λλος οÙδεˆς οÙδένα µετρήσει· Óπερ œδει δε‹ξαι.

Let A, B, C, D, E be any multitude whatsoever of continuously proportional numbers, and let A not measure B. I say that no other (number) will measure any other (number) either. Now, (it is) clear that A, B, C, D, E do not successively measure one another. For A does not even measure B. So I say that no other (number) will measure any other (number) either. For, if possible, let A measure C. And as many (numbers) as are A, B, C, let so many of the least numbers, F , G, H, have been taken of those (numbers) having the same ratio as A, B, C [Prop. 7.33]. And since F , G, H are in the same ratio as A, B, C, and the multitude of A, B, C is equal to the multitude of F , G, H, thus, via equality, as A is to C, so F (is) to H [Prop. 7.14]. And since as A is to B, so F (is) to G, and A does not measure B, F does not measure G either [Def. 7.20]. Thus, F is not a unit. For a unit measures all numbers. And F and H are prime to one another [Prop. 8.3] [and thus F does not measure H]. And as F is to H, so A (is) to C. And thus A does not measure C either [Def. 7.20]. So, similarly, we can show that no other (number) can measure any other (number) either.

234

ΣΤΟΙΧΕΙΩΝ η΄.

ELEMENTS BOOK 8 (Which is) the very thing it was required to show.

ζ΄.

Proposition 7

'Ε¦ν ðσιν Ðποσοιοàν ¢ριθµοˆ [˜ξÁς] ¢νάλογον, Ð δ If there are any multitude whatsoever of [continuπρîτος τÕν œσχατον µετρÍ, κሠτÕν δεύτερον µετρήσει. ously] proportional numbers, and the first measures the last, then (the first) will also measure the second.

Α Β Γ ∆

A B C D

”Εστωσαν Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον οƒ Α, Β, Γ, ∆, Ð δ Α τÕν ∆ µετρείτω· λέγω, Óτι καˆ Ð Α τÕν Β µετρε‹. Ε„ γ¦ρ οÙ µετρε‹ Ð Α τÕν Β, οÙδ ¥λλος οÙδεˆς οÙδένα µετρήσει· µετρε‹ δ Ð Α τÕν ∆. µετρε‹ ¥ρα καˆ Ð Α τÕν Β· Óπερ œδει δε‹ξαι.

Let A, B, C, D be any number whatsoever of continuously proportional numbers. And let A measure D. I say that A also measures B. For if A does not measure B then no other (number) will measure any other (number) either [Prop. 8.6]. But A measures D. Thus, A also measures B. (Which is) the very thing it was required to show.

η΄.

Proposition 8

'Ε¦ν δύο ¢ριθµîν µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπίπτωσιν ¢ριθµοί, Óσοι ε„ς αÙτοÝς µεταξÝ κατ¦ τÕ συνεχς ¢νόλογον ™µπίπτουσιν ¢ριθµοί, τοσοàτοι κሠε„ς τοÝς τÕν αÙτÕν λόγον œχοντας [αÙτο‹ς] µεταξÝ κατ¦ τÕ συνχες ¢νάλογον ™µπεσοàνται

If between two numbers there fall (some) numbers in continued proportion, then as many numbers as fall in between them in continued proportion, so many (numbers) will also fall in between (any two numbers) having the same ratio [as them] in continued proportion.

Α Γ ∆ Β

Ε Μ Ν Ζ

A C D B

Η Θ Κ Λ

E M N F

G H K L

∆ύο γ¦ρ ¢ριθµîν τîν Α, Β µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπιπτέτωσαν ¢ριθµοˆ οƒ Γ, ∆, κሠπεποιήσθω æς Ð Α πρÕς τÕν Β, οÛτως Ð Ε πρÕς τÕν Ζ· λέγω, Óτι Óσοι ε„ς τοÝς Α, Β µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεπτώκασιν ¢ριθµοί, τοσοàτοι κሠε„ς τοÝς Ε, Ζ µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεσοàνται. “Οσοι γάρ ε„σι τù πλήθει οƒ Α, Β, Γ, ∆, τοσοàτοι ε„λήφθωσαν ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον

For let the numbers, C and D, fall between two numbers, A and B, in continued proportion, and let it have been made (so that) as A (is) to B, so E (is) to F . I say that as many numbers as have fallen in between A and B in continued proportion, so many (numbers) will also fall in between E and F in continued proportion. For as many as A, B, C, D are in multitude, let so many of the least numbers, G, H, K, L, having the same

235

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ELEMENTS BOOK 8

™χόντων το‹ς Α, Γ, ∆, Β οƒ Η, Θ, Κ, Λ· οƒ ¥ρα ¥κροι αÙτîν οƒ Η, Λ πρîτοι πρÕς ¢λλήλους ε„σίν. κሠ™πεˆ οƒ Α, Γ, ∆, Β το‹ς Η, Θ, Κ, Λ ™ν τù αÙτù λόγJ ε„σίν, καί ™στιν ‡σον τÕ πλÁθος τîν Α, Γ, ∆, Β τù πλήθει τîν Η, Θ, Κ, Λ, δι' ‡σου ¥ρα ™στˆν æς Ð Α πρÕς τÕν Β, οÛτως Ð Η πρÕς τÕν Λ. æς δ Ð Α πρÕς τÕν Β, οÛτως Ð Ε πρÕς τÕν Ζ· κሠæς ¥ρα Ð Η πρÕς τÕν Λ, οÛτως Ð Ε πρÕς τÕν Ζ. οƒ δ Η, Λ πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι ¢ριθµοˆ µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάσσων τÕν ™λάσσονα, τουτέστιν Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον. „σάκις ¥ρα Ð Η τÕν Ε µετρε‹ καˆ Ð Λ τÕν Ζ. Ðσάκις δ¾ Ð Η τÕν Ε µετρε‹, τοσαυτάκις κሠ˜κάτερος τîν Θ, Κ ˜κάτερον τîν Μ, Ν µετρείτω· οƒ Η, Θ, Κ, Λ ¥ρα τοÝς Ε, Μ, Ν, Ζ „σάκις µετροàσιν. οƒ Η, Θ, Κ, Λ ¥ρα το‹ς Ε, Μ, Ν, Ζ ™ν τù αÙτù λόγJ ε„σίν. ¢λλ¦ οƒ Η, Θ, Κ, Λ το‹ς Α, Γ, ∆, Β ™ν τù αÙτù λόγJ ε„σίν· καˆ οƒ Α, Γ, ∆, Β ¥ρα το‹ς Ε, Μ, Ν, Ζ ™ν τù αÙτù λόγJ ε„σίν. οƒ δ Α, Γ, ∆, Β ˜ξÁς ¢νάλογόν ε„σιν· καˆ οƒ Ε, Μ, Ν, Ζ ¥ρα ˜ξÁς ¢νάλογόν ε„σιν. Óσοι ¥ρα ε„ς τοÝς Α, Β µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεπτώκασιν ¢ριθµοί, τοσοàτοι κሠε„ς τοÝς Ε, Ζ µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεπτώκασιν ¢ριθµοί· Óπερ œδει δε‹ξαι.

ratio as A, B, C, D, have been taken [Prop. 7.33]. Thus, the outermost of them, G and L, are prime to one another [Prop. 8.3]. And since A, B, C, D are in the same ratio as G, H, K, L, and the multitude of A, B, C, D is equal to the multitude of G, H, K, L, thus, via equality, as A is to B. so G (is) to L [Prop. 7.14]. And as A (is) to B, so E (is) to F . And thus as G (is) to L, so E (is) to F . And G and L (are) prime (to one another). And (numbers) prime (to one another are) also the least (numbers having the same ratio as them) [Prop. 7.21]. And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, G measures E the same number of times as L (measures) F . So as many times as G measures E, so many times let H, K also measure M , N , respectively. Thus, G, H, K, L measure E, M , N , F (respectively) an equal number of times. Thus, G, H, K, L are in the same ratio as E, M , N , F [Def. 7.20]. But, G, H, K, L are in the same ratio as A, C, D, B. Thus, A, C, D, B are also in the same ratio as E, M , N , F . And A, C, D, B are continuously proportional. Thus, E, M , N , F are also continuously proportional. Thus, as many numbers as have fallen in between A and B in continued proportion, so many numbers have also fallen in between E and F in continued proportion. (Which is) the very thing it was required to show.

θ΄.

Proposition 9

'Ε¦ν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους ðσιν, κሠε„ς αÙτοÝς µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπίπτωσιν ¢ριθµοί, Óσοι ε„ς αÙτοÝς µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπίπτουσιν ¢ριθµοί, τοσοàτοι κሠ˜κατέρου αÙτîν κሠµονάδος µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεσοàνται.

If two numbers are prime to one another, and there fall in between them (some) numbers in continued proportion, then as many numbers as fall in between them in continued proportion, so many (numbers) will also fall between each of them and a unit in continued proportion.

Α Γ ∆ Β Ε Ζ Η

Θ Κ Λ

A C D B

Μ Ν Ξ Ο

E F G

”Εστωσαν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους οƒ Α, Β, κሠε„ς αÙτοÝς µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον

H K L M N O P

Let A and B be two numbers (which are) prime to one another, and let the (numbers) C and D fall in be-

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ELEMENTS BOOK 8

™µπιπτέτωσαν οƒ Γ, ∆, κሠ™κκείσθω ¹ Ε µονάς· λέγω, Óτι Óσοι ε„ς τοÝς Α, Β µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεπτώκασιν ¢ριθµοί, τοσοàτοι κሠ˜κατέρου τîν Α, Β κሠτÁς µονάδος µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεσοàνται. Ε„λήφθωσαν γ¦ρ δύο µν ¢ριθµοˆ ™λάχιστοι ™ν τù τîν Α, Γ, ∆, Β λόγJ Ôντες οƒ Ζ, Η, τρε‹ς δ οƒ Θ, Κ, Λ, κሠ¢εˆ ˜ξÁς ˜νˆ πλείους, ›ως ¨ν ‡σον γένηται τÕ πλÁθος αÙτîν τù πλήθει τîν Α, Γ, ∆, Β. ε„λήφθωσαν, κሠœστωσαν οƒ Μ, Ν, Ξ, Ο. φανερÕν δή, Óτι Ð µν Ζ ˜αυτÕν πολλαπλασιάσας τÕν Θ πεποίηκεν, τÕν δ Θ πολλαπλασιάσας τÕν Μ πεποίηκεν, καˆ Ð Η ˜αυτÕν µν πολλαπλασιάσας τÕν Λ πεποίηκεν, τÕν δ Λ πολλαπλασιάσας τÕν Ο πεποίηκεν. κሠ™πεˆ οƒ Μ, Ν, Ξ, Ο ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Ζ, Η, ε„σˆ δ καˆ οƒ Α, Γ, ∆, Β ™λάχιστοι τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Ζ, Η, καί ™στιν ‡σον τÕ πλÁθος τîν Μ, Ν, Ξ, Ο τù πλήθει τîν Α, Γ, ∆, Β, ›καστος ¥ρα τîν Μ, Ν, Ξ, Ο ˜κάστJ τîν Α, Γ, ∆, Β ‡σος ™στίν· ‡σος ¥ρα ™στˆν Ð µν Μ τù Α, Ð δ Ο τù Β. κሠ™πεˆ Ð Ζ ˜αυτÕν πολλαπλασιάσας τÕν Θ πεποίηκεν, Ð Ζ ¥ρα τÕν Θ µετρε‹ κατ¦ τ¦ς ™ν τù Ζ µονάδας. µετρε‹ δ κሠ¹ Ε µον¦ς τÕν Ζ κατ¦ τ¦ς ™ν αÙτù µονάδας· „σάκις ¥ρα ¹ Ε µον¦ς τÕν Ζ ¢ριθµÕν µετρε‹ καˆ Ð Ζ τÕν Θ. œστιν ¥ρα æς ¹ Ε µον¦ς πρÕς τÕν Ζ ¢ριθµόν, οÛτως Ð Ζ πρÕς τÕν Θ. πάλιν, ™πεˆ Ð Ζ τÕν Θ πολλαπλασιάσας τÕν Μ πεποίηκεν, Ð Θ ¥ρα τÕν Μ µετρε‹ κατ¦ τ¦ς ™ν τù Ζ µονάδας. µετρε‹ δ κሠ¹ Ε µον¦ς τÕν Ζ ¢ριθµÕν κατ¦ τ¦ς ™ν αÙτù µονάδας· „σάκις ¥ρα ¹ Ε µον¦ς τÕν Ζ ¢ριθµÕν µετρε‹ καˆ Ð Θ τÕν Μ. œστιν ¥ρα æς ¹ Ε µον¦ς πρÕς τÕν Ζ ¢ριθµόν, οÛτως Ð Θ πρÕς τÕν Μ. ™δείχθη δ κሠæς ¹ Ε µον¦ς πρÕς τÕν Ζ ¢ριθµόν, οÛτως Ð Ζ πρÕς τÕν Θ· κሠæς ¥ρα ¹ Ε µον¦ς πρÕς τÕν Ζ ¢ριθµόν, οÛτως Ð Ζ πρÕς τÕν Θ καˆ Ð Θ πρÕς τÕν Μ. ‡σος δ Ð Μ τù Α· œστιν ¥ρα æς ¹ Ε µον¦ς πρÕς τÕν Ζ ¢ριθµόν, οÛτως Ð Ζ πρÕς τÕν Θ καˆ Ð Θ πρÕς τÕν Α. δι¦ τ¦ αÙτ¦ δ¾ κሠæς ¹ Ε µον¦ς πρÕς τÕν Η ¢ριθµόν, οÛτως Ð Η πρÕς τÕν Λ καˆ Ð Λ πρÕς τÕν Β. Óσοι ¥ρα ε„ς τοÝς Α, Β µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεπτώκασιν ¢ριθµοί, τοσοàτοι κሠ˜κατέρου τîν Α, Β κሠµονάδος τÁς Ε µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεπτώκασιν ¢ριθµοί· Óπερ œδει δε‹ξαι.

tween them in continued proportion. And let the unit E be taken. I say that as many numbers as have fallen in between A and B in continued proportion, so many (numbers) will also fall between each of A and B and a unit in continued proportion. For let the least two numbers, F and G, which are in the ratio of A, B, C, D, have been taken [Prop. 7.33]. And the (least) three (numbers), H, K, L. And so on, successively increasing by one, until the multitude of the (least numbers taken) is made equal to the multitude of A, B, C, D [Prop. 8.2]. Let them have been taken, and let them be M , N , O, P . So (it is) clear that F has made H (by) multiplying itself, and has made M (by) multiplying H. And G has made L (by) multiplying itself, and has made P (by) multiplying L [Prop. 8.2 corr.]. And since M , N , O, P are the least of those (numbers) having same ratio as F , G, and A, B, C, D are also the least of those (numbers) having the same ratio as F , G [Prop. 8.2], and the multitude of M , N , O, P is equal to the multitude of A, B, C, D, thus M , N , O, P are equal to A, B, C, D, respectively. Thus, M is equal to A, and P to B. And since F has made H (by) multiplying itself, F thus measures H according to the units in F [Def. 7.15]. And the unit E also measures F according to the units in it. Thus, the unit E measures the number F as many times as F (measures) H. Thus, as the unit E is to the number F , so F (is) to H [Def. 7.20]. Again, since F has made M (by) multiplying H, H thus measures M according to the units in F [Def. 7.15]. And the unit E also measures the number F according to the units in it. Thus, the unit E measures the number F as many times as H (measures) M . Thus, as the unit E is to the number F , so H (is) to M [Prop. 7.20]. And it was shown that as the unit E (is) to the number F , so F (is) to H. And thus as the unit E (is) to the number F , so F (is) to H, and H (is) to M . And M (is) equal to A. Thus, as the unit E is to the number F , so F (is) to H, and H to A. And so, for the same (reasons), as the unit E (is) to the number G, so G (is) to L, and L to B. Thus, as many (numbers) as have fallen in between A and B in continued proportion, so many numbers have also fallen between each of A and B and the unit E in continued proportion. (Which is) the very thing it was required to show.

ι΄.

Proposition 10

'Εάν δύο ¢ριθµîν ˜κατέρου κሠµονάδος µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπίπτωσιν ¢ριθµοί, Óσοι ˜κατέρου αÙτîν κሠµονάδος µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπίπτουσιν ¢ριθµοί, τοσοàτοι κሠε„ς αÙτοÝς µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεσοàνται.

If (some) numbers fall between each of two numbers and a unit in continued proportion, then as many (numbers) as fall between each of the (two numbers) and the unit in continued proportion, so many (numbers) will also fall in between the (two numbers) themselves in con-

237

ΣΤΟΙΧΕΙΩΝ η΄.

ELEMENTS BOOK 8 tinued proportion.

Γ ∆ Ε Α

Γ Ζ Η Β

C D E A

Θ Κ Λ

C F G B

H K L

∆ύο γ¦ρ ¢ριθµîν τîν Α, Β κሠµονάδος τÁς Γ µεταξύ κατ¦ τÕ συνεχς ¢νάλογον ™µπιπτέτωσαν ¢ριθµοˆ ο† τε ∆, Ε καˆ οƒ Ζ, Η· λέγω, Óτι Óσοι ˜κατέρου τîν Α, Β κሠµονάδος τÁς Γ µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεπτώκασιν ¢ριθµοί, τοσοàτοι κሠε„ς τοÝς Α, Β µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπεσοàνται. `Ο ∆ γ¦ρ τÕν Ζ πολλαπλασιάσας τÕν Θ ποιείτω, ˜κάτερος δ τîν ∆, Ζ τÕν Θ πολλαπλασιάσας ˜κάτερον τîν Κ, Λ ποιείτω. Κሠ™πεί ™στιν æς ¹ Γ µον¦ς πρÕς τÕν ∆ ¢ριθµόν, οÛτως Ð ∆ πρÕς τÕν Ε, „σάκις ¥ρα ¹ Γ µον¦ς τÕν ∆ ¢ριθµÕν µετρε‹ καˆ Ð ∆ τÕν Ε. ¹ δ Γ µον¦ς τÕν ∆ ¢ριθµÕν µετρε‹ κατ¦ τ¦ς ™ν τù ∆ µονάδας· καˆ Ð ∆ ¥ρα ¢ριθµÕς τÕν Ε µετρε‹ κατ¦ τ¦ς ™ν τù ∆ µονάδας· Ð ∆ ¥ρα ˜αυτÕν πολλαπλασιάσας τÕν Ε πεποίηκεν. πάλιν, ™πεί ™στιν æς ¹ Γ [µον¦ς] πρÕς τÕν ∆ ¢ριθµÕν, οÛτως Ð Ε πρÕς τÕν Α, „σάκις ¥ρα ¹ Γ µον¦ς τÕν ∆ ¢ριθµÕν µετρε‹ καˆ Ð Ε τÕν Α. ¹ δ Γ µον¦ς τÕν ∆ ¢ριθµÕν µετρε‹ κατ¦ τ¦ς ™ν τù ∆ µονάδας· καˆ Ð Ε ¥ρα τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù ∆ µονάδας· Ð ∆ ¥ρα τÕν Ε πολλαπλασιάσας τÕν Α πεποίηκεν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð µν Ζ ˜αυτÕν πολλαπλασιάσας τÕν Η πεποίηκεν, τÕν δ Η πολλαπλασιάσας τÕν Β πεποίηκεν. κሠ™πεˆ Ð ∆ ˜αυτÕν µν πολλαπλασιάσας τÕν Ε πεποίηκεν, τÕν δ Ζ πολλαπλασιάσας τÕν Θ πεποίηκεν, œστιν ¥ρα æς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Ε πρÕς τÕν Θ. δι¦ τ¦ αÙτ¦ δ¾ κሠæς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Θ πρÕς τÕν Η. κሠæς ¥ρα Ð Ε πρÕς τÕν Θ, οÛτως Ð Θ πρÕς τÕν Η. πάλιν, ™πεˆ Ð ∆ ˜κάτερον τîν Ε, Θ πολλαπλασιάσας ˜κάτερον τîν Α, Κ πεποίηκεν, œστιν ¥ρα æς Ð Ε πρÕς τÕν Θ, οÛτως Ð Α πρÕς τÕν Κ. ¢λλ' æς Ð Ε πρÕς τÕν Θ, οÛτως Ð ∆ πρÕς τÕν Ζ· κሠæς ¥ρα Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Α πρÕς τÕν Κ. πάλιν, ™πεˆ ˜κάτερος τîν ∆, Ζ τÕν Θ πολλαπλασιάσας ˜κάτερον τîν Κ, Λ πεποίηκεν, ›στιν ¥ρα æς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Κ πρÕς τÕν Λ. ¢λλ' æς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Α πρÕς τÕν Κ· κሠæς ¥ρα Ð Α πρÕς τÕν Κ, οÛτως Ð Κ πρÕς τÕν Λ. œτι ™πεˆ Ð Ζ ˜κάτερον τîν Θ, Η πολλαπλασιάσας ˜κάτερον τîν Λ, Β πεποίηκεν, œστιν ¥ρα æς Ð Θ πρÕς τÕν Η, οÛτως Ð Λ

For let the numbers D, E and F , G fall between the numbers A and B (respectively) and the unit C in continued proportion. I say that as many numbers as have fallen between each of A and B and the unit C in continued proportion, so many will also fall in between A and B in continued proportion. For let D make H (by) multiplying F . And let D, F make K, L, respectively, by multiplying H. As since as the unit C is to the number D, so D (is) to E, the unit C thus measures the number D as many times as D (measures) E [Def. 7.20]. And the unit C measures the number D according to the units in D. Thus, the number D also measures E according to the units in D. Thus, D has made E (by) multiplying itself. Again, since as the [unit] C is to the number D, so E (is) to A, the unit C thus measures the number D as many times as E (measures) A [Def. 7.20]. And the unit C measures the number D according to the units in D. Thus, E also measures A according to the units in D. Thus, D has made A (by) multiplying E. And so, for the same (reasons), F has made G (by) multiplying itself, and has made B (by) multiplying G. And since D has made E (by) multiplying itself, and has made H (by) multiplying F , thus as D is to F , so E (is) to H [Prop 7.17]. And so, for the same reasons, as D (is) to F , so H (is) to G [Prop. 7.18]. And thus as E (is) to H, so H (is) to G. Again, since D has made A, K (by) multiplying E, H, respectively, thus as E is to H, so A (is) to K [Prop 7.17]. But, as E (is) to H, so D (is) to F . And thus as D (is) to F , so A (is) to K. Again, since D, F have made K, L, respectively, (by) multiplying H, thus as D is to F , so K (is) to L [Prop. 7.18]. But, as D (is) to F , so A (is) to K. And thus as A (is) to K, so K (is) to L. Further, since F has made L, B (by) multiplying H, G, respectively, thus as H is to G, so L (is) to B [Prop 7.17]. And as H (is) to G, so D (is) to F . And thus as D (is) to F , so L (is) to B. And it was also shown that as D (is) to F , so A (is) to K, and K to L. And thus as A (is) to K, so K (is) to L, and L to B. Thus, A, K, L, B are successively in continued proportion. Thus, as

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πρÕς τÕν Β. æς δ Ð Θ πρÕς τÕν Η, οÛτως Ð ∆ πρÕς τÕν Ζ· κሠæς ¥ρα Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Λ πρÕς τÕν Β. ™δείχθη δ κሠæς Ð ∆ πρÕς τÕν Ζ, οÛτως Ó τε Α πρÕς τÕν Κ καˆ Ð Κ πρÕς τÕν Λ· κሠæς ¥ρα Ð Α πρÕς τÕν Κ, οÛτως Ð Κ πρÕς τÕν Λ καˆ Ð Λ πρÕς τÕν Β. οƒ Α, Κ, Λ, Β ¥ρα κατ¦ τÕ συνεχς ˜ξÁς ε„σιν ¢νάλογον. Óσοι ¥ρα ˜κατέρου τîν Α, Β κሠτÁς Γ µονάδος µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπίπτουσιν ¢ριθµοί, τοσοàτοι κሠε„ς τοÝς Α, Β µεταξÝ κατ¦ τÕ συνεχς ™µπεσοàνται· Óπερ œδει δε‹ξαι.

many numbers as fall between each of A and B and the unit C in continued proportion, so many will also fall in between A and B in continued proportion. (Which is) the very thing it was required to show.

ια΄.

Proposition 11

∆ύο τετραγώνων ¢ριθµîν εŒς µέσος ¢νάλογόν ™στιν There exists one number in mean proportion to two ¢ριθµός, καˆ Ð τετράγωνος πρÕς τÕν τετράγωνον δι- (given) square numbers.† And (one) square (number) πλασίονα λόγον œχει ½περ ¹ πλευρ¦ πρÕς τ¾ν πλευράν. has to the (other) square (number) a squared‡ ratio with respect to (that) the side (of the former has) to the side (of the latter).

Α Β Γ Ε

A B C E

∆

”Εστωσαν τετράγωνοι ¢ριθµοˆ οƒ Α, Β, κሠτοà µν Α πλευρ¦ œστω Ð Γ, τοà δ Β Ð ∆· λέγω, Óτι τîν Α, Β εŒς µέσος ¢νάλογόν ™στιν ¢ριθµός, καˆ Ð Α πρÕς τÕν Β διπλασίονα λόγον œχει ½περ Ð Γ πρÕς τÕν ∆. `Ο Γ γ¦ρ τÕν ∆ πολλαπλασιάσας τÕν Ε ποιείτω. κሠ™πεˆ τετράγωνός ™στιν Ð Α, πλευρ¦ δ αÙτοà ™στιν Ð Γ, Ð Γ ¥ρα ˜αυτÕν πολλαπλασιάσας τÕν Α πεποίηκεν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð ∆ ˜αυτÕν πολλαπλασιάσας τÕν Β πεποίηκεν. ™πεˆ οâν Ð Γ ˜κάτερον τîν Γ, ∆ πολλαπλασιάσας ˜κάτερον τîν Α, Ε πεποίηκεν, œστιν ¥ρα æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Α πρÕς τÕν Ε. δι¦ τ¦ αÙτ¦ δ¾ κሠæς Ð Γ πρÕς τÕν ∆, οÛτως Ð Ε πρÕς τÕν Β. κሠæς ¥ρα Ð Α πρÕς τÕν Ε, οÛτως Ð Ε πρÕς τÕν Β. τîν Α, Β ¥ρα εŒς µέσος ¢νάλογόν ™στιν ¢ριθµός. Λέγω δή, Óτι καˆ Ð Α πρÕς τÕν Β διπλασίονα λόγον œχει ½περ Ð Γ πρÕς τÕν ∆. ™πεˆ γ¦ρ τρε‹ς ¢ριθµοˆ ¢νάλογόν ε„σιν οƒ Α, Ε, Β, Ð Α ¥ρα πρÕς τÕν Β διπλασίονα λόγον œχει ½περ Ð Α πρÕς τÕν Ε. æς δ Ð Α πρÕς τÕν Ε, οÛτως Ð Γ πρÕς τÕν ∆. Ð Α ¥ρα πρÕς τÕν Β διπλασίονα λόγον œχει ½περ ¹ Γ πλευρ¦ πρÕς τ¾ν ∆· Óπερ œδει δε‹ξαι.

D

Let A and B be square numbers, and let C be the side of A, and D (the side) of B. I say that there exists one number in mean proportion to A and B, and that A has to B a squared ratio with respect to (that) C (has) to D. For let C make E (by) multiplying D. And since A is square, and C is its side, C has thus made A (by) multiplying itself. And so, for the same (reasons), D has made B (by) multiplying itself. Therefore, since C has made A, E (by) multiplying C, D, respectively, thus as C is to D, so A (is) to E [Prop. 7.17]. And so, for the same (reasons), as C (is) to D, so E (is) to B [Prop. 7.18]. And thus as A (is) to E, so E (is) to B. Thus, one number (namely, E) is in mean proportion to A and B. So I say that A also has to B a squared ratio with respect to (that) C (has) to D. For since A, E, B are three (continuously) proportional numbers, A thus has to B a squared ratio with respect to (that) A (has) to E [Def. 5.9]. And as A (is) to E, so C (is) to D. Thus, A has to B a squared ratio with respect to (that) side C (has) to (side) D. (Which is) the very thing it was required to show.

†

In other words, between two given square numbers there exists a number in continued proportion.

‡

Literally, “double”.

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ELEMENTS BOOK 8 ιβ΄.

Proposition 12

∆ύο κύβων ¢ριθµîν δύο µέσοι ¢νάλογόν ε„σιν There exist two numbers in mean proportion to two ¢ριθµοί, καˆ Ð κύβος πρÕς τÕν κύβον τριπλασίονα λόγον (given) cube numbers.† And (one) cube (number) has to œχει ½περ ¹ πλευρ¦ πρÕς τ¾ν πλευράν. the (other) cube (number) a cubed‡ ratio with respect to (that) the side (of the former has) to the side (of the latter).

Α Β Γ ∆

Ε Ζ Η Θ Κ

A B C D

”Εστωσαν κύβοι ¢ριθµοˆ οƒ Α, Β κሠτοà µν Α πλευρ¦ œστω Ð Γ, τοà δ Β Ð ∆· λέγω, Óτι τîν Α, Β δύο µέσοι ¢νάλογόν ε„σιν ¢ριθµοί, καˆ Ð Α πρÕς τÕν Β τριπλασίονα λόγον œχει ½περ Ð Γ πρÕς τÕν ∆. `Ο γ¦ρ Γ ˜αυτÕν µν πολλαπλασιάσας τÕν Ε ποιείτω, τÕν δ ∆ πολλαπλασιάσας τÕν Ζ ποιείτω, Ð δ ∆ ˜αυτÕν πολλαπλασιάσας τÕν Η ποιείτω, ˜κάτερος δ τîν Γ, ∆ τÕν Ζ πολλαπλασιάσας ˜κάτερον τîν Θ, Κ ποιείτω. Κሠ™πεˆ κύβος ™στˆν Ð Α, πλευρ¦ δ αÙτοà Ð Γ, καˆ Ð Γ ˜αυτÕν µν πολλαπλασιάσας τÕν Ε πεποίηκεν, Ð Γ ¥ρα ˜αυτÕν µν πολλαπλασιάσας τÕν Ε πεποίηκεν, τÕν δ Ε πολλαπλασιάσας τÕν Α πεποίηκεν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð ∆ ˜αυτÕν µν πολλαπλασιάσας τÕν Η πεποίηκεν, τÕν δ Η πολλαπλασιάσας τÕν Β πεποίηκεν. κሠ™πεˆ Ð Γ ˜κάτερον τîν Γ, ∆ πολλαπλασιάσας ˜κάτερον τîν Ε, Ζ πεποίηκεν, œστιν ¥ρα æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Ε πρÕς τÕν Ζ. δι¦ τ¦ αÙτ¦ δ¾ κሠæς Ð Γ πρÕς τÕν ∆, οÛτως Ð Ζ πρÕς τÕν Η. πάλιν, ™πεˆ Ð Γ ˜κάτερον τîν Ε, Ζ πολλαπλασιάσας ˜κάτερον τîν Α, Θ πεποίηκεν, œστιν ¥ρα æς Ð Ε πρÕς τÕν Ζ, οÛτως Ð Α πρÕς τÕν Θ. æς δ Ð Ε πρÕς τÕν Ζ, οÛτως Ð Γ πρÕς τÕν ∆· κሠæς ¥ρα Ð Γ πρÕς τÕν ∆, οÛτως Ð Α πρÕς τÕν Θ. πάλιν, ™πεˆ ˜κάτερος τîν Γ, ∆ τÕν Ζ πολλαπλασιάσας ˜κάτερον τîν Θ, Κ πεποίηκεν, œστιν ¥ρα æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Θ πρÕς τÕν Κ. πάλιν, ™πεˆ Ð ∆ ˜κάτερον τîν Ζ, Η πολλαπλασιάσας ˜κάτερον τîν Κ, Β πεποίηκεν, œστιν ¥ρα æς Ð Ζ πρÕς τÕν Η, οÛτως Ð Κ πρÕς τÕν Β. æς δ Ð Ζ πρÕς τÕν Η, οÛτως Ð Γ πρÕς τÕν ∆· κሠæς ¥ρα Ð Γ πρÕς τÕν ∆, οÛτως Ó τε Α πρÕς τÕν Θ καˆ Ð Θ πρÕς τÕν Κ καˆ Ð Κ πρÕς τÕν Β. τîν Α, Β ¥ρα δύο µέσοι ¢νάλογόν ε„σιν οƒ Θ, Κ. Λέγω δή, Óτι καˆ Ð Α πρÕς τÕν Β τριπλασίονα λόγον œχει ½περ Ð Γ πρÕς τÕν ∆. ™πεˆ γ¦ρ τέσσαρες ¢ριθµοˆ ¢νάλογόν ε„σιν οƒ Α, Θ, Κ, Β, Ð Α ¥ρα πρÕς τÕν Β τριπλασίονα λόγον œχει ½περ Ð Α πρÕς τÕν Θ. æς δ Ð Α πρÕς τÕν Θ, οÛτως Ð Γ πρÕς τÕν ∆· καˆ Ð Α [¥ρα] πρÕς τÕν Β τριπλασίονα λόγον œχει ½περ Ð Γ πρÕς τÕν ∆· Óπερ œδει δε‹ξαι.

E F G H K

Let A and B be cube numbers, and let C be the side of A, and D (the side) of B. I say that there exist two numbers in mean proportion to A and B, and that A has to B a cubed ratio with respect to (that) C (has) to D. For let C make E (by) multiplying itself, and let it make F (by) multiplying D. And let D make G (by) multiplying itself, and let C, D make H, K, respectively, (by) multiplying F . And since A is cube, and C (is) its side, and C has made E (by) multiplying itself, C has thus made E (by) multiplying itself, and has made A (by) multiplying E. And so, for the same (reasons), D has made G (by) multiplying itself, and has made B (by) multiplying G. And since C has made E, F (by) multiplying C, D, respectively, thus as C is to D, so E (is) to F [Prop. 7.17]. And so, for the same (reasons), as C (is) to D, so F (is) to G [Prop. 7.18]. Again, since C has made A, H (by) multiplying E, F , respectively, thus as E is to F , so A (is) to H [Prop. 7.17]. And as E (is) to F , so C (is) to D. And thus as C (is) to D, so A (is) to H. Again, since C, D have made H, K, respectively, (by) multiplying F , thus as C is to D, so H (is) to K [Prop. 7.18]. Again, since D has made K, B (by) multiplying F , G, respectively, thus as F is to G, so K (is) to B [Prop. 7.17]. And as F (is) to G, so C (is) to D. And thus as C (is) to D, so A (is) to H, and H to K, and K to B. Thus, H and K are two (numbers) in mean proportion to A and B. So I say that A also has to B a cubed ratio with respect to (that) C (has) to D. For since A, H, K, B are four (continuously) proportional numbers, A thus has to B a cubed ratio with respect to (that) A (has) to H [Def. 5.10]. And as A (is) to H, so C (is) to D. And [thus] A has to B a cubed ratio with respect to (that) C (has) to D. (Which is) the very thing it was required to show.

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†

In other words, between two given cube numbers there exist two numbers in continued proportion.

‡

Literally, “triple”.

ιγ΄.

Proposition 13

'Ε¦ν ðσιν Ðσοιδηποτοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον, κሠπολλαπλασιάσας ›καστος ˜αυτÕν ποιÍ τινα, οƒ γενόµενοι ™ξ αÙτîν ¢νάλογον œσονται· κሠ™¦ν οƒ ™ξ ¢ρχÁς τοÝς γενοµένους πολλαπλασιάσαντες ποιîσί τινας, κሠαÙτοˆ ¢νάλογον œσονται [κሠ¢εˆ περˆ τοÝς ¥κρους τοàτο συµβαίνει]. ”Εστωσαν Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον, οƒ Α, Β, Γ, æς Ð Α πρÕς τÕν Β, οÛτως Ð Β πρÕς τÕν Γ, καˆ οƒ Α, Β, Γ ˜αυτοÝς µν πολλαπλασιάσαντες τοÝς ∆, Ε, Ζ ποιείτωσαν, τοÝς δ ∆, Ε, Ζ πολλαπλασιάσαντες τοÝς Η, Θ, Κ ποιείτωσαν· λέγω, Óτι ο† τε ∆, Ε, Ζ καˆ οƒ Η, Θ, Κ ˜ξÁς ¢νάλογον ε„σιν.

If there are any multitude whatsoever of continuously proportional numbers, and each makes some (number by) multiplying itself, then the (numbers) created from them will (also) be (continuously) proportional. And if the original (numbers) make some (more numbers by) multiplying the created (numbers) then these will also be (continuously) proportional [and this always happens with the extremes]. Let A, B, C be any multitude whatsoever of continuously proportional numbers, (such that) as A (is) to B, so B (is) to C. And let A, B, C make D, E, F (by) multiplying themselves, and let them make G, H, K (by) multiplying D, E, F . I say that D, E, F and G, H, K are continuously proportional.

Α

Λ

Β

Ξ

Γ

Μ

A B C

Ν

D E F

∆ Ο Ε Π Ζ

L O M N P Q

G H K

Η Θ Κ `Ο µν γ¦ρ Α τÕν Β πολλαπλασιάσας τÕν Λ ποιείτω, ˜κάτερος δ τîν Α, Β τÕν Λ πολλαπλασιάσας ˜κάτερον τîν Μ, Ν ποιείτω. κሠπάλιν Ð µν Β τÕν Γ πολλαπλασιάσας τÕν Ξ ποιείτω, ˜κάτερος δ τîν Β, Γ τÕν Ξ πολλαπλασιάσας ˜κάτερον τîν Ο, Π ποιείτω. `Οµοίως δ¾ το‹ς ™πάνω δε‹ξοµεν, Óτι οƒ ∆, Λ, Ε καˆ οƒ Η, Μ, Ν, Θ ˜ξÁς ε„σιν ¢νάλογον ™ν τù τοà Α πρÕς τÕν Β λόγJ, κሠœτι οƒ Ε, Ξ, Ζ καˆ οƒ Θ, Ο, Π, Κ ˜ξÁς ε„σιν ¢νάλογον ™ν τù τοà Β πρÕς τÕν Γ λόγJ. καί ™στιν æς Ð Α πρÕς τÕν Β, οÛτως Ð Β πρÕς τÕν Γ· καˆ οƒ ∆, Λ, Ε ¥ρα το‹ς Ε, Ξ, Ζ ™ν τù αÙτù λόγJ ε„σˆ κሠœτι οƒ Η, Μ, Ν, Θ το‹ς Θ, Ο, Π, Κ. καί ™στιν ‡σον τÕ µν τîν ∆, Λ, Ε πλÁθος τù τîν Ε, Ξ, Ζ πλήθει, τÕ δ τîν Η, Μ, Ν, Θ τù τîν Θ, Ο, Π, Κ· δι' ‡σου ¥ρα ™στˆν æς µν Ð ∆ πρÕς τÕν Ε, οÛτως Ð Ε πρÕς τÕν Ζ, æς δ Ð Η πρÕς τÕν Θ, οÛτως Ð Θ πρÕς τÕν Κ· Óπερ œδει δε‹ξαι.

For let A make L (by) multiplying B. And let A, B make M , N , respectively, (by) multiplying L. And, again, let B make O (by) multiplying C. And let B, C make P , Q, respectively, (by) multplying O. So, similarly to the above, we can show that D, L, E and G, M , N , H are continuously proportional in the ratio of A to B, and, further, (that) E, O, F and H, P , Q, K are continuously proportional in the ratio of B to C. And as A is to B, so B (is) to C. And thus D, L, E are in the same ratio as E, O, F , and, further, G, M , N , H (are in the same ratio) as H, P , Q, K. And the multitude of D, L, E is equal to the multitude of E, O, F , and that of G, M , N , H to that of H, P , Q, K. Thus, via equality, as D is to E, so E (is) to F , and as G (is) to H, so H (is) to K [Prop. 7.14]. (Which is) the very thing it was required to show.

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ιδ΄.

Proposition 14

'Ε¦ν τετράγωνος τετράγωνον µετρÍ, κሠ¹ πλευρ¦ If a square (number) measures a(nother) square τ¾ν πλευρ¦ν µετρήσει· κሠ™¦ν ¹ πλευρ¦ τ¾ν πλευρ¦ν (number) then the side (of the former) will also meaµετρÍ, καˆ Ð τετράγωνος τÕν τετράγωνον µετρήσει. sure the side (of the latter). And if the side (of a square number) measures the side (of another square number) then the (former) square (number) will also measure the (latter) square (number).

Α Β

Γ ∆

A B

Ε

C D

E

”Εστωσαν τετράγωνοι ¢ριθµοˆ οƒ Α, Β, πλευραˆ δ αÙτîν œστωσαν οƒ Γ, ∆, Ð δ Α τÕν Β µετρείτω· λέγω, Óτι καˆ Ð Γ τÕν ∆ µετρε‹. `Ο Γ γ¦ρ τÕν ∆ πολλαπλασιάσας τÕν Ε ποιείτω· οƒ Α, Ε, Β ¥ρα ˜ξÁς ¢νάλογόν ε„σιν ™ν τù τοà Γ πρÕς τÕν ∆ λόγJ. κሠ™πεˆ οƒ Α, Ε, Β ™ξÁς ¢νάλογόν ε„σιν, κሠµετρε‹ Ð Α τÕν Β, µετρε‹ ¥ρα καˆ Ð Α τÕν Ε. καί ™στιν æς Ð Α πρÕς τÕν Ε, οÛτως Ð Γ πρÕς τÕν ∆· µετρε‹ ¥ρα καˆ Ð Γ τÕν ∆. Πάλιν δ¾ Ð Γ τÕν ∆ µετρείτω· λέγω, Óτι καˆ Ð Α τÕν Β µετρε‹. Τîν γ¦ρ αÙτîν κατασκευασθέντων еοίως δείξοµεν, Óτι οƒ Α, Ε, Β ˜ξÁς ¢νάλογόν ε„σιν ™ν τù τοà Γ πρÕς τÕν ∆ λόγJ. κሠ™πεί ™στιν æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Α πρÕς τÕν Ε, µετρε‹ δ Ð Γ τÕν ∆, µετρε‹ ¥ρα καˆ Ð Α τÕν Ε. καί ε„σιν οƒ Α, Ε, Β ˜ξÁς ¢νάλογον· µετρε‹ ¥ρα καˆ Ð Α τÕν Β. 'Ε¦ν ¥ρα τετράγωνος τετράγωνον µετρÍ, κሠ¹ πλευρ¦ τ¾ν πλευρ¦ν µετρήσει· κሠ™¦ν ¹ πλευρ¦ τ¾ν πλευρ¦ν µετρÍ, καˆ Ð τετράγωνος τÕν τετράγωνον µετρήσει· Óπερ œδει δε‹ξαι.

Let A and B be square numbers, and let C and D be their sides (respectively). And let A measure B. I say that C also measures D. For let C make E (by) multiplying D. Thus, A, E, B are continuously proportional in the ratio of C to D [Prop. 8.11]. And since A, E, B are continuously proportional, and A measures B, A thus also measures E [Prop. 8.7]. And as A is to E, so C (is) to D. Thus, C also measures D [Def. 7.20]. So, again, let C measure D. I say that A also measures B. For similarly, with the same construction, we can show that A, E, B are continuously proportional in the ratio of C to D. And since as C is to D, so A (is) to E, and C measures D, A thus also measures E [Def. 7.20]. And A, E, B are continuously proportional. Thus, A also measures B. Thus, if a square (number) measures a(nother) square (number) then the side (of the former) will also measure the side (of the latter). And if the side (of a square number) measures the side (of another square number) then the (former) square (number) will also measure the (latter) square (number). (Which is) the very thing it was required to show.

ιε΄.

Proposition 15

'Ε¦ν κύβος ¢ριθµÕς κύβον ¢ριθµÕν µετρÍ, κሠ¹ πλευρ¦ τ¾ν πλευρ¦ν µετρήσει· κሠ™¦ν ¹ πλευρ¦ τ¾ν πλευρ¦ν µετρÍ, καˆ Ð κύβος τÕν κύβον µετρήσει. Κύβος γ¦ρ ¢ριθµÕς Ð Α κύβον τÕν Β µετρείτω, κሠτοà µν Α πλευρ¦ œστω Ð Γ, τοà δ Β Ð ∆· λέγω, Óτι Ð Γ τÕν ∆ µετρε‹.

If a cube number measures a(nother) cube number then the side (of the former) will also measure the side (of the latter). And if the side (of a cube number) measures the side (of another cube number) then the (former) cube (number) will also measure the (latter) cube (number). For let the cube number A measure the cube (number) B, and let C be the side of A, and D (the side) of B.

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ELEMENTS BOOK 8 I say that C measures D.

Α Β

Γ ∆

Ε Η

C D

A B

Θ Κ

H K

E G

Ζ

F

`Ο Γ γ¦ρ ˜αυτÕν πολλαπλασιάσας τÕν Ε ποιείτω, Ð δ ∆ ˜αυτÕν πολλαπλασιάσας τÕν Η ποιείτω, κሠœτι Ð Γ τÕν ∆ πολλαπλασιάσας τÕν Ζ [ποιείτω], ˜κάτερος δ τîν Γ, ∆ τÕν Ζ πολλαπλασιάσας ˜κάτερον τîν Θ, Κ ποιείτω. φανερÕν δή, Óτι οƒ Ε, Ζ, Η καˆ οƒ Α, Θ, Κ, Β ˜ξÁς ¢νάλογόν ε„σιν ™ν τù τοà Γ πρÕς τÕν ∆ λόγJ. κሠ™πεˆ οƒ Α, Θ, Κ, Β ˜ξÁς ¢νάλογόν ε„σιν, κሠµετρε‹ Ð Α τÕν Β, µετρε‹ ¥ρα κሠτÕν Θ. καί ™στιν æς Ð Α πρÕς τÕν Θ, οÛτως Ð Γ πρÕς τÕν ∆· µετρε‹ ¥ρα καˆ Ð Γ τÕν ∆. 'Αλλ¦ δ¾ µετρείτω Ð Γ τÕν ∆· λέγω, Óτι καˆ Ð Α τÕν Β µετρήσει. Τîν γ¦ρ αÙτîν κατασκευασθέντων еοίως δ¾ δείξοµεν, Óτι οƒ Α, Θ, Κ, Β ˜ξÁς ¢νάλογόν ε„σιν ™ν τù τοà Γ πρÕς τÕν ∆ λόγJ. κሠ™πεˆ Ð Γ τÕν ∆ µετρε‹, καί ™στιν æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Α πρÕς τÕν Θ, καˆ Ð Α ¥ρα τÕν Θ µετρε‹· éστε κሠτÕν Β µετρε‹ Ð Α· Óπερ œδει δε‹ξαι.

For let C make E (by) multiplying itself. And let D make G (by) multiplying itself. And, further, [let] C [make] F (by) multiplying D, and let C, D make H, K, respectively, (by) multiplying F . So it is clear that E, F , G and A, H, K, B are continuously proportional in the ratio of C to D [Prop. 8.12]. And since A, H, K, B are continuously proportional, and A measures B, (A) thus also measures H [Prop. 8.7]. And as A is to H, so C (is) to D. Thus, C also measures D [Def. 7.20]. And so let C measure D. I say that A will also measure B. For similarly, with the same construction, we can show that A, H, K, B are continuously proportional in the ratio of C to D. And since C measures D, and as C is to D, so A (is) to H, A thus also measures H [Def. 7.20]. Hence, A also measures B. (Which is) the very thing it was required to show.

ι$΄.

Proposition 16

'Ε¦ν τετράγωνος ¢ριθµÕς τετράγωνον ¢ριθµÕν µ¾ If a square number does not measure a(nother) µετρÍ, οÙδ ¹ πλευρ¦ τ¾ν πλευρ¦ν µετρήσει· κ¨ν ¹ square number then the side (of the former) will not πλευρ¦ τ¾ν πλευρ¦ν µ¾ µετρÍ, οÙδ Ð τετράγωνος τÕν measure the side (of the latter) either. And if the side (of τετράγωνον µετρήσει. a square number) does not measure the side (of another square number) then the (former) square (number) will not measure the (latter) square (number) either.

Α Β

Γ ∆

.

”Εστωσαν τετρ¦γωνοι ¢ριθµοˆ οƒ Α, Β, πλευραˆ δ αÙτîν œστωσαν οƒ Γ, ∆, κሠµ¾ µετρείτω Ð Α τÕν Β· λγω, Óτι οÙδ Ð Γ τÕν ∆ µετρε‹. Ε„ γ¦ρ µετρε‹ Ð Γ τÕν ∆, µετρήσει καˆ Ð Α τÕν Β. οÙ µετρε‹ δ Ð Α τÕν Β· οÙδ ¥ρα Ð Γ τÕν ∆ µετρήσει. Μ¾ µετρείτω [δ¾] πάλιν Ð Γ τÕν ∆· λέγω, Óτι οÙδ Ð Α τÕν Β µετρήσει.

A B

C D

Let A and B be square numbers, and let C and D be their sides (respectively). And let A not measure B. I say that C does not measure D either. For if C measures D then A will also measure B [Prop. 8.14]. And A does not measure B. Thus, C will not measure D either. [So], again, let C not measure D. I say that A will not measure B either. Ε„ γ¦ρ µετρε‹ Ð Α τÕν Β, µετρήσει καˆ Ð Γ τÕν ∆. οÙ For if A measures B then C will also measure D µετρε‹ δ Ð Γ τÕν ∆· οÙδ' ¥ρα Ð Α τÕν Β µετρήσει· [Prop. 8.14]. And C does not measure D. Thus, A will

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ELEMENTS BOOK 8

Óπερ œδει δε‹ξαι.

not measure B either. (Which is) the very thing it was required to show.

ιζ΄.

Proposition 17

'Ε¦ν κύβος ¢ριθµÕς κύβον ¢ριθµÕν µ¾ µετρÍ, οÙδ ¹ πλευρ¦ τ¾ν πλευρ¦ν µετρήσει· κ¨ν ¹ πλευρ¦ τ¾ν πλευρ¦ν µ¾ µετρÍ, οÙδ Ð κύβος τÕν κύβον µετρήσει.

If a cube number does not measure a(nother) cube number then the side (of the former) will not measure the side (of the latter) either. And if the side (of a cube number) does not measure the side (of another cube number) then the (former) cube (number) will not measure the (latter) cube (number) either.

Α Β

Γ ∆

.

A B

C D

Κύβος γ¦ρ ¢ριθµÕς Ð Α κύβον ¢ριθµÕν τÕν Β µ¾ µετρείτω, κሠτοà µν Α πλευρ¦ œστω Ð Γ, τοà δ Β Ð ∆· λέγω, Óτι Ð Γ τÕν ∆ οÙ µετρήσει. Ε„ γ¦ρ µετρε‹ Ð Γ τÕν ∆, καˆ Ð Α τÕν Β µετρήσει. οÙ µετρε‹ δ Ð Α τÕν Β· οÙδ' ¥ρα Ð Γ τÕν ∆ µετρε‹. 'Αλλ¦ δ¾ µ¾ µετρείτω Ð Γ τÕν ∆· λέγω, Óτι οÙδ Ð Α τÕν Β µετρήσει. Ε„ γ¦ρ Ð Α τÕν Β µετρε‹, καˆ Ð Γ τÕν ∆ µετρήσει. οÙ µετρε‹ δ Ð Γ τÕν ∆· οÙδ' ¥ρα Ð Α τÕν Β µετρήσει· Óπερ œδει δε‹ξαι.

For let the cube number A not measure the cube number B. And let C be the side of A, and D (the side) of B. I say that C will not measure D. For if C measures D then A will also measure B [Prop. 8.15]. And A does not measure B. Thus, C does not measure D either. And so let C not measure D. I say that A will not measure B either. For if A measures B then C will also measure D [Prop. 8.15]. And C does not measure D. Thus, A will not measure B either. (Which is) the very thing it was required to show.

ιη΄.

Proposition 18

∆ύο еοίων ™πιπέδων ¢ριθµîν εŒς µέσος ¢νάλογόν ™στιν ¢ριθµός· καˆ Ð ™πίπεδος πρÕς τÕν ™πίπεδον διπλασίονα λόγον œχει ½περ ¹ еόλογος πλευρ¦ πρÕς τ¾ν еόλογον πλευράν.

There exists one number in mean proportion to two similar plane numbers. And (one) plane (number) has to the (other) plane (number) a squared ratio with respect to (that) a corresponding side (of the former has) to a corresponding side (of the latter).

Α Γ ∆

Β Ε Ζ

A C D

Η

B E F

G

”Εστωσαν δύο Óµοιοι ™πίπεδοι ¢ριθµοˆ οƒ Α, Β, κሠτοà µν Α πλευρሠœστωσαν οƒ Γ, ∆ ¢ριθµοί, τοà δ Β οƒ Ε, Ζ. κሠ™πεˆ Óµοιοι ™πίπεδοί ε„σιν οƒ ¢νάλογον œχοντες τ¦ς πλευράς, œστιν ¥ρα æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Ε πρÕς τÕν Ζ. λέγω οâν, Óτι τîν Α, Β εŒς µέσος ¢νάλογόν ™στιν ¢ριθµός, καˆ Ð Α πρÕς τÕν Β διπλασίονα λόγον œχει ½περ Ð Γ πρÕς τÕν Ε À Ð ∆ πρÕς τÕν Ζ, τουτέστιν ½περ ¹ еόλογος πλευρ¦ πρÕς τ¾ν еόλογον [πλευράν]. Κሠ™πεί ™στιν æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Ε πρÕς τÕν Ζ, ™ναλλ¦ξ ¥ρα ™στˆν æς Ð Γ πρÕς τÕν Ε, Ð ∆ πρÕς τÕν Ζ. κሠ™πεˆ ™πίπεδός ™στιν Ð Α, πλευραˆ δ αÙτοà οƒ

Let A and B be two similar plane numbers. And let the numbers C, D be the sides of A, and E, F (the sides) of B. And since similar numbers are those having proportional sides [Def. 7.21], thus as C is to D, so E (is) to F . Therefore, I say that there exists one number in mean proportion to A and B, and that A has to B a squared ratio with respect to that C (has) to E, or D to F —that is to say, with respect to (that) a corresponding side (has) to a corresponding [side]. For since as C is to D, so E (is) to F , thus, alternately, as C is to E, so D (is) to F [Prop. 7.13]. And since A is

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Γ, ∆, Ð ∆ ¥ρα τÕν Γ πολλαπλασιάσας τÕν Α πεποίηκεν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Ε τÕν Ζ πολλαπλασιάσας τÕν Β πεποίηκεν. Ð ∆ δ¾ τÕν Ε πολλαπλασιάσας τÕν Η ποιείτω. κሠ™πεˆ Ð ∆ τÕν µν Γ πολλαπλασιάσας τÕν Α πεποίηκεν, τÕν δ Ε πολλαπλασιάσας τÕν Η πεποίηκεν, œστιν ¥ρα æς Ð Γ πρÕς τÕν Ε, οÛτως Ð Α πρÕς τÕν Η. ¢λλ' æς Ð Γ πρÕς τÕν Ε, [οÛτως] Ð ∆ πρÕς τÕν Ζ· κሠæς ¥ρα Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Α πρÕς τÕν Η. πάλιν, ™πεˆ Ð Ε τÕν µν ∆ πολλαπλασιάσας τÕν Η πεποίηκεν, τÕν δ Ζ πολλαπλασιάσας τÕν Β πεποίηκεν, œστιν ¥ρα æς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Η πρÕς τÕν Β. ™δείχθη δ κሠæς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Α πρÕς τÕν Η· κሠæς ¥ρα Ð Α πρÕς τÕν Η, οÛτως Ð Η πρÕς τÕν Β. οƒ Α, Η, Β ¥ρα ˜ξÁς ¢νάλογόν ε„σιν. τîν Α, Β ¥ρα εŒς µέσος ¢νάλογόν ™στιν ¢ριθµός. Λέγω δή, Óτι καˆ Ð Α πρÕς τÕν Β διπλασίονα λόγον œχει ½περ ¹ еόλογος πλευρ¦ πρÕς τ¾ν еόλογον πλευράν, τουτέστιν ½περ Ð Γ πρÕς τÕν Ε À Ð ∆ πρÕς τÕν Ζ. ™πεˆ γ¦ρ οƒ Α, Η, Β ˜ξÁς ¢νάλογόν ε„σιν, Ð Α πρÕς τÕν Β διπλασίονα λόγον œχει ½περ πρÕς τÕν Η. καί ™στιν æς Ð Α πρÕς τÕν Η, οÛτως Ó τε Γ πρÕς τÕν Ε καˆ Ð ∆ πρÕς τÕν Ζ. καˆ Ð Α ¥ρα πρÕς τÕν Β διπλασίονα λόγον œχει ½περ Ð Γ πρÕς τÕν Ε À Ð ∆ πρÕς τÕν Ζ· Óπερ œδει δε‹ξαι.

plane, and C, D its sides, D has thus made A (by) multiplying C. And so, for the same (reasons), E has made B (by) multiplying F . So let D make G (by) multiplying E. And since D has made A (by) multiplying C, and has made G (by) multiplying E, thus as C is to E, so A (is) to G [Prop. 7.17]. But as C (is) to E, [so] D (is) to F . And thus as D (is) to F , so A (is) to G. Again, since E has made G (by) multiplying D, and has made B (by) multiplying F , thus as D is to F , so G (is) to B [Prop. 7.17]. And it was also shown that as D (is) to F , so A (is) to G. And thus as A (is) to G, so G (is) to B. Thus, A, G, B are continously proportional. Thus, there exists one number (namely, G) in mean proportion to A and B. So I say that A also has to B a squared ratio with respect to (that) a corresponding side (has) to a corresponding side—that is to say, with respect to (that) C (has) to E, or D to F . For since A, G, B are continuously proportional, A has to B a squared ratio with respect to (that A has) to G [Prop. 5.9]. And as A is to G, so C (is) to E, and D to F . And thus A has to B a squared ratio with respect to (that) C (has) to E, or D to F . (Which is) the very thing it was required to show.

ιθ΄.

Proposition 19

∆ύο еοίων στερεîν ¢ριθµîν δύο µέσοι ¢νάλογον ™µπίπτουσιν ¢ριθµοί· καˆ Ð στερεÕς πρÕς τÕν Óµοιον στερεÕν τριπλασίονα λόγον œχει ½περ ¹ еόλογος πλευρ¦ πρÕς τ¾ν еόλογον πλευράν. Γ Α

Two numbers fall (between) two similar solid numbers in mean proportion. And a solid (number) has to a similar solid (number) a cubed† ratio with respect to (that) a corresponding side (has) to a corresponding side. A C

∆ Ε Β

Κ Μ

D E

Ζ Η Θ

B

Ν Λ Ξ ”Εστωσαν δύο Óµοιοι στερεοˆ οƒ Α, Β, κሠτοà µν Α πλευρሠœστωσαν οƒ Γ, ∆, Ε, τοà δ Β οƒ Ζ, Η, Θ. κሠ™πεˆ Óµοιοι στερεοί ε„σιν οƒ ¢νάλογον œχοντες τ¦ς πλευράς, œστιν ¥ρα æς µν Ð Γ πρÕς τÕν ∆, οÛτως Ð Ζ πρÕς τÕν Η, æς δ Ð ∆ πρÕς τÕν Ε, οÛτως Ð Η πρÕς τÕν Θ. λέγω, Óτι τîν Α, Β δύο µέσοι ¢νάλογόν ™µπίπτουσιν ¢ριθµοί, καˆ Ð Α πρÕς τÕν Β τριπλασίονα λόγον œχει ½περ Ð Γ πρÕς τÕν Ζ καˆ Ð ∆ πρÕς τÕν Η κሠœτι Ð Ε

K M L

F G H

N O Let A and B be two similar solid numbers, and let C, D, E be the sides of A, and F , G, H (the sides) of B. And since similar solid (numbers) are those having proportional sides [Def. 7.21], thus as C is to D, so F (is) to G, and as D (is) to E, so G (is) to H. I say that two numbers fall (between) A and B in mean proportion, and (that) A has to B a cubed ratio with respect to (that) C (has) to F , and D to G, and, further, E to H.

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ΣΤΟΙΧΕΙΩΝ η΄.

ELEMENTS BOOK 8

πρÕς τÕν Θ. `Ο Γ γ¦ρ τÕν ∆ πολλαπλασιάσας τÕν Κ ποιείτω, Ð δ Ζ τÕν Η πολλαπλασιάσας τÕν Λ ποιείτω. κሠ™πεˆ οƒ Γ, ∆ τοˆς Ζ, Η ™ν τù αÙτù λόγJ ε„σίν, κሠ™κ µν τîν Γ, ∆ ™στιν Ð Κ, ™κ δ τîν Ζ, Η Ð Λ, οƒ Κ, Λ [¥ρα] Óµοιοι ™πίπεδοί ε„σιν ¢ριθµοί· τîν Κ, Λ ¥ρα εŒς µέσος ¢νάλογόν ™στιν ¢ριθµός. œστω Ð Μ. Ð Μ ¥ρα ™στˆν Ð ™κ τîν ∆, Ζ, æς ™ν τù πρÕ τούτου θεωρήµατι ™δείχθη. κሠ™πεˆ Ð ∆ τÕν µν Γ πολλαπλασιάσας τÕν Κ πεποίηκεν, τÕν δ Ζ πολλαπλασιάσας τÕν Μ πεποίηκεν, œστιν ¥ρα æς Ð Γ πρÕς τÕν Ζ, οÛτως Ð Κ πρÕς τÕν Μ. ¢λλ' æς Ð Κ πρÕς τÕν Μ, Ð Μ πρÕς τÕν Λ. οƒ Κ, Μ, Λ ¥ρα ˜ξÁς ε„σιν ¢νάλογον ™ν τù τοà Γ πρÕς τÕν Ζ λόγù. κሠ™πεί ™στιν æς Ð Γ πρÕς τÕν ∆, οÛτως Ð Ζ πρÕς τÕν Η, ™ναλλ¦ξ ¥ρα ™στˆν æς Ð Γ πρÕς τÕν Ζ, οÛτως Ð ∆ πρÕς τÕν Η. δι¦ τ¦ αÙτ¦ δ¾ κሠæς Ð ∆ πρÕς τÕν Η, οÛτως Ð Ε πρÕς τÕν Θ. οƒ Κ, Μ, Λ ¥ρα ˜ξÁς ε„σιν ¢νάλογον œν τε τù τοà Γ πρÕς τÕν Ζ λόγJ κሠτù τοà ∆ πρÕς τÕν Η κሠœτι τù τοà Ε πρÕς τÕν Θ. ˜κατερος δ¾ τîν Ε, Θ τÕν Μ πολλαπλασιάσας ˜κάτερον τîν Ν, Ξ ποιείτω. κሠ™πεˆ στερεός ™στιν Ð Α, πλευραˆ δ αÙτοà ε„σιν οƒ Γ, ∆, Ε, Ð Ε ¥ρα τÕν ™κ τîν Γ, ∆ πολλαπλασιάσας τÕν Α πεποίηκεν. Ð δ ™κ τîν Γ, ∆ ™στιν Ð Κ· Ð Ε ¥ρα τÕν Κ πολλαπλασιάσας τÕν Α πεποίηκεν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Θ τÕν Λ πολλαπλασιάσας τÕν Β πεποίηκεν. κሠ™πεˆ Ð Ε τÕν Κ πολλαπλασιάσας τÕν Α πεποίηκεν, ¢λλ¦ µ¾ν κሠτÕν Μ πολλαπλασιάσας τÕν Ν πεποίηκεν, œστιν ¥ρα æς Ð Κ πρÕς τÕν Μ, οÛτως Ð Α πρÕς τÕν Ν. æς δ Ð Κ πρÕς τÕν Μ, οÛτως Ó τε Γ πρÕς τÕν Ζ καˆ Ð ∆ πρÕς τÕν Η κሠœτι Ð Ε πρÕς τÕν Θ· κሠæς ¥ρα Ð Γ πρÕς τÕν Ζ καˆ Ð ∆ πρÕς τÕν Η καˆ Ð Ε πρÕς τÕν Θ, οÛτως Ð Α πρÕς τÕν Ν. πάλιν, ™πεˆ ˜κάτερος τîν Ε, Θ τÕν Μ πολλαπλασιάσας ˜κάτερον τîν Ν, Ξ πεποίηκεν, œστιν ¥ρα æς Ð Ε πρÕς τÕν Θ, οÛτως Ð Ν πρÕς τÕν Ξ. ¢λλ' æς Ð Ε πρÕς τÕν Θ, οÛτως Ó τε Γ πρÕς τÕν Ζ καˆ Ð ∆ πρÕς τÕν Η· κሠæς ¥ρα Ð Γ πρÕς τÕν Ζ καˆ Ð ∆ πρÕς τÕν Η καˆ Ð Ε πρÕς τÕν Θ, οÛτως Ó τε Α πρÕς τÕν Ν καˆ Ð Ν πρÕς τÕν Ξ. πάλιν, ™πεˆ Ð Θ τÕν Μ πολλαπλασιάσας τÕν Ξ πεποίηκεν, ¢λλ¦ µ¾ν κሠτÕν Λ πολλαπλασιάσας τÕν Β πεποίηκεν, œστιν ¥ρα æς Ð Μ πρÕς τÕν Λ, οÛτως Ð Ξ πρÕς τÕν Β. ¢λλ' æς Ð Μ πρÕς τÕν Λ, οÛτως Ó τε Γ πρÕς τÕν Ζ καˆ Ð ∆ πρÕς τÕν Η καˆ Ð Ε πρÕς τÕν Θ. κሠæς ¥ρα Ð Γ πρÕς τÕν Ζ καˆ Ð ∆ πρÕς τÕν Η καˆ Ð Ε πρÕς τÕν Θ, οÛτως οÙ µόνον Ð Ξ πρÕς τÕν Β, ¢λλ¦ καˆ Ð Α πρÕς τÕν Ν καˆ Ð Ν πρÕς τÕν Ξ. οƒ Α, Ν, Ξ, Β ¥ρα ˜ξÁς ε„σιν ¢νάλογον ™ν το‹ς ε„ρηµένοις τîν πλευρîν λόγοις. Λέγω, Óτι καˆ Ð Α πρÕς τÕν Β τριπλασίονα λόγον œχει ½περ ¹ еόλογος πλευρ¦ πρÕς τ¾ν еόλογον πλευράν, τουτέστιν ½περ Ð Γ ¢ριθµÕς πρÕς τÕν Ζ À Ð ∆ πρÕς τÕν Η κሠœτι Ð Ε πρÕς τÕν Θ. ™πεˆ γ¦ρ

For let C make K (by) multiplying D, and let F make L (by) multiplying G. And since C, D are in the same ratio as F , G, and K is the (number created) from (multiplying) C, D, and L the (number created) from (multiplying) F , G, [thus] K and L are similar plane numbers [Def. 7.21]. Thus, there exits one number in mean proportion to K and L [Prop. 8.18]. Let it be M . Thus, M is the (number created) from (multiplying) D, F , as shown in the theorem before this (one). And since D has made K (by) multiplying C, and has made M (by) multiplying F , thus as C is to F , so K (is) to M [Prop. 7.17]. But, as K (is) to M , (so) M (is) to L. Thus, K, M , L are continuously proportional in the ratio of C to F . And since as C is to D, so F (is) to G, thus, alternately, as C is to F , so D (is) to G [Prop. 7.13]. And so, for the same (reasons), as D (is) to G, so E (is) to H. Thus, K, M , L are continuously proportional in the ratio of C to F , and of D to G, and, further, of E to H. So let E, H make N , O, respectively, (by) multiplying M . And since A is solid, and C, D, E are its sides, E has thus made A (by) multiplying the (number created) from (multiplying) C, D. And K is the (number created) from (multiplying) C, D. Thus, E has made A (by) multiplying K. And so, for the same (reasons), H has made B (by) multiplying L. And since E has made A (by) multiplying K, but has, in fact, also made N (by) multiplying M , thus as K is to M , so A (is) to N [Prop. 7.17]. And as K (is) to M , so C (is) to F , and D to G, and, further, E to H. And thus as C (is) to F , and D to G, and E to H, so A (is) to N . Again, since E, H have made N , O, respectively, (by) multiplying M , thus as E is to H, so N (is) to O [Prop. 7.18]. But, as E (is) to H, so C (is) to F , and D to G. And thus as C (is) to F , and D to G, and E to H, so (is) A to N , and N to O. Again, since H has made O (by) multiplying M , but has, in fact, also made B (by) multiplying L, thus as M (is) to L, so O (is) to B [Prop. 7.17]. But, as M (is) to L, so C (is) to F , and D to G, and E to H. And thus as C (is) to F , and D to G, and E to H, so not only (is) O to B, but also A to N , and N to O. Thus, A, N , O, B are continuously proportional in the aforementioned ratios of the sides. So I say that A also has to B a cubed ratio with respect to (that) a corresponding side (has) to a corresponding side—that is to say, with respect to (that) the number C (has) to F , or D to G, and, further, E to H. For since A, N , O, B are four continuously proportional numbers, A thus has to B a cubed ratio with respect to (that) A (has) to N [Def. 5.10]. But, as A (is) to N , so it was shown (is) C to F , and D to G, and, further, E to H. And thus A has to B a cubed ratio with respect to (that) a corresponding side (has) to a corresponding side—that is to say, with

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τέσσαρες ¢ριθµοˆ ˜ξÁς ¢νάλογόν ε„σιν οƒ Α, Ν, Ξ, Β, respect to (that) the number C (has) to F , and D to G, Ð Α ¥ρα πρÕς τÕν Β τριπλασίονα λόγον œχει ½περ Ð Α and, further, E to H. (Which is) the very thing it was πρÕς τÕν Ν. ¢λλ' æς Ð Α πρÕς τÕν Ν, οÛτως ™δείχθη Ó required to show. τε Γ πρÕς τÕν Ζ καˆ Ð ∆ πρÕς τÕν Η κሠœτι Ð Ε πρÕς τÕν Θ. καˆ Ð Α ¥ρα πρÕς τÕν Β τριπλασίονα λόγον œχει ½περ ¹ οµόλογος πλευρ¦ πρÕς τ¾ν еόλογον πλευράν, τουτέστιν ½περ Ð Γ ¢ριθµÕς πρÕς τÕν Ζ καˆ Ð ∆ πρÕς τÕν Η κሠœτι Ð Ε πρÕς τÕν Θ· Óπερ œδει δε‹ξαι. †

Literally, “triple”.

κ΄.

Proposition 20

'Ε¦ν δύο ¢ριθµîν εŒς µέσος ¢νάλογον ™µπίπτÍ If one number falls between two numbers in mean ¢ριθµός, Óµοιοι ™πίπεδοι œσονται οƒ ¢ριθµοί. proportion then the numbers will be similar plane (num∆ύο γ¦ρ ¢ριθµîν τîν Α, Β εŒς µέσος ¢νάλογον bers). ™µπιπτέτω ¢ριθµÕς Ð Γ· λέγω, Óτι οƒ Α, Β Óµοιοι For let one number C fall between the two numbers A ™πίπεδοί ε„σιν ¢ριθµοί. and B in mean proportion. I say that A and B are similar plane numbers.

Α Γ Β

∆ Ζ

A C B

Ε Η

D F E G

Ε„λήφθωσαν [γ¦ρ] ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Γ οƒ ∆, Ε· „σάκις ¥ρα Ð ∆ τÕν Α µετρε‹ καˆ Ð Ε τÕν Γ. Ðσάκις δ¾ Ð ∆ τÕν Α µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ζ· Ð Ζ ¥ρα τÕν ∆ πολλαπλασιάσας τÕν Α πεποίηκεν. éστε Ð Α ™πίπεδός ™στιν, πλευραˆ δ αÙτοà οƒ ∆, Ζ. πάλιν, ™πεˆ οƒ ∆, Ε ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Γ, Β, „σάκις ¥ρα Ð ∆ τÕν Γ µετρε‹ καˆ Ð Ε τÕν Β. Ðσάκις δ¾ Ð Ε τÕν Β µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Η. Ð Ε ¥ρα τÕν Β µετρε‹ κατ¦ τ¦ς ™ν τù Η µονάδας· Ð Η ¥ρα τÕν Ε πολλαπλασιάσας τÕν Β πεποίηκεν. Ð Β ¥ρα ™πίπεδος ™στι, πλευραˆ δ αÙτοà ε„σιν οƒ Ε, Η. οƒ Α, Β ¥ρα ™πίπεδοί ε„σιν ¢ριθµοί. λέγω δή, Óτι καˆ Óµοιοι. ™πεˆ γ¦ρ Ð Ζ τÕν µν ∆ πολλαπλασιάσας τÕν Α πεποίηκεν, τÕν δ Ε πολλαπλασιάσας τÕν Γ πεποίηκεν, œστιν ¥ρα æς Ð ∆ πρÕς τÕν Ε, οÛτως Ð Α πρÕς τÕν Γ, τουτέστιν Ð Γ πρÕς τÕν Β. πάλιν, ™πεˆ Ð Ε ˜κάτερον τîν Ζ, Η πολλαπλασιάσας τοÝς Γ, Β πεποίηκεν, œστιν ¥ρα æς Ð Ζ πρÕς τÕν Η, οÛτως Ð Γ πρÕς τÕν Β. æς δ Ð Γ πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Ε· κሠæς ¥ρα Ð ∆ πρÕς τÕν Ε, οÛτως Ð Ζ πρÕς τÕν Η· κሠ™ναλλ¦ξ æς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Ε πρÕς τÕν Η. οƒ Α, Β ¥ρα Óµοιοι ™πίπεδοι ¢ριθµοί ε„σιν· αƒ γ¦ρ πλευρሠαÙτîν ¢νάλογόν ε„σιν· Óπερ œδει δε‹ξαι.

[For] let the least numbers, D and E, having the same ratio as A and C have been taken [Prop. 7.33]. Thus, D measures A as many times as E (measures) C [Prop. 7.20]. So as many times as D measures A, so many units let there be in F . Thus, F has made A (by) multiplying D [Def. 7.15]. Hence, A is plane, and D, F (are) its sides. Again, since D and E are the least of those (numbers) having the same ratio as C and B, D thus measures C as many times as E (measures) B [Prop. 7.20]. So as many times as E measures B, so many units let there be in G. Thus, E measures B according to the units in G. Thus, G has made B (by) multiplying E [Def. 7.15]. Thus, B is plane, and E, G are its sides. Thus, A and B are (both) plane numbers. So I say that (they are) also similar. For since F has made A (by) multiplying D, and has made C (by) multiplying E, thus as D is to E, so A (is) to C—that is to say, C to B [Prop. 7.17].† Again, since E has made C, B (by) multiplying F , G, respectively, thus as F is to G, so C (is) to B [Prop. 7.17]. And as C (is) to B, so D (is) to E. And thus as D (is) to E, so F (is) to G. And, alternately, as D (is) to F , so E (is) to G [Prop. 7.13]. Thus, A and B are similar plane numbers. For their sides are proportional [Def. 7.21]. (Which is) the very thing it was required to show.

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ELEMENTS BOOK 8

This part of the proof is defective, since it is not demonstrated that F × E = C. Furthermore, it is not necessary to show that D : E :: A : C,

because this is true by hypothesis.

κα΄.

Proposition 21

'Ε¦ν δύο ¢ριθµîν δύο µέσοι ¢νάλογον ™µπίπτωσιν If two numbers fall between two numbers in mean ¢ριθµοί, Óµοιοι στερεοί ε„σιν οƒ ¢ριθµοί. proportion then the (latter) are similar solid (numbers).

Α Γ ∆ Β

Θ Κ Ν

A C D B

H K N

Ε Ζ Η

Λ Μ Ξ

E F G

L M O

∆ύο γ¦ρ ¢ριθµîν τîν Α, Β δύο µέσοι ¢νάλογον ™µπιπτέτωσαν ¢ριθµοˆ οƒ Γ, ∆· λέγω, Óτι οƒ Α, Β Óµοιοι στερεοί ε„σιν. Ε„λήφθωσαν γ¦ρ ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Γ, ∆ τρε‹ς οƒ Ε, Ζ, Η· οƒ ¥ρα ¥κροι αÙτîν οƒ Ε, Η πρîτοι πρÕς ¢λλήλους ε„σίν. κሠ™πεˆ τîν Ε, Η εŒς µέσος ¢νάλογον ™µπέπτωκεν ¢ριθµÕς Ð Ζ, οƒ Ε, Η ¥ρα ¢ριθµοˆ Óµοιοι ™πίπεδοί ε„σιν. œστωσαν οâν τοà µν Ε πλευραˆ οƒ Θ, Κ, τοà δ Η οƒ Λ, Μ. φανερÕν ¥ρα ™στˆν ™κ τοà πρÕ τούτου, Óτι οƒ Ε, Ζ, Η ˜ξÁς ε„σιν ¢νάλογον œν τε τù τοà Θ πρÕς τÕν Λ λόγJ κሠτù τοà Κ πρÕς τÕν Μ. κሠ™πεˆ οƒ Ε, Ζ, Η ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Γ, ∆, καί ™στιν ‡σον τÕ πλÁθος τîν Ε, Ζ, Η τù πλήθει τîν Α, Γ, ∆, δι' ‡σου ¥ρα ™στˆν æς Ð Ε πρÕς τÕν Η, οÛτως Ð Α πρÕς τÕν ∆. οƒ δ Ε, Η πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας αÙτο‹ς „σάκις Ó τε µείζων τÕν µείζονα καˆ Ð ™λάσσων τÕν ™λάσσονα, τουτέστιν Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον· „σάκις ¥ρα Ð Ε τÕν Α µετρε‹ καˆ Ð Η τÕν ∆. Ðσάκις δ¾ Ð Ε τÕν Α µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ν. Ð Ν ¥ρα τÕν Ε πολλαπλασιάσας τÕν Α πεποίηκεν. Ð δ Ε ™στιν Ð ™κ τîν Θ, Κ· Ð Ν ¥ρα τÕν ™κ τîν Θ, Κ πολλαπλασιάσας τÕν Α πεποίηκεν. στερεÕς ¥ρα ™στˆν Ð Α, πλευραˆ δ αÙτοà ε„σιν οƒ Θ, Κ, Ν. πάλιν, ™πεˆ οƒ Ε, Ζ, Η ™λάχιστοί ε„σι τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Γ, ∆, Β, „σάκις ¥ρα Ð Ε τÕν Γ µετρε‹ καˆ Ð Η τÕν Β. Ðσάκις δ¾ Ð Ε τÕν Γ µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ξ. Ð Η ¥ρα τÕν Β µετρε‹ κατ¦ τ¦ς ™ν τù Ξ µονάδας· Ð Ξ ¥ρα τÕν Η πολλαπλασιάσας τÕν Β πεποίηκεν. Ð δ Η ™στιν Ð ™κ τîν Λ, Μ· Ð Ξ ¥ρα τÕν ™κ τîν Λ, Μ πολλαπλασιάσας

For let the two numbers C and D fall between the two numbers A and B in mean proportion. I say that A and B are similar solid (numbers). For let the three least numbers E, F , G having the same ratio as A, C, D have been taken [Prop. 8.2]. Thus, the outermost of them, E and G, are prime to one another [Prop. 8.3]. And since one number, F , has fallen (between) E and G in mean proportion, E and G are thus similar plane numbers [Prop. 8.20]. Therefore, let H, K be the sides of E, and L, M (the sides) of G. Thus, it is clear from the (proposition) before this (one) that E, F , G are continuously proportional in the ratio of H to L, and of K to M . And since E, F , G are the least (numbers) having the same ratio as A, C, D, and the multitude of E, F , G is equal to the multitude of A, C, D, thus, via equality, as E is to G, so A (is) to D [Prop. 7.14]. And E and G (are) prime (to one another), and prime (numbers) are also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures A the same number of times as G (measures) D. So as many times as E measures A, so many units let there be in N . Thus, N has made A (by) multiplying E [Def. 7.15]. And E is the (number created) from (multiplying) H and K. Thus, N has made A (by) multiplying the (number created) from (multiplying) H and K. Thus, A is solid, and its sides are H, K, N . Again, since E, F , G are the least (numbers) having the same ratio as C, D, B, thus E measures C the

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τÕν Β πεποίηκεν. στερεÕς ¥ρα ™στˆν Ð Β, πλευραˆ δ αÙτοà ε„σιν οƒ Λ, Μ, Ξ· οƒ Α, Β ¥ρα στερεοί ε„σιν. Λέγω [δή], Óτι καˆ Óµοιοι. ™πεˆ γ¦ρ οƒ Ν, Ξ τÕν Ε πολλαπλασιάσαντες τοÝς Α, Γ πεποιήκασιν, œστιν ¥ρα æς Ð Ν πρÕς τÕν Ξ, Ð Α πρÕς τÕν Γ, τουτέστιν Ð Ε πρÕς τÕν Ζ. ¢λλ' æς Ð Ε πρÕς τÕν Ζ, Ð Θ πρÕς τÕν Λ καˆ Ð Κ πρÕς τÕν Μ· κሠæς ¥ρα Ð Θ πρÕς τÕν Λ, οÛτως Ð Κ πρÕς τÕν Μ καˆ Ð Ν πρÕς τÕν Ξ. καί ε„σιν οƒ µν Θ, Κ, Ν πλευρሠτοà Α, οƒ δ Ξ, Λ, Μ πλευρሠτοà Β. οƒ Α, Β ¥ρα ¢ριθµοˆ Óµοιοι στερεοί ε„σιν· Óπερ œδει δε‹ξαι.

†

same number of times as G (measures) B [Prop. 7.20]. So as many times as E measures C, so many units let there be in O. Thus, G measures B according to the units in O. Thus, O has made B (by) multiplying G. And G is the (number created) from (multiplying) L and M . Thus, O has made B (by) multiplying the (number created) from (multiplying) L and M . Thus, B is solid, and its sides are L, M , O. Thus, A and B are (both) solid. [So] I say that (they are) also similar. For since N , O have made A, C (by) multiplying E, thus as N is to O, so A (is) to C—that is to say, E to F [Prop. 7.18]. But, as E (is) to F , so H (is) to L, and K to M . And thus as H (is) to L, so K (is) to M , and N to O. And H, K, N are the sides of A, and L, M , O the sides of B. Thus, A and B are similar solid numbers [Def. 7.21]. (Which is) the very thing it was required to show.

The Greek text has “O, L, M ”, which is obviously a mistake.

κβ΄.

Proposition 22

'Ε¦ν τρε‹ς ¢ριθµοˆ ˜ξÁς ¢νάλογον ðσιν, Ð δ πρîτος If three numbers are continuously proportional, and τετράγωνος Ï, καˆ Ð τρίτος τετράγωνος œσται. the first is square, then the third will also be square.

Α Β Γ

A B C

”Εστωσαν τρε‹ς ¢ριθµοˆ ˜ξÁς ¢νάλογον οƒ Α, Β, Γ, Ð δ πρîτος Ð Α τετράγωνος œστω· λέγω, Óτι καˆ Ð τρίτος Ð Γ τετράγωνός ™στιν. 'Επεˆ γ¦ρ τîν Α, Γ εŒς µέσος ¢νάλογόν ™στιν ¢ριθµÕς Ð Β, οƒ Α, Γ ¥ρα Óµοιοι ™πίπεδοί ε„σιν. τετράγωνος δ Ð Α· τετράγωνος ¥ρα καˆ Ð Γ· Óπερ œδει δε‹ξαι.

Let A, B, C be three continuously proportional numbers, and let the first A be square. I say that the third C is also square. For since one number, B, is in mean proportion to A and C, A and C are thus similar plane (numbers) [Prop. 8.20]. And A is square. Thus, C is also square [Def. 7.21]. (Which is) the very thing it was required to show.

κγ΄.

Proposition 23

'Ε¦ν τέσσαρες ¢ριθµοˆ ˜ξÁς ¢νάλογον ðσιν, Ð δ If four numbers are continuously proportional, and πρîτος κύβος Ï, καˆ Ð τέταρτος κύβος œσται. the first is cube, then the fourth will also be cube.

Α Β Γ ∆

.

A B C D

”Εστωσαν τέσσαρες ¢ριθµοˆ ˜ξÁς ¢νάλογον οƒ Α, Β, Let A, B, C, D be four continuously proportional Γ, ∆, Ð δ Α κύβος œστω· λέγω, Óτι καˆ Ð ∆ κύβος ™στίν. numbers, and let A be cube. I say that D is also cube. 249

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ELEMENTS BOOK 8

'Επεˆ γ¦ρ τîν Α, ∆ δύο µέσοι ¢νάλογόν ε„σιν For since two numbers, B and C, are in mean propor¢ριθµοˆ οƒ Β, Γ, οƒ Α, ∆ ¥ρα Óµοιοί ε„σι στερεοˆ tion to A and D, A and D are thus similar solid numbers ¢ριθµοί. κύβος δ Ð Α· κύβος ¥ρα καˆ Ð ∆· Óπερ œδει [Prop. 8.21]. And A (is) cube. Thus, D (is) also cube δε‹ξαι. [Def. 7.21]. (Which is) the very thing it was required to show.

κδ΄.

Proposition 24

'Ε¦ν δύο ¢ριθµοˆ πρÕς ¢λλήλους λόγον œχωσιν, Öν If two numbers have to one another the ratio which τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, Ð δ a square number (has) to a(nother) square number, and πρîτος τετράγωνος Ï, καˆ Ð δεύτερος τετράγωνος œσται. the first is square, then the second will also be square.

Α Β

Γ ∆

A B

C D

∆ύο γ¦ρ ¢ριθµοˆ οƒ Α, Β πρÕς ¢λλήλους λόγον ™χέτωσαν, Öν τετράγωνος ¢ριθµÕς Ð Γ πρÕς τετράγωνον ¢ριθµÕν τÕν ∆, Ð δ Α τετράγωνος œστω· λέγω, Óτι καˆ Ð Β τετράγωνός ™στιν. 'Επεˆ γ¦ρ οƒ Γ, ∆ τετράγωνοί ε„σιν, οƒ Γ, ∆ ¥ρα Óµοιοι ™πίπεδοί ε„σιν. τîν Γ, ∆ ¥ρα εŒς µέσος ¢νάλογον ™µπίπτει ¢ριθµός. καί ™στιν æς Ð Γ πρÕς τÕν ∆, Ð Α πρÕς τÕν Β· κሠτîν Α, Β ¥ρα εŒς µέσος ¢νάλογον ™µπίπτει ¢ριθµός. καί ™στιν Ð Α τετράγωνος· καˆ Ð Β ¥ρα τετράγωνός ™στιν· Óπερ œδει δε‹ξαι.

For let two numbers, A and B, have to one another the ratio which the square number C (has) to the square number D. And let A be square. I say that B is also square. For since C and D are square, C and D are thus similar plane (numbers). Thus, one number falls (between) C and D in mean proportion [Prop. 8.18]. And as C is to D, (so) A (is) to B. Thus, one number also falls (between) A and B in mean proportion [Prop. 8.8]. And A is square. Thus, B is also square [Prop. 8.22]. (Which is) the very thing it was required to show.

κε΄.

Proposition 25

'Ε¦ν δύο ¢ριθµοˆ πρÕς ¢λλήλους λόγον œχωσιν, Öν κύβος ¢ριθµÕς πρÕς κύβον ¢ριθµόν, Ð δ πρîτος κύβος Ï, καˆ Ð δεύτερος κύβος œσται.

If two numbers have to one another the ratio which a cube number (has) to a(nother) cube number, and the first is cube, then the second will also be cube.

Α Ε Ζ Β

Γ

A E F B

∆

∆ύο γ¦ρ ¢ριθµοˆ οƒ Α, Β πρÕς ¢λλήλους λόγον ™χέτωσαν, Öν κύβος ¢ριθµÕς Ð Γ πρÕς κύβον ¢ριθµÕν τÕν ∆, κύβος δ œστω Ð Α· λέγω [δή], Óτι καˆ Ð Β κύβος ™στίν. 'Επεˆ γ¦ρ οƒ Γ, ∆ κύβοι ε„σίν, οƒ Γ, ∆ Óµοιοι στερεοί ε„σιν· τîν Γ, ∆ ¥ρα δύο µέσοι ¢νάλογον ™µπίπτουσιν ¢ριθµοί. Óσοι δ ε„ς τοÝς Γ, ∆ µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπίπτουσιν, τοσοàτοι κሠε„ς τοÝς τÕν αÙτÕν λόγον œχοντας αÙτο‹ς· éστε κሠτîν Α, Β δύο µέσοι ¢νάλογον ™µπίπτουσιν ¢ριθµοί. ™µπιπτέτωσαν οƒ Ε, Ζ. ™πεˆ οâν τέσσαρες ¢ριθµοˆ οƒ Α, Ε, Ζ, Β ˜ξÁς ¢νάλογόν ε„σιν, καί ™στι κύβος Ð Α, κύβος ¥ρα καˆ Ð Β· Óπερ œδει δε‹ξαι.

C

D

For let two numbers, A and B, have to one another the ratio which the cube number C (has) to the cube number D. And let A be cube. [So] I say that B is also cube. For since C and D are cube (numbers), C and D are (thus) similar solid (numbers). Thus, two numbers fall (between) C and D in mean proportion [Prop. 8.19]. And as many (numbers) as fall in between C and D in continued proportion, so many also (fall) in (between) those (numbers) having the same ratio as them (in continued proportion) [Prop. 8.8]. And hence two numbers fall (between) A and B in mean proportion. Let E and F (so) fall. Therefore, since the four numbers A, E, F , B are continuously proportional, and A is cube, B (is) thus

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ELEMENTS BOOK 8 also cube [Prop. 8.23]. (Which is) the very thing it was required to show.

κ$΄.

Proposition 26

Οƒ Óµοιοι ™πίπεδοι ¢ριθµοˆ πρÕς ¢λλήλους λόγον œχουσιν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν.

Similar plane numbers have to one another the ratio which (some) square number (has) to a(nother) square number.

Α Γ Β

∆ Ε Ζ

A C B

D E F

”Εστωσαν Óµοιοι ™πίπεδοι ¢ριθµοˆ οƒ Α, Β· λέγω, Óτι Ð Α πρÕς τÕν Β λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. 'Επεˆ γ¦ρ οƒ Α, Β Óµοιοι ™πίπεδοί ε„σιν, τîν Α, Β ¥ρα εŒς µέσος ¢νάλογον ™µπίπτει ¢ριθµός. ™µπιπτέτω κሠœστω Ð Γ, κሠε„λήφθωσαν ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Γ, Β οƒ ∆, Ε, Ζ· οƒ ¥ρα ¥κροι αÙτîν οƒ ∆, Ζ τετράγωνοί ε„σιν. κሠ™πεί ™στιν æς Ð ∆ πρÕς τÕν Ζ, οÛτως Ð Α πρÕς τÕν Β, καί ε„σιν οƒ ∆, Ζ τετράγωνοι, Ð Α ¥ρα πρÕς τÕν Β λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· Óπερ œδει δε‹ξαι.

Let A and B be similar plane numbers. I say that A has to B the ratio which (some) square number (has) to a(nother) square number. For since A and B are similar plane numbers, one number thus falls (between) A and B in mean proportion [Prop. 8.18]. Let it (so) fall, and let it be C. And let the least numbers, D, E, F , having the same ratio as A, C, B have been taken [Prop. 8.2]. The outermost of them, D and F , are thus square [Prop. 8.2 corr.]. And since as D is to F , so A (is) to B, and D and F are square, A thus has to B the ratio which (some) square number (has) to a(nother) square number. (Which is) the very thing it was required to show.

κζ΄.

Proposition 27

Οƒ Óµοιοι στερεοˆ ¢ριθµοˆ πρÕς ¢λλήλους λόγον œχουσιν, Öν κύβος ¢ριθµÕς πρÕς κύβον ¢ριθµόν.

Similar solid numbers have to one another the ratio which (some) cube number (has) to a(nother) cube number.

Α Γ ∆ Β

Ε Ζ Η Θ

A C D B

”Εστωσαν Óµοιοι στερεοˆ ¢ριθµοˆ οƒ Α, Β· λέγω, Óτι Ð Α πρÕς τÕν Β λόγον œχει, Öν κύβος ¢ριθµÕς πρÕς κύβον ¢ριθµόν. 'Επεˆ γ¦ρ οƒ Α, Β Óµοιοι στερεοί ε„σιν, τîν Α, Β ¥ρα δύο µέσοι ¢νάλογον ™µπίπτουσιν ¢ριθµοί. ™µπιπτέτωσαν οƒ Γ, ∆, κሠε„λήφθωσαν ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Γ, ∆, Β ‡σοι αÙτο‹ς τÕ πλÁθος οƒ Ε, Ζ, Η, Θ· οƒ ¥ρα ¥κροι αÙτîν οƒ Ε, Θ κύβοι ε„σίν. καί ™στιν æς Ð Ε πρÕς τÕν Θ, οÛτως Ð Α πρÕς τÕν Β· καˆ Ð Α ¥ρα πρÕς τÕν Β λόγον œχει, Öν κύβος ¢ριθµÕς πρÕς κύβον ¢ριθµόν· Óπερ œδει δε‹ξαι.

E F G H

Let A and B be similar solid numbers. I say that A has to B the ratio which (some) cube number (has) to a(nother) cube number. For since A and B are similar solid (numbers), two numbers thus fall (between) A and B in mean proportion [Prop. 8.19]. Let C and D have (so) fallen. And let the least numbers, E, F , G, H, having the same ratio as A, C, D, B, (and) equal in multitude to them, have been taken [Prop. 8.2]. Thus, the outermost of them, E and H, are cube [Prop. 8.2 corr.]. And as E is to H, so A (is) to B. And thus A has to B the ratio which (some) cube number (has) to a(nother) cube number. (Which is) the very thing it was required to show.

251

252

ELEMENTS BOOK 9 Applications of number theory†

† The

propositions contained in Books 7–9 are generally attributed to the school of Pythagoras.

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ELEMENTS BOOK 9

α΄.

Proposition 1

'Ε¦ν δύο Óµοιοι ™πίπεδοι ¢ριθµοˆ πολλαπλασιάσαντες If two similar plane numbers make some (number by) ¢λλήλους ποιîσί τινα, Ð γενόµενος τετράγωνος œσται. multiplying one another then the created (number) will be square.

Α Β Γ ∆

A B C D

”Εστωσαν δύο Óµοιοι ™πίπεδοι ¢ριθµοˆ οƒ Α, Β, καˆ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ ποιείτω· λέγω, Óτι Ð Γ τετράγωνός ™στιν. `Ο γ¦ρ Α ˜αυτÕν πολλαπλασιάσας τÕν ∆ ποιείτω. Ð ∆ ¥ρα τετράγωνός ™στιν. ™πεˆ οâν Ð Α ˜αυτÕν µν πολλαπλασιάσας τÕν ∆ πεποίηκεν, τÕν δ Β πολλαπλασιάσας τÕν Γ πεποίηκεν, œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Γ. κሠ™πεˆ οƒ Α, Β Óµοιοι ™πίπεδοί ε„σιν ¢ριθµοί, τîν Α, Β ¥ρα εŒς µέσος ¢νάλογον ™µπίπτει ¢ριθµός. ™¦ν δ δύο ¢ριθµîν µεταξÝ κατ¦ τÕ συνεχς ¢νάλογον ™µπίπτωσιν ¢ριθµοί, Óσοι ε„ς αÙτοÝς ™µπίπτουσι, τοσοàτοι κሠε„ς τοÝς τÕν αÙτÕν λόγον œχοντας· éστε κሠτîν ∆, Γ εŒς µέσος ¢νάλογον ™µπίπτει ¢ριθµός. καί ™στι τετράγωνος Ð ∆· τετράγωνος ¥ρα καˆ Ð Γ· Óπερ œδει δε‹ξαι.

Let A and B be two similar plane numbers, and let A make C (by) multiplying B. I say that C is square. For let A make D (by) multiplying itself. D is thus square. Therefore, since A has made D (by) multiplying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17]. And since A and B are similar plane numbers, one number thus falls (between) A and B in mean proportion [Prop. 8.18]. And if (some) numbers fall between two numbers in continued proportion, then as many (numbers) as fall in (between) them (in continued proportion), so many also (fall) in (between numbers) having the same ratio (as them in continued proportion) [Prop. 8.8]. And hence one number falls (between) D and C in mean proportion. And D is square. Thus, C (is) also square [Prop. 8.22]. (Which is) the very thing it was required to show.

β΄.

Proposition 2

'Ε¦ν δύο ¢ριθµοˆ πολλαπλασιάσαντες ¢λλήλους ποιîσι τετράγωνον, Óµοιοι ™πίπεδοί ε„σιν ¢ριθµοί.

If two numbers make a square (number by) multiplying one another then they are similar plane numbers.

Α Β Γ ∆

A B C D

”Εστωσαν δύο ¢ριθµοˆ οƒ Α, Β, καˆ Ð Α τÕν Β πολλαπλασιάσας τετράγωνον τÕν Γ ποιείτω· λέγω, Óτι οƒ Α, Β Óµοιοι ™πίπεδοί ε„σιν ¢ριθµοί. `Ο γ¦ρ Α ˜αυτÕν πολλαπλασιάσας τÕν ∆ ποιείτω· Ð ∆ ¥ρα τετράγωνός ™στιν. κሠ™πεˆ Ð Α ˜αυτÕν µν πολλαπλασιάσας τÕν ∆ πεποίηκεν, τÕν δ Β πολλαπλασιάσας τÕν Γ πεποίηκεν, œστιν ¥ρα æς Ð Α πρÕς τÕν Β, Ð ∆ πρÕς τÕν Γ. κሠ™πεˆ Ð ∆ τετράγωνός ™στιν, ¢λλ¦ καˆ Ð Γ, οƒ ∆, Γ ¥ρα Óµοιοι ™πίπεδοί ε„σιν. τîν ∆, Γ ¥ρα

Let A and B be two numbers, and let A make the square (number) C (by) multiplying B. I say that A and B are similar plane numbers. For let A make D (by) multiplying itself. Thus, D is square. And since A has made D (by) multiplying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17]. And since D is square, and also C, D and C are thus similar plane numbers. Thus, one (number) falls (between) D and C in mean proportion

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ELEMENTS BOOK 9

εŒς µέσος ¢νάλογον ™µπίπτει. καί ™στιν æς Ð ∆ πρÕς τÕν Γ, οÛτως Ð Α πρÕς τÕν Β· κሠτîν Α, Β ¥ρα εŒς µέσος ¢νάλογον ™µπίπτει. ™¦ν δ δύο ¢ριθµîν εŒς µέσος ¢νάλογον ™µπίπτV, Óµοιοι ™πίπεδοί ε„σιν [οƒ] ¢ριθµοί· οƒ ¥ρα Α, Β Óµοιοί ε„σιν ™πίπεδοι· Óπερ œδει δε‹ξαι.

[Prop. 8.18]. And as D is to C, so A (is) to B. Thus, one (number) also falls (between) A and B in mean proportion [Prop. 8.8]. And if one (number) falls (between) two numbers in mean proportion then [the] numbers are similar plane (numbers) [Prop. 8.20]. Thus, A and B are similar plane (numbers). (Which is) the very thing it was required to show.

γ΄.

Proposition 3

'Ε¦ν κύβος ¢ριθµÕς ˜αυτÕν πολλαπλασιάσας ποιÍ τινα, Ð γενόµενος κύβος œσται.

If a cube number makes some (number by) multiplying itself then the created (number) will be cube.

Α Β Γ ∆

A B C D

Κύβος γ¦ρ ¢ριθµÕς Ð Α ˜αυτÕν πολλαπλασιάσας τÕν Β ποιείτω· λέγω, Óτι Ð Β κύβος ™στίν. Ε„λήφθω γ¦ρ τοà Α πλευρ¦ Ð Γ, καˆ Ð Γ ˜αυτÕν πολλαπλασιάσας τÕν ∆ ποιείτω. φανερÕν δή ™στιν, Óτι Ð Γ τÕν ∆ πολλαπλασιάσας τÕν Α πεποίηκεν. κሠ™πεˆ Ð Γ ˜αυτÕν πολλαπλασιάσας τÕν ∆ πεποίηκεν, Ð Γ ¥ρα τÕν ∆ µετρε‹ κατ¦ τ¦ς ™ν αØτù µονάδας. ¢λλ¦ µ¾ν κሠ¹ µον¦ς τÕν Γ µετρε‹ κατ¦ τ¦ς ™ν αÙτù µονάδας· œστιν ¥ρα æς ¹ µον¦ς πρÕς τÕν Γ, Ð Γ πρÕς τÕν ∆. πάλιν, ™πεˆ Ð Γ τÕν ∆ πολλαπλασιάσας τÕν Α πεποίηκεν, Ð ∆ ¥ρα τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù Γ µονάδας. µετρε‹ δ κሠ¹ µον¦ς τÕν Γ κατ¦ τ¦ς ™ν αÙτù µονάδας· œστιν ¥ρα æς ¹ µον¦ς πρÕς τÕν Γ, Ð ∆ πρÕς τÕν Α. ¢λλ' æς ¹ µον¦ς πρÕς τÕν Γ, Ð Γ πρÕς τÕν ∆· κሠæς ¥ρα ¹ µον¦ς πρÕς τÕν Γ, οÛτως Ð Γ πρÕς τÕν ∆ καˆ Ð ∆ πρÕς τÕν Α. τÁς ¥ρα µονάδος κሠτοà Α ¢ριθµοà δύο µέσοι ¢νάλογον κατ¦ τÕ συνεχς ™µπεπτώκασιν ¢ριθµοˆ οƒ Γ, ∆. πάλιν, ™πεˆ Ð Α ˜αυτÕν πολλαπλασιάσας τÕν Β πεποίηκεν, Ð Α ¥ρα τÕν Β µετρε‹ κατ¦ τ¦ς ™ν αÙτù µονάδας· µετρε‹ δ κሠ¹ µον¦ς τÕν Α κατ¦ τ¦ς ™ν αÙτù µονάδας· œστιν ¥ρα æς ¹ µον¦ς πρÕς τÕν Α, Ð Α πρÕς τÕν Β. τÁς δ µονάδος κሠτοà Α δύο µέσοι ¢νάλογον ™µπεπτώκασιν ¢ριθµοί· κሠτîν Α, Β ¥ρα δύο µέσοι ¢νάλογον ™µπεσοàνται ¢ριθµοί. ™¦ν δ δύο ¢ριθµîν δύο µέσοι ¢νάλογον ™µπίπτωσιν, Ð δ πρîτος κύβος Ï, καˆ Ð δεύτερος κύβος œσται. καί ™στιν Ð Α κύβος· καˆ Ð Β ¥ρα κύβος ™στίν· Óπερ œδει δε‹ξαι.

For let the cube number A make B (by) multiplying itself. I say that B is cube. For let the side C of A have been taken. And let C make D by multiplying itself. So it is clear that C has made A (by) multiplying D. And since C has made D (by) multiplying itself, C thus measures D according to the units in it [Def. 7.15]. But, in fact, a unit also measures C according to the units in it [Def. 7.20]. Thus, as a unit is to C, so C (is) to D. Again, since C has made A (by) multiplying D, D thus measures A according to the units in C. And a unit also measures C according to the units in it. Thus, as a unit is to C, so D (is) to A. But, as a unit (is) to C, so C (is) to D. And thus as a unit (is) to C, so C (is) to D, and D to A. Thus, two numbers, C and D, have fallen (between) a unit and the number A in successive mean proportion. Again, since A has made B (by) multiplying itself, A thus measures B according to the units in it. And a unit also measures A according to the units in it. Thus, as a unit is to A, so A (is) to B. And two numbers have fallen (between) a unit and A in mean proportion. Thus two numbers will also fall (between) A and B in mean proportion [Prop. 8.8]. And if two (numbers) fall (between) two numbers in mean proportion, and the first (number) is cube, then the second will also be cube [Prop. 8.23]. And A is cube. Thus, B is also cube. (Which is) the very thing it was required to show.

δ΄.

Proposition 4

'Ε¦ν κύβος ¢ριθµÕς κύβον ¢ριθµÕν πολλαπλασιάσας ποιÍ τινα, Ð γενόµενος κύβος œσται.

If a cube number makes some (number by) multiplying a(nother) cube number then the created (number)

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ELEMENTS BOOK 9 will be cube.

Α Β Γ ∆

A B C D

Κύβος γ¦ρ ¢ριθµÕς Ð Α κύβον ¢ριθµÕν τÕν Β πολλαπλασιάσας τÕν Γ ποιείτω· λέγω, Óτι Ð Γ κύβος ™στίν. `Ο γ¦ρ Α ˜αυτÕν πολλαπλασιάσας τÕν ∆ ποιείτω· Ð ∆ ¥ρα κύβος ™στίν. κሠ™πεˆ Ð Α ˜αυτÕν µν πολλαπλασιάσας τÕν ∆ πεποίηκεν, τÕν δ Β πολλαπλασιάσας τÕν Γ πεποίηκεν, œστιν ¥ρα æς Ð Α πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Γ. κሠ™πεˆ οƒ Α, Β κύβοι ε„σίν, Óµοιοι στερεοί ε„σιν οƒ Α, Β. τîν Α, Β ¥ρα δύο µέσοι ¢νάλογον ™µπίπτουσιν ¢ριθµοί· éστε κሠτîν ∆, Γ δύο µέσοι ¢νάλογον ™µπεσοàνται ¢ριθµοί. καί ™στι κύβος Ð ∆· κύβος ¥ρα καˆ Ð Γ· Óπερ œδει δε‹ξαι.

For let the cube number A make C (by) multiplying the cube number B. I say that C is cube. For let A make D (by) multiplying itself. Thus, D is cube [Prop. 9.3]. And since A has made D (by) multiplying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17]. And since A and B are cube, A and B are similar solid (numbers). Thus, two numbers fall (between) A and B in mean proportion [Prop. 8.19]. Hence, two numbers will also fall (between) D and C in mean proportion [Prop. 8.8]. And D is cube. Thus, C (is) also cube [Prop. 8.23]. (Which is) the very thing it was required to show.

ε΄.

Proposition 5

'Ε¦ν κύβος ¢ριθµÕς ¢ριθµόν τινα πολλαπλασιάσας κύβον ποιÍ, καˆ Ð πολλαπλασιασθεˆς κύβος œσται.

If a cube number makes a(nother) cube number (by) multiplying some (number) then the (number) multiplied will also be cube.

Α Β Γ ∆

A B C D

Κύβος γ¦ρ ¢ριθµος Ð Α ¢ριθµόν τινα τÕν Β πολλαπλασιάσας κύβον τÕν Γ ποιείτω· λέγω, Óτι Ð Β κύβος ™στίν. `Ο γ¦ρ Α ˜αυτÕν πολλαπλασιάσας τÕν ∆ ποιείτω· κύβος ¥ρα ™στίν Ð ∆. κሠ™πεˆ Ð Α ˜αυτÕν µν πολλαπλασιάσας τÕν ∆ πεποίηκεν, τÕν δ Β πολλαπλασιάσας τÕν Γ πεποίηκεν, œστιν ¢ρα æς Ð Α πρÕς τÕν Β, Ð ∆ πρÕς τÕν Γ. κሠ™πεˆ οƒ ∆, Γ κύβοι ε„σίν, Óµοιοι στερεοί ε„σιν. τîν ∆, Γ ¥ρα δύο µέσοι ¢νάλογον ™µπίπτουσιν ¢ριθµοί. καί ™στιν æς Ð ∆ πρÕς τÕν Γ, οÛτως Ð Α πρÕς τÕν Β· κሠτîν Α, Β ¥ρα δύο µέσοι ¢νάλογον ™µπίπτουσιν ¢ριθµοί. καί ™στι κύβος Ð Α· κύβος ¥ρα ™στˆ καˆ Ð Β· Óπερ œδει δε‹ξαι.

For let the cube number A make the cube (number) C (by) multiplying some number B. I say that B is cube. For let A make D (by) multiplying itself. D is thus cube [Prop. 9.3]. And since A has made D (by) multiplying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17]. And since D and C are (both) cube, they are similar solid (numbers). Thus, two numbers fall (between) D and C in mean proportion [Prop. 8.19]. And as D is to C, so A (is) to B. Thus, two numbers also fall (between) A and B in mean proportion [Prop. 8.8]. And A is cube. Thus, B is also cube [Prop. 8.23]. (Which is) the very thing it was required to show.

$΄.

Proposition 6

'Ε¦ν ¢ριθµÕς ˜αυτÕν πολλαπλασιάσας κύβον ποιÍ,

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If a number makes a cube (number by) multiplying

ΣΤΟΙΧΕΙΩΝ θ΄.

ELEMENTS BOOK 9

κሠαÙτÕς κύβος œσται.

itself then it itself will also be cube.

Α Β Γ

A B C

'ΑριθµÕς γ¦ρ Ð Α ˜αυτÕν πολλαπλασιάσας κύβον τÕν Β ποιείτω· λέγω, Óτι καˆ Ð Α κύβος ™στίν. `Ο γ¦ρ Α τÕν Β πολλαπλασιάσας τÕν Γ ποιείτω. ™πεˆ οâν Ð Α ˜αυτÕν µν πολλαπλασιάσας τÕν Β πεποίηκεν, τÕν δ Β πολλαπλασιάσας τÕν Γ πεποίηκεν, Ð Γ ¥ρα κύβος ™στίν. κሠ™πεˆ Ð Α ˜αυτÕν πολλαπλασιάσας τÕν Β πεποίηκεν, Ð Α ¥ρα τÕν Β µετρε‹ κατ¦ τ¦ς ™ν αØτù µονάδας. µετρε‹ δ κሠ¹ µον¦ς τÕν Α κατ¦ τ¦ς ™ν αÙτù µονάδας. œστιν ¥ρα æς ¹ µον¦ς πρÕς τÕν Α, οÛτως Ð Α πρÕς τÕν Β. κሠ™πεˆ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν, Ð Β ¥ρα τÕν Γ µετρε‹ κατ¦ τ¦ς ™ν τù Α µονάδας. µετρεˆ δ κሠ¹ µον¦ς τÕν Α κατ¦ τ¦ς ™ν αÙτù µονάδας. œστιν ¥ρα æς ¹ µον¦ς πρÕς τÕν Α, οÛτως Ð Β πρÕς τÕν Γ. ¢λλ' æς ¹ µον¦ς πρÕς τÕν Α, οÛτως Ð Α πρÕς τÕν Β· κሠæς ¥ρα Ð Α πρÕς τÕν Β, Ð Β πρÕς τÕν Γ. κሠ™πεˆ οƒ Β, Γ κύβοι ε„σίν, Óµοιοι στερεοί ε„σιν. τîν Β, Γ ¥ρα δύο µέσοι ¢νάλογόν ε„σιν ¢ριθµοί. καί ™στιν æς Ð Β πρÕς τÕν Γ, Ð Α πρÕς τÕν Β. κሠτîν Α, Β ¥ρα δύο µέσοι ¢νάλογόν ε„σιν ¢ριθµοί. καί ™στιν κύβος Ð Β· κύβος ¥ρα ™στˆ καˆ Ð Α· Óπερ œδει δεˆξαι.

For let the number A make the cube (number) B (by) multiplying itself. I say that A is also cube. For let A make C (by) multiplying B. Therefore, since A has made B (by) multiplying itself, and has made C (by) multiplying B, C is thus cube. And since A has made B (by) multiplying itself, A thus measures B according to the units in (A). And a unit also measures A according to the units in it. Thus, as a unit is to A, so A (is) to B. And since A has made C (by) multiplying B, B thus measures C according to the units in A. And a unit also measures A according to the units in it. Thus, as a unit is to A, so B (is) to C. But, as a unit (is) to A, so A (is) to B. And thus as A (is) to B, (so) B (is) to C. And since B and C are cube, they are similar solid (numbers). Thus, there exist two numbers in mean proportion (between) B and C [Prop. 8.19]. And as B is to C, (so) A (is) to B. Thus, there also exist two numbers in mean proportion (between) A and B [Prop. 8.8]. And B is cube. Thus, A is also cube [Prop. 8.23]. (Which is) the very thing it was required to show.

ζ΄.

Proposition 7

'Ε¦ν σύνθετος ¢ριθµÕς ¢ριθµόν τινα πολλαπλασιάσας ποιÍ τινα, Ð γενόµενος στερεÕς œσται.

If a composite number makes some (number by) multiplying some (other) number then the created (number) will be solid.

Α Β Γ ∆ Ε

A B C D E

Σύνθετος γ¦ρ ¢ριθµÕς Ð Α ¢ριθµόν τινα τÕν Β πολλαπλασιάσας τÕν Γ ποιείτω· λέγω, Óτι Ð Γ στερεός ™στιν. 'Επεˆ γ¦ρ Ð Α σύνθετός ™στιν, ØπÕ ¢ριθµοà τινος µετρηθήσεται. µετρείσθω ØπÕ τοà ∆, κሠÐσάκις Ð ∆ τÕν Α µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ε. ™πεˆ οâν Ð ∆ τÕν Α µετρε‹ κατ¦ τ¦ς ™ν τù Ε µονάδας, Ð Ε ¥ρα τÕν ∆ πολλαπλασιάσας τÕν Α πεποίηκεν. κሠ™πεˆ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν, Ð δ Α ™στιν Ð ™κ τîν ∆, Ε, Ð ¥ρα ™κ τîν ∆, Ε τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν. Ð Γ ¥ρα στερεός ™στιν, πλευρሠδ

For let the composite number A make C (by) multiplying some number B. I say that C is solid. For since A is a composite (number), it will be measured by some number. Let it be measured by D, and as many times as D measures A, so many units let there be in E. Therefore, since D measures A according to the units in E, E has thus made A (by) multiplying D [Def. 7.15]. And since A has made C (by) multiplying B, and A is the (number created) from (multiplying) D, E, the (number created) from (multiplying) D, E has thus

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αÙτοà ε„σιν οƒ ∆, Ε, Β· Óπερ œδει δε‹ξαι.

made C (by) multiplying B. Thus, C is solid, and its sides are D, E, B. (Which is) the very thing it was required to show.

η΄.

Proposition 8

'Ε¦ν ¢πÕ µονάδος Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον ðσιν, Ð µν τρίτος ¢πÕ τÁς µονάδος τετράγωνος œσται καˆ οƒ ›να διαλείποντες, Ð δ τέταρτος κύβος καˆ οƒ δύο διαλείποντες πάντες, Ð δ ›βδοµος κύβος ¤µα κሠτετράγωνος καˆ οƒ πέντε διαλείποντες.

If any multitude whatsoever of numbers is continuously proportional, (starting) from a unit, then the third from the unit will be square, and (all) those (numbers after that) which leave an interval of one (number), and the fourth (will be) cube, and all those (numbers after that) which leave an interval of two (numbers), and the seventh (will be) both cube and square, and (all) those (numbers after that) which leave an interval of five (numbers).

Α Β Γ ∆ Ε Ζ

A B C D E F

”Εστωσαν ¢πÕ µονάδος Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον οƒ Α, Β, Γ, ∆, Ε, Ζ· λέγω, Óτι Ð µν τρίτος ¢πÕ τÁς µονάδος Ð Β τετράγωνός ™στι καˆ οƒ ›να διαλείποντες πάντες, Ð δ τέταρτος Ð Γ κύβος καˆ οƒ δύο διαλείποντες πάντες, Ð δ ›βδοµος Ð Ζ κύβος ¤µα κሠτετράγωνος καˆ οƒ πέντε διαλείποντες πάντες. 'Επεˆ γάρ ™στιν æς ¹ µον¦ς πρÕς τÕν Α, οÛτως Ð Α πρÕς τÕν Β, „σάκις ¥ρα ¹ µον¦ς τÕν Α ¢ριθµÕν µετρε‹ καˆ Ð Α τÕν Β. ¹ δ µον¦ς τÕν Α ¢ριθµÕν µετρε‹ κατ¦ τ¦ς ™ν αÙτù µονάδας· καˆ Ð Α ¥ρα τÕν Β µετρε‹ κατ¦ τ¦ς ™ν τù Α µονάδας. Ð Α ¥ρα ˜αυτÕν πολλαπλασιάσας τÕν Β πεποίηκεν· τετράγωνος ¥ρα ™στˆν Ð Β. κሠ™πεˆ οƒ Β, Γ, ∆ ˜ξÁς ¢νάλογόν ε„σιν, Ð δ Β τετράγωνός ™στιν, καˆ Ð ∆ ¥ρα τετράγωνός ™στιν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Ζ τετράγωνός ™στιν. еοίως δ¾ δείξοµεν, Óτι καˆ οƒ ›να διαλείποντες πάντες τετράγωνοί ε„σιν. λέγω δή, Óτι καˆ Ð τέταρτος ¢πÕ τÁς µονάδος Ð Γ κύβος ™στˆ καˆ οƒ δύο διαλείποντες πάντες. ™πεˆ γάρ ™στιν æς ¹ µον¦ς πρÕς τÕν Α, οÛτως Ð Β πρÕς τÕν Γ, „σάκις ¥ρα ¹ µον¦ς τÕν Α ¢ριθµÕν µετρε‹ καˆ Ð Β τÕν Γ. ¹ δ µον¦ς τÕν Α ¢ριθµÕν µετρε‹ κατ¦ τ¦ς ™ν τù Α µονάδας· καˆ Ð Β ¥ρα τÕν Γ µετρε‹ κατ¦ τ¦ς ™ν τù Α µονάδας· Ð Α ¥ρα τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν. ™πεˆ οâν Ð Α ˜αυτÕν µν πολλαπλασιάσας τÕν Β πεποίηκεν, τÕν δ Β πολλαπλασιάσας τÕν Γ πεποίηκεν, κύβος ¥ρα ™στˆν

Let any multitude whatsoever of numbers, A, B, C, D, E, F , be continuously proportional, (starting) from a unit. I say that the third from the unit, B, is square, and all those (numbers after that) which leave an interval of one (number). And the fourth (from the unit), C, (is) cube, and all those (numbers after that) which leave an interval of two (numbers). And the seventh (from the unit), F , (is) both cube and square, and all those (numbers after that) which leave an interval of five (numbers). For since as the unit is to A, so A (is) to B, the unit thus measures the number A the same number of times as A (measures) B [Def. 7.20]. And the unit measures the number A according to the units in it. Thus, A also measures B according to the units in A. A has thus made B (by) multiplying itself [Def. 7.15]. Thus, B is square. And since B, C, D are continuously proportional, and B is square, D is thus also square [Prop. 8.22]. So, for the same (reasons), F is also square. So, similarly, we can also show that all those (numbers after that) which leave an interval of one (number) are square. So I also say that the fourth (number) from the unit, C, is cube, and all those (numbers after that) which leave an interval of two (numbers). For since as the unit is to A, so B (is) to C, the unit thus measures the number A the same number of times that B (measures) C. And the unit measures the

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ELEMENTS BOOK 9

Ð Γ. κሠ™πεˆ οƒ Γ, ∆, Ε, Ζ ˜ξÁς ¢νάλογόν ε„σιν, Ð δ Γ κύβος ™στˆν, καˆ Ð Ζ ¥ρα κύβος ™στίν. ™δείχθη δ κሠτετράγωνος· Ð ¥ρα ›βδοµος ¢πÕ τÁς µονάδος κύβος τέ ™στι κሠτετράγωνος. еοίως δ¾ δείξοµεν, Óτι καˆ οƒ πέντε διαλείποντες πάντες κύβοι τέ ε„σι κሠτετράγωνοι· Óπερ œδει δε‹ξαι.

number A according to the units in A. And thus B measures C according to the units in A. A has thus made C (by) multiplying B. Therefore, since A has made B (by) multiplying itself, and has made C (by) multiplying B, C is thus cube. And since C, D, E, F are continuously proportional, and C is cube, F is thus also cube [Prop. 8.23]. And it was also shown (to be) square. Thus, the seventh (number) from the unit is (both) cube and square. So, similarly, we can show that all those (numbers after that) which leave an interval of five (numbers) are (both) cube and square. (Which is) the very thing it was required to show.

θ΄.

Proposition 9

'Ε¦ν ¢πÕ µονάδος Ðποσοιοàν ˜ξÁς κατ¦ τÕ συνεχς ¢ριθµοˆ ¢νάλογον ðσιν, Ð δ µετ¦ τ¾ν µονάδα τετράγωνος Ï, καˆ οƒ λοιποˆ πάντες τετράγωνοι œσονται. κሠ™¦ν Ð µετ¦ τ¾ν µονάδα κύβος Ï, καˆ οƒ λοιποˆ πάντες κύβοι œσονται.

If any multitude whatsoever of numbers is continuously proportional, (starting) from a unit, and the (one) after the unit is square, then all the remaining (numbers) will also be square. And if the (one) after the unit is cube, then all the remaining (numbers) will also be cube.

Α Β Γ ∆ Ε Ζ

A B C D E F

”Εστωσαν ¢πÕ µονάδος ˜ξÁς ¢νάλογον Ðσοιδηποτοàν ¢ριθµοˆ οƒ Α, Β, Γ, ∆, Ε, Ζ, Ð δ µετ¦ τ¾ν µονάδα Ð Α τετράγωνος œστω· λέγω, Óτι καˆ οƒ λοιποˆ πάντες τετράγωνοι œσονται. “Οτι µν οâν Ð τρίτος ¢πÕ τÁς µονάδος Ð Β τετράγωνός ™στι καˆ οƒ ›να διαπλείποντες πάντες, δέδεικται· λέγω [δή], Óτι καˆ οƒ λοιποˆ πάντες τετράγωνοί ε„σιν. ™πεˆ γ¦ρ οƒ Α, Β, Γ ˜ξÁς ¢νάλογόν ε„σιν, καί ™στιν Ð Α τετράγωνος, καˆ Ð Γ [¥ρα] τετράγωνος ™στιν. πάλιν, ™πεˆ [καˆ] οƒ Β, Γ, ∆ ˜ξÁς ¢νάλογόν ε„σιν, καί ™στιν Ð Β τετράγωνος, καˆ Ð ∆ [¥ρα] τετράγωνός ™στιν. еοίως δ¾ δείξοµεν, Óτι καˆ οƒ λοιποˆ πάντες τετράγωνοί ε„σιν. 'Αλλ¦ δ¾ œστω Ð Α κύβος· λέγω, Óτι καˆ οƒ λοιποˆ πάντες κύβοι ε„σίν. “Οτι µν οâν Ð τέταρτος ¢πÕ τÁς µονάδος Ð Γ κύβος ™στˆ καˆ οƒ δύο διαλείποντες πάντες, δέδεικται· λέγω [δή], Óτι καˆ οƒ λοιποˆ πάντες κύβοι ε„σίν. ™πεˆ γάρ ™στιν æς ¹ µον¦ς πρÕς τÕν Α, οÛτως Ð Α πρÕς τÕν Β, „σάκις ¢ρα ¹ µον¦ς τÕν Α µετρε‹ καˆ Ð Α τÕν Β. ¹ δ µον¦ς τÕν Α µετρε‹ κατ¦ τ¦ς ™ν αÙτù µονάδας· καˆ Ð Α ¥ρα τÕν

Let any multitude whatsoever of numbers, A, B, C, D, E, F , be continuously proportional, (starting) from a unit. And let the (one) after the unit, A, be square. I say that all the remaining (numbers) will also be square. In fact, it has (already) been shown that the third (number) from the unit, B, is square, and all those (numbers after that) which leave an interval of one (number) [Prop. 9.8]. [So] I say that all the remaining (numbers) are also square. For since A, B, C are continuously proportional, and A (is) square, C is [thus] also square [Prop. 8.22]. Again, since B, C, D are [also] continuously proportional, and B is square, D is [thus] also square [Prop. 8.22]. So, similarly, we can show that all the remaining (numbers) are also square. And so let A be cube. I say that all the remaining (numbers) are also cube. In fact, it has (already) been shown that the fourth (number) from the unit, C, is cube, and all those (numbers after that) which leave an interval of two (numbers) [Prop. 9.8]. [So] I say that all the remaining (numbers)

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ELEMENTS BOOK 9

Β µετρε‹ κατ¦ τ¦ς ™ν αØτù µονάδας· Ð Α ¥ρα ˜αυτÕν πολλαπλασιάσας τÕν Β πεποίηκεν. καί ™στιν Ð Α κύβος. ™¦ν δ κύβος ¢ριθµÕς ˜αυτÕν πολλαπλασιάσας ποιÍ τινα, Ð γενόµενος κύβος ™στίν· καˆ Ð Β ¥ρα κύβος ™στίν. κሠ™πεˆ τέσσαρες ¢ριθµοˆ οƒ Α, Β, Γ, ∆ ˜ξÁς ¢νάλογόν ε„σιν, καί ™στιν Ð Α κύβος, καˆ Ð ∆ ¥ρα κύβος ™στίν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Ε κύβος ™στίν, καˆ Ðµοίως οƒ λοιποˆ πάντες κύβοι ε„σίν· Óπερ œδει δε‹ξαι.

are also cube. For since as the unit is to A, so A (is) to B, the unit thus measures A the same number of times as A (measures) B. And the unit measures A according to the units in it. Thus, A also measures B according to the units in (A). A has thus made B (by) multiplying itself. And A is cube. And if a cube number makes some (number by) multiplying itself then the created (number) is cube [Prop. 9.3]. Thus, B is also cube. And since the four numbers A, B, C, D are continuously proportional, and A is cube, D is thus also cube [Prop. 8.23]. So, for the same (reasons), E is also cube, and, similarly, all the remaining (numbers) are cube. (Which is) the very thing it was required to show.

ι΄.

Proposition 10

'Ε¦ν ¢πÕ µονάδος Ðποσοιοàν ¢ριθµοˆ [˜ξÁς] ¢νάλογον ðσιν, Ð δ µετ¦ τ¾ν µονάδα µ¾ Ï τετράγωνος, οÙδ' ¥λλος οÙδεˆς τετράγωνος œσται χωρˆς τοà τρίτου ¢πÕ τÁς µονάδος κሠτîν ›να διαλειπόντων πάντων. κሠ™¦ν Ð µετ¦ τ¾ν µονάδα κύβος µ¾ Ï, οÙδ ¥λλος οÙδεˆς κύβος œσται χωρˆς τοà τετάρτου ¢πÕ τÁς µονάδος κሠτîν δύο διαλειπόντων πάντων.

If any multitude whatsoever of numbers is [continuously] proportional, (starting) from a unit, and the (one) after the unit is not square, then no other (number) will be square either, apart from the third from the unit, and all those (numbers after that) which leave an interval of one (number). And if the (number) after the unit is not cube, then no other (number) will be cube either, apart from the fourth from the unit, and all those (numbers after that) which leave an interval of two (numbers).

Α Β Γ ∆

A B C D

”Εστωσαν ¢πÕ µονάδος ˜ξÁς ¢νάλογον Ðσοιδηποτοàν ¢ριθµοˆ οƒ Α, Β, Γ, ∆, Ε, Ζ, Ð µετ¦ τ¾ν µονάδα Ð Α µ¾ œστω τετράγωνος· λέγω, Óτι οÙδ ¥λλος οÙδεˆς τετράγωνος œσται χωρˆς τοà τρίτου ¢πÕ τ¾ς µονάδος [κሠτîν ›να διαλειπόντων]. Ε„ γ¦ρ δυνατόν, œστω Ð Γ τετράγωνος. œστι δ καˆ Ð Β τετράγωνος· οƒ Β, Γ ¥ρα πρÕς ¢λλήλους λόγον œχουσιν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. καί ™στιν æς Ð Β πρÕς τÕν Γ, Ð Α πρÕς τÕν Β· οƒ Α, Β ¥ρα πρÕς ¢λλήλους λόγον œχουσιν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· éστε οƒ Α, Β Óµοιοι ™πίπεδοί ε„σιν. καί ™στι τετράγωνος Ð Β· τετράγωνος ¥ρα ™στˆ καˆ Ð Α· Óπερ οÙχ Øπέκειτο. οÙκ ¥ρα Ð Γ τετράγωνός ™στιν. еοίως δ¾ δείξοµεν, Óτι οÙδ' ¥λλος οÙδεˆς τετράγωνός ™στι χωρˆς τοà τρίτου ¢πÕ τÁς µονάδος κሠτîν ›να διαλειπόντων. 'Αλλ¦ δ¾ µ¾ œστω Ð Α κύβος. λέγω, Óτι οÙδ' ¥λλος οÙδεˆς κύβος œσται χωρˆς τοà τετάρτου ¢πÕ τÁς µονάδος κሠτîν δύο διαλειπόντων.

Let any multitude whatsoever of numbers, A, B, C, D, E, F , be continuously proportional, (starting) from a unit. And let the (number) after the unit, A, not be square. I say that no other (number) will be square either, apart from the third from the unit [and (all) those (numbers after that) which leave an interval of one (number)]. For, if possible, let C be square. And B is also square [Prop. 9.8]. Thus, B and C have to one another (the) ratio which (some) square number (has) to (some other) square number. And as B is to C, (so) A (is) to B. Thus, A and B have to one another (the) ratio which (some) square number has to (some other) square number. Hence, A and B are similar plane (numbers) [Prop. 8.26]. And B is square. Thus, A is also square. The very opposite thing was assumed. C is thus not square. So, similarly, we can show that no other (number is) square either, apart from the third from the unit, and (all) those (numbers after that) which leave an interval

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ELEMENTS BOOK 9

Ε„ γ¦ρ δυνατόν, œστω Ð ∆ κύβος. œστι δ καˆ Ð Γ κύβος· τέταρτος γάρ ™στιν ¢πÕ τÁς µονάδος. καί ™στιν æς Ð Γ πρÕς τÕν ∆, Ð Β πρÕς τÕν Γ· καˆ Ð Β ¥ρα πρÕς τÕν Γ λόγον œχει, Öν κύβος πρÕς κύβον. καί ™στιν Ð Γ κύβος· καˆ Ð Β ¥ρα κύβος ™στίν. κሠ™πεί ™στιν æς ¹ µον¦ς πρÕς τÕν Α, Ð Α πρÕς τÕν Β, ¹ δ µον¦ς τÕν Α µετρε‹ κατ¦ τ¦ς ™ν αÙτù µονάδας, καˆ Ð Α ¥ρα τÕν Β µετρε‹ κατ¦ τ¦ς ™ν αØτù µονάδας· Ð Α ¥ρα ˜αυτÕν πολλαπλασιάσας κύβον τÕν Β πεποίηκεν. ™¦ν δ ¢ριθµÕς ˜αυτÕν πολλαπλασιάσας κύβον ποιÍ, κሠαÙτÕς κύβος œσται. κύβος ¥ρα καˆ Ð Α· Óπερ οÙχ Øπόκειται. οÙχ ¥ρα Ð ∆ κύβος ™στίν. еοίως δ¾ δείξοµεν, Óτι οÙδ' ¥λλος οÙδεˆς κύβος ™στˆ χωρˆς τοà τετάρτου ¢πÕ τÁς µονάδος κሠτîν δύο διαλειπόντων· Óπερ œδει δε‹ξαι.

of one (number). And so let A not be cube. I say that no other (number) will be cube either, apart from the fourth from the unit, and (all) those (numbers after that) which leave an interval of two (numbers). For, if possible, let D be cube. And C is also cube [Prop. 9.8]. For it is the fourth (number) from the unit. And as C is to D, (so) B (is) to C. And B thus has to C the ratio which (some) cube (number has) to (some other) cube (number). And C is cube. Thus, B is also cube [Props. 7.13, 8.25]. And since as the unit is to A, (so) A (is) to B, and the unit measures A according to the units in it, A thus also measures B according to the units in (A). Thus, A has made the cube (number) B (by) multiplying itself. And if a number makes a cube (number by) multiplying itself then it itself will be cube [Prop. 9.6]. Thus, A (is) also cube. The very opposite thing was assumed. Thus, D is not cube. So, similarly, we can show that no other (number) is cube either, apart from the fourth from the unit, and (all) those (numbers after that) which leave an interval of two (numbers). (Which is) the very thing it was required to show.

ια΄.

Proposition 11

'Ε¦ν ¢πÕ µονάδος Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον If any multitude whatsoever of numbers is continuðσιν, Ð ™λάττων τÕν µείζονα µετρε‹ κατά τινα τîν ously proportional, (starting) from a unit, then a lesser Øπαρχόντων ™ν το‹ς ¢νάλογον ¢ριθµο‹ς. (number) measures a greater according to some existing (number) among the proportional numbers.

Α Β Γ ∆ Ε

A B C D E

”Εστωσαν ¢πÕ µονάδος τÁς Α Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον οƒ Β, Γ, ∆, Ε· λέγω, Óτι τîν Β, Γ, ∆, Ε Ð ™λάχιστος Ð Β τÕν Ε µετρε‹ κατά τινα τîν Γ, ∆. 'Επεˆ γάρ ™στιν æς ¹ Α µον¦ς πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Ε, „σάκις ¥ρα ¹ Α µον¦ς τÕν Β ¢ριθµÕν µετρε‹ καˆ Ð ∆ τÕν Ε· ™ναλλ¦ξ ¥ρα „σάκις ¹ Α µον¦ς τÕν ∆ µετρε‹ καˆ Ð Β τÕν Ε. ¹ δ Α µον¦ς τÕν ∆ µετρε‹ κατ¦ τ¦ς ™ν αÙτù µονάδας· καˆ Ð Β ¥ρα τÕν Ε µετρε‹ κατ¦ τ¦ς ™ν τù ∆ µονάδας· éστε Ð ™λάσσων Ð Β τÕν µείζονα τÕν Ε µετρε‹ κατά τινα ¢ριθµÕν τîν Øπαρχόντων ™ν το‹ς ¢νάλογον ¢ριθµο‹ς.

Let any multitude whatsoever of numbers, B, C, D, E, be continuously proportional, (starting) from the unit A. I say that, for B, C, D, E, the least (number), B, measures E according to some (one) of C, D. For since as the unit A is to B, so D (is) to E, the unit A thus measures the number B the same number of times as D (measures) E. Thus, alternately, the unit A measures D the same number of times as B (measures) E [Prop. 7.15]. And the unit A measures D according to the units in it. Thus, B also measures E according to the units in D. Hence, the lesser (number) B measures the greater E according to some existing number among the

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ΣΤΟΙΧΕΙΩΝ θ΄.

ELEMENTS BOOK 9 proportional numbers (namely, D).

Πόρισµα.

Corollary

Κሠφανερόν, Óτι ¿ν œχει τάξιν Ð µετρîν ¢πÕ And (it is) clear that what(ever relative) place the µονάδος, τ¾ν αÙτ¾ν œχει καˆ Ð καθ' Öν µετρε‹ ¢πÕ τοà measuring (number) has from the unit, the (number) µετρουµένου ™πˆ τÕ πρÕ αÙτοà. Óπερ œδει δε‹ξαι. according to which it measures has the same (relative) place from the measured (number), in (the direction of the number) before it. (Which is) the very thing it was required to show.

ιβ΄.

Proposition 12

'Ε¦ν ¢πÕ µονάδος Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον ðσιν, Øφ' Óσων ¨ν Ð œσχατος πρώτων ¢ριθµîν µετρÁται, ØπÕ τîν αÙτîν καˆ Ð παρ¦ τ¾ν µονάδα µετρηθήσεται.

If any multitude whatsoever of numbers is continuously proportional, (starting) from a unit, then however many prime numbers the last (number) is measured by, the (number) next to the unit will also be measured by the same (prime numbers).

Α Β Γ ∆

Ε Ζ Η Θ

”Εστωσαν ¢πÕ µονάδος Ðποσοιδηποτοàν ¢ριθµοˆ ¢νάλογον οƒ Α, Β, Γ, ∆· λέγω, Óτι Øφ' Óσων ¨ν Ð ∆ πρώτων ¢ριθµîν µετρÁται, ØπÕ τîν αÙτîν καˆ Ð Α µετρηθήσεται. Μετρείσθω γ¦ρ Ð ∆ Øπό τινος πρώτου ¢ριθµοà τοà Ε· λέγω, Óτι Ð Ε τÕν Α µετρε‹. µ¾ γάρ· καί ™στιν Ð Ε πρîτος, ¤πας δ πρîτος ¢ριθµÕς πρÕς ¤παντα, Öν µ¾ µετρε‹, πρîτός ™στιν· οƒ Ε, Α ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. κሠ™πεˆ Ð Ε τÕν ∆ µετρε‹, µετρείτω αÙτÕν κατ¦ τÕν Ζ· Ð Ε ¥ρα τÕν Ζ πολλαπλασιάσας τÕν ∆ πεποίηκεν. πάλιν, ™πεˆ Ð Α τÕν ∆ µετρε‹ κατ¦ τ¦ς ™ν τù Γ µονάδας, Ð Α ¥ρα τÕν Γ πολλαπλασιάσας τÕν ∆ πεποίηκεν. ¢λλ¦ µ¾ν καˆ Ð Ε τÕν Ζ πολλαπλασιάσας τÕν ∆ πεποίηκεν· Ð ¥ρα ™κ τîν Α, Γ ‡σος ™στˆ τù ™κ τîν Ε, Ζ. œστιν ¥ρα æς Ð Α πρÕς τÕν Ε, Ð Ζ πρÕς τÕν Γ. οƒ δ Α, Ε πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον· µετρε‹ ¥ρα Ð Ε τÕν Γ. µετρείτω αÙτÕν κατ¦ τÕν Η· Ð Ε ¥ρα τÕν Η πολλαπλασιάσας τÕν Γ πεποίηκεν. ¢λλ¦ µ¾ν δι¦ τÕ πρÕ τούτου καˆ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν. Ð ¥ρα ™κ τîν Α, Β ‡σος ™στˆ τù ™κ τîν Ε, Η. œστιν ¥ρα æς Ð Α πρÕς τÕν Ε, Ð Η πρÕς τÕν Β. οƒ δ Α, Ε πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι ¢ριθµοˆ µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας αÙτο‹ς „σάκις Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον· µετρε‹

A

E

B

F

C

G

D

H

Let any multitude whatsoever of numbers, A, B, C, D, be (continuously) proportional, (starting) from a unit. I say that however many prime numbers D is measured by, A will also be measured by the same (prime numbers). For let D be measured by some prime number E. I say that E measures A. For (suppose it does) not. E is prime, and every prime number is prime to every number which it does not measure [Prop. 7.29]. Thus, E and A are prime to one another. And since E measures D, let it measure it according to F . Thus, E has made D (by) multiplying F . Again, since A measures D according to the units in C [Prop. 9.11 corr.], A has thus made D (by) multiplying C. But, in fact, E has also made D (by) multiplying F . Thus, the (number created) from (multiplying) A, C is equal to the (number created) from (multiplying) E, F . Thus, as A is to E, (so) F (is) to C [Prop. 7.19]. And A and E (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures C. Let it measure it according to G. Thus, E has made C (by) multiplying G. But, in fact, via the (proposition) before this, A has also made C (by) multiplying B [Prop. 9.11 corr.]. Thus, the (number created)

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ELEMENTS BOOK 9

¥ρα Ð Ε τÕν Β. µετρείτω αÙτÕν κατ¦ τÕν Θ· Ð Ε ¥ρα τÕν Θ πολλαπλασιάσας τÕν Β πεποίηκεν. ¢λλ¦ µ¾ν καˆ Ð Α ˜αυτÕν πολλαπλασιάσας τÕν Β πεποίηκεν· Ð ¥ρα ™κ τîν Ε, Θ ‡σος ™στˆ τù ¢πÕ τοà Α. œστιν ¥ρα æς Ð Ε πρÕς τÕν Α, Ð Α πρÕς τÕν Θ. οƒ δ Α, Ε πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον· µετρε‹ ¥ρα Ð Ε τÕν Α æς ¹γούµενος ¹γούµενον. ¢λλ¦ µ¾ν κሠοÙ µετρε‹· Óπερ ¢δύνατον. οÙκ ¥ρα οƒ Ε, Α πρîτοι πρÕς ¢λλήλους ε„σίν. σύνθετοι ¥ρα. οƒ δ σύνθετοι ØπÕ [πρώτου] ¢ριθµοà τινος µετροàνται. κሠ™πεˆ Ð Ε πρîτος Øπόκειται, Ð δ πρîτος ØπÕ ˜τέρου ¢ριθµοà οÙ µετρε‹ται À Øφ' ˜αυτοà, Ð Ε ¥ρα τοÝς Α, Ε µετρε‹· éστε Ð Ε τÕν Α µετρε‹. µετρε‹ δ κሠτÕν ∆· Ð Ε ¥ρα τοÝς Α, ∆ µετρε‹. еοίως δ¾ δείξοµεν, Óτι Øφ' Óσων ¨ν Ð ∆ πρώτων ¢ριθµîν µετρÁται, ØπÕ τîν αÙτîν καˆ Ð Α µετρηθήσεται· Óπερ œδει δε‹ξαι.

from (multiplying) A, B is equal to the (number created) from (multiplying) E, G. Thus, as A is to E, (so) G (is) to B [Prop. 7.19]. And A and E (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures B. Let it measure it according to H. Thus, E has made B (by) multiplying H. But, in fact, A has also made B (by) multiplying itself [Prop. 9.8]. Thus, the (number created) from (multiplying) E, H is equal to the (square) on A. Thus, as E is to A, (so) A (is) to H [Prop. 7.19]. And A and E are prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures A, as the leading (measuring the) leading. But, in fact, (E) also does not measure (A). The very thing (is) impossible. Thus, E and A are not prime to one another. Thus, (they are) composite (to one another). And (numbers) composite (to one another) are (both) measured by some [prime] number [Def. 7.14]. And since E is assumed (to be) prime, and a prime (number) is not measured by another number (other) than itself [Def. 7.11], E thus measures (both) A and E. Hence, E measures A. And it also measures D. Thus, E measures (both) A and D. So, similarly, we can show that however many prime numbers D is measured by, A will also be measured by the same (prime numbers). (Which is) the very thing it was required to show.

ιγ΄.

Proposition 13

'Ε¦ν ¢πÕ µονάδος Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον If any multitude whatsoever of numbers is continuðσιν, Ð δ µετ¦ τ¾ν µονάδα πρîτος Ï, Ð µέγιστος Øπ' ously proportional, (starting) from a unit, and the (numοÙδενÕς [¥λλου] µετρηθήσεται παρξ τîν Øπαρχόντων ber) after the unit is prime, then the greatest (number) ™ν το‹ς ¢νάλογον ¢ριθµο‹ς. will be measured by no [other] (numbers) except (numbers) existing among the proportional numbers.

Α Β Γ ∆

Ε Ζ Η Θ

A B C D

”Εστωσαν ¢πÕ µονάδος Ðποσοιοàν ¢ριθµοˆ ˜ξÁς

263

E F G H

Let any multitude whatsoever of numbers, A, B, C,

ΣΤΟΙΧΕΙΩΝ θ΄.

ELEMENTS BOOK 9

¢νάλογον οƒ Α, Β, Γ, ∆, Ð δ µετ¦ τ¾ν µονάδα Ð Α πρîτος œστω· λέγω, Óτι Ð µέγιστος αÙτîν Ð ∆ Øπ' οÙδενÕς ¥λλου µετρηθήσεται παρξ τîν Α, Β, Γ. Ε„ γ¦ρ δυνατόν, µετρείσθω ØπÕ τοà Ε, καˆ Ð Ε µηδενˆ τîν Α, Β, Γ œστω Ð αÙτός. φανερÕν δή, Óτι Ð Ε πρîτος οÜκ ™στιν. ε„ γ¦ρ Ð Ε πρîτός ™στι κሠµετρε‹ τÕν ∆, κሠτÕν Α µετρήσει πρîτον Ôντα µ¾ íν αÙτù Ð αÙτός· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα Ð Ε πρîτός ™στιν. σύνθετος ¥ρα. π©ς δ σύνθετος ¢ριθµÕς ØπÕ πρώτου τινÕς ¢ριθµοà µετρε‹ται. Ð Ε ¥ρα ØπÕ πρώτου τινÕς ¢ριθµοà µετρε‹ται. λέγω δή, Óτι Øπ' οÙδενÕς ¥λλου πρώτου µετρηθήσεται πλ¾ν τοà Α. ε„ γ¦ρ Øφ' ˜τέρου µετρε‹ται Ð Ε, Ð δ Ε τÕν ∆ µετρε‹, κ¢κε‹νος ¥ρα τÕν ∆ µετρήσει· éστε κሠτÕν Α µετρήσει πρîτον Ôντα µ¾ íν αÙτù Ð αÙτός· Óπερ ™στˆν ¢δύνατον. Ð Α ¥ρα τÕν Ε µετρε‹. κሠ™πεˆ Ð Ε τÕν ∆ µετρε‹, µετρείτω αÙτÕν κατ¦ τÕν Ζ. λέγω, Óτι Ð Ζ οÙδενˆ τîν Α, Β, Γ ™στιν Ð αÙτός. ε„ γ¦ρ Ð Ζ ˜νˆ τîν Α, Β, Γ ™στιν Ð αÙτÕς κሠµετρε‹ τÕν ∆ κατ¦ τÕν Ε, κሠεŒς ¥ρα τîν Α, Β, Γ τÕν ∆ µετρε‹ κατά τÕν Ε. ¢λλ¦ εŒς τîν Α, Β, Γ τÕν ∆ µετρε‹ κατά τινα τîν Α, Β, Γ· καˆ Ð Ε ¥ρα ˜νˆ τîν Α, Β, Γ ™στιν Ð αÙτός· Óπερ οÙχ Øπόκειται. οÙκ ¥ρα Ð Ζ ˜νˆ τîν Α, Β, Γ ™στιν Ð αÙτός. еοίως δ¾ δείξοµεν, Óτι µετρε‹ται Ð Ζ ØπÕ τοà Α, δεικνύντες πάλιν, Óτι Ð Ζ οÜκ ™στι πρîτος. ε„ γ¦ρ, κሠµετρε‹ τÕν ∆, κሠτÕν Α µετρήσει πρîτον Ôντα µ¾ íν αÙτù Ð αÙτός· Óπερ ™στˆν ¢δύνατον· οÙκ ¥ρα πρîτός ™στιν Ð Ζ· σύνθετος ¥ρα. ¤πας δ σύνθετος ¢ριθµÕς ØπÕ πρώτου τινÕς ¢ριθµοà µετρε‹ται· Ð Ζ ¥ρα ØπÕ πρώτου τινÕς ¢ριθµοà µετρε‹ται. λέγω δή, Óτι Øφ' ˜τέρου πρώτου οÙ µετρηθήσεται πλ¾ν τοà Α. ε„ γ¦ρ ›τερός τις πρîτος τÕν Ζ µετρε‹, Ð δ Ζ τÕν ∆ µετρε‹, κ¢κε‹νος ¥ρα τÕν ∆ µετρήσει· éστε κሠτÕν Α µετρήσει πρîτον Ôντα µ¾ íν αÙτù Ð αÙτός· Óπερ ™στˆν ¢δύνατον. Ð Α ¥ρα τÕν Ζ µετρε‹. κሠ™πεˆ Ð Ε τÕν ∆ µετρε‹ κατ¦ τÕν Ζ, Ð Ε ¥ρα τÕν Ζ πολλαπλασιάσας τÕν ∆ πεποίηκεν. ¢λλ¦ µ¾ν καˆ Ð Α τÕν Γ πολλαπλασιάσας τÕν ∆ πεποίηκεν· Ð ¥ρα ™κ τîν Α, Γ ‡σος ™στˆ τù ™κ τîν Ε, Ζ. ¢νάλογον ¥ρα ™στˆν æς Ð Α πρÕς τÕν Ε, οÛτως Ð Ζ πρÕς τÕν Γ. Ð δ Α τÕν Ε µετρε‹· καˆ Ð Ζ ¥ρα τÕν Γ µετρε‹. µετρείτω αÙτÕν κατ¦ τÕν Η. еοίως δ¾ δείξοµεν, Óτι Ð Η οÙδενˆ τîν Α, Β ™στιν Ð αÙτός, κሠÓτι µετρε‹ται ØπÕ τοà Α. κሠ™πεˆ Ð Ζ τÕν Γ µετρε‹ κατ¦ τÕν Η, Ð Ζ ¥ρα τÕν Η πολλαπλασιάσας τÕν Γ πεποίηκεν. ¢λλ¦ µ¾ν καˆ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν· Ð ¥ρα ™κ τîν Α, Β ‡σος ™στˆ τù ™κ τîν Ζ, Η. ¢νάλογον ¥ρα æς Ð Α πρÕς τÕν Ζ, Ð Η πρÕς τÕν Β. µετρε‹ δ Ð Α τÕν Ζ· µετρε‹ ¥ρα καˆ Ð Η τÕν Β. µετρείτω αÙτÕν κατ¦ τÕν Θ. еοίως δ¾ δείξοµεν, Óτι Ð Θ τù Α οÙκ œστιν Ð αÙτός. κሠ™πεˆ Ð Η τÕν Β µετρε‹ κατ¦ τÕν Θ, Ð Η ¥ρα τÕν Θ πολλαπλασιάσας τÕν Β πεποίηκεν. ¢λλ¦ µ¾ν καˆ Ð Α ˜αυτÕν πολλαπλασιάσας τÕν Β πεποίηκεν· Ð ¥ρα ØπÕ

D, be continuously proportional, (starting) from a unit. And let the (number) after the unit, A, be prime. I say that the greatest of them, D, will be measured by no other (numbers) except A, B, C. For, if possible, let it be measured by E, and let E not be the same as one of A, B, C. So it is clear that E is not prime. For if E is prime, and measures D, then it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12]. The very thing is impossible. Thus, E is not prime. Thus, (it is) composite. And every composite number is measured by some prime number [Prop. 7.31]. Thus, E is measured by some prime number. So I say that it will be measured by no other prime number than A. For if E is measured by another (prime number), and E measures D, then this (prime number) will thus also measure D. Hence, it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12]. The very thing is impossible. Thus, A measures E. And since E measures D, let it measure it according to F . I say that F is not the same as one of A, B, C. For if F is the same as one of A, B, C, and measures D according to E, then one of A, B, C thus also measures D according to E. But one of A, B, C (only) measures D according to some (one) of A, B, C [Prop. 9.11]. And thus E is the same as one of A, B, C. The very opposite thing was assumed. Thus, F is not the same as one of A, B, C. Similarly, we can show that F is measured by A, (by) again showing that F is not prime. For if (F is prime), and measures D, then it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12]. The very thing is impossible. Thus, F is not prime. Thus, (it is) composite. And every composite number is measured by some prime number [Prop. 7.31]. Thus, F is measured by some prime number. So I say that it will be measured by no other prime number than A. For if some other prime (number) measures F , and F measures D, then this (prime number) will thus also measure D. Hence, it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12]. The very thing is impossible. Thus, A measures F . And since E measures D according to F , E has thus made D (by) multiplying F . But, in fact, A has also made D (by) multiplying C [Prop. 9.11 corr.]. Thus, the (number created) from (multiplying) A, C is equal to the (number created) from (multiplying) E, F . Thus, proportionally as A is to E, so F (is) to C [Prop. 7.19]. And A measures E. Thus, F also measures C. Let it measure it according to G. So, similarly, we can show that G is not the same as one of A, B, and that it is measured by A. And since F measures C according to G, F has thus made C (by) multiplying G. But, in fact, A has also made C (by) mul-

264

ΣΤΟΙΧΕΙΩΝ θ΄.

ELEMENTS BOOK 9

Θ, Η ‡σος ™στˆ τù ¢πÕ τοà Α τετραγώνJ· œστιν ¥ρα æς Ð Θ πρÕς τÕν Α, Ð Α πρÕς τÕν Η. µετρε‹ δ Ð Α τÕν Η· µετρε‹ ¥ρα καˆ Ð Θ τÕν Α πρîτον Ôντα µ¾ íν αÙτù Ð αÙτός· Óπερ ¥τοπον. οÙκ ¥ρα Ð µέγιστος Ð ∆ ØπÕ ˜τέρου ¢ριθµοà µετρηθήσεται παρξ τîν Α, Β, Γ· Óπερ œδει δε‹ξαι.

tiplying B [Prop. 9.11 corr.]. Thus, the (number created) from (multiplying) A, B is equal to the (number created) from (multiplying) F , G. Thus, proportionally, as A (is) to F , so G (is) to B [Prop. 7.19]. And A measures F . Thus, G also measures B. Let it measure it according to H. So, similarly, we can show that H is not the same as A. And since G measures B according to H, G has thus made B (by) multiplying H. But, in fact, A has also made B (by) multiplying itself [Prop. 9.8]. Thus, the (number created) from (multiplying) H, G is equal to the square on A. Thus, as H is to A, (so) A (is) to G [Prop. 7.19]. And A measures G. Thus, H also measures A, (despite A) being prime (and) not being the same as it. The very thing (is) absurd. Thus, the greatest (number) D cannot be measured by another (number) except (one of) A, B, C. (Which is) the very thing it was required to show.

ιδ΄.

Proposition 14

'Ε¦ν ™λάχιστος ¢ριθµÕς ØπÕ πρώτων ¢ριθµîν µετρÁται, Øπ' οØδενÕς ¥λλου πρώτου ¢ριθµοà µετρηθήσεται παρξ τîν ™ξ ¢ρχÁς µετρούντων.

If a least number is measured by (some) prime numbers then it will not be measured by any other prime number except (one of) the original measuring (numbers).

Α Ε Ζ

A E F

Β Γ ∆

'Ελάχιστος γ¦ρ ¢ριθµÕς Ð Α ØπÕ πρώτων ¢ριθµîν τîν Β, Γ, ∆ µετρείσθω· λέγω, Óτι Ð Α Øπ' οÙδενÕς ¥λλου πρώτου ¢ριθµοà µετρηθήσεται παρξ τîν Β, Γ, ∆. Ε„ γ¦ρ δυνατόν, µετρείσθω ØπÕ πρώτου τοà Ε, καˆ Ð Ε µηδενˆ τîν Β, Γ, ∆ œστω Ð αÙτός. κሠ™πεˆ Ð Ε τÕν Α µετρε‹, µετρείτω αÙτÕν κατ¦ τÕν Ζ· Ð Ε ¥ρα τÕν Ζ πολλαπλασιάσας τÕν Α πεποίηκεν. κሠµετρε‹ται Ð Α ØπÕ πρώτων ¢ριθµîν τîν Β, Γ, ∆. ™¦ν δ δύο ¢ριθµοˆ πολλαπλασιάσαντες ¢λλήλους ποιîσί τινα, τÕν δ γενόµενον ™ξ αÙτîν µετρÍ τις πρîτος ¢ριθµός, κሠ›να τîν ™ξ ¢ρχÁς µετρήσει· οƒ Β, Γ, ∆ ¥ρα ›να τîν Ε, Ζ µετρήσουσιν. τÕν µν οâν Ε οÙ µετρήσουσιν· Ð γ¦ρ Ε πρîτός ™στι κሠοÙδενˆ τîν Β, Γ, ∆ Ð αÙτός. τÕν Ζ ¥ρα µετροàσιν ™λάσσονα Ôντα τοà Α· Óπερ ¢δύνατον. Ð γ¦ρ Α Øπόκειται ™λάχιστος ØπÕ τîν Β, Γ, ∆ µετρούµενος. οÙκ ¥ρα τÕν Α µετρήσει πρîτος ¢ριθµÕς παρξ τîν Β, Γ, ∆· Óπερ œδει δε‹ξαι.

B C D

For let A be the least number measured by the prime numbers B, C, D. I say that A will not be measured by any other prime number except (one of) B, C, D. For, if possible, let it be measured by the prime (number) E. And let E not be the same as one of B, C, D. And since E measures A, let it measure it according to F . Thus, E has made A (by) multiplying F . And A is measured by the prime numbers B, C, D. And if two numbers make some (number by) multiplying one another, and some prime number measures the number created from them, then (the prime number) will also measure one of the original (numbers) [Prop. 7.30]. Thus, B, C, D will measure one of E, F . In fact, they do not measure E. For E is prime, and not the same as one of B, C, D. Thus, they (all) measure F , which is less than A. The very thing (is) impossible. For A was assumed (to be) the least (number) measured by B, C, D. Thus, no prime number can measure A except (one of) B, C, D. (Which is) the very thing it was required to show.

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ΣΤΟΙΧΕΙΩΝ θ΄.

ELEMENTS BOOK 9

ιε΄.

Proposition 15

'Ε¦ν τρε‹ς ¢ριθµοˆ ˜ξÁς ¢νάλογον ðσιν ™λάχιστοι If three continuously proportional numbers are the τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς, δύο Ðποιοιοàν least of those (numbers) having the same ratio as them, συντεθέντες πρÕς τÕν λοιπÕν πρîτοί ε„σιν. then two (of them) added together in any way are prime to the remaining (one).

Α Β Γ

∆

Ε

Ζ

D

E

F

A B C

”Εστωσαν τρε‹ς ¢ριθµοˆ ˜ξÁς ¢νάλογον ™λάχιστοι τîν τÕν αÙτÕν λόγον ™χόντων αÙτο‹ς οƒ Α, Β, Γ· λέγω, Óτι τîν Α, Β, Γ δύο Ðποιοιοàν συντεθέντες πρÕς τÕν λοιπÕν πρîτοι ε„σιν, οƒ µν Α, Β πρÕς τÕν Γ, οƒ δ Β, Γ πρÕς τÕν Α κሠœτι οƒ Α, Γ πρÕς τÕν Β. Ε„λήφθωσαν γ¦ρ ™λάχιστοι ¢ριθµοˆ τîν τÕν αÙτÕν λόγον ™χόντων το‹ς Α, Β, Γ δύο οƒ ∆Ε, ΕΖ. φανερÕν δή, Óτι Ð µν ∆Ε ˜αυτÕν πολλαπλασιάσας τÕν Α πεποίηκεν, τÕν δ ΕΖ πολλαπλασιάσας τÕν Β πεποίηκεν, κሠœτι Ð ΕΖ ˜αυτÕν πολλαπλασιάσας τÕν Γ πεποίηκεν. κሠ™πεˆ οƒ ∆Ε, ΕΖ ™λάχιστοί ε„σιν, πρîτοι πρÕς ¢λλήλους ε„σιν. ™¦ν δ δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους ðσιν, κሠσυναµφότερος πρÕς ˜κάτερον πρîτός ™στιν· καˆ Ð ∆Ζ ¥ρα πρÕς ˜κάτερον τîν ∆Ε, ΕΖ πρîτός ™στιν. ¢λλ¦ µ¾ν καˆ Ð ∆Ε πρÕς τÕν ΕΖ πρîτός ™στιν· οƒ ∆Ζ, ∆Ε ¥ρα πρÕς τÕν ΕΖ πρîτοί ε„σιν. ™¦ν δ δύο ¢ριθµοˆ πρός τινα ¢ριθµÕν πρîτοι ðσιν, καˆ Ð ™ξ αÙτîν γενόµενος πρÕς τÕν λοιπÕν πρîτός ™στιν· éστε Ð ™κ τîν Ζ∆, ∆Ε πρÕς τÕν ΕΖ πρîτός ™στιν· éστε καˆ Ð ™κ τîν Ζ∆, ∆Ε πρÕς τÕν ¢πÕ τοà ΕΖ πρîτός ™στιν. [™¦ν γ¦ρ δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους ðσιν, Ð ™κ τοà ˜νÕς αÙτîν γενόµενος πρÕς τÕν λοιπÕν πρîτός ™στιν]. ¢λλ' Ð ™κ τîν Ζ∆, ∆Ε Ð ¢πÕ τοà ∆Ε ™στι µετ¦ τοà ™κ τîν ∆Ε, ΕΖ· Ð ¥ρα ¢πÕ τοà ∆Ε µετ¦ τοà ™κ τîν ∆Ε, ΕΖ πρÕς τÕν ¢πÕ τοà ΕΖ πρîτός ™στιν. καί ™στιν Ð µν ¢πÕ τοà ∆Ε Ð Α, Ð δ ™κ τîν ∆Ε, ΕΖ Ð Β, Ð δ ¢πÕ τοà ΕΖ Ð Γ· οƒ Α, Β ¥ρα συντεθέντες πρÕς τÕν Γ πρîτοί ε„σιν. еοίως δ¾ δείξοµεν, Óτι καˆ οƒ Β, Γ πρÕς τÕν Α πρîτοί ε„σιν. λέγω δή, Óτι καˆ οƒ Α, Γ πρÕς τÕν Β πρîτοί ε„σιν. ™πεˆ γ¦ρ Ð ∆Ζ πρÕς ˜κάτερον τîν ∆Ε, ΕΖ πρîτός ™στιν, καˆ Ð ¢πÕ τοà ∆Ζ πρÕς τÕν ™κ τîν ∆Ε, ΕΖ πρîτός ™στιν. ¢λλ¦ τù ¢πÕ τοà ∆Ζ ‡σοι ε„σˆν οƒ ¢πÕ τîν ∆Ε, ΕΖ µετ¦ τοà δˆς ™κ τîν ∆Ε, ΕΖ· καˆ οƒ ¢πÕ τîν ∆Ε, ΕΖ ¥ρα µετ¦ τοà δˆς ØπÕ τîν ∆Ε, ΕΖ πρÕς τÕν ØπÕ τîν ∆Ε, ΕΖ πρîτοί [ε„σι]. διελόντι οƒ ¢πÕ τîν ∆Ε, ΕΖ µετ¦ τοà ¤παξ ØπÕ ∆Ε, ΕΖ πρÕς τÕν ØπÕ ∆Ε, ΕΖ πρîτοί ε„σιν. œτι διελόντι οƒ ¢πÕ τîν ∆Ε, ΕΖ

Let A, B, C be three continuously proportional numbers (which are) the least of those (numbers) having the same ratio as them. I say that two of A, B, C added together in any way are prime to the remaining (one), (that is) A and B (prime) to C, B and C to A, and, further, A and C to B. Let the two least numbers, DE and EF , having the same ratio as A, B, C, have been taken [Prop. 8.2]. So it is clear that DE has made A (by) multiplying itself, and has made B (by) multiplying EF , and, further, EF has made C (by) multiplying itself [Prop. 8.2]. And since DE, EF are the least (of those numbers having the same ratio as them), they are prime to one another [Prop. 7.22]. And if two numbers are prime to one another then the sum (of them) is also prime to each [Prop. 7.28]. Thus, DF is also prime to each of DE, EF . But, in fact, DE is also prime to EF . Thus, DF , DE are (both) prime to EF . And if two numbers are (both) prime to some number then the (number) created from (multiplying) them is also prime to the remaining (number) [Prop. 7.24]. Hence, the (number created) from (multiplying) F D, DE is prime to EF . Hence, the (number created) from (multiplying) F D, DE is also prime to the (square) on EF [Prop. 7.25]. [For if two numbers are prime to one another then the (number) created from (squaring) one of them is prime to the remaining (number).] But the (number created) from (multiplying) F D, DE is the (square) on DE plus the (number created) from (multiplying) DE, EF [Prop. 2.3]. Thus, the (square) on DE plus the (number created) from (multiplying) DE, EF is prime to the (square) on EF . And the (square) on DE is A, and the (number created) from (multiplying) DE, EF (is) B, and the (square) on EF (is) C. Thus, A, B summed is prime to C. So, similarly, we can show that B, C (summed) is also prime to A. So I say that A, C (summed) is also prime to B. For since DF is prime to each of DE, EF then the (square) on DF

266

ΣΤΟΙΧΕΙΩΝ θ΄.

ELEMENTS BOOK 9

¥ρα πρÕς τÕν ØπÕ ∆Ε, ΕΖ πρîτοί ε„σιν. καί ™στιν Ð µν ¢πÕ τοà ∆Ε Ð Α, Ð δ ØπÕ τîν ∆Ε, ΕΖ Ð Β, Ð δ ¢πÕ τοà ΕΖ Ð Γ. οƒ Α, Γ ¥ρα συντεθέντες πρÕς τÕν Β πρîτοί ε„σιν· Óπερ œδει δε‹ξαι.

†

is also prime to the (number created) from (multiplying) DE, EF [Prop. 7.25]. But, the (sum of the squares) on DE, EF plus twice the (number created) from (multiplying) DE, EF is equal to the (square) on DF [Prop. 2.4]. And thus the (sum of the squares) on DE, EF plus twice the (rectangle contained) by DE, EF [is] prime to the (rectangle contained) by DE, EF . By separation, the (sum of the squares) on DE, EF plus once the (rectangle contained) by DE, EF is prime to the (rectangle contained) by DE, EF .† Again, by separation, the (sum of the squares) on DE, EF is prime to the (rectangle contained) by DE, EF . And the (square) on DE is A, and the (rectangle contained) by DE, EF (is) B, and the (square) on EF (is) C. Thus, A, C summed is prime to B. (Which is) the very thing it was required to show.

Since if α β measures α2 + β 2 + 2 α β then it also measures α2 + β 2 + α β, and vice versa.

ι$΄.

Proposition 16

'Ε¦ν δύο ¢ριθµοˆ πρîτοι πρÕς ¢λλήλους ðσιν, οÙκ œσται æς Ð πρîτος πρÕς τÕν δεύτερον, οÛτως Ð δεύτερος πρÕς ¥λλον τινά.

If two numbers are prime to one another then as the first is to the second, so the second (will) not (be) to some other (number).

Α Β Γ

A B C

∆ύο γ¦ρ ¢ριθµοˆ οƒ Α, Β πρîτοι πρÕς ¢λλήλους œστωσαν· λέγω, Óτι οÙκ œστιν æς Ð Α πρÕς τÕν Β, οÛτως Ð Β πρÕς ¥λλον τινά. Ε„ γ¦ρ δυνατόν, œστω æς Ð Α πρÕς τÕν Β, Ð Β πρÕς τÕν Γ. οƒ δ Α, Β πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι ¢ριθµοˆ µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον· µετρε‹ ¥ρα Ð Α τÕν Β æς ¹γούµενος ¹γούµενον. µετρε‹ δ κሠ˜αυτόν· Ð Α ¥ρα τοÝς Α, Β µετρε‹ πρώτους Ôντας πρÕς ¢λλήλους· Óπερ ¥τοπον. οÙκ ¥ρα œσται æς Ð Α πρÕς τÕν Β, οÛτως Ð Β πρÕς τÕν Γ· Óπερ œδει δε‹ξαι.

For let the two numbers A and B be prime to one another. I say that as A is to B, so B is not to some other (number). For, if possible, let it be that as A (is) to B, (so) B (is) to C. And A and B (are) prime (to one another). And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21]. And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, A measures B, as the leading (measuring) the leading. And (A) also measures itself. Thus, A measures A and B, which are prime to one another. The very thing (is) absurd. Thus, as A (is) to B, so B cannot be to C. (Which is) the very thing it was required to show.

ιζ΄.

Proposition 17

'Ε¦ν ðσιν Ðσοιδηποτοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον, οƒ If any multitude whatsoever of numbers is continuδ ¥κροι αÙτîν πρîτοι πρÕς ¢λλήλους ðσιν, οÙκ œσται ously proportional, and the outermost of them are prime æς Ð πρîτος πρÕς τÕν δεύτερον, οÛτως Ð œσχατος πρÕς to one another, then as the first (is) to the second, so the ¥λλον τινά. last will not be to some other (number). 267

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”Εστωσαν Ðσοιδηποτοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον οƒ Let A, B, C, D be any multitude whatsoever of conΑ, Β, Γ, ∆, οƒ δ ¥κροι αÙτîν οƒ Α, ∆ πρîτοι πρÕς tinuously proportional numbers. And let the outermost ¢λλήλους œστωσαν· λέγω, Óτι οÙκ œστιν æς Ð Α πρÕς τÕν of them, A and D, be prime to one another. I say that as Β, οÛτως Ð ∆ πρÕς ¥λλον τινά. A is to B, so D (is) not to some other (number).

Α Β Γ ∆ Ε

A B C D E

Ε„ γ¦ρ δυνατόν, œστω æς Ð Α πρÕς τÕν Β, οÛτως Ð ∆ πρÕς τÕν Ε. ™ναλλ¦ξ ¥ρα ™στˆν æς Ð Α πρÕς τÕν ∆, Ð Β πρÕς τÕν Ε. οƒ δ Α, ∆ πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι ¢ριθµοˆ µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον. µετρε‹ ¥ρα Ð Α τÕν Β. καί ™στιν æς Ð Α πρÕς τÕν Β, Ð Β πρÕς τÕν Γ. καˆ Ð Β ¥ρα τÕν Γ µετρε‹΄· éστε καˆ Ð Α τÕν Γ µετρε‹. κሠ™πεί ™στιν æς Ð Β πρÕς τÕν Γ, Ð Γ πρÕς τÕν ∆, µετρε‹ δ Ð Β τÕν Γ, µετρε‹ ¥ρα καˆ Ð Γ τÕν ∆. ¢λλ' Ð Α τÕν Γ ™µέτρει· éστε Ð Α κሠτÕν ∆ µετρε‹. µετρε‹ δ κሠ˜αυτόν. Ð Α ¥ρα τοÝς Α, ∆ µετρε‹ πρώτους Ôντας πρÕς ¢λλήλους· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα œσται æς Ð Α πρÕς τÕν Β, οÛτως Ð ∆ πρÕς ¥λλον τινά· Óπερ œδει δεˆξαι.

For, if possible, let it be that as A (is) to B, so D (is) to E. Thus, alternately, as A is to D, (so) B (is) to E [Prop. 7.13]. And A and D are prime (to one another). And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21]. And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, A measures B. And as A is to B, (so) B (is) to C. Thus, B also measures C. And hence A measures C [Def. 7.20]. And since as B is to C, (so) C (is) to D, and B measures C, C thus also measures D [Def. 7.20]. But, A was measuring C. And hence A measures D. And (A) also measures itself. Thus, A measures A and D, which are prime to one another. The very thing is impossible. Thus, as A (is) to B, so D cannot be to some other (number). (Which is) the very thing it was required to show.

ιη΄.

Proposition 18

∆ύο ¢ριθµîν δοθέντων ™πισκέψασθαι, ε„ δυνατόν ™στιν αÙτο‹ς τρίτον ¢νάλογον προσευρε‹ν.

For two given numbers, to investigate whether it is possible to find a third (number) proportional to them.

Α Β

Γ ∆

A B

”Εστωσαν οƒ δοθέντες δύο ¢ριθµοˆ οƒ Α, Β, κሠδέον œστω ™πισκέψασθαι, ε„ δυνατόν ™στιν αÙτο‹ς τρίτον ¢νάλογον προσευρε‹ν. Οƒ δ¾ Α, Β ½τοι πρîτοι πρÕς ¢λλήλους ε„σˆν À οÜ. καˆ ε„ πρîτοι πρÕς ¢λλήλους ε„σίν, δέδεικται, Óτι ¢δύνατόν ™στιν αÙτο‹ς τρίτον ¢νάλογον προσευρε‹ν. 'Αλλ¦ δ¾ µ¾ œστωσαν οƒ Α, Β πρîτοι πρÕς ¢λλήλους, καˆ Ð Β ˜αυτον πολλαπλασιάσας τÕν Γ ποιείτω. Ð Α δ¾ τÕν Γ ½τοι µετρε‹ À οÙ µετρε‹. µετρείτω πρότερον κατ¦ τÕν ∆· Ð Α ¥ρα τÕν ∆ πολλαπλασιάσας τÕν Γ πεποίηκεν. ¢λλα µ¾ν καˆ Ð Β ˜αυτÕν πολλαπλασιάσας τÕν Γ πεποίηκεν· Ð ¥ρα ™κ τîν Α, ∆ ‡σος ™στˆ τù ¢πÕ τοà Β. œστιν ¥ρα æς Ð Α πρÕς τÕν Β, Ð Β πρÕς τÕν ∆·

C D

Let A and B be the two given numbers. And let it be required to investigate whether it is possible to find a third (number) proportional to them. So A and B are either prime to one another or not. And if they are prime to one another it has (already) been show that it is impossible to find a third (number) proportional to them [Prop. 9.16]. And so let A and B not be prime to one another. And let B make C (by) multiplying itself. So A either measures, or does not measure, C. Let it first of all measure (C) according to D. Thus, A has made C (by) multiplying D. But, in fact, B has also made C (by) multiplying itself. Thus, the (number created) from (multiplying) A,

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το‹ς Α, Β ¥ρα τρίτος ¢ριθµÕς ¢νάλογον προσηύρηται Ð ∆. 'Αλλ¦ δ¾ µ¾ µετρείτω Ð Α τÕν Γ· λέγω, Óτι το‹ς Α, Β ¢δύνατόν ™στι τρίτον ¢νάλογον προσευρε‹ν ¢ριθµόν. ε„ γ¦ρ δυνατόν, προσηυρήσθω Ð ∆. Ð ¥ρα ™κ τîν Α, ∆ ‡σος ™στˆ τù ¢πÕ τοà Β. Ð δ ¢πÕ τοà Β ™στιν Ð Γ· Ð ¥ρα ™κ τîν Α, ∆ ‡σος ™στˆ τù Γ. éστε Ð Α τÕν ∆ πολλαπλασιάσας τÕν Γ πεποίηκεν· Ð Α ¥ρα τÕν Γ µετρε‹ κατ¦ τÕν ∆. ¢λλα µ¾ν Øπόκειται κሠµ¾ µετρîν· Óπερ ¥τοπον. οÙκ ¥ρα δυνατόν ™στι το‹ς Α, Β τρίτον ¢νάλογον προσευρε‹ν ¢ριθµÕν, Óταν Ð Α τÕν Γ µ¾ µετρÍ· Óπερ œδει δε‹ξαι.

D is equal to the (square) on B. Thus, as A is to B, (so) B (is) to D [Prop. 7.19]. Thus, a third number has been found proportional to A, B, (namely) D. And so let A not measure C. I say that it is impossible to find a third number proportional to A, B. For, if possible, let it have been found, (and let it be) D. Thus, the (number created) from (multiplying) A, D is equal to the (square) on B [Prop. 7.19]. And the (square) on B is C. Thus, the (number created) from (multiplying) A, D is equal to C. Hence, A has made C (by) multiplying D. Thus, A measures C according to D. But (A) was, in fact, also assumed (to be) not measuring (C). The very thing (is) absurd. Thus, it is not possible to find a third number proportional to A, B when A does not measure C. (Which is) the very thing it was required to show.

ιθ΄.

Proposition 19†

Τριîν ¢ριθµîν δοθέντων ™πισκέψασθαι, πότε δυνατόν ™στιν αÙτο‹ς τέταρτον ¢νάλογον προσευρε‹ν.

For three given numbers, to investigate when it is possible to find a fourth (number) proportional to them.

Α Β Γ ∆ Ε

A B C D E

”Εστωσαν οƒ δοθέντες τρε‹ς ¢ριθµοˆ οƒ Α, Β, Γ, κሠδέον œστω επισκέψασθαι, πότε δυνατόν ™στιν αÙτο‹ς τέταρτον ¢νάλογον προσευρε‹ν. ”Ητοι οâν οÜκ ε„σιν ˜ξÁς ¢νάλογον, καˆ οƒ ¥κροι αÙτîν πρîτοι πρÕς ¢λλήλους ε„σίν, À ˜ξÁς ε„σιν ¢νάλογον, καˆ οƒ ¥κροι αÙτîν οÜκ ε„σι πρîτοι πρÕς ¢λλήλους, À οÛτε ˜ξÁς ε„σιν ¢νάλογον, οÜτε οƒ ¥κροι αÙτîν πρîτοι πρÕς ¢λλήλους ε„σίν, À κሠ˜ξÁς ε„σιν ¢νάλογον, καˆ οƒ ¥κροι αÙτîν πρîτοι πρÕς ¢λλήλους ε„σίν. Ε„ µν οâν οƒ Α, Β, Γ ˜ξÁς ε„σιν ¢νάλογον, καˆ οƒ ¥κροι αÙτîν οƒ Α, Γ πρîτοι πρÕς ¢λλήλους ε„σίν, δέδεικται, Óτι ¢δύνατόν ™στιν αÙτο‹ς τέταρτον ¢νάλογον προσευρε‹ν ¢ριθµόν. µ¾ œστωσαν δ¾ οƒ Α, Β, Γ ˜ξÁς ¢νάλογον τîν ¢κρîν πάλιν Ôντων πρώτων πρÕς ¢λλήλους. λέγω, Óτι κሠοÛτως ¢δύνατόν ™στιν αÙτο‹ς τέταρτον ¢νάλογον προσευρε‹ν. ε„ γ¦ρ δυνατόν, προσευρήσθω Ð ∆, éστε εναι æς τÕν Α πρÕς τÕν Β, τÕν Γ πρÕς τÕν ∆, κሠγεγονέτω æς Ð Β πρÕς τÕν Γ, Ð ∆ πρÕς τÕν Ε. κሠ™πεί ™στιν æς µν Ð Α πρÕς τÕν Β, Ð Γ πρÕς τÕν ∆, æς δ Ð Β πρÕς τÕν Γ, Ð ∆ πρÕς τÕν Ε, δι' ‡σου ¥ρα æς Ð Α πρÕς τÕν Γ, Ð Γ πρÕς τÕν Ε. οƒ δ Α, Γ πρîτοι, οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ

Let A, B, C be the three given numbers. And let it be required to investigate when it is possible to find a fourth (number) proportional to them. In fact, (A, B, C) are either not continuously proportional and the outermost of them are prime to one another, or are continuously proportional and the outermost of them are not prime to one another, or are neither continuously proportional nor are the outermost of them prime to one another, or are continuously proportional and the outermost of them are prime to one another. In fact, if A, B, C are continuously proportional, and the outermost of them, A and C, are prime to one another, (then) it has (already) been shown that it is impossible to find a fourth number proportional to them [Prop. 9.17]. So let A, B, C not be continuously proportional, (with) the outermost of them again being prime to one another. I say that, in this case, it is also impossible to find a fourth (number) proportional to them. For, if possible, let it have been found, (and let it be) D. Hence, it will be that as A (is) to B, (so) C (is) to D. And let it be contrived that as B (is) to C, (so) D (is) to E. And since as A is to B, (so) C (is) to D, and as B (is) to C, (so) D (is) to E, thus, via equality, as A (is) to C, (so) C (is) to E

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™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον. µετρε‹ ¥ρα Ð Α τÕν Γ æς ¹γούµενος ¹γούµενον. µετρε‹ δ κሠ˜αυτόν· Ð Α ¥ρα τοÝς Α, Γ µετρε‹ πρώτους Ôντας πρÕς ¢λλήλους· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα το‹ς Α, Β, Γ δυνατόν ™στι τέταρτον ¢νάλογον προσευρε‹ν. 'Αλλά δ¾ πάλιν œστωσαν οƒ Α, Β, Γ ˜ξÁς ¢νάλογον, οƒ δ Α, Γ µ¾ œστωσαν πρîτοι πρÕς ¢λλήλους. λέγω, Óτι δυνατόν ™στιν αÙτο‹ς τέταρτον ¢νάλογον προσευρε‹ν. Ð γ¦ρ Β τÕν Γ πολλαπλασιάσας τÕν ∆ ποιείτω· Ð Α ¥ρα τÕν ∆ ½τοι µετρε‹ À οÙ µετρε‹. µετρείτω αÙτÕν πρότερον κατ¦ τÕν Ε· Ð Α ¥ρα τÕν Ε πολλαπλασιάσας τÕν ∆ πεποίηκεν. ¢λλ¦ µ¾ν καˆ Ð Β τÕν Γ πολλαπλασιάσας τÕν ∆ πεποίηκεν· Ð ¥ρα ™κ τîν Α, Ε ‡σος ™στˆ τù ™κ τîν Β, Γ. ¢νάλογον ¥ρα [™στˆν] æς Ð Α πρÕς τÕν Β, Ð Γ πρÕς τÕν Ε· τοˆς Α, Β, Γ ¥ρα τέταρτος ¢νάλογον προσηύρηται Ð Ε. 'Αλλ¦ δ¾ µ¾ µετρείτω Ð Α τÕν ∆· λέγω, Óτι ¢δύνατόν ™στι το‹ς Α, Β, Γ τέταρτον ¢νάλογον προσευρε‹ν ¢ριθµόν. ε„ γ¦ρ δυνατόν, προσευρήσθω Ð Ε· Ð ¥ρα ™κ τîν Α, Ε ‡σος ™στˆ τù ™κ τîν Β, Γ. ¢λλ¦ Ð ˜κ τîν Β, Γ ™στιν Ð ∆· καˆ Ð ™κ τîν Α, Ε ¥ρα ‡σος ™στˆ τù ∆. Ð Α ¥ρα τÕν Ε πολλαπλασιάσας τÕν ∆ πεποίηκεν· Ð Α ¥ρα τÕν ∆ µετρε‹ κατ¦ τÕν Ε· éστε µετρε‹ Ð Α τÕν ∆. ¢λλ¦ κሠοÙ µετρε‹· Óπερ ¥τοπον. οÙκ ¥ρα δυνάτον ™στι το‹ς Α, Β, Γ τέταρτον ¢νάλογον προσευρε‹ν ¢ριθµόν, Óταν Ð Α τÕν ∆ µ¾ µετρÍ. ¢λλ¦ δ¾ οƒ Α, Β, Γ µήτε ˜ξÁς œστωσαν ¢νάλογον µήτε οƒ ¥κροι πρîτοι πρÕς ¢λλήλους. καˆ Ð Β τÕν Γ πολλαπλασιάσας τÕν ∆ ποιείτω. еοίως δ¾ δειχθήσεται, Óτι ε„ µν µετρε‹ Ð Α τÕν ∆, δυνατόν ™στιν αÙτο‹ς ¢νάλογον προσευρε‹ν, ε„ δ οÙ µετρε‹, ¢δύνατον· Óπερ œδει δε‹ξαι.

†

[Prop. 7.14]. And A and C (are) prime (to one another). And (numbers) prime (to one another are) also the least (numbers having the same ratio as them) [Prop. 7.21]. And the least (numbers) measure those numbers having the same ratio as them (the same number of times), the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, A measures C, (as) the leading (measuring) the leading. And it also measures itself. Thus, A measures A and C, which are prime to one another. The very thing is impossible. Thus, it is not possible to find a fourth (number) proportional to A, B, C. And so let A, B, C again be continuously proportional, and let A and C not be prime to one another. I say that it is possible to find a fourth (number) proportional to them. For let B make D (by) multiplying C. Thus, A either measures or does not measure D. Let it, first of all, measure (D) according to E. Thus, A has made D (by) multiplying E. But, in fact, B has also made D (by) multiplying C. Thus, the (number created) from (multiplying) A, E is equal to the (number created) from (multiplying) B, C. Thus, proportionally, as A [is] to B, (so) C (is) to E [Prop. 7.19]. Thus, a fourth (number) proportional to A, B, C has been found, (namely) E. And so let A not measure D. I say that it is impossible to find a fourth number proportional to A, B, C. For, if possible, let it have been found, (and let it be) E. Thus, the (number created) from (multiplying) A, E is equal to the (number created) from (multiplying) B, C. But, the (number created) from (multiplying) B, C is D. And thus the (number created) from (multiplying) A, E is equal to D. Thus, A has made D (by) multiplying E. Thus, A measures D according to E. Hence, A measures D. But, it also does not measure (D). The very thing (is) absurd. Thus, it is not possible to find a fourth number proportional to A, B, C when A does not measure D. And so (let) A, B, C (be) neither continuously proportional, nor (let) the outermost of them (be) prime to one another. And let B make D (by) multiplying C. So, similarly, it can be show that if A measures D then it is possible to find a fourth (number) proportional to (A, B, C), and impossible if (A) does not measure (D). (Which is) the very thing it was required to show.

The proof of this proposition is incorrect. There are, in fact, only two cases. Either A, B, C are continuously proportional, with A and C prime

to one another, or not. In the first case, it is impossible to find a fourth proportional number. In the second case, it is possible to find a fourth proportional number provided that A measures B times C. Of the four cases considered by Euclid, the proof given in the second case is incorrect, since it only demonstrates that if A : B :: C : D then a number E cannot be found such that B : C :: D : E. The proofs given in the other three cases are correct.

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κ΄.

Proposition 20

Οƒ πρîτοι ¢ριθµοˆ πλείους ε„σˆ παντÕς τοà προThe (set of all) prime numbers is more numerous than τεθέντος πλήθους πρώτων ¢ριθµîν. any assigned multitude of prime numbers.

Α Β Γ Ε

Η

A B C E

∆ Ζ

G

D F

”Εστωσαν οƒ προτεθέντες πρîτοι ¢ριθµοˆ οƒ Α, Β, Γ· λέγω, Óτι τîν Α, Β, Γ πλείους ε„σˆ πρîτοι ¢ριθµοί. Ε„λήφθω γ¦ρ Ð ØπÕ τîν Α, Β, Γ ™λάχιστος µετρούµενος κሠœστω ∆Ε, κሠπροσκείσθω τù ∆Ε µον¦ς ¹ ∆Ζ. Ð δ¾ ΕΖ ½τοι πρîτός ™στιν À οÜ. œστω πρότερον πρîτος· εÙρηµένοι ¥ρα ε„σˆ πρîτοι ¢ριθµοˆ οƒ Α, Β, Γ, ΕΖ πλείους τîν Α, Β, Γ. 'Αλλ¦ δ¾ µ¾ œστω Ð ΕΖ πρîτος· ØπÕ πρώτου ¥ρα τινÕς ¢ριθµοà µετρε‹ται. µετρείσθω ØπÕ πρώτου τοà Η· λέγω, Óτι Ð Η οÙδενˆ τîν Α, Β, Γ ™στιν Ð αÙτός. ε„ γ¦ρ δυνατόν, œστω. οƒ δ Α, Β, Γ τÕν ∆Ε µετροàσιν· καˆ Ð Η ¥ρα τÕν ∆Ε µετρήσει. µετρε‹ δ κሠτÕν ΕΖ· κሠλοιπ¾ν τ¾ν ∆Ζ µονάδα µετρήσει Ð Η ¢ριθµÕς êν· Ôπερ ¥τοπον. οÙκ ¥ρα Ð Η ˜νˆ τîν Α, Β, Γ ™στιν Ð αÙτός. κሠØπόκειται πρîτος. εØρηµένοι ¥ρα ε„σˆ πρîτοι ¢ριθµοˆ πλείους τοà προτεθέντος πλήθους τîν Α, Β, Γ οƒ Α, Β, Γ, Η· Óπερ œδει δε‹ξαι.

Let A, B, C be the assigned prime numbers. I say that the (set of all) primes numbers is more numerous than A, B, C. For let the least number measured by A, B, C have been taken, and let it be DE [Prop. 7.36]. And let the unit DF have been added to DE. So EF is either prime or not. Let it, first of all, be prime. Thus, the (set of) prime numbers A, B, C, EF , (which is) more numerous than A, B, C, has been found. And so let EF not be prime. Thus, it is measured by some prime number [Prop. 7.31]. Let it be measured by the prime (number) G. I say that G is not the same as any of A, B, C. For, if possible, let it be (the same). And A, B, C (all) measure DE. Thus, G will also measure DE. And it also measures EF . (So) G will also measure the remainder, unit DF , (despite) being a number [Prop. 7.28]. The very thing (is) absurd. Thus, G is not the same as one of A, B, C. And it was assumed (to be) prime. Thus, the (set of) prime numbers A, B, C, G, (which is) more numerous than the assigned multitude (of prime numbers), A, B, C, has been found. (Which is) the very thing it was required to show.

κα΄.

Proposition 21

'Ε¦ν ¥ρτιοι ¢ριθµοˆ Ðποσοιοàν συντεθîσιν, Ð Óλος ¥ρτιός ™στιν.

If any multitude whatsoever of even numbers is added together then the whole is even.

Α

Β

Γ

∆

Ε

A

Συγκείσθωσαν γ¦ρ ¥ρτιοι ¢ριθµοˆ Ðποσοιοàν οƒ ΑΒ, ΒΓ, Γ∆, ∆Ε· λέγω, Óτι Óλος Ð ΑΕ ¥ρτιός ™στιν. 'Επεˆ γ¦ρ ›καστος τîν ΑΒ, ΒΓ, Γ∆, ∆Ε ¥ρτιός ™στιν, œχει µέρος ¼µισυ· éστε κሠÓλος Ð ΑΕ œχει µέρος ¼µισυ. ¥ρτιος δ ¢ριθµός ™στιν Ð δίχα διαιρούµενος· ¥ρτιος ¥ρα ™στˆν Ð ΑΕ· Óπερ œδει δε‹ξαι.

B

C

D

E

For let any multitude whatsoever of even numbers, AB, BC, CD, DE, lie together. I say that the whole, AE, is even. For since everyone of AB, BC, CD, DE is even, it has a half part [Def. 7.6]. And hence the whole AE has a half part. And an even number is one (which can be) divided in two [Def. 7.6]. Thus, AE is even. (Which is)

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ELEMENTS BOOK 9 the very thing it was required to show.

κβ΄.

Proposition 22

'Ε¦ν περισσοˆ ¢ριθµοˆ Ðποσοιοàν συντεθîσιν, τÕ δ If any multitude whatsoever of odd numbers is added πλÁθος αÙτîν ¥ρτιον Ï, Ð Óλος ¥ρτιος œσται. together, and the multitude of them is even, then the whole will be even.

Α

Β

Γ

∆

Ε

A

B

C

D

E

Συγκείσθωσαν γ¦ρ περισσοˆ ¢ριθµοˆ Ðσοιδηποτοàν ¥ρτιοι τÕ πλÁθος οƒ ΑΒ, ΒΓ, Γ∆, ∆Ε· λέγω, Óτι Óλος Ð ΑΕ ¥ρτιός ™στιν. 'Επεˆ γ¦ρ ›καστος τîν ΑΒ, ΒΓ, Γ∆, ∆Ε περιττός ™στιν, ¢φαιρεθείσης µονάδος ¢φ' ˜κάστου ›καστος τîν λοιπîν ¥ρτιος œσται· éστε καˆ Ð συγκείµενος ™ξ αÙτîν ¥ρτιος œσται. œστι δ κሠτÕ πλÁθος τîν µονάδων ¥ρτιον. κሠÓλος ¥ρα Ð ΑΕ ¥ρτιός ™στιν· Óπερ œδει δε‹ξαι.

For let any even multitude whatsoever of odd numbers, AB, BC, CD, DE, lie together. I say that the whole, AE, is even. For since everyone of AB, BC, CD, DE is odd then, a unit being subtracted from each, everyone of the remainders will be (made) even [Def. 7.7]. And hence the sum of them will be even [Prop. 9.21]. And the multitude of the units is even. Thus, the whole AE is also even [Prop. 9.21]. (Which is) the very thing it was required to show.

κγ΄.

Proposition 23

'Ε¦ν περισσοˆ ¢ριθµοˆ Ðποσοιοàν συντεθîσιν, τÕ δ If any multitude whatsoever of odd numbers is added πλÁθος αÙτîν περισσÕν Ï, καˆ Ð Óλος περισσÕς œσται. together, and the multitude of them is odd, then the

Α

Β Γ

Ε ∆

whole will also be odd.

A Συγκείσθωσαν γ¦ρ Ðποσοιοàν περισσοˆ ¢ριθµοί, ïν τÕ πλÁθος περισσÕν œστω, οƒ ΑΒ, ΒΓ, Γ∆· λέγω, Óτι κሠÓλος Ð Α∆ περισσός ™στιν. 'ΑφVρήσθω ¢πÕ τοà Γ∆ µον¦ς ¹ ∆Ε· λοιπÕς ¥ρα Ð ΓΕ ¥ρτιός ™στιν. œστι δ καˆ Ð ΓΑ ¥ρτιος· κሠÓλος ¥ρα Ð ΑΕ ¥ρτιός ™στιν. καί ™στι µον¦ς ¹ ∆Ε. περισσÕς ¥ρα ™στˆν Ð Α∆· Óπερ œδει δε‹ξαι.

B C

E D

For let any multitude whatsoever of odd numbers, AB, BC, CD, lie together, and let the multitude of them be odd. I say that the whole, AD, is also odd. For let the unit DE have been subtracted from CD. The remainder CE is thus even [Def. 7.7]. And CA is also even [Prop. 9.22]. Thus, the whole AE is also even [Prop. 9.21]. And DE is a unit. Thus, AD is odd [Def. 7.7]. (Which is) the very thing it was required to show.

κδ΄.

Proposition 24

'Ε¦ν ¢πÕ ¢ρτίου ¢ριθµοà ¥ρτιος ¢φαιρεθÍ, Ð λοιπÕς ¥ρτιος œσται.

If an even (number) is subtracted from an(other) even number then the remainder will be even.

Α

Γ

Β

A

'ΑπÕ γ¦ρ ¢ρτίου τοà ΑΒ ¥ρτιος ¢φVρήσθω Ð ΒΓ· λέγω, Óτι Ð λοιπÕς Ð ΓΑ ¥ρτιός ™στιν. 'Επεˆ γ¦ρ Ð ΑΒ ¥ρτιός ™στιν, œχει µέρος ¼µισυ. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð ΒΓ œχει µέρος ¼µισυ· éστε κሠλοιπÕς [Ð ΓΑ œχει µέρος ¼µισυ] ¥ρτιος [¥ρα] ™στˆν Ð ΑΓ· Óπερ œδει δε‹ξαι.

C

B

For let the even (number) BC have been subtracted from the even number AB. I say that the remainder CA is even. For since AB is even, it has a half part [Def. 7.6]. So, for the same (reasons), BC also has a half part. And hence the remainder [CA has a half part]. [Thus,] AC is

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κε΄.

Proposition 25

'Ε¦ν ¢πÕ ¢ρτίου ¢ριθµοà περισσÕς ¢φαιρεθÍ, Ð If an odd (number) is subtracted from an even numλοιπÕς περισσÕς œσται. ber then the remainder will be odd.

Α

Γ

∆

Β

A

C

D

B

'ΑπÕ γ¦ρ ¢ρτίου τοà ΑΒ περισσÕς ¢φVρήσθω Ð ΒΓ· λέγω, Óτι Ð λοιπÕς Ð ΓΑ περισσός ™στιν. 'ΑφVρήσθω γ¦ρ ¢πÕ τοà ΒΓ µον¦ς ¹ Γ∆· Ð ∆Β ¥ρα ¥ρτιός ™στιν. œστι δ καˆ Ð ΑΒ ¥ρτιος· κሠλοιπÕς ¥ρα Ð Α∆ ¥ρτιός ™στιν. καί ™στι µον¦ς ¹ Γ∆· Ð ΓΑ περισσός ™στιν· Óπερ œδει δε‹ξαι.

For let the odd (number) BC have been subtracted from the even number AB. I say that the remainder CA is odd. For let the unit CD have been subtracted from BC. DB is thus even [Def. 7.7]. And AB is also even. And thus the remainder AD is even [Prop. 9.24]. And CD is a unit. Thus, CA is odd [Def. 7.7]. (Which is) the very thing it was required to show.

κ$΄.

Proposition 26

'Ε¦ν ¢πÕ περισσοà ¢ριθµοà περισσÕς ¢φαιρεθÍ, Ð If an odd (number) is subtracted from an odd number λοιπÕς ¥ρτιος œσται. then the remainder will be even.

Α

Γ

∆ Β

A

C

D B

'ΑπÕ γ¦ρ περισσοà τοà ΑΒ περισσÕς ¢φVρήσθω Ð ΒΓ· λέγω, Óτι Ð λοιπÕς Ð ΓΑ ¥ρτιός ™στιν. 'Επεˆ γ¦ρ Ð ΑΒ περισσός ™στιν, ¢φVρήσθω µον¦ς ¹ Β∆· λοιπÕς ¥ρα Ð Α∆ ¥ρτιός ™στιν. δι¦ τ¦ αÙτ¦ δ¾ καˆ Ð Γ∆ ¥ρτιός ™στιν· éστε κሠλοιπÕς Ð ΓΑ ¥ρτιός ™στιν· Óπερ œδει δε‹ξαι.

For let the odd (number) BC have been subtracted from the odd (number) AB. I say that the remainder CA is even. For since AB is odd, let the unit BD have been subtracted (from it). Thus, the remainder AD is even [Def. 7.7]. So, for the same (reasons), CD is also even. And hence the remainder CA is even [Prop. 9.24]. (Which is) the very thing it was required to show.

κζ΄.

Proposition 27

'Ε¦ν ¢πÕ περισσοà ¢ριθµοà ¥ρτιος ¢φαιρεθÍ, Ð If an even (number) is subtracted from an odd numλοιπÕς περισσÕς œσται. ber then the remainder will be odd.

Α ∆

Γ

Β

A D

C

B

'ΑπÕ γ¦ρ περισσοà τοà ΑΒ ¥ρτιος ¢φVρήσθω Ð ΒΓ· For let the even (number) BC have been subtracted λέγω, Óτι Ð λοιπÕς Ð ΓΑ περισσός ™στιν. from the odd (number) AB. I say that the remainder CA 'ΑφVρήσθω [γ¦ρ] µον¦ς ¹ Α∆· Ð ∆Β ¥ρα ¥ρτιός is odd. ™στιν. œστι δ καˆ Ð ΒΓ ¥ρτιος· κሠλοιπÕς ¥ρα Ð Γ∆ [For] let the unit AD have been subtracted (from AB). ¥ρτιός ™στιν. περισσÕς ¥ρα Ð ΓΑ· Óπερ œδει δε‹ξαι. DB is thus even [Def. 7.7]. And BC is also even. Thus, the remainder CD is also even [Prop. 9.24]. CA (is) thus odd [Def. 7.7]. (Which is) the very thing it was required to show.

κη΄.

Proposition 28

'Ε¦ν περισσÕς ¢ριθµÕς ¥ρτιον πολλαπλασιάσας ποιÍ τινα, Ð γενόµενος ¥ρτιος œσται.

If an odd number makes some (number by) multiplying an even (number) then the created (number) will be even.

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ELEMENTS BOOK 9

Α Β Γ

A B C

ΠερισσÕς γ¦ρ ¢ριθµÕς Ð Α ¥ρτιον τÕν Β πολλαπλασιάσας τÕν Γ ποιείτω· λέγω, Óτι Ð Γ ¥ρτιός ™στιν. 'Επεˆ γ¦ρ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν, Ð Γ ¥ρα σύγκειται ™κ τοσούτων ‡σων τù Β, Óσαι ε„σˆν ™ν τù Α µονάδες. καί ™στιν Ð Β ¥ρτιος· Ð Γ ¥ρα σύγκειται ™ξ ¢ρτίων. ™¦ν δ ¥ρτιοι ¢ριθµοˆ Ðποσοιοàν συντεθîσιν, Ð Óλος ¥ρτιός ™στιν. ¥ρτιος ¥ρα ™στˆν Ð Γ· Óπερ œδει δε‹ξαι.

For let the odd number A make C (by) multiplying the even (number) B. I say that C is even. For since A has made C (by) multiplying B, C is thus composed out of so many (magnitudes) equal to B, as many as (there) are units in A [Def. 7.15]. And B is even. Thus, C is composed out of even (numbers). And if any multitude whatsoever of even numbers is added together then the whole is even [Prop. 9.21]. Thus, C is even. (Which is) the very thing it was required to show.

κθ΄.

Proposition 29

'Ε¦ν περισσÕς ¢ριθµÕς περισσÕν ¢ριθµÕν πολλαIf an odd number makes some (number by) multiplyπλασιάσας ποιÍ τινα, Ð γενόµενος περισσÕς œσται. ing an odd (number) then the created (number) will be odd.

Α Β Γ

A B C

ΠερισσÕς γ¦ρ ¢ριθµÕς Ð Α περισσÕν τÕν Β πολλαπλασιάσας τÕν Γ ποιείτω· λέγω, Óτι Ð Γ περισσός ™στιν. 'Επεˆ γ¦ρ Ð Α τÕν Β πολλαπλασιάσας τÕν Γ πεποίηκεν, Ð Γ ¥ρα σύγκειται ™κ τοσούτων ‡σων τù Β, Óσαι ε„σˆν ™ν τù Α µονάδες. καί ™στιν ˜κάτερος τîν Α, Β περισσός· Ð Γ ¥ρα σύγκειται ™κ περισσîν ¢ριθµîν, ïν τÕ πλÁθος περισσόν ™στιν. éστε Ð Γ περισσός ™στιν· Óπερ œδει δε‹ξαι.

For let the odd number A make C (by) multiplying the odd (number) B. I say that C is odd. For since A has made C (by) multiplying B, C is thus composed out of so many (magnitudes) equal to B, as many as (there) are units in A [Def. 7.15]. And each of A, B is odd. Thus, C is composed out of odd (numbers), (and) the multitude of them is odd. Hence C is odd [Prop. 9.23]. (Which is) the very thing it was required to show.

λ΄.

Proposition 30

'Ε¦ν περισσÕς ¢ριθµÕς ¥ρτιον ¢ριθµÕν µετρÍ, κሠτÕν ¼µισυν αÙτοà µετρήσει.

If an odd number measures an even number then it will also measure (one) half of it.

Α Β Γ

A B C

ΠερισσÕς γ¦ρ ¢ριθµÕς Ð Α ¥ρτιον τÕν Β µετρείτω· λέγω, Óτι κሠτÕν ¼µισυν αÙτοà µετρήσει. 'Επεˆ γ¦ρ Ð Α τÕν Β µετρε‹, µετρείτω αÙτÕν κατ¦ τÕν Γ· λέγω, Óτι Ð Γ οÙκ œστι περισσός. ε„ γ¦ρ δυνατόν, œστω. κሠ™πεˆ Ð Α τÕν Β µετρε‹ κατ¦ τÕν Γ, Ð Α ¥ρα τÕν Γ πολλαπλασιάσας τÕν Β πεποίηκεν. Ð Β ¥ρα σύγκειται ™κ περισσîν ¢ριθµîν, ïν τÕ πλÁθος περισσόν

For let the odd number A measure the even (number) B. I say that (A) will also measure (one) half of (B). For since A measures B, let it measure it according to C. I say that C is not odd. For, if possible, let it be (odd). And since A measures B according to C, A has thus made B (by) multiplying C. Thus, B is composed out of odd numbers, (and) the multitude of them is odd. B is thus

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™στιν. Ð Β ¥ρα περισσός ™στιν· Óπερ ¥τοπον· Øπόκειται γ¦ρ ¥ρτιος. οÙκ ¥ρα Ð Γ περισσός ™στιν· ¥ρτιος ¥ρα ™στˆν Ð Γ. éστε Ð Α τÕν Β µετρε‹ ¢ρτιάκις. δι¦ δ¾ τοàτο κሠτÕν ¼µισυν αÙτοà µετρήσει· Óπερ œδει δε‹ξαι.

odd [Prop. 9.23]. The very thing (is) absurd. For (B) was assumed (to be) even. Thus, C is not odd. Thus, C is even. Hence, A measures B an even number of times. So, on account of this, (A) will also measure (one) half of (B). (Which is) the very thing it was required to show.

λα΄.

Proposition 31

'Ε¦ν περισσÕς ¢ριθµÕς πρός τινα ¢ριθµÕν πρîτος Ï, κሠπρÕς τÕν διπλασίονα αÙτοà πρîτος œσται.

If an odd number is prime to some number then it will also be prime to its double.

Α Β Γ ∆

A B C D

ΠερισσÕς γ¦ρ ¢ριθµÕς Ð Α πρός τινα ¢ριθµÕν τÕν Β πρîτος œστω, τοà δ Β διπλασίων œστω Ð Γ· λέγω, Óτι Ð Α [καˆ] πρÕς τÕν Γ πρîτός ™στιν. Ε„ γ¦ρ µή ε„σιν [οƒ Α, Γ] πρîτοι, µετρήσει τις αÙτοÝς ¢ριθµός. µετρείτω, κሠœστω Ð ∆. καί ™στιν Ð Α περισσός· περισσÕς ¥ρα καˆ Ð ∆. κሠ™πεˆ Ð ∆ περισσÕς íν τÕν Γ µετρε‹, καί ™στιν Ð Γ ¥ρτιος, κሠτÕν ¼µισυν ¥ρα τοà Γ µετρήσει [Ð ∆]. τοà δ Γ ¼µισύ ™στιν Ð Β· Ð ∆ ¥ρα τÕν Β µετρε‹. µετρε‹ δ κሠτÕν Α. Ð ∆ ¥ρα τοÝς Α, Β µετρε‹ πρώτους Ôντας πρÕς ¢λλήλους· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα Ð Α πρÕς τÕν Γ πρîτος οÜκ ™στιν. οƒ Α, Γ ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν· Óπερ œδει δε‹ξαι.

For let the odd number A be prime to some number B. And let C be double B. I say that A is [also] prime to C. For if [A and C] are not prime (to one another) then some number will measure them. Let it measure (them), and let it be D. And A is odd. Thus, D (is) also odd. And since D, which is odd, measures C, and C is even, [D] will thus also measure half of C [Prop. 9.30]. And B is half of C. Thus, D measures B. And it also measures A. Thus, D measures (both) A and B, (despite) them being prime to one another. The very thing is impossible. Thus, A is not unprime to C. Thus, A and C are prime to one another. (Which is) the very thing it was required to show.

λβ΄.

Proposition 32

Τîν ¢πÕ δύαδος διπλασιαζοµένων ¢ριθµων ›καστος ¢ρτιάκις ¥ρτιός ™στι µόνον.

Each of the numbers (which is continually) doubled, (starting) from a dyad, is an even-times-even (number) only.

Α Β Γ ∆

A B C D

'ΑπÕ γ¦ρ δύαδος τÁς Α δεδιπλασιάσθωσαν Ðσοιδηποτοàν ¢ριθµοˆ οƒ Β, Γ, ∆· λέγω, Óτι οƒ Β, Γ, ∆ ¢ρτιάκις ¥ρτιοί ε„σι µόνον. “Οτι µν οâν ›καστος [τîν Β, Γ, ∆] ¢ρτιάκις ¥ρτιός ™στιν, φανερόν· ¢πÕ γ¦ρ δυάδος ™στˆ διπλασιασθείς. λέγω, Óτι κሠµόνον. ™κκείσθω γ¦ρ µονάς. ™πεˆ οâν ¢πÕ µονάδος Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ¢νάλογόν ε„σιν, Ð

For let any multitude of numbers whatsoever, B, C, D, have been (continually) doubled, (starting) from the dyad A. I say that B, C, D are even-times-even (numbers) only. In fact, (it is) clear that each [of B, C, D] is an eventimes-even (number). For they are doubled from a dyad [Def. 7.8]. I also say that (they are even-times-even num-

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δ µετ¦ τ¾ν µονάδα Ð Α πρîτός ™στιν, Ð µέγιστος τîν Α, Β, Γ, ∆ Ð ∆ Øπ' οÙδενÕς ¥λλου µετρηθήσεται παρξ τîν Α, Β, Γ. καί ™στιν ›καστος τîν Α, Β, Γ ¥ρτιος· Ð ∆ ¥ρα ¢ρτιάκις ¥ρτιός ™στι µόνον. еοίως δ¾ δε‹ξοµεν, Óτι [καˆ] ˜κάτερος τîν Β, Γ ¢ρτιάκις ¥ρτιός ™στι µόνον· Óπερ œδει δε‹ξαι.

bers) only. For let a unit be laid down. Therefore, since any multitude of numbers whatsoever are continuously proportional, starting from a unit, and the (number) A after the unit is prime, the greatest of A, B, C, D, (namely) D, will not be measured by any other (numbers) except A, B, C [Prop. 9.13]. And each of A, B, C is even. Thus, D is an even-time-even (number) only [Def. 7.8]. So, similarly, we can show that each of B, C is [also] an eventime-even (number) only. (Which is) the very thing it was required to show.

λγ΄.

Proposition 33

'Ε¦ν ¢ριθµÕς τÕν ¼µισυν œχV περισσόν, ¢ρτιάκις πεIf a number has an odd half then it is an even-timeρισσός ™στι µόνον. odd (number) only.

Α

A

'ΑριθµÕς γ¦ρ Ð Α τÕν ¼µισυν ™χέτω περισσόν· λέγω, Óτι Ð Α ¢ρτιάκις περισσός ™στι µόνον. “Οτι µν οâν ¢ρτιάκις περισσός ™στιν, φανερόν· Ð γ¦ρ ¼µισυς αÙτοà περισσÕς íν µετρε‹ αÙτÕν ¢ρτιάκις, λέγω δή, Óτι κሠµόνον. ε„ γ¦ρ œσται Ð Α κሠ¢ρτιάκις ¥ρτιος, µετρηθήσεται ØπÕ ¢ρτίου κατ¦ ¥ρτιον ¢ριθµόν· éστε καˆ Ð ¼µισυς αÙτοà µετρηθήσεται ØπÕ ¢ρτίου ¢ριθµοà περισσÕς êν· Óπερ ™στˆν ¥τοπον. Ð Α ¥ρα ¢ρτιάκις περισσός ™στι µόνον· Óπερ œδει δε‹ξαι.

For let the number A have an odd half. I say that A is an even-times-odd (number) only. In fact, (it is) clear that (A) is an even-times-odd (number). For its half, being odd, measures it an even number of times [Def. 7.9]. So I also say that (it is an even-times-odd number) only. For if A is also an even-times-even (number) then it will be measured by an even (number) according to an even number [Def. 7.8]. Hence, its half will also be measured by an even number, (despite) being odd. The very thing is absurd. Thus, A is an even-times-odd (number) only. (Which is) the very thing it was required to show.

λδ΄.

Proposition 34

'Ε¦ν ¢ριθµÕς µήτε τîν ¢πÕ δυάδος διπλασιαζοµένων If a number is neither (one) of the (numbers) doubled Ï, µήτε τÕν ¼µισυν œχV περισσόν, ¢ρτιάκις τε ¥ρτιός ™στι from a dyad, nor has an odd half, then it is (both) an κሠ¢ρτιάκις περισσός. even-times-even and an even-times-odd (number).

Α

A

'ΑριθµÕς γ¦ρ Ð Α µήτε τîν ¢πÕ δυάδος διπλασιαζοµένων œστω µήτε τÕν ¼µισυν ™χέτω περισσόν· λέγω, Óτι Ð Α ¢ρτιάκις τέ ™στιν ¥ρτιος κሠ¢ρτιάκις περισσός. “Οτι µν οâν Ð Α ¢ρτιάκις ™στˆν ¥ρτιος, φανερόν· τÕν γ¦ρ ¼µισυν οÙκ œχει περισσόν. λέγω δή, Óτι κሠ¢ρτιάκις περισσός ™στιν. ™¦ν γ¦ρ τÕν Α τέµνωµεν δίχα κሠτÕν ¼µισυν αÙτοà δίχα κሠτοàτο ¢εˆ ποιîµεν, καταντήσοµεν ε‡ς τινα ¢ριθµÕν περισσόν, Öς µετρήσει τÕν Α κατ¦ ¥ρτιον ¢ριθµόν. ε„ γ¦ρ οÜ, καταντήσοµεν ε„ς δυάδα, κሠœσται Ð Α τîν ¢πÕ δυάδος διπλασιαζοµένων· Óπερ οÙχ Øπόκειται. éστε Ð Α ¢ρτιάκις περισσόν ™στιν. ™δείχθη δ κሠ¢ρτιάκις ¥ρτιος. Ð Α ¥ρα ¢ρτιάκις τε ¥ρτιός ™στι κሠ¢ρτιάκις περισσός· Óπερ œδει δε‹ξαι.

For let the number A neither be (one) of the (numbers) doubled from a dyad, nor let it have an odd half. I say that A is (both) an even-times-even and an eventimes-odd (number). In fact, (it is) clear that A is an even-times-even (number) [Def. 7.8]. For it does not have an odd half. So I say that it is also an even-times-odd (number). For if we cut A in half, and (then cut) its half in half, and we do this continually, then we will arrive at some odd number which will measure A according to an even number. For if not, we will arrive at a dyad, and A will be (one) of the (numbers) doubled from a dyad. The very opposite thing (was) assumed. Hence, A is an even-times-odd (number) [Def. 7.9]. And it was also shown (to be) an even-times-even (number). Thus, A is (both) an eventimes-even and an even-times-odd (number). (Which is)

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ELEMENTS BOOK 9 the very thing it was required to show.

λε΄.

Proposition 35†

'Ε¦ν ðσιν Ðσοιδηποτοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον, ¢φαιρεθîσι δ ¢πό τε τοà δευτέρου κሠτοà ™σχάτου ‡σοι τù πρώτJ, œσται æς ¹ τοà δευτέρου Øπεροχ¾ πρÕς τÕν πρîτον, οÛτως ¹ τοà ™σχάτου Øπεροχ¾ πρÕς τοÝς πρÕ ˜αυτοà πάντας.

If there is any multitude whatsoever of continually proportional numbers, and (numbers) equal to the first are subtracted from (both) the second and the last, then as the excess of the second (number is) to the first, so the excess of the last will be to (the sum of) all those (numbers) before it.

Α Β

A B

Η Γ

∆ Ε Λ Κ Θ

G C

D E L K H

Ζ

”Εστωσαν Ðποσοιδηποτοàν ¢ριθµοˆ ˜ξÁς ¢νάλογον οƒ Α, ΒΓ, ∆, ΕΖ ¢φχόµενοι ¢πÕ ™λαχίστου τοà Α, κሠ¢φVρήσθω ¢πÕ τοà ΒΓ κሠτοà ΕΖ τö Α ‡σος ˜κάτερος τîν ΒΗ, ΖΘ· λέγω, Óτι ™στˆν æς Ð ΗΓ πρÕς τÕν Α, οÛτως Ð ΕΘ πρÕς τοÝς Α, ΒΓ, ∆. Κείσθω γ¦ρ τù µν ΒΓ ‡σος Ð ΖΚ, τù δ ∆ ‡σος Ð ΖΛ. κሠ™πεˆ Ð ΖΚ τù ΒΓ ‡σος ™στίν, ïν Ð ΖΘ τù ΒΗ ‡σος ™στίν, λοιπÕς ¥ρα Ð ΘΚ λοιπù τù ΗΓ ™στιν ‡σος. κሠ™πεί ™στιν æς Ð ΕΖ πρÕς τÕν ∆, οÛτως Ð ∆ πρÕς τÕν ΒΓ καˆ Ð ΒΓ πρÕς τÕν Α, ‡σος δ Ð µν ∆ τù ΖΛ, Ð δ ΒΓ τù ΖΚ, Ð δ Α τù ΖΘ, œστιν ¥ρα æς Ð ΕΖ πρÕς τÕν ΖΛ, οÛτως Ð ΛΖ πρÕς τÕν ΖΚ καˆ Ð ΖΚ πρÕς τÕν ΖΘ. διελόντι, æς Ð ΕΛ πρÕς τÕν ΛΖ, οÛτως Ð ΛΚ πρÕς τÕν ΖΚ καˆ Ð ΚΘ πρÕς τÕν ΖΘ. œστιν ¥ρα κሠæς εŒς τîν ¹γουµένων πρÕς ›να τîν ˜ποµένων, οÛτως ¤παντες οƒ ¹γούµενοι πρÕς ¤παντας τοÝς ˜ποµένους· œστιν ¥ρα æς Ð ΚΘ πρÕς τÕν ΖΘ, οÛτως οƒ ΕΛ, ΛΚ, ΚΘ πρÕς τοÝς ΛΖ, ΖΚ, ΘΖ. ‡σος δ Ð µν ΚΘ τù ΓΗ, Ð δ ΖΘ τù Α, οƒ δ ΛΖ, ΖΚ, ΘΖ τοˆς ∆, ΒΓ, Α· œστιν ¥ρα æς Ð ΓΗ πρÕς τÕν Α, οÛτως Ð ΕΘ πρÕς τοÝς ∆, ΒΓ, Α. œστιν ¥ρα æς ¹ τοà δευτέρου Øπεροχ¾ πρÕς τÕν πρîτον, οÛτως ¹ τοà ™σχάτου Øπεροχ¾ πρÕς τοÝς πρÕ ˜αυτοà πάντας· Óπερ œδει δε‹ξαι.

F

Let A, BC, D, EF be any multitude whatsoever of continuously proportional numbers, beginning from the least A. And let BG and F H, each equal to A, have been subtracted from BC and EF (respectively). I say that as GC is to A, so EH is to A, BC, D. For let F K be made equal to BC, and F L to D. And since F K is equal to BC, of which F H is equal to BG, the remainder HK is thus equal to the remainder GC. And since as EF is to D, so D (is) to BC, and BC to A [Prop. 7.13], and D (is) equal to F L, and BC to F K, and A to F H, thus as EF is to F L, so LF (is) to F K, and F K to F H. By separation, as EL (is) to LF , so LK (is) to F K, and KH to F H [Props. 7.11, 7.13]. And thus as one of the leading (numbers) is to one of the following, so (the sum of) all of the leading (numbers is) to (the sum of) all of the following [Prop. 7.12]. Thus, as KH is to F H, so EL, LK, KH (are) to LF , F K, HF . And KH (is) equal to CG, and F H to A, and LF , F K, HF to D, BC, A. Thus, as CG is to A, so EH (is) to D, BC, A. Thus, as the excess of the second (number) is to the first, so the excess of the last (is) to (the sum of) all those (numbers) before it. (Which is) the very thing it was required to show.

†

This proposition allows us to sum a geometric series of the form a, a r, a r 2 , a r 3 , · · · a r n−1 . According to Euclid, the sum Sn satisfies (a r − a)/a = (a r n − a)/Sn . Hence, Sn = a (r n − 1)/(r − 1).

λ$΄.

Proposition 36†

'Ε¦ν ¢πÕ µονάδος Ðποσοιοàν ¢ριθµοˆ ˜ξÁς ™κτεθîσιν If any multitude whatsoever of numbers is set out con™ν τÍ διπλασίονι ¢ναλογίv, ›ως οá Ð σύµπας συντεθεˆς tinuously in a double proportion, (starting) from a unit, πρîτος γένηται, καˆ Ð σύµπας ™πˆ τÕν œσχατον πολλα- until the whole sum added together becomes prime, and 277

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πλασιασθεˆς ποιÍ τινα, Ð γενόµενος τέλειος œσται. 'ΑπÕ γ¦ρ µονάδος ™κκείσθωσαν Ðσοιδηποτοàν ¢ριθµοˆ ™ν τÍ διπλασίονι ¢ναλογίv, ›ως οá Ð σύµπας συντεθεˆς πρîτος γένηται, οƒ Α, Β, Γ, ∆, κሠτù σύµπαντι ‡σος œστω Ð Ε, καˆ Ð Ε τÕν ∆ πολλαπλασιάσας τÕν ΖΗ ποιείτω. λέγω, Óτι Ð ΖΗ τέλειός ™στιν.

the sum multiplied into the last (number) makes some (number), then the (number so) created will be perfect. For let any multitude of numbers, A, B, C, D, be set out (continuouly) in a double proportion, until the whole sum added together is made prime. And let E be equal to the sum. And let E make F G (by) multiplying D. I say that F G is a perfect (number).

A B C D

Α Β Γ ∆ Ε Θ Ν Κ

E H N K

Λ Μ Ζ Ξ

L M F O

Η

Ο Π

G

P Q

“Οσοι γάρ ε„σιν οƒ Α, Β, Γ, ∆ τù πλήθει, τοσοàτοι ¢πÕ τοà Ε ε„λήφθωσαν ™ν τÍ διπλασίονι ¢ναλογίv οƒ Ε, ΘΚ, Λ, Μ· δι' ‡σου ¥ρα ™στˆν æς Ð Α πρÕς τÕν ∆, οÛτως Ð Ε πρÕς τÕν Μ. Ð ¥ρα ™κ τîν Ε, ∆ ‡σος ™στˆ τù ™κ τîν Α, Μ. καί ™στιν Ð ™κ τîν Ε, ∆ Ð ΖΗ· καˆ Ð ™κ τîν Α, Μ ¥ρα ™στˆν Ð ΖΗ. Ð Α ¥ρα τÕν Μ πολλαπλασιάσας τÕν ΖΗ πεποίηκεν· Ð Μ ¥ρα τÕν ΖΗ µετρε‹ κατ¦ τ¦ς ™ν τù Α µονάδας. καί ™στι δυ¦ς Ð Α· διπλάσιος ¥ρα ™στˆν Ð ΖΗ τοà Μ. ε„σˆ δ καˆ οƒ Μ, Λ, ΘΚ, Ε ˜ξÁς διπλάσιοι ¢λλήλων· οƒ Ε, ΘΚ, Λ, Μ, ΖΗ ¥ρα ˜ξÁς ¢νάλογόν ε„σιν ™ν τÍ διπλασίονι ¢ναλογίv. ¢φVρήσθω δ¾ ¢πÕ τοà δευτέρου τοà ΘΚ κሠτοà ™σχάτου τοà ΖΗ τù πρώτJ τù Ε ‡σος ˜κάτερος τîν ΘΝ, ΖΞ· œστιν ¥ρα æς ¹ τοà δευτέρου ¢ριθµοà Øπεροχ¾ πρÕς τÕν πρîτον, οÛτως ¹ τοà ™σχάτου Øπεροχ¾ πρÕς τοÝς πρÕ ˜αυτοà πάντας. œστιν ¥ρα æς Ð ΝΚ πρÕς τÕν Ε, οÛτως Ð ΞΗ πρÕς τοÝς Μ, Λ, ΚΘ, Ε. καί ™στιν Ð ΝΚ ‡σος τù Ε· καˆ Ð ΞΗ ¥ρα ‡σος ™στˆ το‹ς Μ, Λ, ΘΚ, Ε. œστι δ καˆ Ð ΖΞ τù Ε ‡σος, Ð δ Ε το‹ς Α, Β, Γ, ∆ κሠτÍ µονάδι. Óλος ¥ρα Ð ΖΗ ‡σος ™στˆ το‹ς τε Ε, ΘΚ, Λ, Μ κሠτο‹ς Α, Β, Γ, ∆ κሠτÍ µονάδι· κሠµετρε‹ται Øπ' αÙτîν. λέγω, Óτι καˆ Ð ΖΗ Ùπ' οÙδενÕς ¥λλου µετρηθήσεται παρξ τîν Α, Β, Γ, ∆, Ε, ΘΚ, Λ, Μ κሠτÁς µονάδος. ε„ γ¦ρ δυνατόν, µετρείτω τις τÕν ΖΗ Ð Ο, καˆ Ð Ο µηδενˆ τîν

For as many as is the multitude of A, B, C, D, let so many (numbers), E, HK, L, M , have been taken in a double proportion, (starting) from E. Thus, via equality, as A is to D, so E (is) to M [Prop. 7.14]. Thus, the (number created) from (multiplying) E, D is equal to the (number created) from (multiplying) A, M . And F G is the (number created) from (multiplying) E, D. Thus, F G is also the (number created) from (multiplying) A, M [Prop. 7.19]. Thus, A has made F G (by) multiplying M . Thus, M measures F G according to the units in A. And A is a dyad. Thus, F G is double M . And M , L, HK, E are also continuously double one another. Thus, E, HK, L, M , F G are continuously proportional in a double proportion. So let HN and F O, each equal to the first (number) E, have been subtracted from the second (number) HK and the last F G (respectively). Thus, as the excess of the second number is to the first, so the excess of the last (is) to (the sum of) all those (numbers) before it [Prop. 9.35]. Thus, as N K is to E, so OG (is) to M , L, KH, E. And N K is equal to E. And thus OG is equal to M , L, HK, E. And F O is also equal to E, and E to A, B, C, D, and a unit. Thus, the whole of F G is equal to E, HK, L, M , and A, B, C, D, and a unit. And it is measured by them. I also say that F G will be

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Α, Β, Γ, ∆, Ε, ΘΚ, Λ, Μ œστω Ð αÙτός. κሠÐσάκις Ð Ο τÕν ΖΗ µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Π· Ð Π ¥ρα τÕν Ο πολλαπλασιάσας τÕν ΖΗ πεποίηκεν. ¢λλ¦ µ¾ν καˆ Ð Ε τÕν ∆ πολλαπλασιάσας τÕν ΖΗ πεποίηκεν· œστιν ¥ρα æς Ð Ε πρÕς τÕν Π, Ð Ο πρÕς τÕν ∆. κሠ™πεˆ ¢πÕ µονάδος ˜ξÁς ¢νάλογόν ε„σιν οƒ Α, Β, Γ, ∆, Ð ∆ ¥ρα Øπ' οÙδενÕς ¥λλου ¢ριθµοà µετρηθήσεται παρξ τîν Α, Β, Γ. κሠØπόκειται Ð Ο οÙδενˆ τîν Α, Β, Γ Ð αÙτός· οÙκ ¥ρα µετρήσει Ð Ο τÕν ∆. ¢λλ' æς Ð Ο πρÕς τÕν ∆, Ð Ε πρÕς τÕν Π· οÙδ Ð Ε ¥ρα τÕν Π µετρε‹. καί ™στιν Ð Ε πρîτος· π©ς δ πρîτος ¢ριθµÕς πρÕς ¤παντα, Öν µ¾ µετρε‹, πρîτός [™στιν]. οƒ Ε, Π ¥ρα πρîτοι πρÕς ¢λλήλους ε„σίν. οƒ δ πρîτοι κሠ™λάχιστοι, οƒ δ ™λάχιστοι µετροàσι τοÝς τÕν αÙτÕν λόγον œχοντας „σάκις Ó τε ¹γούµενος τÕν ¹γούµενον καˆ Ð ˜πόµενος τÕν ˜πόµενον· καί ™στιν æς Ð Ε πρÕς τÕν Π, Ð Ο πρÕς τÕν ∆. „σάκις ¥ρα Ð Ε τÕν Ο µετρε‹ καˆ Ð Π τÕν ∆. Ð δ ∆ Øπ' οÙδενÕς ¥λλου µετρε‹ται παρξ τîν Α, Β, Γ· Ð Π ¥ρα ˜νˆ τîν Α, Β, Γ ™στιν Ð αÙτός. œστω τù Β Ð αÙτός. κሠÓσοι ε„σˆν οƒ Β, Γ, ∆ τù πλήθει τοσοàτοι ε„λήφθωσαν ¢πÕ τοà Ε οƒ Ε, ΘΚ, Λ. καί ε„σιν οƒ Ε, ΘΚ, Λ το‹ς Β, Γ, ∆ ™ν τù αÙτù λόγω· δι' ‡σου ¥ρα ™στˆν æς Ð Β πρÕς τÕν ∆, Ð Ε πρÕς τÕν Λ. Ð ¥ρα ™κ τîν Β, Λ ‡σος ™στˆ τù ™κ τîν ∆, Ε· ¢λλ' Ð ™κ τîν ∆, Ε ‡σος ™στˆ τù ™κ τîν Π, Ο· καˆ Ð ™κ τîν Π, Ο ¥ρα ‡σος ™στˆ τù ™κ τîν Β, Λ. œστιν ¥ρα æς Ð Π πρÕς τÕν Β, Ð Λ πρÕς τÕν Ο. καί ™στιν Ð Π τù Β Ð αÙτός· καˆ Ð Λ ¥ρα τJ Ο ™στιν Ð αÙτός· Óπερ ¢δύνατον· Ð γ¦ρ Ο Øπόκειται µηδενˆ τîν ™κκειµένων Ð αÙτός· οÙκ ¥ρα τÕν ΖΗ µετρήσει τις ¢ριθµÕς παρξ τîν Α, Β, Γ, ∆, Ε, ΘΚ, Λ, Μ κሠτÁς µονάδος. κሠ™δείχη Ð ΖΗ το‹ς Α, Β, Γ, ∆, Ε, ΘΚ, Λ, Μ κሠτÍ µονάδι ‡σος. τέλειος δ ¢ριθµός ™στιν Ð το‹ς ˜αυτοà µέρεσιν ‡σος êν· τέλειος ¥ρα ™στˆν Ð ΖΗ· Óπερ œδει δε‹ξαι.

measured by no other (numbers) except A, B, C, D, E, HK, L, M , and a unit. For, if possible, let some (number) P measure F G, and let P not be the same as any of A, B, C, D, E, HK, L, M . And as many times as P measures F G, so many units let there be in Q. Thus, Q has made F G (by) multiplying P . But, in fact, E has also made F G (by) multiplying D. Thus, as E is to Q, so P (is) to D [Prop. 7.19]. And since A, B, C, D are continually proportional, (starting) from a unit, D will thus not be measured by any other numbers except A, B, C [Prop. 9.13]. And P was assumed not (to be) the same as any of A, B, C. Thus, P does not measure D. But, as P (is) to D, so E (is) to Q. Thus, E does not measure Q either [Def. 7.20]. And E is a prime (number). And every prime number [is] prime to every (number) which it does not measure [Prop. 7.29]. Thus, E and Q are prime to one another. And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. And as E is to Q, (so) P (is) to D. Thus, E measures P the same number of times as Q (measures) D. And D is not measured by any other (numbers) except A, B, C. Thus, Q is the same as one of A, B, C. Let it be the same as B. And as many as is the multitude of B, C, D, let so many (of the set out numbers) have been taken, (starting) from E, (namely) E, HK, L. And E, HK, L are in the same ratio as B, C, D. Thus, via equality, as B (is) to D, (so) E (is) to L [Prop. 7.14]. Thus, the (number created) from (multiplying) B, L is equal to the (number created) from multiplying D, E [Prop. 7.19]. But, the (number created) from (multiplying) D, E is equal to the (number created) from (multiplying) Q, P . Thus, the (number created) from (multiplying) Q, P is equal to the (number created) from (multiplying) B, L. Thus, as Q is to B, (so) L (is) to P [Prop. 7.19]. And Q is the same as B. Thus, L is also the same as P . The very thing (is) impossible. For P was assumed not (to be) the same as any of the (numbers) set out. Thus, F G cannot be measured by any number except A, B, C, D, E, HK, L, M , and a unit. And F G was shown (to be) equal to (the sum of) A, B, C, D, E, HK, L, M , and a unit. And a perfect number is one which is equal to (the sum of) its own parts [Def. 7.22]. Thus, F G is a perfect (number). (Which is) the very thing it was required to show.

†

This proposition demonstrates that perfect numbers take the form 2n−1 (2n − 1) provided 2n − 1 is a prime number. The ancient Greeks knew of four perfect numbers: 6, 28, 496, and 8128, which correspond to n = 2, 3, 5, and 7, respectively.

279

280

ELEMENTS BOOK 10 Incommensurable magnitudes†

† The theory of incommensurable magntidues set out in this book is generally attributed to Theaetetus of Athens. In the footnotes throughout this book, k, k ′ , etc. stand for distinct ratios of positive integers.

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“Οροι.

Definitions

α΄. Σύµµετρα µεγέθη λέγεται τ¦ τù αÙτù µετρJ µετρούµενα, ¢σύµµετρα δέ, ïν µηδν ™νδέχεται κοινÕν µέτρον γενέσθαι. β΄. ΕÙθε‹αι δυνάµει σύµµετροί ε„σιν, Óταν τ¦ ¢π' αÙτîν τετράγωνα τù αÙτù χωρίJ µετρÁται, ¢σύµµετροι δέ, Óταν το‹ς ¢π' αÙτîν τετραγώνοις µηδν ™νδέχηται χωρίον κοινÕν µέτρον γενέσθαι. γ΄. Τούτων Øποκειµένων δείκνυται, Óτι τÍ προτεθείσV εÙθείv Øπάρχουσιν εÙθε‹αι πλήθει ¥πειροι σύµµετροί τε κሠ¢σύµµετροι αƒ µν µήκει µόνον, αƒ δ κሠδυνάµει. καλείσθω οâν ¹ µν προτεθε‹σα εÙθε‹α ·ητή, καˆ αƒ ταύτV σύµµετροι ε‡τε µήκει κሠδυνάµει ε‡τε δυνάµει µόνον ·ηταί, αƒ δ ταύτV ¢σύµµετροι ¥λογοι καλείσθωσαν. δ΄. ΚሠτÕ µν ¢πÕ τÁς προτεθείσης εÙθείας τετράγωνον ·ητόν, κሠτ¦ τούτJ σύµµετρα ·ητά, τ¦ δ τούτJ ¢σύµµετρα ¥λογα καλείσθω, καˆ αƒ δυνάµεναι αÙτ¦ ¥λογοι, ε„ µν τετράγωνα ε‡η, αÙταˆ αƒ πλευραί, ε„ δ ›τερά τινα εÙθύγραµµα, αƒ ‡σα αÙτο‹ς τετράγωνα ¢ναγράφουσαι.

1. Those magnitudes measured by the same measure are said (to be) commensurable, but (those) of which no (magnitude) admits to be a common measure (are said to be) incommensurable.† 2. (Two) straight-lines are commensurable in square‡ when the squares on them are measured by the same area, but (are) incommensurable (in square) when no area admits to be a common measure of the squares on them.§ 3. These things being assumed, it is proved that there exist an infinite multitude of straight-lines commensurable and incommensurable with an assigned straightline—those (incommensurable) in length only, and those also (commensurable or incommensurable) in square.¶ Therefore, let the assigned straight-line be called rational. And (let) the (straight-lines) commensurable with it, either in length and square, or in square only, (also be called) rational. But let the (straight-lines) incommensurable with it be called irrational.∗ 4. And let the square on the assigned straight-line be called rational. And (let areas) commensurable with it (also be called) rational. But (let areas) incommensurable with it (be called) irrational, and (let) their squareroots$ (also be called) irrational—the sides themselves, if the (areas) are squares, and the (straight-lines) describing squares equal to them, if the (areas) are some other rectilinear (figure).k

†

In other words, two magnitudes α and β are commensurable if α : β :: 1 : k, and incommensurable otherwise.

‡

Literally, “in power”.

§

In other words, two straight-lines of length α and β are commensurable in square if α : β :: 1 : k 1/2 , and incommensurable in square otherwise.

Likewise, the straight-lines are commensurable in length if α : β :: 1 : k, and incommensurable in length otherwise. ¶

To be more exact, straight-lines can either be commensurable in square only, incommensurable in length only, or commenusrable/incommensurable

in both length and square, with an assigned straight-line. ∗

Let the length of the assigned straight-line be unity. Then rational straight-lines have lengths expressible as k or k 1/2 , depending on whether the lengths are commensurable in length, or in square only, respectively, with unity. All other straight-lines are irrational. $

The square-root of an area is the length of the side of an equal area square.

k

The area of the square on the assigned straight-line is unity. Rational areas are expressible as k. All other areas are irrational. Thus, squares

whose sides are of rational length have rational areas, and vice versa.

α΄.

Proposition 1†

∆ύο µεγεθîν ¢νίσων ™κκειµένων, ™¦ν ¢πÕ τοà µείζονος ¢φαιρεθÍ µε‹ζον À τÕ ¼µισυ κሠτοà καταλειποµένου µε‹ζον À τÕ ¼µισυ, κሠτοàτο ¢εˆ γίγνηται, λειφθήσεταί τι µέγεθος, Ö œσται œλασσον τοà ™κκειµένου ™λάσσονος µεγέθους.

If, from the greater of two unequal magnitudes (which are) laid out, (a part) greater than half is subtracted, and (if from) the remainder (a part) greater than half (is subtracted), and (if) this happens continually, then some magnitude will (eventually) be left which will

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ELEMENTS BOOK 10

”Εστω δύο µεγέθη ¥νισα τ¦ ΑΒ, Γ, ïν µε‹ζον τÕ ΑΒ· λέγω, Óτι, ™αν ¢πÕ τοà ΑΒ ¢φαιρεθÍ µε‹ζον À τÕ ¼µισυ κሠτοà καταλειποµένου µε‹ζον À τÕ ¼µισυ, κሠτοàτο ¢εˆ γίγνηται, λειφθήσεταί τι µέγεθος, Ö œσται œλασσον τοà Γ µεγέθους.

ΑΚ Θ

be less than the lesser laid out magnitude. Let AB and C be two unequal magnitudes, of which (let) AB (be) the greater. I say that if (an amount) greater than half is subtracted from AB, and (if) (an amount) greater than half (is subtracted) from the remainder, and (if) this happens continually, then some magnitude will (eventually) be left which will be less than the magnitude C.

Β

A K

Γ ∆

B

C Ζ

Η

Ε

D

ΤÕ Γ γ¦ρ πολλαπλασιαζόµενον œσται ποτ τοà ΑΒ µε‹ζον. πεπολλαπλασιάσθω, κሠœστω τÕ ∆Ε τοà µν Γ πολλαπλάσιον, τοà δ ΑΒ µε‹ζον, κሠδιVρήσθω τÕ ∆Ε ε„ς τ¦ τù Γ ‡σα τ¦ ∆Ζ, ΖΗ, ΗΕ, κሠ¢φVρήσθω ¢πÕ µν τοà ΑΒ µε‹ζον À τÕ ¼µισυ τÕ ΒΘ, ¢πÕ δ τοà ΑΘ µε‹ζον À τÕ ¼µισυ τÕ ΘΚ, κሠτοàτο ¢εˆ γιγνέσθω, ›ως ¨ν αƒ ™ν τù ΑΒ διαιρέσεις „σοπληθε‹ς γένωνται τα‹ς ™ν τù ∆Ε διαιρέσεσιν. ”Εστωσαν οâν αƒ ΑΚ, ΚΘ, ΘΒ διαιρέσεις „σοπληθε‹ς οâσαι τα‹ς ∆Ζ, ΖΗ, ΗΕ· κሠ™πεˆ µε‹ζόν ™στι τÕ ∆Ε τοà ΑΒ, κሠ¢φÇρηται ¢πÕ µν τοà ∆Ε œλασσον τοà ¹µίσεως τÕ ΕΗ, ¢πÕ δ τοà ΑΒ µε‹ζον À τÕ ¼µισυ τÕ ΒΘ, λοιπÕν ¥ρα τÕ Η∆ λοιποà τοà ΘΑ µε‹ζόν ™στιν. κሠ™πεˆ µε‹ζόν ™στι τÕ Η∆ τοà ΘΑ, κሠ¢φÇρηται τοà µν Η∆ ¼µισυ τÕ ΗΖ, τοà δ ΘΑ µε‹ζον À τÕ ¼µισυ τÕ ΘΚ, λοιπÕν ¥ρα τÕ ∆Ζ λοιποà τοà ΑΚ µε‹ζόν ™στιν. ‡σον δ τÕ ∆Ζ τù Γ· κሠτÕ Γ ¥ρα τοà ΑΚ µε‹ζόν ™στιν. œλασσον ¥ρα τÕ ΑΚ τοà Γ. Καταλείπεται ¥ρα ¢πÕ τοà ΑΒ µεγέθους τÕ ΑΚ µέγεθος œλασσον ×ν τοà ™κκειµένου ™λάσσονος µεγέθους τοà Γ· Óπερ œδει δε‹ξαι. — еοίως δ δειχθήσεται, κ¨ν ¹µίση Ï τ¦ ¢φαιρούµενα.

†

H

F

G

E

For C, when multiplied (by some number), will sometimes be greater than AB [Def. 5.4]. Let it have been (so) multiplied. And let DE be (both) a multiple of C, and greater than AB. And let DE have been divided into the (divisions) DF , F G, GE, equal to C. And let BH, (which is) greater than half, have been subtracted from AB. And (let) HK, (which is) greater than half, (have been subtracted) from AH. And let this happen continually, until the divisions in AB become equal in number to the divisions in DE. Therefore, let the divisions (in AB) be AK, KH, HB, being equal in number to DF , F G, GE. And since DE is greater than AB, and EG, (which is) less than half, has been subtracted from DE, and BH, (which is) greater than half, from AB, the remainder GD is thus greater than the remainder HA. And since GD is greater than HA, and the half GF has been subtracted from GD, and HK, (which is) greater than half, from HA, the remainder DF is thus greater than the remainder AK. And DF (is) equal to C. C is thus also greater than AK. Thus, AK (is) less than C. Thus, the magnitude AK, which is less than the lesser laid out magnitude C, is left over from the magnitude AB. (Which is) the very thing it was required to show. — (The theorem) can similarly be proved even if the (parts) subtracted are halves.

This theorem is the basis of the so-called method of exhaustion, and is generally attributed to Eudoxus of Cnidus.

β΄.

Proposition 2

'Ε¦ν δύο µεγεθîν [™κκειµένων] ¢νίσων ¢νθυφαιρουµένου ¢εˆ τοà ™λάσσονος ¢πÕ τοà µείζονος τÕ καταλειπόµενον µηδέποτε καταµετρÍ τÕ πρÕ ˜αυτοà, ¢σύµµετρα œσται τ¦ µεγέθη. ∆ύο γ¦ρ µεγεθîν Ôντων ¢νίσων τîν ΑΒ, Γ∆ κሠ™λάσσονος τοà ΑΒ ¢νθυφαιρουµένου ¢εˆ τοà ™λάσσονος

If the remainder of two unequal magnitudes (which are) [laid out] never measures the (magnitude) before it, (when) the lesser (magnitude is) continually subtracted in turn from the greater, then the (original) magnitudes will be incommensurable. For, AB and CD being two unequal magnitudes, and

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¢πÕ τοà µείζονος τÕ περιλειπόµενον µηδέποτε καταµετρείτω τÕ πρÕ ˜αυτοà· λέγω, Óτι ¢σύµµετρά ™στι τ¦ ΑΒ, Γ∆ µεγέθη.

Α Η

AB (being) the lesser, let the remainder never measure the (magnitude) before it, (when) the lesser (magnitude is) continually subtracted in turn from the greater. I say that the magnitudes AB and CD are incommensurable.

Β

A G

Ε Γ

E Ζ

∆

C

Ε„ γάρ ™στι σύµµετρα, µετρήσει τι αÙτ¦ µέγεθος. µετρείτω, ε„ δυνατόν, κሠœστω τÕ Ε· κሠτÕ µν ΑΒ τÕ Ζ∆ καταµετροàν λειπέτω ˜αυτοà œλασσον τÕ ΓΖ, τÕ δ ΓΖ τÕ ΒΗ καταµετροàν λειπέτω ˜αυτοà œλασσον τÕ ΑΗ, κሠτοàτο ¢εˆ γινέσθω, ›ως οá λειφθÍ τι µέγεθος, Ó ™στιν œλασσον τοà Ε. γεγονέτω, κሠλελείφθω τÕ ΑΗ œλασσον τοà Ε. ™πεˆ οâν τÕ Ε τÕ ΑΒ µετρε‹, ¢λλ¦ τÕ ΑΒ τÕ ∆Ζ µετρε‹, κሠτÕ Ε ¥ρα τÕ Ζ∆ µετρήσει. µετρε‹ δ κሠÓλον τÕ Γ∆· κሠλοιπÕν ¥ρα τÕ ΓΖ µετρήσει. ¢λλ¦ τÕ ΓΖ τÕ ΒΗ µετρε‹· κሠτÕ Ε ¥ρα τÕ ΒΗ µετρε‹. µετρε‹ δ κሠÓλον τÕ ΑΒ· κሠλοιπÕν ¥ρα τÕ ΑΗ µετρήσει, τÕ µε‹ζον τÕ œλασσον· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τ¦ ΑΒ, Γ∆ µεγέθη µετρήσει τι µέγεθος· ¢σύµµετρα ¥ρα ™στˆ τ¦ ΑΒ, Γ∆ µεγέθη. 'Ε¦ν ¥ρα δύο µεγεθîν ¢νίσων, κሠτ¦ ˜ξÁς.

†

B

F

D

For if they are commensurable then some magnitude will measure them (both). If possible, let it (so) measure (them), and let it be E. And let AB leave CF less than itself (in) measuring F D, and let CF leave AG less than itself (in) measuring BG, and let this happen continually, until some magnitude which is less than E is left. Let (this) have occurred,† and let AG, (which is) less than E, have been left. Therefore, since E measures AB, but AB measures DF , E will thus also measure F D. And it also measures the whole (of) CD. Thus, it will also measure the remainder CF . But, CF measures BG. Thus, E also measures BG. And it also measures the whole (of) AB. Thus, it will also measure the remainder AG, the greater (measuring) the lesser. The very thing is impossible. Thus, some magnitude cannot measure (both) the magnitudes AB and CD. Thus, the magnitudes AB and CD are incommensurable [Def. 10.1]. Thus, if . . . of two unequal magnitudes, and so on . . . .

The fact that this will eventually occur is guaranteed by Prop. 10.1.

γ΄.

Proposition 3

∆ύο µεγεθîν συµµέτρων δοθέντων τÕ µέγιστον αÙτîν κοινÕν µέτρον εØρε‹ν.

To find the greatest common measure of two given commensurable magnitudes.

ΑΖ Γ Ε Η

Β

A F ∆

C E G

”Εστω τ¦ δοθέντα δύο µεγέθη σύµµετρα τ¦ ΑΒ, Γ∆, ïν œλασσον τÕ ΑΒ· δε‹ δ¾ τîν ΑΒ, Γ∆ τÕ µέγιστον κοινÕν µέτρον εØρε‹ν. ΤÕ ΑΒ γ¦ρ µέγεθος ½τοι µετρε‹ τÕ Γ∆ À οÜ. ε„ µν οâν µετρε‹, µετρε‹ δ κሠ˜αυτό, τÕ ΑΒ ¥ρα τîν ΑΒ, Γ∆ κοινÕν µέτρον ™στίν· κሠφανερόν, Óτι κሠµέγιστον. µε‹ζον γ¦ρ τοà ΑΒ µεγέθους τÕ ΑΒ οÙ µετρήσει. Μ¾ µετρείτω δ¾ τÕ ΑΒ τÕ Γ∆. κሠ¢νθυφαιρουµένου ¢εˆ τοà ™λάσσονος ¢πÕ τοà µείζονος, τÕ περιλειπόµενον µετρήσει ποτ τÕ πρÕ ˜αυτοà δι¦ τÕ µ¾ εναι ¢σύµµετρα

B D

Let AB and CD be the two given magnitudes, of which (let) AB (be) the lesser. So, it is required to find the greatest common measure of AB and CD. For the magnitude AB either measures, or (does) not (measure), CD. Therefore, if it measures (CD), and (since) it also measures itself, AB is thus a common measure of AB and CD. And (it is) clear that (it is) also (the) greatest. For a (magnitude) greater than magnitude AB cannot measure AB. So let AB not measure CD. And continually subtract-

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τ¦ ΑΒ, Γ∆· κሠτÕ µν ΑΒ τÕ Ε∆ καταµετροàν λειπέτω ˜αυτοà œλασσον τÕ ΕΓ, τÕ δ ΕΓ τÕ ΖΒ καταµετροàν λειπέτω ˜αυτοà œλασσον τÕ ΑΖ, τÕ δ ΑΖ τÕ ΓΕ µετρείτω. 'Επεˆ οâν τÕ ΑΖ τÕ ΓΕ µετρε‹, ¢λλ¦ τÕ ΓΕ τÕ ΖΒ µετρε‹, κሠτÕ ΑΖ ¥ρα τÕ ΖΒ µετρήσει. µετρε‹ δ κሠ˜αυτό· κሠÓλον ¥ρα τÕ ΑΒ µετρήσει τÕ ΑΖ. ¢λλ¦ τÕ ΑΒ τÕ ∆Ε µετρε‹· κሠτÕ ΑΖ ¥ρα τÕ Ε∆ µετρήσει. µετρε‹ δ κሠτÕ ΓΕ· καί Óλον ¥ρα τÕ Γ∆ µετρε‹· τÕ ΑΖ ¥ρα τîν ΑΒ, Γ∆ κοινÕν µέτρον ™στίν. λέγω δή, Óτι κሠµέγιστον. ε„ γ¦ρ µή, œσται τι µέγεθος µε‹ζον τοà ΑΖ, Ö µετρήσει τ¦ ΑΒ, Γ∆. œστω τÕ Η. ™πεˆ οâν τÕ Η τÕ ΑΒ µετρε‹, ¢λλ¦ τÕ ΑΒ τÕ Ε∆ µετρε‹, κሠτÕ Η ¥ρα τÕ Ε∆ µετρήσει. µετρε‹ δ κሠÓλον τÕ Γ∆· κሠλοιπÕν ¥ρα τÕ ΓΕ µετρήσει τÕ Η. ¢λλ¦ τÕ ΓΕ τÕ ΖΒ µετρε‹· κሠτÕ Η ¥ρα τÕ ΖΒ µετρήσει. µετρε‹ δ κሠÓλον τÕ ΑΒ, κሠλοιπÕν τÕ ΑΖ µετρήσει, τÕ µε‹ζον τÕ œλασσον· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα µε‹ζόν τι µέγεθος τοà ΑΖ τ¦ ΑΒ, Γ∆ µετρήσει· τÕ ΑΖ ¥ρα τîν ΑΒ, Γ∆ τÕ µέγιστον κοινÕν µέτρον ™στίν. ∆ύο ¥ρα µεγεθîν συµµέτρων δοθέντων τîν ΑΒ, Γ∆ τÕ µέγιστον κοινÕν µέτρον ηÛρηται· Óπερ œδει δε‹ξαι.

ing in turn the lesser (magnitude) from the greater, the remaining (magnitude) will (at) some time measure the (magnitude) before it, on account of AB and CD not being incommensurable [Prop. 10.2]. And let AB leave EC less than itself (in) measuring ED, and let EC leave AF less than itself (in) measuring F B, and let AF measure CE. Therefore, since AF measures CE, but CE measures F B, AF will thus also measure F B. And it also measures itself. Thus, AF will also measure the whole (of) AB. But, AB measures DE. Thus, AF will also measure ED. And it also measures CE. Thus, it also measures the whole of CD. Thus, AF is a common measure of AB and CD. So I say that (it is) also (the) greatest (common measure). For, if not, there will be some magnitude, greater than AF , which will measure (both) AB and CD. Let it be G. Therefore, since G measures AB, but AB measures ED, G will thus also measure ED. And it also measures the whole of CD. Thus, G will also measure the remainder CE. But CE measures F B. Thus, G will also measure F B. And it also measures the whole (of) AB. And (so) it will measure the remainder AF , the greater (measuring) the lesser. The very thing is impossible. Thus, some magnitude greater than AF cannot measure (both) AB and CD. Thus, AF is the greatest common measure of AB and CD. Thus, the greatest common measure of two given commensurable magnitudes, AB and CD, has been found. (Which is) the very thing it was required to show.

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι, ™¦ν µέγεθος δύο µεγέθη µετρÍ, κሠτÕ µέγιστον αÙτîν κοινÕν µέτρον µετρήσει.

So (it is) clear, from this, that if a magnitude measures two magnitudes then it will also measure their greatest common measure.

δ΄.

Proposition 4

Τριîν µεγεθîν συµµέτρων δοθέντων τÕ µέγιστον αÙτîν κοινÕν µέτρον εØρε‹ν.

To find the greatest common measure of three given commensurable magnitudes.

Α Β Γ

A B C ∆

Ε

Ζ

D

E

F

”Εστω τ¦ δοθέντα τρία µεγέθη σύµµετρα τ¦ Α, Β, Let A, B, C be the three given commensurable magΓ· δε‹ δ¾ τîν Α, Β, Γ τÕ µέγιστον κοινÕν µέτρον εØρε‹ν. nitudes. So it is required to find the greatest common Ε„λήφθω γ¦ρ δύο τîν Α, Β τÕ µέγιστον κοινÕν measure of A, B, C. µέτρον, κሠœστω τÕ ∆· τÕ δ¾ ∆ τÕ Γ ½τοι µετρε‹ À For let the greatest common measure of the two (mag285

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οÜ [µετρε‹]. µετρείτω πρότερον. ™πεˆ οâν τÕ ∆ τÕ Γ µετρε‹, µετρε‹ δ κሠτ¦ Α, Β, τÕ ∆ ¥ρα τ¦ Α, Β, Γ µετρε‹· τÕ ∆ ¥ρα τîν Α, Β, Γ κοινÕν µέτρον ™στίν. κሠφανερόν, Óτι κሠµέγιστον· µε‹ζον γ¦ρ τοà ∆ µεγέθους τ¦ Α, Β οÙ µετρε‹. Μ¾ µετρείτω δ¾ τÕ ∆ τÕ Γ. λέγω πρîτον, Óτι σύµµετρά ™στι τ¦ Γ, ∆. ™πεˆ γ¦ρ σύµµετρά ™στι τ¦ Α, Β, Γ, µετρήσει τι αÙτ¦ µέγεθος, Ö δηλαδ¾ κሠτ¦ Α, Β µετρήσει· éστε κሠτÕ τîν Α, Β µέγιστον κοινÕν µέτρον τÕ ∆ µετρήσει. µετρε‹ δ κሠτÕ Γ· éστε τÕ ε„ρηµένον µέγεθος µετρήσει τ¦ Γ, ∆· σύµµετρα ¥ρα ™στˆ τ¦ Γ, ∆. ε„λήφθω οâν αÙτîν τÕ µέγιστον κοινÕν µέτρον, κሠœστω τÕ Ε. ™πεˆ οâν τÕ Ε τÕ ∆ µετρε‹, ¥λλ¦ τÕ ∆ τ¦ Α, Β µετρε‹, κሠτÕ Ε ¥ρα τ¦ Α, Β µετρήσει. µετρε‹ δ κሠτÕ Γ. τÕ Ε ¥ρα τ¦ Α, Β, Γ µετρε‹· τÕ Ε ¥ρα τîν Α, Β, Γ κοινόν ™στι µέτρον. λέγω δή, Óτι κሠµέγιστον. ε„ γ¦ρ δυνατόν, œστω τι τοà Ε µε‹ζον µέγεθος τÕ Ζ, κሠµετρείτω τ¦ Α, Β, Γ. κሠ™πεˆ τÕ Ζ τ¦ Α, Β, Γ µετρε‹, κሠτ¦ Α, Β ¥ρα µετρήσει κሠτÕ τîν Α, Β µέγιστον κοινÕν µέτρον µετρήσει. τÕ δ τîν Α, Β µέγιστον κοινÕν µέτρον ™στˆ τÕ ∆· τÕ Ζ ¥ρα τÕ ∆ µετρε‹. µετρε‹ δ κሠτÕ Γ· τÕ Ζ ¥ρα τ¦ Γ, ∆ µετρε‹· κሠτÕ τîν Γ, ∆ ¥ρα µέγιστον κοινÕν µέτρον µετρήσει τÕ Ζ. œστι δ τÕ Ε· τÕ Ζ ¥ρα τÕ Ε µετρήσει, τÕ µε‹ζον τÕ œλασσον· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα µε‹ζόν τι τοà Ε µεγέθους [µέγεθος] τ¦ Α, Β, Γ µετρε‹· τÕ Ε ¥ρα τîν Α, Β, Γ τÕ µέγιστον κοινÕν µέτρον ™στίν, ™¦ν µ¾ µετρÍ τÕ ∆ τÕ Γ, ™¦ν δ µετρÍ, αÙτÕ τÕ ∆. Τριîν ¥ρα µεγεθîν συµµέτρων δοθέντων τÕ µέγιστον κοινÕν µέτρον ηÛρηται [Óπερ œδει δε‹ξαι].

nitudes) A and B have been taken [Prop. 10.3], and let it be D. So D either measures, or [does] not [measure], C. Let it, first of all, measure C. Therefore, since D measures C, and it also measures A and B, D thus measures A, B, C. Thus, D is a common measure of A, B, C. And (it is) clear that (it is) also (the) greatest (common measure). For no magnitude larger than D measures (both) A and B. So let D not measure C. I say, first, that C and D are commensurable. For if A, B, C are commensurable then some magnitude will measure them which will clearly also measure A and B. Hence, it will also measure D, the greatest common measure of A and B [Prop. 10.3 corr.]. And it also measures C. Hence, the aforementioned magnitude will measure (both) C and D. Thus, C and D are commensurable [Def. 10.1]. Therefore, let their greatest common measure have been taken [Prop. 10.3], and let it be E. Therefore, since E measures D, but D measures (both) A and B, E will thus also measure A and B. And it also measures C. Thus, E measures A, B, C. Thus, E is a common measure of A, B, C. So I say that (it is) also (the) greatest (common measure). For, if possible, let F be some magnitude greater than E, and let it measure A, B, C. And since F measures A, B, C, it will thus also measure A and B, and will (thus) measure the greatest common measure of A and B [Prop. 10.3 corr.]. And D is the greatest common measure of A and B. Thus, F measures D. And it also measures C. Thus, F measures (both) C and D. Thus, F will also measure the greatest common measure of C and D [Prop. 10.3 corr.]. And it is E. Thus, F will measure E, the greater (measuring) the lesser. The very thing is impossible. Thus, some [magnitude] greater than the magnitude E cannot measure A, B, C. Thus, if D does not measure C, then E is the greatest common measure of A, B, C. And if it does measure (C), then D itself (is the greatest common measure). Thus, the greatest common measure of three given commensurable magnitudes has been found. [(Which is) the very thing it was required to show.]

Πόρισµα.

Corollary

'Εκ δ¾ τούτου µεγέθη µετρÍ, κሠµετρήσει. `Οµοίως δ¾ κሠµέτρον ληφθήσεται, œδει δε‹ξαι.

φανερόν, Óτι, ™¦ν µέγεθος τρία So (it is) clear, from this, that if a magnitude measures τÕ µέγιστον αÙτîν κοινÕν µέτρον three magnitudes then it will also measure their greatest common measure. ™πˆ πλειόνων τÕ µέγιστον κοινÕν So, similarly, the greatest common measure of more κሠτÕ πόρισµα προχωρήσει. Óπερ (magnitudes) can also be taken, and the (above) corollary will go forward. (Which is) the very thing it was required to show.

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ε΄.

Proposition 5

Τ¦ σύµµετρα µεγέθη πρÕς ¥λληλα λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν.

Commensurable magnitudes have to one another the ratio which (some) number (has) to (some) number.

Α

Β

∆

Γ

A

Ε

D

”Εστω σύµµετρα µεγέθη τ¦ Α, Β· λέγω, Óτι τÕ Α πρÕς τÕ Β λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν. 'Επεˆ γ¦ρ σύµµετρά ™στι τ¦ Α, Β, µετρήσει τι αÙτ¦ µέγεθος. µετρείτω, κሠœστω τÕ Γ. κሠÐσάκις τÕ Γ τÕ Α µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù ∆, Ðσάκις δ τÕ Γ τÕ Β µετρε‹, τοσαàται µονάδες œστωσαν ™ν τù Ε. 'Επεˆ οâν τÕ Γ τÕ Α µετρε‹ κατ¦ τ¦ς ™ν τù ∆ µονάδας, µετρε‹ δ κሠ¹ µον¦ς τÕν ∆ κατ¦ τ¦ς ™ν αÙτù µονάδας, „σάκις ¥ρα ¹ µον¦ς τÕν ∆ µετρε‹ ¢ριθµÕν κሠτÕ Γ µέγεθος τÕ Α· œστιν ¥ρα æς τÕ Γ πρÕς τÕ Α, οÛτως ¹ µον¦ς πρÕς τÕν ∆· ¢νάπαλιν ¥ρα, æς τÕ Α πρÕς τÕ Γ, οÛτως Ð ∆ πρÕς τ¾ν µονάδα. πάλιν ™πεˆ τÕ Γ τÕ Β µετρε‹ κατ¦ τ¦ς ™ν τù Ε µονάδας, µετρε‹ δ κሠ¹ µον¦ς τÕν Ε κατ¦ τ¦ς ™ν αÙτù µονάδας, „σάκις ¥ρα ¹ µον¦ς τÕν Ε µετρε‹ κሠτÕ Γ τÕ Β· œστιν ¥ρα æς τÕ Γ πρÕς τÕ Β, οÛτως ¹ µον¦ς πρÕς τÕν Ε. ™δείχθη δ κሠæς τÕ Α πρÕς τÕ Γ, Ð ∆ πρÕς τ¾ν µονάδα· δι' ‡σου ¥ρα ™στˆν æς τÕ Α πρÕς τÕ Β, οÛτως Ð ∆ ¢ριθµÕς πρÕς τÕν Ε. Τ¦ ¥ρα σύµµετρα µεγέθη τ¦ Α, Β πρÕς ¥λληλα λόγον œχει, Öν ¢ριθµÕς Ð ∆ πρÕς ¢ριθµÕν τÕν Ε· Óπερ œδει δε‹ξαι.

†

B

C

E

Let A and B be commensurable magnitudes. I say that A has to B the ratio which (some) number (has) to (some) number. For if A and B are commensurable (magnitudes) then some magnitude will measure them. Let it (so) measure (them), and let it be C. And as many times as C measures A, so many units let there be in D. And as many times as C measures B, so many units let there be in E. Therefore, since C measures A according to the units in D, and a unit also measures D according to the units in it, a unit thus measures the number D as many times as the magnitude C (measures) A. Thus, as C is to A, so a unit (is) to D [Def. 7.20].† Thus, inversely, as A (is) to C, so D (is) to a unit [Prop. 5.7 corr.]. Again, since C measures B according to the units in E, and a unit also measures E according to the units in it, a unit thus measures E the same number of times that C (measures) B. Thus, as C is to B, so a unit (is) to E [Def. 7.20]. And it was also shown that as A (is) to C, so D (is) to a unit. Thus, via equality, as A is to B, so the number D (is) to the (number) E [Prop. 5.22]. Thus, the commensurable magnitudes A and B have to one another the ratio which the number D (has) to the number E. (Which is) the very thing it was required to show.

There is a slight logical gap here, since Def. 7.20 applies to four numbers, rather than two number and two magnitudes.

$΄.

Proposition 6

'Ε¦ν δύο µεγέθη πρÕς ¥λληλα λόγον œχV, Öν ¢ριθµÕς πρÕς ¢ριθµόν, σύµµετρα œσται τ¦ µεγέθη.

If two magnitudes have to one another the ratio which (some) number (has) to (some) number, then the magnitudes will be commensurable.

Α ∆ Γ

Β Ε Ζ

A D C

B E F

∆ύο γ¦ρ µεγέθη τ¦ Α, Β πρÕς ¥λληλα λόγον For let the two magnitudes A and B have to one an™χέτω, Öν ¢ριθµÕς Ð ∆ πρÕς ¢ριθµÕν τÕν Ε· λέγω, Óτι other the ratio which the number D (has) to the number

287

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

σύµµετρά ™στι τ¦ Α, Β µεγέθη. “Οσαι γάρ ε„σιν ™ν τù ∆ µονάδες, ε„ς τοσαàτα ‡σα διVρήσθω τÕ Α, κሠ˜νˆ αÙτîν ‡σον œστω τÕ Γ· Óσαι δέ ε„σιν ™ν τù Ε µονάδες, ™κ τοσούτων µεγεθîν ‡σων τù Γ συγκείσθω τÕ Ζ. 'Επεˆ οâν, Óσαι ε„σˆν ™ν τù ∆ µονάδες, τοσαàτά ε„σι κሠ™ν τù Α µεγέθη ‡σα τù Γ, Ö ¥ρα µέρος ™στˆν ¹ µον¦ς τοà ∆, τÕ αÙτÕ µέρος ™στˆ κሠτÕ Γ τοà Α· œστιν ¥ρα æς τÕ Γ πρÕς τÕ Α, οÛτως ¹ µον¦ς πρÕς τÕν ∆. µετρε‹ δ ¹ µον¦ς τÕν ∆ ¢ριθµόν· µετρε‹ ¥ρα κሠτÕ Γ τÕ Α. κሠ™πεί ™στιν æς τÕ Γ πρÕς τÕ Α, οÛτως ¹ µον¦ς πρÕς τÕν ∆ [¢ριθµόν], ¢νάπαλιν ¥ρα æς τÕ Α πρÕς τÕ Γ, οÛτως Ð ∆ ¢ριθµÕς πρÕς τ¾ν µονάδα. πάλιν ™πεί, Óσαι ε„σˆν ™ν τù Ε µονάδες, τοσαàτά ε„σι κሠ™ν τù Ζ ‡σα τù Γ, œστιν ¥ρα æς τÕ Γ πρÕς τÕ Ζ, οÛτως ¹ µον¦ς πρÕς τÕν Ε [¢ριθµόν]. ™δείχθη δ κሠæς τÕ Α πρÕς τÕ Γ, οÛτως Ð ∆ πρÕς τ¾ν µονάδα· δι' ‡σου ¥ρα ™στˆν æς τÕ Α πρÕς τÕ Ζ, οÛτως Ð ∆ πρÕς τÕν Ε. ¢λλ' æς Ð ∆ πρÕς τÕν Ε, οÛτως ™στˆ τÕ Α πρÕς τÕ Β· κሠæς ¥ρα τÕ Α πρÕς τÕ Β, οÛτως κሠπρÕς τÕ Ζ. τÕ Α ¥ρα πρÕς ˜κάτερον τîν Β, Ζ τÕν αÙτÕν œχει λόγον· ‡σον ¥ρα ™στˆ τÕ Β τù Ζ. µετρε‹ δ τÕ Γ τÕ Ζ· µετρε‹ ¥ρα κሠτÕ Β. ¢λλ¦ µ¾ν κሠτÕ Α· τÕ Γ ¥ρα τ¦ Α, Β µετρε‹. σύµµετρον ¥ρα ™στˆ τÕ Α τù Β. 'Ε¦ν ¥ρα δύο µεγέθη πρÕς ¥λληλα, κሠτ¦ ˜ξÁς.

E. I say that the magnitudes A and B are commensurable. For, as many units as there are in D, let A have been divided into so many equal (divisions). And let C be equal to one of them. And as many units as there are in E, let F be the sum of so many magnitudes equal to C. Therefore, since as many units as there are in D, so many magnitudes equal to C are also in A, therefore whichever part a unit is of D, C is also the same part of A. Thus, as C is to A, so a unit (is) to D [Def. 7.20]. And a unit measures the number D. Thus, C also measures A. And since as C is to A, so a unit (is) to the [number] D, thus, inversely, as A (is) to C, so the number D (is) to a unit [Prop. 5.7 corr.]. Again, since as many units as there are in E, so many (magnitudes) equal to C are also in F , thus as C is to F , so a unit (is) to the [number] E [Def. 7.20]. And it was also shown that as A (is) to C, so D (is) to a unit. Thus, via equality, as A is to F , so D (is) to E [Prop. 5.22]. But, as D (is) to E, so A is to B. And thus as A (is) to B, so (it) also is to F [Prop. 5.11]. Thus, A has the same ratio to each of B and F . Thus, B is equal to F [Prop. 5.9]. And C measures F . Thus, it also measures B. But, in fact, (it) also (measures) A. Thus, C measures (both) A and B. Thus, A is commensurable with B [Def. 10.1]. Thus, if two magnitudes . . . to one another, and so on ....

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι, ™¦ν ðσι δύο ¢ριθµοί, æς οƒ ∆, Ε, κሠεÙθε‹α, æς ¹ Α, δύνατόν ™στι ποιÁσαι æς Ð ∆ ¢ριθµÕς πρÕς τÕν Ε ¢ριθµόν, οÛτως τ¾ν εÙθε‹αν πρÕς εÙθε‹αν. ™¦ν δ κሠτîν Α, Ζ µέση ¢νάλογον ληφθÍ, æς ¹ Β, œσται æς ¹ Α πρÕς τ¾ν Ζ, οÛτως τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς Β, τουτέστιν æς ¹ πρώτη πρÕς τ¾ν τρίτην, οÛτως τÕ ¢πÕ τÁς πρώτης πρÕς τÕ ¢πÕ τÁς δευτέρας τÕ Óµοιον καˆ Ðµοίως ¢ναγραφόµενον. ¢λλ' æς ¹ Α πρÕς τ¾ν Ζ, οÛτως ™στˆν Ð ∆ ¢ριθµος πρÕς τÕν Ε ¢ριθµόν· γέγονεν ¥ρα κሠæς Ð ∆ ¢ριθµÕς πρÕς τÕν Ε ¢ριθµόν, οÛτως τÕ ¢πÕ τÁς Α εÙθείας πρÕς τÕ ¢πÕ τÁς Β εÙθείας· Óπερ œδει δε‹ξαι.

So it is clear, from this, that if there are two numbers, like D and E, and a straight-line, like A, then it is possible to contrive that as the number D (is) to the number E, so the straight-line (is) to (another) straight-line (i.e., F ). And if the mean proportion, (say) B, is taken of A and F , then as A is to F , so the (square) on A (will be) to the (square) on B. That is to say, as the first (is) to the third, so the (figure) on the first (is) to the similar, and similarly described, (figure) on the second [Prop. 6.19 corr.]. But, as A (is) to F , so the number D is to the number E. Thus, it has also been contrived that as the number D (is) to the number E, so the (figure) on the straight-line A (is) to the (similar figure) on the straight-line B. (Which is) the very thing it was required to show.

ζ΄.

Proposition 7

Τ¦ ¢σύµµετρα µεγέθη πρÕς ¥λληλα λόγον οÙκ œχει, Incommensurable magnitudes do not have to one anÖν ¢ριθµÕς πρÕς ¢ριθµόν. other the ratio which (some) number (has) to (some) ”Εστω ¢σύµµετρα µεγέθη τ¦ Α, Β· λέγω, Óτι τÕ Α number. πρÕς τÕ Β λόγον οÙκ œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν. Let A and B be incommensurable magnitudes. I say 288

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 that A does not have to B the ratio which (some) number (has) to (some) number.

Α Β

A B

Ε„ γ¦ρ œχει τÕ Α πρÕς τÕ Β λόγον, Öν ¢ριθµÕς πρÕς ¢ριθµόν, σύµµετρον œσται τÕ Α τù Β. οÙκ œστι δέ· οÙκ ¥ρα τÕ Α πρÕς τÕ Β λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν. Τ¦ ¥ρα ¢σύµµετρα µεγέθη πρÕς ¥λληλα λόγον οÙκ œχει, κሠτ¦ ˜ξÁς.

For if A has to B the ratio which (some) number (has) to (some) number, then A will be commensurable with B [Prop. 10.6]. But it is not. Thus, A does not have to B the ratio which (some) number (has) to (some) number. Thus, incommensurable numbers do not have to one another, and so on . . . .

η΄.

Proposition 8

'Ε¦ν δύο µεγέθη πρÕς ¥λληλα λόγον µ¾ œχV, Öν ¢ριθµÕς πρÕς ¢ριθµόν, ¢σύµµετρα œσται τ¦ µεγέθη.

If two magnitudes do not have to one another the ratio which (some) number (has) to (some) number, then the magnitudes will be incommensurable.

Α Β

A B

∆ύο γ¦ρ µεγέθη τ¦ Α, Β πρÕς ¥λληλα λόγον µ¾ ™χέτω, Öν ¢ριθµÕς πρÕς ¢ριθµόν· λέγω, Óτι ¢σύµµετρά ™στι τ¦ Α, Β µεγέθη. Ε„ γ¦ρ œσται σύµµετρα, τÕ Α πρÕς τÕ Β λόγον ›ξει, Öν ¢ριθµÕς πρÕς ¢ριθµόν. οÙκ œχει δέ. ¢σύµµετρα ¥ρα ™στˆ τ¦ Α, Β µεγέθη. 'Ε¦ν ¥ρα δύο µεγέθη πρÕς ¥λληλα, κሠτ¦ ˜ξÁς.

For let the two magnitudes A and B not have to one another the ratio which (some) number (has) to (some) number. I say that the magnitudes A and B are incommensurable. For if they are commensurable, A will have to B the ratio which (some) number (has) to (some) number [Prop. 10.5]. But it does not have (such a ratio). Thus, the magnitudes A and B are incommensurable. Thus, if two magnitudes . . . to one another, and so on ....

θ΄.

Proposition 9

Τ¦ ¢πÕ τîν µήκει συµµέτρων εÙθειîν τετράγωνα πρÕς ¥λληλα λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· κሠτ¦ τετράγωνα τ¦ πρÕς ¥λληλα λόγον œχοντα, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, κሠτ¦ς πλευρ¦ς ›ξει µήκει συµµέτρους. τ¦ δ ¢πÕ τîν µήκει ¢συµµέτρων εÙθειîν τετράγωνα πρÕς ¥λληλα λόγον οÙκ œχει, Óνπερ τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· κሠτ¦ τετράγωνα τ¦ πρÕς ¥λληλα λόγον µ¾ œχοντα, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ τ¦ς πλευρ¦ς ›ξει µήκει συµµέτρους.

Squares on straight-lines (which are) commensurable in length have to one another the ratio which (some) square number (has) to (some) square number. And squares having to one another the ratio which (some) square number (has) to (some) square number will also have sides (which are) commensurable in length. But squares on straight-lines (which are) incommensurable in length do not have to one another the ratio which (some) square number (has) to (some) square number. And squares not having to one another the ratio which (some) square number (has) to (some) square number will not have sides (which are) commensurable in length either.

Α Γ

Β ∆

A C

”Εστωσαν γ¦ρ αƒ Α, Β µήκει σύµµετροι· λέγω, Óτι τÕ

289

B D

For let A and B be (straight-lines which are) commen-

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς Β τετράγωνον λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. 'Επεˆ γ¦ρ σύµµετρός ™στιν ¹ Α τÍ Β µήκει, ¹ Α ¥ρα πρÕς τ¾ν Β λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν. ™χέτω, Öν Ð Γ πρÕς τÕν ∆. ™πεˆ οâν ™στιν æς ¹ Α πρÕς τ¾ν Β, οÛτως Ð Γ πρÕς τÕν ∆, ¢λλ¦ τοà µν τÁς Α πρÕς τ¾ν Β λόγου διπλασίων ™στˆν Ð τοà ¢πÕ τÁς Α τετραγώνου πρÕς τÕ ¢πÕ τÁς Β τετράγωνον· τ¦ γ¦ρ Óµοια σχήµατα ™ν διπλασίονι λόγJ ™στˆ τîν еολόγων πλευρîν· τοà δ τοà Γ [¢ριθµοà] πρÕς τÕν ∆ [¢ριθµÕν] λόγου διπλασίων ™στˆν Ð τοà ¢πÕ τοà Γ τετραγώνου πρÕς τÕν ¢πÕ τοà ∆ τετράγωνον· δύο γ¦ρ τετραγώνων ¢ριθµîν εŒς µέσος ¢νάλογόν ™στιν ¢ριθµός, καί Ð τετράγωνος πρÕς τÕν τετράγωνον [¢ριθµÕν] διπλασίονα λόγον œχει, ½περ ¹ πλευρ¦ πρÕς τ¾ν πλευράν· œστιν ¥ρα κሠæς τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς Β τετράγωνον, οÛτως Ð ¢πÕ τοà Γ τετράγωνος [¢ριθµÕς] πρÕς τÕν ¢πÕ τοà ∆ [¢ριθµοà] τετράγωνον [¢ριθµόν]. 'Αλλ¦ δ¾ œστω æς τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς Β, οÛτως Ð ¢πÕ τοà Γ τετράγωνος πρÕς τÕν ¢πÕ τοà ∆ [τετράγωνον]· λέγω, Óτι σύµµετρός ™στιν ¹ Α τÍ Β µήκει. 'Επεˆ γάρ ™στιν æς τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς Β [τετράγωνον], οÛτως Ð ¢πÕ τοà Γ τετράγωνος πρÕς τÕν ¢πÕ τοà ∆ [τετράγωνον], ¢λλ' Ð µν τοà ¢πÕ τ¾ς Α τετραγώνου πρÕς τÕ ¢πÕ τÁς Β [τετράγωνον] λόγος διπλασίων ™στˆ τοà τÁς Α πρÕς τ¾ν Β λόγου, Ð δ τοà ¢πÕ τοà Γ [¢ριθµοà] τετραγώνου [¢ριθµοà] πρÕς τÕν ¢πÕ τοà ∆ [¢ριθµοà] τετράγωνον [¢ριθµÕν] λόγος διπλασίων ™στˆ τοà τοà Γ [¢ριθµοà] πρÕς τÕν ∆ [¢ριθµÕν] λόγου, œστιν ¥ρα κሠæς ¹ Α πρÕς τ¾ν Β, οÛτως Ð Γ [¢ριθµÕς] πρÕς τÕν ∆ [¢ριθµόν]. ¹ Α ¥ρα πρÕς τ¾ν Β λόγον œχει, Öν ¢ριθµÕς Ð Γ πρÕς ¢ριθµÕν τÕν ∆· σύµµετρος ¥ρα ™στˆν ¹ Α τÍ Β µήκει. 'Αλλ¦ δ¾ ¢σύµµετρος œστω ¹ Α τÍ Β µήκει· λέγω, Óτι τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς Β [τετράγωνον] λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. Ε„ γ¦ρ œχει τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς Β [τετράγωνον] λόγον, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, σύµµετρος œσται ¹ Α τÍ Β. οÙκ œστι δέ· οÙκ ¥ρα τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς Β [τετράγωνον] λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. Πάλιν δ¾ τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς Β [τετράγωνον] λόγον µ¾ ™χέτω, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· λέγω, Óτι ¢σύµµετρός ™στιν ¹ Α τÍ Β µήκει. Ε„ γάρ ™στι σύµµετρος ¹ Α τÍ Β, ›ξει τÕ ¢πÕ τÁς Α

surable in length. I say that the square on A has to the square on B the ratio which (some) square number (has) to (some) square number. For since A is commensurable in length with B, A thus has to B the ratio which (some) number (has) to (some) number [Prop. 10.5]. Let it have (that) which C (has) to D. Therefore, since as A is to B, so C (is) to D, but the (ratio) of the square on A to the square on B is the square of the ratio of A to B. For similar figures are in the squared ratio of (their) corresponding sides [Prop. 6.20 corr.]. And the (ratio) of the square on C to the square on D is the square of the ratio of the [number] C to the [number] D. For there exits one number in mean proportion to two square numbers, and (one) square (number) has to the (other) square [number] a squared ratio with respect to (that) the side (of the former has) to the side (of the latter) [Prop. 8.11]. And, thus, as the square on A is to the square on B, so the square [number] on the (number) C (is) to the square [number] on the [number] D.† And so let the square on A be to the (square) on B as the square (number) on C (is) to the [square] (number) on D. I say that A is commensurable in length with B. For since as the square on A is to the [square] on B, so the square (number) on C (is) to the [square] (number) on D. But, the ratio of the square on A to the (square) on B is the square of the (ratio) of A to B [Prop. 6.20 corr.]. And the (ratio) of the square [number] on the [number] C to the square [number] on the [number] D is the square of the ratio of the [number] C to the [number] D [Prop. 8.11]. Thus, as A is to B, so the [number] C also (is) to the [number] D. A, thus, has to B the ratio which the number C has to the number D. Thus, A is commensurable in length with B [Prop. 10.6].‡ And so let A be incommensurable in length with B. I say that the square on A does not have to the [square] on B the ratio which (some) square number (has) to (some) square number. For if the square on A has to the [square] on B the ratio which (some) square number (has) to (some) square number then A will be commensurable (in length) with B. But it is not. Thus, the square on A does not have to the [square] on the B the ratio which (some) square number (has) to (some) square number. So, again, let the square on A not have to the [square] on B the ratio which (some) square number (has) to (some) square number. I say that A is incommensurable in length with B. For if A is commensurable (in length) with B then the (square) on A will have to the (square) on B the ra-

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πρÕς τÕ ¢πÕ τÁς Β λόγον, Öν τετράγωνος ¢ριθµÕς πρÕς tio which (some) square number (has) to (some) square τετράγωνον ¢ριθµόν. οÙκ œχει δέ· οÙκ ¥ρα σύµµετρός number. But it does not have (such a ratio). Thus, A is ™στιν ¹ Α τÍ Β µήκει. not commensurable in length with B. Τ¦ ¥ρα ¢πÕ τîν µήκει συµµέτρων, κሠτ¦ ˜ξÁς. Thus, (squares) on (straight-lines which are) commensurable in length, and so on . . . .

Πόρισµα.

Corollary

ΚሠφανερÕν ™κ τîν δεδειγµένων œσται, Óτι αƒ µήκει σύµµετροι πάντως κሠδυνάµει, αƒ δ δυνάµει οÙ πάντως κሠµήκει.

And it will be clear, from (what) has been demonstrated, that (straight-lines) commensurable in length (are) always also (commensurable) in square, but (straightlines commensurable) in square (are) not always (commensurable) in length.

†

There is an unstated assumption here that if α : β :: γ : δ then α2 : β 2 :: γ 2 : δ2 .

‡

There is an unstated assumption here that if α2 : β 2 :: γ 2 : δ2 then α : β :: γ : δ.

ι΄.

Proposition 10†

ΤÍ προτεθείσV εÙθείv προσευρε‹ν δύο εÙθείας ¢συµµέτρους, τ¾ν µν µήκει µόνον, τ¾ν δ κሠδυνάµει.

To find two straight-lines incommensurable with a given straight-line, the one (incommensurable) in length only, the other also (incommensurable) in square.

Α Ε ∆ Β Γ

A E D B C

”Εστω ¹ προτεθε‹σα εÙθε‹α ¹ Α· δε‹ δ¾ τÍ Α προσευρε‹ν δύο εÙθείας ¢συµµέτρους, τ¾ν µν µήκει µόνον, τ¾ν δ κሠδυνάµει. 'Εκκείσθωσαν γ¦ρ δύο αριθµοˆ οƒ Β, Γ πρÕς ¢λλήλους λόγον µ¾ œχοντες, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, τουτέστι µ¾ Óµοιοι ™πίπεδοι, κሠγεγονέτω æς Ð Β πρÕς τÕν Γ, οÛτως τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς ∆ τετράγωνον· ™µάθοµεν γάρ· σύµµετρον ¥ρα τÕ ¢πÕ τÁς Α τù ¢πÕ τÁς ∆. κሠ™πεˆ Ð Β πρÕς τÕν Γ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ' ¥ρα τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς ∆ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ Α τÍ ∆ µήκει. ε„λήφθω τîν Α, ∆ µέση ¢νάλογον ¹ Ε· œστιν ¥ρα æς ¹ Α πρÕς τ¾ν ∆, οÛτως τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς Ε. ¢σύµµετρος δέ ™στιν ¹ Α τÍ ∆ µήκει· ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς Α τετράγωνον τù ¢πÕ τÁς Ε τετραγώνJ· ¢σύµµετρος ¥ρα ™στˆν ¹ Α τÍ Ε δυνάµει. ΤÊ ¥ρα προτεθείσV εÙθείv τÍ Α προσεύρηνται δύο

Let A be the given straight-line. So it is required to find two straight-lines incommensurable with A, the one (incommensurable) in length only, the other also (incommensurable) in square. For let two numbers, B and C, not having to one another the ratio which (some) square number (has) to (some) square number—that is to say, not (being) similar plane (numbers)—have been taken. And let it be contrived that as B (is) to C, so the square on A (is) to the square on D. For we learned (how to do this) [Prop. 10.6 corr.]. Thus, the (square) on A (is) commensurable with the (square) on D [Prop. 10.6]. And since B does not have to C the ratio which (some) square number (has) to (some) square number, the (square) on A thus does not have to the (square) on D the ratio which (some) square number (has) to (some) square number either. Thus, A is incommensurable in length with D [Prop. 10.9]. Let the (straight-line) E (which is) in mean proportion to A and D have been taken [Prop. 6.13]. Thus, as A is to D, so the square on A (is) to the (square) on E [Def. 5.9].

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εÙθε‹αι ¢σύµµετροι αƒ ∆, Ε, µήκει µν µόνον ¹ ∆, And A is incommensurable in length with D. Thus, the δυνάµει δ κሠµήκει δηλαδ¾ ¹ Ε [Óπερ œδει δε‹ξαι]. square on A is also incommensurble with the square on E [Prop. 10.11]. Thus, A is incommensurable in square with E. Thus, two straight-lines, D and E, (which are) incommensurable with the given straight-line A, have been found, the one, D, (incommensurable) in length only, the other, E, (incommensurable) in square, and, clearly, also in length. [(Which is) the very thing it was required to show.] †

This whole proposition is regarded by Heiberg as an interpolation into the original text.

ια΄.

Proposition 11

'Ε¦ν τέσσαρα µεγέθη ¢νάλογον Ï, τÕ δ πρîτον If four magnitudes are proportional, and the first is τù δευτέρJ σύµµετρον Ï, κሠτÕ τρίτον τù τετάρτJ commensurable with the second, then the third will also σύµµετρον œσται· κ¨ν τÕ πρîτον τù δευτέρJ ¢σύµµετρον be commensurable with the fourth. And if the first is inÏ, κሠτÕ τρίτον τù τετάρτJ ¢σύµµετρον œσται. commensurable with the second, then the third will also be incommensurable with the fourth.

Α Γ

Β ∆

A C

B D

”Εστωσαν τέσσαρα µεγέθη ¢νάλογον τ¦ Α, Β, Γ, ∆, æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆, τÕ Α δ τù Β σύµµετρον œστω· λέγω, Óτι κሠτÕ Γ τù ∆ σύµµετρον œσται. 'Επεˆ γ¦ρ σύµµετρόν ™στι τÕ Α τù Β, τÕ Α ¥ρα πρÕς τÕ Β λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν. καί ™στιν æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆· κሠτÕ Γ ¥ρα πρÕς τÕ ∆ λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν· σύµµετρον ¥ρα ™στˆ τÕ Γ τù ∆. 'Αλλ¦ δ¾ τÕ Α τù Β ¢σύµµετρον œστω· λέγω, Óτι κሠτÕ Γ τù ∆ ¢σύµµετρον œσται. ™πεˆ γ¦ρ ¢σύµµετρόν ™στι τÕ Α τù Β, τÕ Α ¥ρα πρÕς τÕ Β λόγον οÙκ œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν. καί ™στιν æς τÕ Α πρÕς τÕ Β, οÛτως τÕ Γ πρÕς τÕ ∆· οÙδ τÕ Γ ¥ρα πρÕς τÕ ∆ λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν· ¢σύµµετρον ¥ρα ™στˆ τÕ Γ τù ∆. 'Ε¦ν ¥ρα τέσσαρα µεγέθη, κሠτ¦ ˜ξÁς.

Let A, B, C, D be four proportional magnitudes, (such that) as A (is) to B, so C (is) to D. And let A be commensurable with B. I say that C will also be commensurable with D. For since A is commensurable with B, A thus has to B the ratio which (some) number (has) to (some) number [Prop. 10.5]. And as A is to B, so C (is) to D. Thus, C also has to D the ratio which (some) number (has) to (some) number. Thus, C is commensurable with D [Prop. 10.6]. And so let A be incommensurable with B. I say that C will also be incommensurable with D. For since A is incommensurable with B, A thus does not have to B the ratio which (some) number (has) to (some) number [Prop. 10.7]. And as A is to B, so C (is) to D. Thus, C does not have to D the ratio which (some) number (has) to (some) number either. Thus, C is incommensurable with D [Prop. 10.8]. Thus, if four magnitudes, and so on . . . .

ιβ΄.

Proposition 12

Τ¦ τù αÙτù µεγέθει σύµµετρα κሠ¢λλήλοις ™στˆ (Magnitudes) commensurable with the same magniσύµµετρα. tude are also commensurable with one another. For let A and B each be commensurable with C. I say `Εκάτερον γ¦ρ τîν Α, Β τù Γ œστω σύµµετρον. λέγω, that A is also commensurable with B. Óτι κሠτÕ Α τù Β ™στι σύµµετρον. For since A is commensurable with C, A thus has 'Επεˆ γ¦ρ σύµµετρόν ™στι τÕ Α τù Γ, τÕ Α ¥ρα πρÕς to C the ratio which (some) number (has) to (some)

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τÕ Γ λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν. ™χέτω, Öν Ð ∆ πρÕς τÕν Ε. πάλιν, ™πεˆ σύµµετρόν ™στι τÕ Γ τù Β, τÕ Γ ¥ρα πρÕς τÕ Β λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν. ™χέτω, Öν Ð Ζ πρÕς τÕν Η. κሠλόγων δοθέντων Ðποσωνοàν τοà τε, Öν œχει Ð ∆ πρÕς τÕν Ε, καˆ Ð Ζ πρÕς τÕν Η ε„λήφθωσαν ¢ριθµοˆ ˜ξÁς ™ν το‹ς δοθε‹σι λόγοις οƒ Θ, Κ, Λ· éστε εναι æς µν τÕν ∆ πρÕς τÕν Ε, οÛτως τÕν Θ πρÕς τÕν Κ, æς δ τÕν Ζ πρÕς τÕν Η, οÛτως τÕν Κ πρÕς τÕν Λ.

Γ Θ Κ Λ

Α ∆ Ε Ζ Η

number [Prop. 10.5]. Let it have (the ratio) which D (has) to E. Again, since C is commensurable with B, C thus has to B the ratio which (some) number (has) to (some) number [Prop. 10.5]. Let it have (the ratio) which F (has) to G. And for any multitude whatsoever of given ratios—(namely,) those which D has to E, and F to G—let the numbers H, K, L (which are) continuously (proportional) in the(se) given ratios have been taken [Prop. 8.4]. Hence, as D is to E, so H (is) to K, and as F (is) to G, so K (is) to L.

Β

C H K L

A D E F G

B

'Επεˆ οâν ™στιν æς τÕ Α πρÕς τÕ Γ, οÛτως Ð ∆ πρÕς τÕν Ε, ¢λλ' æς Ð ∆ πρÕς τÕν Ε, οÛτως Ð Θ πρÕς τÕν Κ, œστιν ¥ρα κሠæς τÕ Α πρÕς τÕ Γ, οÛτως Ð Θ πρÕς τÕν Κ. πάλιν, ™πεί ™στιν æς τÕ Γ πρÕς τÕ Β, οÛτως Ð Ζ πρÕς τÕν Η, ¢λλ' æς Ð Ζ πρÕς τÕν Η, [οÛτως] Ð Κ πρÕς τÕν Λ, κሠæς ¥ρα τÕ Γ πρÕς τÕ Β, οÛτως Ð Κ πρÕς τÕν Λ. œστι δ κሠæς τÕ Α πρÕς τÕ Γ, οÛτως Ð Θ πρÕς τÕν Κ· δι' ‡σου ¥ρα ™στˆν æς τÕ Α πρÕς τÕ Β, οÛτως Ð Θ πρÕς τÕν Λ. τÕ Α ¥ρα πρÕς τÕ Β λόγον œχει, Öν ¢ριθµÕς Ð Θ πρÕς ¢ριθµÕν τÕν Λ· σύµµετρον ¥ρα ™στˆ τÕ Α τù Β. Τ¦ ¥ρα τù αÙτù µεγέθει σύµµετρα κሠ¢λλήλοις ™στˆ σύµµετρα· Óπερ œδει δε‹ξαι.

Therefore, since as A is to C, so D (is) to E, but as D (is) to E, so H (is) to K, thus also as A is to C, so H (is) to K [Prop. 5.11]. Again, since as C is to B, so F (is) to G, but as F (is) to G, [so] K (is) to L, thus also as C (is) to B, so K (is) to L [Prop. 5.11]. And also as A is to C, so H (is) to K. Thus, via equality, as A is to B, so H (is) to L [Prop. 5.22]. Thus, A has to B the ratio which the number H (has) to the number L. Thus, A is commensurable with B [Prop. 10.6]. Thus, (magnitudes) commensurable with the same magnitude are also commensurable with one another. (Which is) the very thing it was required to show.

ιγ΄.

Proposition 13

'Ε¦ν Ï δύο µεγέθη σύµµετρα, τÕ δ ›τερον αÙτîν µεγέθει τινˆ ¢σύµµετρον Ï, κሠτÕ λοιπÕν τù αÙτù ¢σύµµετρον œσται.

If two magnitudes are commensurable, and one of them is incommensurable with some magnitude, then the remaining (magnitude) will also be incommensurable with it.

Α Γ Β

A C B

”Εστω δύο µεγέθη σύµµετρα τ¦ Α, Β, τÕ δ ›τερον αÙτîν τÕ Α ¥λλJ τινˆ τù Γ ¢σύµµετρον œστω· λέγω, Óτι κሠτÕ λοιπÕν τÕ Β τù Γ ¢σύµµετρόν ™στιν. Ε„ γάρ ™στι σύµµετρον τÕ Β τù Γ, ¢λλ¦ κሠτÕ Α τù Β σύµµετρόν ™στιν, κሠτÕ Α ¥ρα τù Γ σύµµετρόν ™στιν. ¢λλ¦ κሠ¢σύµµετρον· Óπερ ¢δύνατον. οÙκ ¥ρα σύµµετρόν ™στι τÕ Β τù Γ· ¢σύµµετρον ¥ρα. 'Ε¦ν ¥ρα Ï δύο µεγέθη σύµµετρα, κሠτ¦ ˜ξÁς.

Let A and B be two commensurable magnitudes, and let one of them, A, be incommensurable with some other (magnitude), C. I say that the remaining (magnitude), B, is also incommensurable with C. For if B is commensurable with C, but A is also commensurable with B, A is thus also commensurable with C [Prop. 10.12]. But, (it is) also incommensurable (with C). The very thing (is) impossible. Thus, B is not commensurable with C. Thus, (it is) incommensurable. Thus, if two magnitudes are commensurable, and so on . . . .

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ΛÁµµα.

Lemma

∆ύο δοθεισîν εÙθειîν ¢νίσων εØρε‹ν, τίνι µε‹ζον δύναται ¹ µείζων τÁς ™λάσσονος.

For two given unequal straight-lines, to find by (the square on) which (straight-line) the square on the greater (straight-line is) larger than (the square on) the lesser.†

Γ

∆

Α

C D

Β

A

”Εστωσαν αƒ δοθε‹σαι δύο ¥νισοι εÙθε‹αι αƒ ΑΒ, Γ, ïν µείζων œστω ¹ ΑΒ· δε‹ δ¾ εØρε‹ν, τίνι µε‹ζον δύναται ¹ ΑΒ τÁς Γ. Γεγράφθω ™πˆ τÁς ΑΒ ¹µικύκλιον τÕ Α∆Β, κሠε„ς αÙτÕ ™νηρµόσθω τÍ Γ ‡ση ¹ Α∆, κሠ™πεζεύχθω ¹ ∆Β. φανερÕν δή, Óτι Ñρθή ™στιν ¹ ØπÕ Α∆Β γωνία, κሠÓτι ¹ ΑΒ τÁς Α∆, τουτέστι τÁς Γ, µε‹ζον δύναται τÍ ∆Β. `Οµοίως δ κሠδύο δοθεισîν εÙθειîν ¹ δυναµένη αÙτ¦ς εØρίσκεται οÛτως. ”Εστωσαν αƒ δοθε‹σαι δύο εÙθε‹αι αƒ Α∆, ∆Β, κሠδέον œστω εØρε‹ν τ¾ν δυναµένην αÙτάς. κείσθωσαν γάρ, éστε Ñρθ¾ν γωνίαν περιέχειν τ¾ν ØπÕ Α∆, ∆Β, κሠ™πεζεύχθω ¹ ΑΒ· φανερÕν πάλιν, Óτι ¹ τ¦ς Α∆, ∆Β δυναµένη ™στˆν ¹ ΑΒ· Óπερ œδει δε‹ξαι.

†

B

Let AB and C be the two given unequal straight-lines, and let AB be the greater of them. So it is required to find by (the square on) which (straight-line) the square on AB (is) greater than (the square on) C. Let the semi-circle ADB have been described on AB. And let AD, equal to C, have been inserted into it [Prop. 4.1]. And let DB have been joined. So (it is) clear that the angle ADB is a right-angle [Prop. 3.31], and that the square on AB (is) greater than (the square on) AD—that is to say, (the square on) C—by (the square on) DB [Prop. 1.47]. And, similarly, the square-root of (the sum of the squares on) two given straight-lines is also found likeso. Let AD and DB be the two given straight-lines. And let it be necessary to find the square-root of (the sum of the squares on) them. For let them have been laid down such as to encompass a right-angle—(namely), that (angle encompassed) by AD and DB. And let AB have been joined. (It is) again clear that AB is the square-root of (the sum of the squares on) AD and DB [Prop. 1.47]. (Which is) the very thing it was required to show.

That is, if α and β are the lengths of two given straight-lines, with α being greater than β, to find a straight-line of length γ such that

α2 = β 2 + γ 2 . Similarly, we can also find γ such that γ 2 = α2 + β 2 .

ιδ΄.

Proposition 14

'Ε¦ν τέσσαρες εÙθε‹αι ¢νάλογον ðσιν, δύνηται δ ¹ πρώτη τÁς δευτέρας µε‹ζον τù ¢πÕ συµµέτρου ˜αυτÍ [µήκει], κሠ¹ τρίτη τÁς τετάρτης µε‹ζον δυνήσεται τù ¢πÕ συµµέτρου ˜αυτÍ [µήκει]. κሠ™¦ν ¹ πρώτη τÁς δευτέρας µε‹ζον δύνηται τù ¢πÕ ¢συµµέτρου ˜αυτÍ [µήκει], κሠ¹ τρίτη τÁς τετάρτης µε‹ζον δυνήσεται τù ¢πÕ ¢συµµέτρου ˜αυτÍ [µήκει]. ”Εστωσαν τέσσαρες εÙθε‹αι ¢νάλογον αƒ Α, Β, Γ, ∆, æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Γ πρÕς τ¾ν ∆, κሠ¹ Α µν τÁς Β µε‹ζον δυνάσθω τù ¢πÕ τÁς Ε, ¹ δ Γ τÁς ∆ µε‹ζον δυνάσθω τù ¢πÕ τÁς Ζ· λέγω, Óτι, ε‡τε

If four straight-lines are proportional, and the square on the first is greater than (the square on) the second by the (square) on (some straight-line) commensurable [in length] with the first, then the square on the third will also be greater than (the square on) the fourth by the (square) on (some straight-line) commensurable [in length] with the third. And if the square on the first is greater than (the square on) the second by the (square) on (some straight-line) incommensurable [in length] with the first, then the square on the third will also be greater than (the square on) the fourth by

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ELEMENTS BOOK 10

σύµµετρός ™στιν ¹ Α τÍ Ε, σύµµετρός ™στι κሠ¹ Γ τÍ the (square) on (some straight-line) incommensurable Ζ, ε‡τε ¢σύµµετρός ™στιν ¹ Α τÍ Ε, ¢σύµµετρός ™στι [in length] with the third. καˆ Ð Γ τÍ Ζ. Let A, B, C, D be four proportional straight-lines, (such that) as A (is) to B, so C (is) to D. And let the square on A be greater than (the square on) B by the (square) on E, and let the square on C be greater than (the square on) D by the (square) on F . I say that A is either commensurable (in length) with E, and C is also commensurable with F , or A is incommensurable (in length) with E, and C is also incommensurable with F.

Α Β Ε Γ ∆ Ζ

A B E C D F

'Επεˆ γάρ ™στιν æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Γ πρÕς τ¾ν ∆, œστιν ¥ρα κሠæς τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς Β, οÛτως τÕ ¢πÕ τÁς Γ πρÕς τÕ ¢πÕ τÁς ∆. ¢λλ¦ τù µν ¢πÕ τÁς Α ‡σα ™στˆ τ¦ ¢πÕ τîν Ε, Β, τù δ ¢πÕ τÁς Γ ‡σα ™στˆ τ¦ ¢πÕ τîν ∆, Ζ. œστιν ¥ρα æς τ¦ ¢πÕ τîν Ε, Β πρÕς τÕ ¢πÕ τÁς Β, οÛτως τ¦ ¢πÕ τîν ∆, Ζ πρÕς τÕ ¢πÕ τÁς ∆· διελόντι ¥ρα ™στˆν æς τÕ ¢πÕ τÁς Ε πρÕς τÕ ¢πÕ τÁς Β, οÛτως τÕ ¢πÕ τÁς Ζ πρÕς τÕ ¢πÕ τÁς ∆· œστιν ¥ρα κሠæς ¹ Ε πρÕς τ¾ν Β, οÛτως ¹ Ζ πρÕς τ¾ν ∆· ¢νάπαλιν ¥ρα ™στˆν æς ¹ Β πρÕς τ¾ν Ε, οÛτως ¹ ∆ πρÕς τ¾ν Ζ. œστι δ κሠæς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Γ πρÕς τ¾ν ∆· δι' ‡σου ¥ρα ™στˆν æς ¹ Α πρÕς τ¾ν Ε, οÛτως ¹ Γ πρÕς τ¾ν Ζ. ε‡τε οâν σύµµετρός ™στιν ¹ Α τÍ Ε, συµµετρός ™στι κሠ¹ Γ τÍ Ζ, ε‡τε ¢σύµµετρός ™στιν ¹ Α τÍ Ε, ¢σύµµετρός ™στι κሠ¹ Γ τÍ Ζ. 'Ε¦ν ¥ρα, κሠτ¦ ˜ξÁς.

For since as A is to B, so C (is) to D, thus as the (square) on A is to the (square) on B, so the (square) on C also (is) to the (square) on D [Prop. 6.22]. But the (sum of the squares) on E and B is equal to the (square) on A, and the (sum of the squares) on D and F is equal to the (square) on C. Thus, as the (sum of the squares) on E and B is to the (square) on B, so the (sum of the squares) on D and F (is) to the (square) on D. Thus, via separation, as the (square) on E is to the (square) on B, so the (square) on F (is) to the (square) on D [Prop. 5.17]. Thus, also, as E is to B, so F (is) to D [Prop. 6.22]. Thus, inversely, as B is to E, so D (is) to F [Prop. 5.7 corr.]. But, as A is to B, so C also (is) to D. Thus, via equality, as A is to E, so C (is) to F [Prop. 5.22]. Therefore, A is either commensurable (in length) with E, and C is commensurable with F , or A is incommensurable (in length) with E, and C is incommensurable with F [Prop. 10.11]. Thus, if, and so on . . . .

ιε΄.

Proposition 15

'Ε¦ν δύο µεγέθη σύµµετρα συντεθÍ, κሠτÕ Óλον ˜κατέρJ αÙτîν σύµµετρον œσται· κ¨ν τÕ Óλον ˜νˆ αÙτîν σύµµετρον Ï, κሠτ¦ ™ξ ¢ρχÁς µεγέθη σύµµετρα œσται. Συγκείσθω γ¦ρ δύο µεγέθη σύµµετρα τ¦ ΑΒ, ΒΓ· λέγω, Óτι κሠÓλον τÕ ΑΓ ˜κατέρJ τîν ΑΒ, ΒΓ ™στι σύµµετρον.

If two commensurable magnitudes are added together, then the whole will also be commensurable with each of them. And if the whole is commensurable with one of them, then the original magnitudes will also be commensurable (with one another). For let the two commensurable magnitudes AB and BC be laid down together. I say that the whole AC is

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ELEMENTS BOOK 10 also commensurable with each of AB and BC.

Α

Β

Γ

A

∆

B

C

D

'Επεˆ γ¦ρ σύµµετρά ™στι τ¦ ΑΒ, ΒΓ, µετρήσει τι αÙτ¦ µέγεθος. µετρείτω, κሠœστω τÕ ∆. ™πεˆ οâν τÕ ∆ τ¦ ΑΒ, ΒΓ µετρε‹, κሠÓλον τÕ ΑΓ µετρήσει. µετρε‹ δ κሠτ¦ ΑΒ, ΒΓ. τÕ ∆ ¥ρα τ¦ ΑΒ, ΒΓ, ΑΓ µετρε‹· σύµµετρον ¥ρα ™στˆ τÕ ΑΓ ˜κατέρJ τîν ΑΒ, ΒΓ. 'Αλλ¦ δ¾ τÕ ΑΓ œστω σύµµετρον τù ΑΒ· λέγω δή, Óτι κሠτ¦ ΑΒ, ΒΓ σύµµετρά ™στιν. 'Επεˆ γ¦ρ σύµµετρά ™στι τ¦ ΑΓ, ΑΒ, µετρήσει τι αÙτ¦ µέγεθος. µετρείτω, κሠœστω τÕ ∆. ™πεˆ οâν τÕ ∆ τ¦ ΓΑ, ΑΒ µετρε‹, κሠλοιπÕν ¥ρα τÕ ΒΓ µετρήσει. µετρε‹ δ κሠτÕ ΑΒ· τÕ ∆ ¥ρα τ¦ ΑΒ, ΒΓ µετρήσει· σύµµετρα ¥ρα ™στˆ τ¦ ΑΒ, ΒΓ. 'Ε¦ν ¥ρα δύο µεγέθη, κሠτ¦ ˜ξÁς.

For since AB and BC are commensurable, some magnitude will measure them. Let it (so) measure (them), and let it be D. Therefore, since D measures (both) AB and BC, it will also measure the whole AC. And it also measures AB and BC. Thus, D measures AB, BC, and AC. Thus, AC is commensurable with each of AB and BC [Def. 10.1]. And so let AC be commensurable with AB. I say that AB and BC are also commensurable. For since AC and AB are commensurable, some magnitude will measure them. Let it (so) measure (them), and let it be D. Therefore, since D measures (both) CA and AB, it will thus also measure the remainder BC. And it also measures AB. Thus, D will measure (both) AB and BC. Thus, AB and BC are commensurable [Def. 10.1]. Thus, if two magnitudes, and so on . . . .

ι$΄.

Proposition 16

'Ε¦ν δύο µεγέθη ¢σύµµετρα συντεθÍ, κሠτÕ Óλον ˜κατέρJ αÙτîν ¢σύµµετρον œσται· κ¨ν τÕ Óλον ˜νˆ αÙτîν ¢σύµµετρον Ï, κሠτ¦ ™ξ ¢ρχÁς µεγέθη ¢σύµµετρα œσται.

If two incommensurable magnitudes are added together, then the whole will also be incommensurable with each of them. And if the whole is incommensurable with one of them, then the original magnitudes will also be incommensurable (with one another).

Α

Β

Γ

A

∆

B

C

D

Συγκείσθω γ¦ρ δύο µεγέθη ¢σύµµετρα τ¦ ΑΒ, ΒΓ· λέγω, Óτι κሠÓλον τÕ ΑΓ ˜κατέρJ τîν ΑΒ, ΒΓ ¢σύµµετρόν ™στιν. Ε„ γ¦ρ µή ™στιν ¢σύµµετρα τ¦ ΓΑ, ΑΒ, µετρήσει τι [αÙτ¦] µέγεθος. µετρείτω, ε„ δυνατόν, κሠœστω τÕ ∆. ™πεˆ οâν τÕ ∆ τ¦ ΓΑ, ΑΒ µετρε‹, κሠλοιπÕν ¥ρα τÕ ΒΓ µετρήσει. µετρε‹ δ κሠτÕ ΑΒ· τÕ ∆ ¥ρα τ¦ ΑΒ, ΒΓ µετρε‹. σύµµετρα ¥ρα ™στˆ τ¦ ΑΒ, ΒΓ· Øπέκειντο δ κሠ¢σύµµετρα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τ¦ ΓΑ, ΑΒ µετρήσει τι µέγεθος· ¢σύµµετρα ¥ρα ™στˆ τ¦ ΓΑ, ΑΒ. еοίως δ¾ δείξοµεν, Óτι κሠτ¦ ΑΓ, ΓΒ ¢σύµµετρά ™στιν. τÕ ΑΓ ¥ρα ˜κατέρJ τîν ΑΒ, ΒΓ ¢σύµµετρόν ™στιν. 'Αλλ¦ δ¾ τÕ ΑΓ ˜νˆ τîν ΑΒ, ΒΓ ¢σύµµετρον œστω. œστω δ¾ πρότερον τù ΑΒ· λέγω, Óτι κሠτ¦ ΑΒ, ΒΓ ¢σύµµετρά ™στιν. ε„ γ¦ρ œσται σύµµετρα, µετρήσει τι αÙτ¦ µέγεθος. µετρείτω, κሠœστω τÕ ∆. ™πεˆ οâν τÕ ∆ τ¦

For let the two incommensurable magnitudes AB and BC be laid down together. I say that that the whole AC is also incommensurable with each of AB and BC. For if CA and AB are not incommensurable, then some magnitude will measure [them]. If possible, let it (so) measure (them), and let it be D. Therefore, since D measures (both) CA and AB, it will thus also measure the remainder BC. And it also measures AB. Thus, D measures (both) AB and BC. Thus, AB and BC are commensurable [Def. 10.1]. But they were also assumed (to be) incommensurable. The very thing is impossible. Thus, some magnitude cannot measure (both) CA and AB. Thus, CA and AB are incommensurable [Def. 10.1]. So, similarly, we can show that AC and CB are also incommensurable. Thus, AC is incommensurable with each of AB and BC. And so let AC be incommensurable with one of AB

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ELEMENTS BOOK 10

ΑΒ, ΒΓ µετρε‹, κሠÓλον ¥ρα τÕ ΑΓ µετρήσει. µετρε‹ δ κሠτÕ ΑΒ· τÕ ∆ ¥ρα τ¦ ΓΑ, ΑΒ µετρε‹. σύµµετρα ¥ρα ™στˆ τ¦ ΓΑ, ΑΒ· Øπέκειτο δ κሠ¢σύµµετρα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τ¦ ΑΒ, ΒΓ µετρήσει τι µέγεθος· ¢σύµµετρα ¥ρα ™στˆ τ¦ ΑΒ, ΒΓ. 'Ε¦ν ¥ρα δύο µεγέθη, κሠτ¦ ˜ξÁς.

and BC. So let it, first of all, be incommensurable with AB. I say that AB and BC are also incommensurable. For if they are commensurable, then some magnitude will measure them. Let it (so) measure (them), and let it be D. Therefore, since D measures (both) AB and BC, it will thus also measure the whole AC. And it also measures AB. Thus, D measures (both) CA and AB. Thus, CA and AB are commensurable [Def. 10.1]. But they were also assumed (to be) incommensurable. The very thing is impossible. Thus, some magnitude cannot measure (both) AB and BC. Thus, AB and BC are incommensurable [Def. 10.1]. Thus, if two. . . magnitudes, and so on . . . .

ΛÁµµα.

Lemma †

'Ε¦ν παρά τινα εÙθε‹αν παραβληθÍ παραλληλόγραµIf a parallelogram, falling short by a square figure, is µον ™λλε‹πον ε‡δει τετραγώνJ, τÕ παραβληθν ‡σον ™στˆ applied to some straight-line, then the applied (parallelτù ØπÕ τîν ™κ τÁς παραβολής γενοµένων τµηµάτων τÁς ogram) is equal (in area) to the (rectangle contained) by εÙθείας. the pieces of the straight-line created via the application (of the parallelogram).

∆

Α

Γ

D

Β

A

Παρ¦ γ¦ρ εÙθε‹αν τ¾ν ΑΒ παραβεβλήσθω παραλληλόγραµµον τÕ Α∆ ™λλε‹πον ε‡δει τετραγώνJ τù ∆Β· λέγω, Óτι ‡σον ™στˆ τÕ Α∆ τù ØπÕ τîν ΑΓ, ΓΒ. Καί ™στιν αÙτόθεν φανερόν· ™πεˆ γ¦ρ τετράγωνόν ™στι τÕ ∆Β, ‡ση ™στˆν ¹ ∆Γ τÍ ΓΒ, καί ™στι τÕ Α∆ τÕ ØπÕ τîν ΑΓ, Γ∆, τουτέστι τÕ ØπÕ τîν ΑΓ, ΓΒ. 'Ε¦ν ¥ρα παρά τινα εÙθε‹αν, κሠτ¦ ˜ξÁς.

†

C

B

For let the parallelogram AD, falling short by the square figure DB, have been applied to the straight-line AB. I say that AD is equal to the (rectangle contained) by AC and CB. And it is immediately obvious. For since DB is a square, DC is equal to CB. And AD is the (rectangle contained) by AC and CD—that is to say, by AC and CB. Thus, if . . . to some straight-line, and so on . . . .

Note that this lemma only applies to rectangular parallelograms.

ιζ΄.

Proposition 17†

'Ε¦ν ðσι δύο εÙθε‹αι ¥νισοι, τù δ τετράτJ µέρει τοà ¢πÕ τÁς ™λάσσονος ‡σον παρ¦ τ¾ν µείζονα παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ κሠε„ς σύµµετρα αÙτ¾ν διαιρÍ µήκει, ¹ µείζων τÁς ™λάσσονος µε‹ζον δυνήσεται τù ¢πÕ συµµέτου ˜αυτÍ [µήκει]. κሠ™¦ν ¹ µείζων τÁς ™λάσσονος µε‹ζον δύνηται τù ¢πÕ συµµέτρου ˜αυτV [µήκει], τù δ τετράρτJ τοà ¢πÕ τÁς ™λάσσονος ‡σον παρ¦ τ¾ν µείζονα παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, ε„ς σύµµετρα αÙτ¾ν διαιρε‹ µήκει.

If there are two unequal straight-lines, and a (rectangle) equal to the fourth part of the (square) on the lesser, falling short by a square figure, is applied to the greater, and divides it into (parts which are) commensurable in length, then the square on the greater will be larger than (the square on) the lesser by the (square) on (some straight-line) commensurable [in length] with the greater. And if the square on the greater is larger than (the square on) the lesser by the (square) on

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”Εστωσαν δύο εÙθε‹αι ¥νισοι αƒ Α, ΒΓ, ïν µείζων ¹ ΒΓ, τù δ τετράρτJ µέρει τοà ¢πÕ ™λάσσονος τÁς Α, τουτέστι τù ¢πÕ τÁς ¹µισείας τÁς Α, ‡σον παρ¦ τ¾ν ΒΓ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ, κሠœστω τÕ ØπÕ τîν Β∆, ∆Γ, σύµµετρος δ œστω ¹ Β∆ τÍ ∆Γ µήκει· λέγω, Öτι ¹ ΒΓ τÁς Α µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ.

(some straight-line) commensurable [in length] with the greater, and a (rectangle) equal to the fourth (part) of the (square) on the lesser, falling short by a square figure, is applied to the greater, then it divides it into (parts which are) commensurable in length. Let A and BC be two unequal straight-lines, of which (let) BC (be) the greater. And let a (rectangle) equal to the fourth part of the (square) on the lesser, A—that is, (equal) to the (square) on half of A—falling short by a square figure, have been applied to BC. And let it be the (rectangle contained) by BD and DC [see previous lemma]. And let BD be commensurable in length with DC. I say that that the square on BC is greater than the (square on) A by (the square on some straight-line) commensurable (in length) with (BC).

Α

Β Ζ

A

Ε

∆ Γ

B F

Τετµήσθω γ¦ρ ¹ ΒΓ δίχα κατ¦ τÕ Ε σηµε‹ον, κሠκείσθω τÍ ∆Ε ‡ση ¹ ΕΖ. λοιπ¾ ¥ρα ¹ ∆Γ ‡ση ™στˆ τÍ ΒΖ. κሠ™πεˆ εÙθε‹α ¹ ΒΓ τέτµηται ε„ς µν ‡σα κατ¦ τÕ Ε, ε„ς δ ¥νισα κατ¦ τÕ ∆, τÕ ¥ρα ØπÕ Β∆, ∆Γ περειχόµενον Ñρθογώνιον µετ¦ τοà ¢πÕ τÁς Ε∆ τετραγώνου ‡σον ™στˆ τù ¢πÕ τÁς ΕΓ τετραγώνJ· κሠτ¦ τετραπλάσια· τÕ ¥ρα τετράκις ØπÕ τîν Β∆, ∆Γ µετ¦ τοà τετραπλασίου τοà ¢πÕ τÁς ∆Ε ‡σον ™στˆ τù τετράκις ¢πÕ τÁς ΕΓ τετραγώνJ. ¢λλ¦ τù µν τετραπλασίJ τοà ØπÕ τîν Β∆, ∆Γ ‡σον ™στˆ τÕ ¢πÕ τÁς Α τετράγωνον, τù δ τετραπλασίJ τοà ¢πÕ τÁς ∆Ε ‡σον ™στˆ τÕ ¢πÕ τÁς ∆Ζ τετράγωνον· διπλασίων γάρ ™στιν ¹ ∆Ζ τÁς ∆Ε. τù δ τετραπλασίJ τοà ¢πÕ τÁς ΕΓ ‡σον ™στˆ τÕ ¢πÕ τÁς ΒΓ τετράγωνον· διπλασίων γάρ ™στι πάλιν ¹ ΒΓ τÁς ΓΕ. τ¦ ¥ρα ¢πÕ τîν Α, ∆Ζ τετράγωνα ‡σα ™στˆ τù ¢πÕ τÁς ΒΓ τετράγωνJ· éστε τÕ ¢πÕ τÁς ΒΓ τοà ¢πÕ τÁς Α µε‹ζόν ™στι τù ¢πÕ τÁς ∆Ζ· ¹ ΒΓ ¥ρα τÁς Α µε‹ζον δύναται τÍ ∆Ζ. δεικτέον, Óτι κሠσύµµετρός ™στιν ¹ ΒΓ τÍ ∆Ζ. ™πεˆ γ¦ρ σύµµετρός ™στιν ¹ Β∆ τÍ ∆Γ µήκει, σύµµετρος ¥ρα ™στˆ κሠ¹ ΒΓ τÍ Γ∆ µήκει. ¢λλ¦ ¹ Γ∆ τα‹ς Γ∆, ΒΖ ™στι σύµµετρος µήκει· ‡ση γάρ ™στιν ¹ Γ∆ τÍ ΒΖ. κሠ¹ ΒΓ ¥ρα σύµµετρός ™στι τα‹ς ΒΖ, Γ∆ µήκει· éστε κሠλοιπÍ τÍ Ζ∆ σύµµετρός ™στιν ¹ ΒΓ µήκει· ¹ ΒΓ ¥ρα τÁς Α µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. 'Αλλ¦ δ¾ ¹ ΒΓ τÁς Α µε‹ζον δυνάσθω τù ¢πÕ συµµέτρου ˜αυτÍ, τù δ τετράτρJ τοà ¢πÕ τÁς Α ‡σον παρ¦ τ¾ν ΒΓ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ, κሠœστω τÕ ØπÕ τîν Β∆, ∆Γ. δεικτέον, Óτι σύµµετρός ™στιν ¹ Β∆ τÍ ∆Γ µήκει.

E

D C

For let BC have been cut in half at the point E [Prop. 1.10]. And let EF be made equal to DE [Prop. 1.3]. Thus, the remainder DC is equal to BF . And since the straight-line BC has been cut into equal (pieces) at E, and into unequal (pieces) at D, the rectangle contained by BD and DC, plus the square on ED, is thus equal to the square on EC [Prop. 2.5]. (The same) also (for) the quadruples. Thus, four times the (rectangle contained) by BD and DC, plus the quadruple of the (square) on DE, is equal to four times the square on EC. But, the square on A is equal to the quadruple of the (rectangle contained) by BD and DC, and the square on DF is equal to the quadruple of the (square) on DE. For DF is double DE. And the square on BC is equal to the quadruple of the (square) on EC. For, again, BC is double CE. Thus, the (sum of the) squares on A and DF is equal to the square on BC. Hence, the (square) on BC is greater than the (square) on A by the (square) on DF . Thus, BC is greater in square than A by DF . It must also be shown that BC is commensurable (in length) with DF . For since BD is commensurable in length with DC, BC is thus also commensurable in length with CD [Prop. 10.15]. But, CD is commensurable in length with CD plus BF . For CD is equal to BF [Prop. 10.6]. Thus, BC is also commensurable in length with BF plus CD [Prop. 10.12]. Hence, BC is also commensurable in length with the remainder F D [Prop. 10.15]. Thus, the square on BC is greater than (the square on) A by the (square) on (some straight-line) commensurable (in

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Τîν γ¦ρ αÙτîν κατασκευασθέντων еοίως δείξοµεν, Óτι ¹ ΒΓ τÁς Α µε‹ζον δύναται τù ¢πÕ τÁς Ζ∆. δύναται δ ¹ ΒΓ τÁς Α µε‹ζον τù ¢πÕ συµµέτρου ˜αυτÍ. σύµµετρος ¥ρα ™στˆν ¹ ΒΓ τÍ Ζ∆ µήκει· éστε κሠλοιπÍ συναµφοτέρJ τÍ ΒΖ, ∆Γ σύµµετρός ™στιν ¹ ΒΓ µήκει. ¢λλ¦ συναµφότερος ¹ ΒΖ, ∆Γ σύµµετρός ™στι τÍ ∆Γ [µήκει]. éστε κሠ¹ ΒΓ τÍ Γ∆ σύµµετρός ™στι µήκει· κሠδιελόντι ¥ρα ¹ Β∆ τÍ ∆Γ ™στι σύµµετρος µήκει. 'Ε¦ν ¥ρα ðσι δύο εÙθε‹αι ¥νισοι, κሠτ¦ ˜ξÁς.

length) with (BC). And so let the square on BC be greater than the (square on) A by the (square) on (some straight-line) commensurable (in length) with (BC). And let a (rectangle) equal to the fourth (part) of the (square) on A, falling short by a square figure, have been applied to BC. And let it be the (rectangle contained) by BD and DC. It must be shown that BD is commensurable in length with DC. For, similarly, by the same construction, we can show that the square on BC is greater than the (square on) A by the (square) on F D. And the square on BC is greater than the (square on) A by the (square) on (some straightline) commensurable (in length) with (BC). Thus, BC is commensurable in length with F D. Hence, BC is also commensurable in length with the remaining sum of BF and DC [Prop. 10.15]. But, the sum of BF and DC is commensurable [in length] with DC [Prop. 10.6]. Hence, BC is also commensurable in length with CD [Prop. 10.12]. Thus, via separation, BD is also commensurable in length with DC [Prop. 10.15]. Thus, if there are two unequal straight-lines, and so on . . . .

†

This proposition states that if α x − x2 = β 2 /4 (where α = BC, x = DC, and β = A) then α and are x are commensurable, and vice versa.

p

α2 − β 2 are commensurable when α − x

ιη΄.

Proposition 18†

'Ε¦ν ðσι δύο εÙθε‹αι ¥νισοι, τù δ τετάρτJ µέρει τοà ¢πÕ τÁς ™λάσσονος ‡σον παρ¦ τ¾ν µείζονα παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, κሠε„ς ¢συµµετρα αÙτ¾ν διαιρÍ [µήκει], ¹ µείζων τÁς ™λάσσονος µε‹ζον δυνήσεται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. κሠ™¦ν ¹ µείζων τÁς ™λάσσονος µε‹ζον δύνηται τù ¢πÕ ¢συµµέτρου ˜αυτÍ, τù δ τετράρτJ τοà ¢πÕ τÁς ™λάσσονος ‡σον παρ¦ τ¾ν µείζονα παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, ε„ς ¢σύµµετρα αÙτ¾ν διαιρε‹ [µήκει].

If there are two unequal straight-lines, and a (rectangle) equal to the fourth part of the (square) on the lesser, falling short by a square figure, is applied to the greater, and divides it into (parts which are) incommensurable [in length], then the square on the greater will be larger than the (square on the) lesser by the (square) on (some straight-line) incommensurable (in length) with the greater. And if the square on the greater is larger than the (square on the) lesser by the (square) on (some straight-line) incommensurable (in length) with the greater, and a (rectangle) equal to the fourth (part) of the (square) on the lesser, falling short by a square figure, is applied to the greater, then it divides it into (parts which are) incommensurable [in length].

Α Β

A Ζ

Ε

∆

Γ

B

F

E

D

C

”Εστωσαν δύο εÙθε‹αι ¥νισοι αƒ Α, ΒΓ, ïν µείζων Let A and BC be two unequal straight-lines, of which ¹ ΒΓ, τù δ τετάρτJ [µέρει] τοà ¢πÕ τÁς ™λάσσονος (let) BC (be) the greater. And let a (rectangle) equal to τÁς Α ‡σον παρ¦ τ¾ν ΒΓ παραβεβλήσθω ™λλε‹πον ε‡δει the fourth [part] of the (square) on the lesser, A, falling

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τετραγώνJ, κሠœστω τÕ ØπÕ τîν Β∆Γ, ¢σύµµετρος δ œστω ¹ Β∆ τÍ ∆Γ µήκει· λέγω, Óτι ¹ ΒΓ τÁς Α µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. Τîν γ¦ρ αÙτîν κατασκευασθέντων τù πρότερον еοίως δείξοµεν, Óτι ¹ ΒΓ τÁς Α µε‹ζον δύναται τù ¢πÕ τÁς Ζ∆. δεικτέον [οâν], Óτι ¢σύµµετρός ™στιν ¹ ΒΓ τÍ ∆Ζ µήκει. ™πεˆ γ¦ρ ¢σύµµετρός ™στιν ¹ Β∆ τÍ ∆Γ µήκει, ¢σύµµετρος ¥ρα ™στˆ κሠ¹ ΒΓ τÍ Γ∆ µήκει. ¢λλ¦ ¹ ∆Γ σύµµετρός ™στι συναµφοτέραις τα‹ς ΒΖ, ∆Γ· κሠ¹ ΒΓ ¥ρα ¢σύµµετρός ™στι συναµφοτέραις τα‹ς ΒΖ, ∆Γ. éστε κሠλοιπÍ τÍ Ζ∆ ¢σύµµετρός œστιν ¹ ΒΓ µήκει. κሠ¹ ΒΓ τÁς Α µε‹ζον δύναται τù ¢πÕ τÁς Ζ∆· ¹ ΒΓ ¥ρα τÁς Α µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. ∆υνάσθω δ¾ πάλιν ¹ ΒΓ τÁς Α µε‹ζον τù ¢πÕ ¢συµµέτρου ˜αυτÍ, τù δ τετάρτJ τοà ¢πÕ τÁς Α ‡σον παρ¦ τ¾ν ΒΓ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ, κሠœστω τÕ ØπÕ τîν Β∆, ∆Γ. δεικτέον, Óτι ¢σύµµετρός ™στιν ¹ Β∆ τÍ ∆Γ µήκει. Τîν γ¦ρ αÙτîν κατασκευασθέντων еοίως δείξοµεν, Óτι ¹ ΒΓ τÁς Α µε‹ζον δύναται τù ¢πÕ τÁς Ζ∆. ¢λλ¦ ¹ ΒΓ τÁς Α µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. ¢σύµµετρος ¥ρα ™στˆν ¹ ΒΓ τÍ Ζ∆ µήκει· éστε κሠλοιπÍ συναµφοτέρJ τÍ ΒΖ, ∆Γ ¢σύµµετρός ™στιν ¹ ΒΓ. ¢λλ¦ συναµφότερος ¹ ΒΖ, ∆Γ τÍ ∆Γ σύµµετρός ™στι µήκει· κሠ¹ ΒΓ ¥ρα τÍ ∆Γ ¢σύµµετρός ™στι µήκει· éστε κሠδιελόντι ¹ Β∆ τÍ ∆Γ ¢σύµµετρός ™στι µήκει. 'Ε¦ν ¥ρα ðσι δύο εÙθε‹αι, κሠτ¦ ˜ξÁς.

†

short by a square figure, have been applied to BC. And let it be the (rectangle contained) by BDC. And let BD be incommensurable in length with DC. I say that that the square on BC is greater than the (square on) A by the (square) on (some straight-line) incommensurable (in length) with (BC). For, similarly, by the same construction as before, we can show that the square on BC is greater than the (square on) A by the (square) on F D. [Therefore] it must be shown that BC is incommensurable in length with DF . For since BD is incommensurable in length with DC, BC is thus also incommensurable in length with CD [Prop. 10.16]. But, DC is commensurable (in length) with the sum of BF and DC [Prop. 10.6]. And, thus, BC is incommensurable (in length) with the sum of BF and DC [Prop. 10.13]. Hence, BC is also incommensurable in length with the remainder F D [Prop. 10.16]. And the square on BC is greater than the (square on) A by the (square) on F D. Thus, the square on BC is greater than the (square on) A by the (square) on (some straight-line) incommensurable (in length) with (BC). So, again, let the square on BC be greater than the (square on) A by the (square) on (some straight-line) incommensurable (in length) with (BC). And let a (rectangle) equal to the fourth [part] of the (square) on A, falling short by a square figure, have been applied to BC. And let it be the (rectangle contained) by BD and DC. It must be shown that BD is incommensurable in length with DC. For, similarly, by the same construction, we can show that the square on BC is greater than the (square) on A by the (square) on F D. But, the square on BC is greater than the (square) on A by the (square) on (some straight-line) incommensurable (in length) with (BC). Thus, BC is incommensurable in length with F D. Hence, BC is also incommensurable (in length) with the remaining sum of BF and DC [Prop. 10.16]. But, the sum of BF and DC is commensurable in length with DC [Prop. 10.6]. Thus, BC is also incommensurable in length with DC [Prop. 10.13]. Hence, via separation, BD is also incommensurable in length with DC [Prop. 10.16]. Thus, if there are two . . . straight-lines, and so on . . . .

This proposition states that if α x − x2 = β 2 /4 (where α = BC, x = DC, and β = A) then α and

α − x are x are incommensurable, and vice versa.

ιθ΄.

p

α2 − β 2 are incommensurable when

Proposition 19

ΤÕ ØπÕ ·ητîν µήκει συµµέτρων εÙθειîν πεThe rectangle contained by rational straight-lines ριεχόµενον Ñρθογώνιον ·ητόν ™στιν. (which are) commensurable in length is rational.

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`ΥπÕ γ¦ρ ·ητîν µήκει συµµέτρων εÙθειîν τîν ΑΒ, For let the rectangle AC have been enclosed by the ΒΓ Ñρθογώνιον περιεχέσθω τÕ ΑΓ· λέγω, Óτι ·ητόν ™στι rational straight-lines AB and BC (which are) commenτÕ ΑΓ. surable in length. I say that AC is rational.

Α

∆

D

Γ

C

Β

A

B

'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ Α∆· ·ητÕν ¥ρα ™στˆ τÕ Α∆. κሠ™πεˆ σύµµετρός ™στιν ¹ ΑΒ τÍ ΒΓ µήκει, ‡ση δέ ™στιν ¹ ΑΒ τÍ Β∆, σύµµετρος ¥ρα ™στˆν ¹ Β∆ τÍ ΒΓ µήκει. καί ™στιν æς ¹ Β∆ πρÕς τ¾ν ΒΓ, οÛτως τÕ ∆Α πρÕς τÕ ΑΓ. σύµµετρον ¥ρα ™στˆ τÕ ∆Α τù ΑΓ. ·ητÕν δ τÕ ∆Α· ·ητÕν ¥ρα ™στˆ κሠτÕ ΑΓ. ΤÕ ¥ρα ØπÕ ·ητîν µήκει συµµέτρων, κሠτ¦ ˜ξÁς.

For let the square AD have been described on AB. AD is thus rational [Def. 10.4]. And since AB is commensurable in length with BC, and AB is equal to BD, BD is thus commensurable in length with BC. And as BD is to BC, so DA (is) to AC [Prop. 6.1]. Thus, DA is commensurable with AC [Prop. 10.11]. And DA (is) rational. Thus, AC is also rational [Def. 10.4]. Thus, the . . . by rational straight-lines . . . commensurable, and so on . . . .

κ΄.

Proposition 20

'Ε¦ν ·ητÕν παρ¦ ·ητ¾ν παραβληθÍ, πλάτος ποιε‹ If a rational (area) is applied to a rational (straight·ητ¾ν κሠσύµµετρον τÍ, παρ' ¿ν παράκειται, µήκει. line) then it produces as breadth a (straight-line which is) rational, and commensurable in length with the (straightline) to which it is applied.

∆

Β

D

Α

B

Γ

A

C

`ΡητÕν γ¦ρ τÕ ΑΓ παρ¦ ·ητ¾ν τ¾ν ΑΒ παραFor let the rational (area) AC have been applied to the βεβλήσθω πλάτος ποιοàν τ¾ν ΒΓ· λέγω, Óτι ·ητή ™στιν rational (straight-line) AB, producing the (straight-line) ¹ ΒΓ κሠσύµµετρος τÍ ΒΑ µήκει. BC as breadth. I say that BC is rational, and commen'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ Α∆· surable in length with BA.

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·ητÕν ¥ρα ™στˆ τÕ Α∆. ·ητÕν δ κሠτÕ ΑΓ· σύµµετρον ¥ρα ™στˆ τÕ ∆Α τù ΑΓ. καί ™στιν æς τÕ ∆Α πρÕς τÕ ΑΓ, οÛτως ¹ ∆Β πρÕς τ¾ν ΒΓ. σύµµετρος ¥ρα ™στˆ κሠ¹ ∆Β τÍ ΒΓ· ‡ση δ ¹ ∆Β τÍ ΒΑ· σύµµετρος ¥ρα κሠ¹ ΑΒ τÍ ΒΓ. ·ητ¾ δέ ™στιν ¹ ΑΒ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΒΓ κሠσύµµετρος τÍ ΑΒ µήκει. 'Ε¦ν ¥ρα ·ητÕν παρ¦ ·ητ¾ν παραβληθÍ, κሠτ¦ ˜ξÁς.

For let the square AD have been described on AB. AD is thus rational [Def. 10.4]. And AC (is) also rational. DA is thus commensurable with AC. And as DA is to AC, so DB (is) to BC [Prop. 6.1]. Thus, DB is also commensurable (in length) with BC [Prop. 10.11]. And DB (is) equal to BA. Thus, AB (is) also commensurable (in length) with BC. And AB is rational. Thus, BC is also rational, and commensurable in length with AB [Def. 10.3]. Thus, if a rational (area) is applied to a rational (straight-line), and so on . . . .

κα΄.

Proposition 21

ΤÕ ØπÕ ·ητîν δυνάµει µόνον συµµέτρων εÙθειîν περιεχόµενον Ñρθογώνιον ¥λογόν ™στιν, κሠ¹ δυναµένη αÙτÕ ¥λογός ™στιν, καλείσθω δ µέση.

The rectangle contained by rational straight-lines (which are) commensurable in square only is irrational, and its square-root is irrational—let it be called medial.†

∆

Β

D

Α

B

Γ

C

`ΥπÕ γ¦ρ ·ητîν δυνάµει µόνον συµµέτρων εÙθειîν τîν ΑΒ, ΒΓ Ñρθογώνιον περιεχέσθω τÕ ΑΓ· λέγω, Óτι ¥λογόν ™στι τÕ ΑΓ, κሠ¹ δυναµένη αÙτÕ ¥λογός ™στιν, καλείσθω δ µέση. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ Α∆· ·ητÕν ¥ρα ™στˆ τÕ Α∆. κሠ™πεˆ ¢σύµµετρός ™στιν ¹ ΑΒ τÍ ΒΓ µήκει· δυνάµει γ¦ρ µόνον Øπόκεινται σύµµετροι· ‡ση δ ¹ ΑΒ τÍ Β∆, ¢σύµµετρος ¥ρα ™στˆ κሠ¹ ∆Β τÍ ΒΓ µήκει. καί ™στιν æς ¹ ∆Β πρÕς τ¾ν ΒΓ, οÛτως τÕ Α∆ πρÕς τÕ ΑΓ· ¢σύµµετρον ¥ρα [™στˆ] τÕ ∆Α τù ΑΓ. ·ητÕν δ τÕ ∆Α· ¥λογον ¥ρα ™στˆ τÕ ΑΓ· éστε κሠ¹ δυναµένη τÕ ΑΓ [τουτέστιν ¹ ‡σον αÙτù τετράγωνον δυναµένη] ¥λογός ™στιν, καλείσθω δε µέση· Óπερ œδει δε‹ξαι.

†

A

For let the rectangle AC be contained by the rational straight-lines AB and BC (which are) commensurable in square only. I say that AC is irrational, and its squareroot is irrational—let it be called medial. For let the square AD have been described on AB. AD is thus rational [Def. 10.4]. And since AB is incommensurable in length with BC. For they were assumed to be commensurable in square only. And AB (is) equal to BD. DB is thus also incommensurable in length with BC. And as DB is to BC, so AD (is) to AC [Prop. 6.1]. Thus, DA [is] incommensurable with AC [Prop. 10.11]. And DA (is) rational. Thus, AC is irrational [Def. 10.4]. Hence, its square-root [that is to say, the square-root of the square equal to it] is also irrational [Def. 10.4]—let it be called medial. (Which is) the very thing it was required to show.

Thus, a medial straight-line has a length expressible as k 1/4 .

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ΛÁµµα.

Lemma

'Ε¦ν ðσι δύο εÙθε‹αι, œστιν æς ¹ πρώτη πρÕς τ¾ν If there are two straight-lines then as the first is to the δευτέραν, οÛτως τÕ ¢πÕ τÁς πρώτης πρÕς τÕ ØπÕ τîν second, so the (square) on the first (is) to the (rectangle δύο εÙθειîν. contained) by the two straight-lines.

Ζ

Ε

Η

F

E

∆

G

D

”Εστωσαν δύο εÙθε‹αι αƒ ΖΕ, ΕΗ. λέγω, Óτι ™στˆν æς ¹ ΖΕ πρÕς τ¾ν ΕΗ, οÛτως τÕ ¢πÕ τÁς ΖΕ πρÕς τÕ ØπÕ τîν ΖΕ, ΕΗ. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΖΕ τετράγωνον τÕ ∆Ζ, κሠσυµπεπληρώσθω τÕ Η∆. ™πεˆ οâν ™στιν æς ¹ ΖΕ πρÕς τ¾ν ΕΗ, οÛτως τÕ Ζ∆ πρÕς τÕ ∆Η, καί ™στι τÕ µν Ζ∆ τÕ ¢πÕ τÁς ΖΕ, τÕ δ ∆Η τÕ ØπÕ τîν ∆Ε, ΕΗ, τουτέστι τÕ ØπÕ τîν ΖΕ, ΕΗ, œστιν ¥ρα æς ¹ ΖΕ πρÕς τ¾ν ΕΗ, οÛτως τÕ ¢πÕ τÁς ΖΕ πρÕς τÕ ØπÕ τîν ΖΕ, ΕΗ. еοίως δ κሠæς τÕ ØπÕ τîν ΗΕ, ΕΖ πρÕς τÕ ¢πÕ τÁς ΕΖ, τουτέστιν æς τÕ Η∆ πρÕς τÕ Ζ∆, οÛτως ¹ ΗΕ πρÕς τ¾ν ΕΖ· Óπερ œδει δε‹ξαι.

Let F E and EG be two straight-lines. I say that as F E is to EG, so the (square) on F E (is) to the (rectangle contained) by F E and EG. For let the square DF have been described on F E. And let GD have been completed. Therefore, since as F E is to EG, so F D (is) to DG [Prop. 6.1], and F D is the (square) on F E, and DG the (rectangle contained) by DE and EG—that is to say, the (rectangle contained) by F E and EG—thus as F E is to EG, so the (square) on F E (is) to the (rectangle contained) by F E and EG. And also, similarly, as the (rectangle contained) by GE and EF is to the (square on) EF —that is to say, as GD (is) to F D—so GE (is) to EF . (Which is) the very thing it was required to show.

κβ΄.

Proposition 22

ΤÕ ¢πÕ µέσης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ·ητ¾ν κሠ¢σύµµετρον τÍ, παρ' ¿ν παράκειται, µήκει.

The square on a medial (straight-line), being applied to a rational (straight-line), produces as breadth a (straight-line which is) rational, and incommensurable in length with the (straight-line) to which it is applied.

Β

B Η

Α

Γ

∆

Ε

G

Ζ

A

”Εστω µέση µν ¹ Α, ·ητ¾ δ ¹ ΓΒ, κሠτù ¢πÕ τÁς Α ‡σον παρ¦ τ¾ν ΒΓ παραβεβλήσθω χωρίον Ñρθογώνιον τÕ Β∆ πλάτος ποιοàν τ¾ν Γ∆· λέγω, Óτι ·ητή ™στιν ¹ Γ∆ κሠ¢σύµµετρος τÍ ΓΒ µήκει. 'Επεˆ γ¦ρ µέση ™στˆν ¹ Α, δύναται χωρίον πε-

C

D

E

F

Let A be a medial (straight-line), and CB a rational (straight-line), and let the rectangular area BD, equal to the (square) on A, have been applied to BC, producing CD as breadth. I say that CD is rational, and incommensurable in length with CB.

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ELEMENTS BOOK 10

ριεχόµενον ØπÕ ·ητîν δυνάµει µόνον συµµέτρων. δυνάσθω τÕ ΗΖ. δύναται δ κሠτÕ Β∆· ‡σον ¥ρα ™στˆ τÕ Β∆ τù ΗΖ. œστι δ αÙτù κሠ„σογώνιον· τîν δ ‡σων τε κሠ„σογωνίων παραλληλογράµµων ¢ντιπεπόνθασιν αƒ πλευραˆ αƒ περˆ τ¦ς ‡σας γωνίας· ¢νάλογον ¥ρα ™στˆν æς ¹ ΒΓ πρÕς τ¾ν ΕΗ, οÛτως ¹ ΕΖ πρÕς τ¾ν Γ∆. œστιν ¥ρα κሠæς τÕ ¢πÕ τÁς ΒΓ πρÕς τÕ ¢πÕ τÁς ΕΗ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς Γ∆. σύµµετρον δέ ™στι τÕ ¢πÕ τÁς ΓΒ τù ¢πÕ τÁς ΕΗ· ·ητ¾ γάρ ™στιν ˜κατέρα αÙτîν· σύµµετρον ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς ΕΖ τù ¢πÕ τÁς Γ∆. ·ητÕν δέ ™στι τÕ ¢πÕ τÁς ΕΖ· ·ητÕν ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς Γ∆· ·ητ¾ ¥ρα ™στˆν ¹ Γ∆. κሠ™πεˆ ¢σύµµετρός ™στιν ¹ ΕΖ τÍ ΕΗ µήκει· δυνάµει γ¦ρ µόνον ε„σˆ σύµµετροι· æς δ ¹ ΕΖ πρÕς τ¾ν ΕΗ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ØπÕ τîν ΖΕ, ΕΗ, ¢σύµµετρον ¥ρα [™στˆ] τÕ ¢πÕ τÁς ΕΖ τù ØπÕ τîν ΖΕ, ΕΗ. ¢λλ¦ τù µν ¢πÕ τÁς ΕΖ σύµµετρόν ™στι τÕ ¢πÕ τÁς Γ∆· ·ητሠγάρ ε„σι δυνάµει· τù δ ØπÕ τîν ΖΕ, ΕΗ σύµµετρόν ™στι τÕ ØπÕ τîν ∆Γ, ΓΒ· ‡σα γάρ ™στι τù ¢πÕ τÁς Α· ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς Γ∆ τù ØπÕ τîν ∆Γ, ΓΒ. æς δ τÕ ¢πÕ τÁς Γ∆ πρÕς τÕ ØπÕ τîν ∆Γ, ΓΒ, οÛτως ™στˆν ¹ ∆Γ πρÕς τ¾ν ΓΒ· ¢σύµµετρος ¥ρα ™στˆν ¹ ∆Γ τÍ ΓΒ µήκει. ·ητ¾ ¥ρα ™στˆν ¹ Γ∆ κሠ¢σύµµετρος τÍ ΓΒ µήκει· Óπερ œδει δε‹ξαι.

†

For since A is medial, the square on it is equal to a (rectangular) area contained by rational (straight-lines which are) commensurable in square only [Prop. 10.21]. Let the square on (A) be equal to GF . And the square on (A) is also equal to BD. Thus, BD is equal to GF . And (BD) is also equiangular with (GF ). And for equal and equiangular parallelograms, the sides about the equal angles are reciprocally proportional [Prop. 6.14]. Thus, proportionally, as BC is to EG, so EF (is) to CD. And, also, as the (square) on BC is to the (square) on EG, so the (square) on EF (is) to the (square) on CD [Prop. 6.22]. And the (square) on CB is commensurable with the (square) on EG. For they are each rational. Thus, the (square) on EF is also commensurable with the (square) on CD [Prop. 10.11]. And the (square) on EF is rational. Thus, the (square) on CD is also rational [Def. 10.4]. Thus, CD is rational. And since EF is incommensurable in length with EG. For they are commensurable in square only. And as EF (is) to EG, so the (square) on EF (is) to the (rectangle contained) by F E and EG [see previous lemma]. The (square) on EF [is] thus incommensurable with the (rectangle contained) by F E and EG [Prop. 10.11]. But, the (square) on CD is commensurable with the (square) on EF . For they are commensurable in square. And the (rectangle contained) by DC and CB is commensurable with the (rectangle contained) by F E and EG. For they are (both) equal to the (square) on A. Thus, the (square) on CD is also incommensurable with the (rectangle contained) by DC and CB [Prop. 10.13]. And as the (square) on CD (is) to the (rectangle contained) by DC and CB, so DC is to CB [see previous lemma]. Thus, DC is incommensurable in length with CB [Prop. 10.11]. Thus, CD is rational, and incommensurable in length with CB. (Which is) the very thing it was required to show.

Literally, “rational”.

κγ΄.

Proposition 23

`Η τÍ µέσV σύµµετρος µέση ™στίν. ”Εστω µέση ¹ Α, κሠτÍ Α σύµµετρος œστω ¹ Β· λέγω, Óτι κሠ¹ Β µέση ™στίν. 'Εκκείσθω γ¦ρ ·ητ¾ ¹ Γ∆, κሠτù µν ¢πÕ τÁς Α ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω χωρίον Ñρθογώνιον τÕ ΓΕ πλάτος ποιοàν τ¾ν Ε∆· ·ητ¾ ¥ρα ™στˆν ¹ Ε∆ κሠ¢σύµµετρος τÊ Γ∆ µήκει. τù δ ¢πÕ τÁς Β ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω χωρίον Ñρθογώνιον τÕ ΓΖ πλάτος ποιοàν τ¾ν ∆Ζ. ™πεˆ οâν σύµµετρός ™στιν ¹ Α τÍ Β, σύµµετρόν ™στι κሠτÕ ¢πÕ τÁς Α τù ¢πÕ τÁς Β. ¢λλ¦ τù µν ¢πÕ τÁς Α ‡σον ™στˆ τÕ ΕΓ, τù δ ¢πÕ τÁς Β ‡σον ™στˆ τÕ ΓΖ· σύµµετρον ¥ρα ™στˆ τÕ ΕΓ τù ΓΖ.

A (straight-line) commensurable with a medial (straightline) is medial. Let A be a medial (straight-line), and let B be commensurable with A. I say that B is also a medial (staightline). Let the rational (straight-line) CD be set out, and let the rectangular area CE, equal to the (square) on A, have been applied to CD, producing ED as width. ED is thus rational, and incommensurable in length with CD [Prop. 10.22]. And let the rectangular area CF , equal to the (square) on B, have been applied to CD, producing DF as width. Therefore, since A is commensurable

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ELEMENTS BOOK 10

καί ™στιν æς τÕ ΕΓ πρÕς τÕ ΓΖ, οÛτως ¹ Ε∆ πρÕς τ¾ν ∆Ζ· σύµµετρος ¥ρα ™στˆν ¹ Ε∆ τÍ ∆Ζ µήκει. ·ητ¾ δέ ™στιν ¹ Ε∆ κሠ¢σύµµετρος τÍ ∆Γ µήκει· ·ητ¾ ¥ρα ™στˆ κሠ¹ ∆Ζ κሠ¢σύµµετρος τÍ ∆Γ µήκει· αƒ Γ∆, ∆Ζ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. ¹ δ τÕ ØπÕ ·ητîν δυνάµει µόνον συµµέτρων δυναµένη µέση ™στίν. ¹ ¥ρα τÕ ØπÕ τîν Γ∆, ∆Ζ δυναµένη µέση ™στίν· κሠδύναται τÕ ØπÕ τîν Γ∆, ∆Ζ ¹ Β· µέση ¥ρα ™στˆν ¹ Β.

Α

with B, the (square) on A is also commensurable with the (square) on B. But, EC is equal to the (square) on A, and CF is equal to the (square) on B. Thus, EC is commensurable with CF . And as EC is to CF , so ED (is) to DF [Prop. 6.1]. Thus, ED is commensurable in length with DF [Prop. 10.11]. And ED is rational, and incommensurable in length with CD. DF is thus also rational [Def. 10.3], and incommensurable in length with DC [Prop. 10.13]. Thus, CD and DF are rational, and commensurable in square only. And the square-root of a (rectangle contained) by rational (straight-lines which are) commensurable in square only is medial [Prop. 10.21]. Thus, the square-root of the (rectangle contained) by CD and DF is medial. And the square on B is equal to the (rectangle contained) by CD and DF . Thus, B is a medial (straight-line).

Β

A

Γ

Ε

∆

B C

Ζ

E

Πόρισµα.

D

F

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι τÕ τù µέσJ χωρίJ And (it is) clear, from this, that an (area) commensuσύµµετρον µέσον ™στίν. rable with a medial area† is medial. †

A medial area is equal to the square on some medial straight-line. Hence, a medial area is expressible as k 1/2 .

κδ΄.

Proposition 24

ΤÕ ØπÕ µέσων µήκει συµµέτρων εÙθειîν περιεχόµενον Ñρθογώνιον µέσον ™στίν. `ΥπÕ γ¦ρ µέσων µήκει συµµέτρων εÙθειîν τîν ΑΒ, ΒΓ περιεχέσθω Ñρθογώνιον τÕ ΑΓ· λέγω, Óτι τÕ ΑΓ µέσον ™στίν. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ Α∆· µέσον ¥ρα ™στˆ τÕ Α∆. κሠ™πεˆ σύµµετρός ™στιν ¹ ΑΒ τÍ ΒΓ µήκει, ‡ση δ ¹ ΑΒ τÍ Β∆, σύµµετρος ¥ρα ™στˆ κሠ¹ ∆Β τÍ ΒΓ µήκει· éστε κሠτÕ ∆Α τù ΑΓ σύµµετρόν ™στιν. µέσον δ τÕ ∆Α· µέσον ¥ρα κሠτÕ ΑΓ· Óπερ œδει δε‹ξαι.

A rectangle contained by medial straight-lines (which are) commensurable in length is medial. For let the rectangle AC be contained by the medial straight-lines AB and BC (which are) commensurable in length. I say that AC is medial. For let the square AD have been described on AB. AD is thus medial [see previous footnote]. And since AB is commensurable in length with BC, and AB (is) equal to BD, DB is thus also commensurable in length with BC. Hence, DA is also commensurable with AC [Props. 6.1, 10.11]. And DA (is) medial. Thus, AC (is) also medial [Prop. 10.23 corr.]. (Which is) the very thing it was required to show.

305

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

Γ

Α

C

Β

A

B

∆

D

κε΄.

Proposition 25

ΤÕ ØπÕ µέσων δυνάµει µόνον συµµέτρων εÙθειîν περιεχόµενον Ñρθογώνιον ½τοι ·ητÕν À µέσον ™στίν.

The rectangle contained by medial straight-lines (which are) commensurable in square only is either rational or medial.

Α

∆

Β Ξ

Γ Ε

Ζ

Η

Θ

Μ

Κ Λ

Ν

A

D

B O

`ΥπÕ γ¦ρ µέσων δυνάµει µόνον συµµέτρων εÙθειîν τîν ΑΒ, ΒΓ Ñρθογώνιον περιεχέσθω τÕ ΑΓ· λέγω, Óτι τÕ ΑΓ ½τοι ·ητÕν À µέσον ™στίν. 'Αναγεγράφθω γ¦ρ ¢πÕ τîν ΑΒ, ΒΓ τετράγωνα τ¦ Α∆, ΒΕ· µέσον ¥ρα ™στˆν ˜κάτερον τîν Α∆, ΒΕ. κሠ™κκείσθω ·ητ¾ ¹ ΖΗ, κሠτù µν Α∆ ‡σον παρ¦ τ¾ν ΖΗ παραβεβλήσθω Ñρθογώνιον παραλληλόγραµµον τÕ ΗΘ πλάτος ποιοàν τ¾ν ΖΘ, τù δ ΑΓ ‡σον παρ¦ τ¾ν ΘΜ παραβεβλήσθω Ñρθογώνιον παραλληλόγραµµον τÕ ΜΚ πλάτος ποιοàν τ¾ν ΘΚ, κሠœτι τù ΒΕ ‡σον еοίως παρ¦ τ¾ν ΚΝ παραβεβλήσθω τÕ ΝΛ πλάτος ποιοàν τ¾ν ΚΛ· ™π' εÙθείας ¥ρα ε„σˆν αƒ ΖΘ, ΘΚ, ΚΛ. ™πεˆ οâν µέσον ™στˆν ˜κάτερον τîν Α∆, ΒΕ, καί ™στιν ‡σον τÕ µν Α∆ τù ΗΘ, τÕ δ ΒΕ τù ΝΛ, µέσον ¥ρα κሠ˜κάτερον τîν ΗΘ, ΝΛ. κሠπαρ¦ ·ητ¾ν τ¾ν ΖΗ παράκειται· ·ητ¾ ¥ρα ™στˆν ˜κατέρα τîν ΖΘ, ΚΛ κሠ¢σύµµετρος τÍ ΖΗ µήκει. κሠ™πεˆ σύµµετρόν ™στι τÕ Α∆ τù ΒΕ, σύµµετρον ¥ρα ™στˆ κሠτÕ ΗΘ τù ΝΛ. καί ™στιν æς τÕ ΗΘ πρÕς τÕ ΝΛ, οÛτως ¹ ΖΘ πρÕς τ¾ν ΚΛ· σύµµετρος ¥ρα ™στˆν ¹ ΖΘ τÍ ΚΛ µήκει. αƒ ΖΘ, ΚΛ ¥ρα ·ηταί

F

G

H

M

K L

N

C

E

For let the rectangle AC be contained by the medial straight-lines AB and BC (which are) commensurable in square only. I say that AC is either rational or medial. For let the squares AD and BE have been described on (the straight-lines) AB and BC (respectively). AD and BE are thus each medial. And let the rational (straight-line) F G be laid out. And let the rectangular parallelogram GH, equal to AD, have been applied to F G, producing F H as breadth. And let the rectangular parallelogram M K, equal to AC, have been applied to HM , producing HK as breadth. And, finally, let N L, equal to BE, have similarly been applied to KN , producing KL as breadth. Thus, F H, HK, and KL are in a straight-line. Therefore, since AD and BE are each medial, and AD is equal to GH, and BE to N L, GH and N L (are) thus each also medial. And they are applied to the rational (straight-line) F G. F H and KL are thus each rational, and incommensurable in length with F G [Prop. 10.22]. And since AD is commensurable with BE, GH is thus also commensurable with N L. And as

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ε„σι µήκει σύµµετροι· ·ητÕν ¥ρα ™στˆ τÕ ØπÕ τîν ΖΘ, ΚΛ. κሠ™πεˆ ‡ση ™στˆν ¹ µν ∆Β τÍ ΒΑ, ¹ δ ΞΒ τÍ ΒΓ, œστιν ¥ρα æς ¹ ∆Β πρÕς τ¾ν ΒΓ, οÛτως ¹ ΑΒ πρÕς τ¾ν ΒΞ. ¢λλ' æς µν ¹ ∆Β πρÕς τ¾ν ΒΓ, οÛτως τÕ ∆Α πρÕς τÕ ΑΓ· æς δ ¹ ΑΒ πρÕς τ¾ν ΒΞ, οÛτως τÕ ΑΓ πρÕς τÕ ΓΞ· œστιν ¥ρα æς τÕ ∆Α πρÕς τÕ ΑΓ, οÛτως τÕ ΑΓ πρÕς τÕ ΓΞ. ‡σον δέ ™στι τÕ µν Α∆ τù ΗΘ, τÕ δ ΑΓ τù ΜΚ, τÕ δ ΓΞ τù ΝΛ· œστιν ¥ρα æς τÕ ΗΘ πρÕς τÕ ΜΚ, οÛτως τÕ ΜΚ πρÕς τÕ ΝΛ· œστιν ¥ρα κሠæς ¹ ΖΘ πρÕς τ¾ν ΘΚ, οÛτως ¹ ΘΚ πρÕς τ¾ν ΚΛ· τÕ ¥ρα ØπÕ τîν ΖΘ, ΚΛ ‡σον ™στˆ τù ¢πÕ τÁς ΘΚ. ·ητÕν δ τÕ ØπÕ τîν ΖΘ, ΚΛ· ·ητÕν ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς ΘΚ· ·ητ¾ ¥ρα ™στˆν ¹ ΘΚ. καˆ ε„ µν σύµµετρός ™στι τÍ ΖΗ µήκει, ·ητόν ™στι τÕ ΘΝ· ε„ δ ¢σύµµετρός ™στι τÍ ΖΗ µήκει, αƒ ΚΘ, ΘΜ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· µέσον ¥ρα τÕ ΘΝ. τÕ ΘΝ ¥ρα ½τοι ·ητÕν À µέσον ™στίν. ‡σον δ τÕ ΘΝ τù ΑΓ· τÕ ΑΓ ¥ρα ½τοι ·ητÕν À µέσον ™στίν. ΤÕ ¥ρα ØπÕ µέσων δυνάµει µόνον συµµέτρων, κሠτ¦ εξÁς.

GH is to N L, so F H (is) to KL [Prop. 6.1]. Thus, F H is commensurable in length with KL [Prop. 10.11]. Thus, F H and KL are rational (straight-lines which are) commensurable in length. Thus, the (rectangle contained) by F H and KL is rational [Prop. 10.19]. And since DB is equal to BA, and OB to BC, thus as DB is to BC, so AB (is) to BO. But, as DB (is) to BC, so DA (is) to AC [Props. 6.1]. And as AB (is) to BO, so AC (is) to CO [Prop. 6.1]. Thus, as DA is to AC, so AC (is) to CO. And AD is equal to GH, and AC to M K, and CO to N L. Thus, as GH is to M K, so M K (is) to N L. Thus, also, as F H is to HK, so HK (is) to KL [Props. 6.1, 5.11]. Thus, the (rectangle contained) by F H and KL is equal to the (square) on HK [Prop. 6.17]. And the (rectangle contained) by F H and KL (is) rational. Thus, the (square) on HK is also rational. Thus, HK is rational. And if it is commensurable in length with F G, then HN is rational [Prop. 10.19]. And if it is incommensurable in length with F G, then KH and HM are rational (straight-lines which are) commensurable in square only: thus, HN is medial [Prop. 10.21]. Thus, HN is either rational or medial. And HN (is) equal to AC. Thus, AC is either rational or medial. Thus, the . . . by medial straight-lines (which are) commensurable in square only, and so on . . . .

κ$΄.

Proposition 26

Μέσον µέσου οÙχ Øπερέχει ·ητù.

Α ∆

Ζ

A medial (area) does not exceed a medial (area) by a rational (area).†

Ε

A

Γ Β Κ

D Η

E

K

G

C B

Θ

F

H

Ε„ γ¦ρ δυνατόν, µέσον τÕ ΑΒ µέσου τοà ΑΓ Øπερεχέτω ·ητù τù ∆Β, κሠ™κκείσθω ·ητ¾ ¹ ΕΖ, κሠτù ΑΒ ‡σον παρ¦ τ¾ν ΕΖ παραβεβλήσθω παραλληλόγραµµον Ñρθογώνιον τÕ ΖΘ πλάτος ποιοàν τ¾ν ΕΘ, τù δ ΑΓ ‡σον ¢φVρήσθω τÕ ΖΗ· λοιπÕν ¥ρα τÕ Β∆ λοιπù τù ΚΘ ™στιν ‡σον. ·ητÕν δέ ™στι τÕ ∆Β· ·ητÕν ¥ρα ™στˆ κሠτÕ ΚΘ. ™πεˆ οâν µέσον ™στˆν ˜κάτερον τîν ΑΒ, ΑΓ, καί ™στι τÕ µν ΑΒ τù ΖΘ ‡σον, τÕ δ ΑΓ τù ΖΗ, µέσον ¥ρα κሠ˜κάτερον τîν ΖΘ, ΖΗ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται· ·ητ¾ ¥ρα ™στˆν

For, if possible, let the medial (area) AB exceed the medial (area) AC by the rational (area) DB. And let the rational (straight-line) EF be laid down. And let the rectangular parallelogram F H, equal to AB, have been applied to to EF , producing EH as breadth. And let F G, equal to AC, have been cut off (from F H). Thus, the remainder BD is equal to the remainder KH. And DB is rational. Thus, KH is also rational. Therefore, since AB and AC are each medial, and AB is equal to F H, and AC to F G, F H and F G are thus each also medial.

307

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˜κατέρα τîν ΘΕ, ΕΗ κሠ¢σύµµετρος τÍ ΕΖ µήκει. κሠ™πεˆ ·ητόν ™στι τÕ ∆Β καί ™στιν ‡σον τù ΚΘ, ·ητÕν ¥ρα ™στˆ κሠτÕ ΚΘ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται· ·ητ¾ ¥ρα ™στˆν ¹ ΗΘ κሠσύµµετρος τÍ ΕΖ µήκει. ¢λλά κሠ¹ ΕΗ ·ητή ™στι κሠ¢σύµµετρος τÍ ΕΖ µήκει· ¢σύµµετρος ¥ρα ™στˆν ¹ ΕΗ τÍ ΗΘ µήκει. καί ™στιν æς ¹ ΕΗ πρÕς τ¾ν ΗΘ, οÛτως τÕ ¢πÕ τÁς ΕΗ πρÕς τÕ ØπÕ τîν ΕΗ, ΗΘ· ¢σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΕΗ τù ØπÕ τîν ΕΗ, ΗΘ. ¢λλ¦ τù µν ¢πÕ τÁς ΕΗ σύµµετρά ™στι τ¦ ¢πÕ τîν ΕΗ, ΗΘ τετράγωνα· ·ητ¦ γ¦ρ ¢µφότερα· τù δ ØπÕ τîν ΕΗ, ΗΘ σύµµετρόν ™στι τÕ δˆς ØπÕ τîν ΕΗ, ΗΘ· διπλάσιον γάρ ™στιν αÙτοà· ¢σύµµετρα ¥ρα ™στˆ τ¦ ¢πÕ τîν ΕΗ, ΗΘ τù δˆς ØπÕ τîν ΕΗ, ΗΘ· κሠσυναµφότερα ¥ρα τά τε ¢πÕ τîν ΕΗ, ΗΘ κሠτÕ δˆς ØπÕ τîν ΕΗ, ΗΘ, Óπερ ™στˆ τÕ ¢πÕ τÁς ΕΘ, ¢σύµµετρόν ™στι το‹ς ¢πÕ τîν ΕΗ, ΗΘ. ·ητ¦ δ τ¦ ¢πÕ τîν ΕΗ, ΗΘ· ¥λογον ¥ρα τÕ ¢πÕ τÁς ΕΘ. ¥λογος ¥ρα ™στˆν ¹ ΕΘ. ¢λλ¦ κሠ·ηρή· Óπερ ™στˆν ¢δύνατον. Μέσον ¥ρα µέσου οÙχ Øπερέχει ·ητù· Óπερ œδει δε‹ξαι.

†

In other words,

And they are applied to the rational (straight-line) EF . Thus, HE and EG are each rational, and incommensurable in length with EF [Prop. 10.22]. And since DB is rational, and is equal to KH, KH is thus also rational. And (KH) is applied to the rational (straight-line) EF . GH is thus rational, and commensurable in length with EF [Prop. 10.20]. But, EG is also rational, and incommensurable in length with EF . Thus, EG is incommensurable in length with GH [Prop. 10.13]. And as EG is to GH, so the (square) on EG (is) to the (rectangle contained) by EG and GH [Prop. 10.13 lem.]. Thus, the (square) on EG is incommensurable with the (rectangle contained) by EG and GH [Prop. 10.11]. But, the (sum of the) squares on EG and GH is commensurable with the (square) on EG. For (EG and GH are) both rational. And twice the (rectangle contained) by EG and GH is commensurable with the (rectangle contained) by EG and GH [Prop. 10.6]. For (the former) is double the latter. Thus, the (sum of the squares) on EG and GH is incommensurable with twice the (rectangle contained) by EG and GH [Prop. 10.13]. And thus the sum of the (squares) on EG and GH plus twice the (rectangle contained) by EG and GH, that is the (square) on EH [Prop. 2.4], is incommensurable with the (sum of the squares) on EG and GH [Prop. 10.16]. And the (sum of the squares) on EG and GH (is) rational. Thus, the (square) on EH is irrational [Def. 10.4]. Thus, EH is irrational [Def. 10.4]. But, (it is) also rational. The very thing is impossible. Thus, a medial (area) does not exceed a medial (area) by a rational (area). (Which is) the very thing it was required to show.

p p book10eps/k − book10eps/k ′ 6= k ′′ .

κζ΄.

Proposition 27

Μέσας εØρε‹ν δυνάµει µόνον συµµέτρους ·ητÕν πεTo find (two) medial (straight-lines), containing a raριεχούσας. tional (area), (which are) commensurable in square only.

Α Β Γ ∆

A B C D

'Εκκείσθωσαν δύο ·ητሠδυνάµει µόνον σύµµετροι Let the two rational (straight-lines) A and B, (which αƒ Α, Β, κሠε„λήφθω τîν Α, Β µέση ¢νάλογον ¹ Γ, κሠare) commensurable in square only, be laid down. And let

308

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ELEMENTS BOOK 10

γεγονέτω æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Γ πρÕς τ¾ν ∆. Κሠ™πεˆ αƒ Α, Β ·ηταί ε„σι δυνάµει µόνον σύµµετροι, τÕ ¥ρα ØπÕ τîν Α, Β, τουτέστι τÕ ¢πÕ τÁς Γ, µέσον ™στίν. µέση ¥ρα ¹ Γ. κሠ™πεί ™στιν æς ¹ Α πρÕς τ¾ν Β, [οÛτως] ¹ Γ πρÕς τ¾ν ∆, αƒ δ Α, Β δυνάµει µόνον [ε„σˆ] σύµµετροι, καˆ αƒ Γ, ∆ ¥ρα δυνάµει µόνον ε„σˆ σύµµετροι. καί ™στι µέση ¹ Γ· µέση ¥ρα κሠ¹ ∆. αƒ Γ, ∆ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι. λέγω, Óτι κሠ·ητÕν περιέχουσιν. ™πεˆ γάρ ™στιν æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Γ πρÕς τ¾ν ∆, ™ναλλ¦ξ ¥ρα ™στˆν æς ¹ Α πρÕς τ¾ν Γ, ¹ Β πρÕς τ¾ν ∆. ¢λλ' æς ¹ Α πρÕς τ¾ν Γ, ¹ Γ πρÕς τ¾ν Β· κሠæς ¥ρα ¹ Γ πρÕς τ¾ν Β, οÛτως ¹ Β πρÕς τ¾ν ∆· τÕ ¥ρα ØπÕ τîν Γ, ∆ ‡σον ™στˆ τù ¢πÕ τÁς Β. ·ητÕν δ τÕ ¢πÕ τÁς Β· ·ητÕν ¥ρα [™στˆ] κሠτÕ ØπÕ τîν Γ, ∆. ΕÛρηνται ¥ρα µέσαι δυνάµει µόνον σύµµετροι ·ητÕν περιέχουσαι· Óπερ œδει δε‹ξαι.

†

C—the mean proportional (straight-line) to A and B— have been taken [Prop. 6.13]. And let it be contrived that as A (is) to B, so C (is) to D [Prop. 6.12]. And since the rational (straight-lines) A and B are commensurable in square only, the (rectangle contained) by A and B—that is to say, the (square) on C [Prop. 6.17]—is thus medial [Prop 10.21]. Thus, C is medial [Prop. 10.21]. And since as A is to B, [so] C (is) to D, and A and B [are] commensurable in square only, C and D are thus also commensurable in square only [Prop. 10.11]. And C is medial. Thus, D is also medial [Prop. 10.23]. Thus, C and D are medial (straight-lines which are) commensurable in square only. I say that they also contain a rational (area). For since as A is to B, so C (is) to D, thus, alternately, as A is to C, so B (is) to D [Prop. 5.16]. But, as A (is) to C, (so) C (is) to B. And thus as C (is) to B, so B (is) to D [Prop. 5.11]. Thus, the (rectangle contained) by C and D is equal to the (square) on B [Prop. 6.17]. And the (square) on B (is) rational. Thus, the (rectangle contained) by C and D [is] also rational. Thus, (two) medial (straight-lines, C and D), containing a rational (area), (which are) commensurable in square only, have been found.† (Which is) the very thing it was required to show.

C and D have lengths k 1/4 and k 3/4 times that of A, respectively, where the length of B is k 1/2 times that of A.

κη΄.

Proposition 28

Μέσας εØρε‹ν δυνάµει µόνον συµµέτρους µέσον πειTo find (two) medial (straight-lines), containing a meριεχούσας. dial (area), (which are) commensurable in square only.

Α Β Γ

∆ Ε

A B C

'Εκκείσθωσαν [τρε‹ς] ·ητሠδυνάµει µόνον σύµµετροι αƒ Α, Β, Γ, κሠε„λήφθω τîν Α, Β µέση ¢νάλογον ¹ ∆, κሠγεγονέτω æς ¹ Β πρÕς τ¾ν Γ, ¹ ∆ πρÕς τ¾ν Ε. 'Επεˆ αƒ Α, Β ·ηταί ε„σι δυνάµει µόνον σύµµετροι, τÕ ¥ρα ØπÕ τîν Α, Β, τουτέστι τÕ ¢πÕ τÁς ∆, µέσον ™στˆν. µέση ¥ρα ¹ ∆. κሠ™πεˆ αƒ Β, Γ δυνάµει µόνον ε„σˆ σύµµετροι, καί ™στιν æς ¹ Β πρÕς τ¾ν Γ, ¹ ∆ πρÕς τ¾ν Ε, καˆ αƒ ∆, Ε ¥ρα δυνάµει µόνον ε„σˆ σύµµετροι. µέση δ ¹ ∆· µέση ¥ρα κሠ¹ Ε· αƒ ∆, Ε ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι. λέγω δή, Óτι κሠµέσον περιέχουσιν. ™πεˆ γάρ ™στιν æς ¹ Β πρÕς τ¾ν Γ, ¹ ∆ πρÕς τ¾ν Ε, ™ναλλ¦ξ ¥ρα æς ¹ Β πρÕς τ¾ν ∆, ¹ Γ πρÕς τ¾ν Ε. æς δ ¹ Β πρÕς τ¾ν ∆, ¹ ∆ πρÕς τ¾ν Α· κሠæς ¤ρα ¹ ∆ πρÕς τ¾ν Α, ¹ Γ πρÕς τ¾ν Ε· τÕ ¥ρα ØπÕ τîν Α, Γ ‡σον ™στˆ τù ØπÕ τîν ∆, Ε. µέσον δ τÕ ØπÕ τîν

D E

Let the [three] rational (straight-lines) A, B, and C, (which are) commensurable in square only, be laid down. And let, D, the mean proportional (straight-line) to A and B, have been taken [Prop. 6.13]. And let it be contrived that as B (is) to C, (so) D (is) to E [Prop. 6.12]. Since the rational (straight-lines) A and B are commensurable in square only, the (rectangle contained) by A and B—that is to say, the (square) on D [Prop. 6.17]— is medial [Prop. 10.21]. Thus, D (is) medial [Prop. 10.21]. And since B and C are commensurable in square only, and as B is to C, (so) D (is) to E, D and E are thus commensurable in square only [Prop. 10.11]. And D (is) medial. E (is) thus also medial [Prop. 10.23]. Thus, D and E are medial (straight-lines which are) commensurable in square only. So, I say that they also enclose a medial

309

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ELEMENTS BOOK 10

Α, Γ· µέσον ¥ρα κሠτÕ ØπÕ τîν ∆, Ε. (area). For since as B is to C, (so) D (is) to E, thus, 'ΕÛρηνται ¥ρα µέσαι δυνάµει µόνον σύµµετροι alternately, as B (is) to D, (so) C (is) to E [Prop. 5.16]. µέσον περιέχουσαι· Óπερ œδει δε‹ξαι. And as B (is) to D, (so) D (is) to A. And thus as D (is) to A, (so) C (is) to E. Thus, the (rectangle contained) by A and C is equal to the (rectangle contained) by D and E [Prop. 6.16]. And the (rectangle contained) by A and C is medial [Prop. 10.21]. Thus, the (rectangle contained) by D and E (is) also medial. Thus, (two) medial (straight-lines, D and E), containing a medial (area), (which are) commensurable in square only, have been found. (Which is) the very thing it was required to show. †

D and E have lengths k 1/4 and k ′1/2 /k 1/4 times that of A, respectively, where the lengths of B and C are k 1/2 and k ′1/2 times that of A, respectively.

ΛÁµµα α΄.

Lemma I

ΕØρειν δύο τετραγώνους ¢ριθµούς, éστε κሠτÕν συγκείµενον ™ξ αÙτîν εναι τετράγωνον.

To find two square numbers such that the sum of them is also square.

Α

∆

Γ

Β

A

'Εκκείσθωσαν δύο ¢ριθµοˆ οƒ ΑΒ, ΒΓ, œστωσαν δ ½τοι ¥ρτιοι À περιττοί. κሠ™πεˆ, ™άν τε ¢πÕ ¢ρτίου ¥ρτιος ¢φαιρεθÍ, ™άν τε ¢πÕ περισσοà περισσός, Ð λοιπÕς ¥ρτιός ™στιν, Ð λοιπÕς ¥ρα Ð ΑΓ ¥ρτιός ™στιν. τετµήσθω Ð ΑΓ δίχα κατ¦ τÕ ∆. œστωσαν δ καˆ οƒ ΑΒ, ΒΓ ½τοι Óµοιοι ™πίπεδοι À τετράγωνοι, ο‰ κሠαÙτοˆ Óµοιοί ε„σιν ™πίπεδοι· Ð ¥ρα ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ [τοà] Γ∆ τετραγώνου ‡σος ™στˆ τù ¢πÕ τοà Β∆ τετραγώνJ. καί ™στι τετράγωνος Ð ™κ τîν ΑΒ, ΒΓ, ™πειδήπερ ™δείχθη, Óτι, ™¦ν δύο Óµοιοι ™πίπεδοι πολλαπλασιάσαντες ¢λλήλους ποιîσι τινα, Ð γενόµενος τετράγωνός ™στιν. εÛρηνται ¥ρα δύο τετράγωνοι ¢ριθµοˆ Ó τε ™κ τîν ΑΒ, ΒΓ καˆ Ð ¢πÕ τοà Γ∆, ο‰ συντεθέντες ποιοàσι τÕν ¢πÕ τοà Β∆ τετράγωνον. Κሠφανερόν, Óτι εÛρηνται πάλιν δύο τετράγωνοι Ó τε ¢πÕ τοà Β∆ καˆ Ð ¢πÕ τοà Γ∆, éστε τ¾ν Øπεροχ¾ν αÙτîν τÕν ØπÕ ΑΒ, ΒΓ εναι τετράγωνον, Óταν οƒ ΑΒ, ΒΓ Óµοιοι ðσιν ™πίπεδοι. Óταν δ µ¾ ðσιν Óµοιοι ™πίπεδοι, εÛρηνται δύο τετράγωνοι Ó τε ¢πÕ τοà Β∆ καˆ Ð ¢πÕ τοà ∆Γ, ïν ¹ Øπεροχ¾ Ð ØπÕ τîν ΑΒ, ΒΓ οÙκ œστι τετράγωνος· Óπερ œδει δε‹ξαι.

D

C

B

Let the two numbers AB and BC be laid down. And let them be either (both) even or (both) odd. And since, if an even (number) is subtracted from an even (number), or if an odd (number is subtracted) from an odd (number), then the remainder is even [Props. 9.24, 9.26], the remainder AC is thus even. Let AC have been cut in half at D. And let AB and BC also be either similar plane (numbers), or square (numbers)—which are themselves also similar plane (numbers). Thus, the (number created) from (multiplying) AB and BC, plus the square on CD, is equal to the square on BD [Prop. 2.6]. And the (number created) from (multiplying) AB and BC is square—in as much as it was shown that if two similar plane (numbers) make some (number) by multiplying one another, then the (number so) created is square [Prop. 9.1]. Thus, two square numbers have been found—(namely,) the (number created) from (multiplying) AB and BC, and the (square) on CD—which, (when) added (together), make the square on BD. And (it is) clear that two square (numbers) have again been found—(namely,) the (square) on BD, and the (square) on CD—such that their difference—(namely,) the (rectangle) contained by AB and BC—is square whenever AB and BC are similar plane (numbers). But when they are not similar plane numbers, two square (numbers) have been found—(namely,) the (square) on BD, and the (square) on DC—between which the difference—(namely,) the (rectangle) contained by AB and BC—is not square. (Which is) the very thing it was

310

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 required to show.

ΛÁµµα β΄.

Lemma II

ΕØρε‹ν δύο τετραγώνους ¢ριθµούς, éστε τÕν ™ξ αÙτîν συγκείµενον µ¾ εναι τετράγωνον.

To find two square numbers such that the sum of them is not square.

Α Η Θ

∆ Ζ Ε

Γ

Β

A G H

”Εστω γ¦ρ Ð ™κ τîν ΑΒ, ΒΓ, æς œφαµεν, τετράγωνος, κሠ¥ρτιος Ð ΓΑ, κሠτετµήσθω Ð ΓΑ δίχα τù ∆. φανερÕν δή, Óτι Ð ™κ τîν ΑΒ, ΒΓ τετράγωνος µετ¦ τοà ¢πÕ [τοà] Γ∆ τετραγώνου ‡σος ™στˆ τù ¢πÕ [τοà] Β∆ τετραγώνJ. ¢φVρήσθω µον¦ς ¹ ∆Ε· Ð ¥ρα ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ [τοà] ΓΕ ™λάσσων ™στˆ τοà ¢πÕ [τοà] Β∆ τετραγώνου. λέγω οâν, Óτι Ð ™κ τîν ΑΒ, ΒΓ τετράγωνος µετ¦ τοà ¢πÕ [τοà] ΓΕ οÙκ œσται τετράγωνος. Ε„ γ¦ρ œσται τετράγωνος, ½τοι ‡σος ™στˆ τù ¢πÕ [τοà] ΒΕ À ™λάσσων τοà ¢πÕ [τοà] ΒΕ, οÙκέτι δ κሠµείζων, †να µ¾ τµηθÍ ¹ µονάς. œστω, ε„ δυνατόν, πρότερον Ð ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΕ ‡σος τù ¢πÕ ΒΕ, κሠœστω τÁς ∆Ε µονάδος διπλασίων Ð ΗΑ. ™πεˆ οâν Óλος Ð ΑΓ Óλου τοà Γ∆ ™στι διπλασίων, ïν Ð ΑΗ τοà ∆Ε ™στι διπλασίων, κሠλοιπÕς ¥ρα Ð ΗΓ λοιποà τοà ΕΓ ™στι διπλασίων· δίχα ¥ρα τέτµηται Ð ΗΓ τù Ε. Ð ¥ρα ™κ τîν ΗΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΕ ‡σος ™στˆ τù ¢πÕ ΒΕ τετραγώνJ. ¢λλ¦ καˆ Ð ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΕ ‡σος Øπόκειται τù ¢πÕ [τοà] ΒΕ τετραγώνJ· Ð ¥ρα ™κ τîν ΗΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΕ ‡σος ™στˆ τù ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΕ. κሠκοινοà ¢φαιρεθέντος τοà ¢πÕ ΓΕ συνάγεται Ð ΑΒ ‡σος τù ΗΒ· Óπερ ¥τοπον. οÙκ ¥ρα Ð ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ [τοà] ΓΕ ‡σος ™στˆ τù ¢πÕ ΒΕ. λέγω δή, Óτι οÙδ ™λάσσων τοà ¢πÕ ΒΕ. ε„ γ¦ρ δυνατόν, œστω τù ¢πÕ ΒΖ ‡σος, κሠτοà ∆Ζ διπλασίων Ð ΘΑ. κሠσυναχθήσεται πάλιν διπλασίων Ð ΘΓ τοà ΓΖ· éστε κሠτÕν ΓΘ δίχα τετµÁσθαι κατ¦ τÕ Ζ, κሠδι¦ τοàτο τÕν ™κ τîν ΘΒ, ΒΓ µετ¦ τοà ¢πÕ ΖΓ ‡σον γίνεσθαι τù ¢πÕ ΒΖ. Øπόκειται δ καˆ Ð ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΕ ‡σος τù ¢πÕ ΒΖ. éστε καˆ Ð ™κ τîν ΘΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΖ ‡σος œσται τù ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΕ· Óπερ ¥τοπον. οÙκ ¥ρα Ð ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΕ ‡σος ™στˆ [τù] ™λάσσονι τοà ¢πÕ ΒΕ. ™δείχθη δέ, Óτι οÙδ [αÙτù] τù ¢πÕ ΒΕ. οÙκ ¥ρα Ð ™κ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΕ τετράγωνός ™στιν. Óπερ œδει δε‹ξαι.

D F E

C

B

For let the (number created) from (multiplying) AB and BC, as we said, be square. And (let) CA (be) even. And let CA have been cut in half at D. So it is clear that the square (number created) from (multiplying) AB and BC, plus the square on CD, is equal to the square on BD [see previous lemma]. Let the unit DE have been subtracted (from BD). Thus, the (number created) from (multiplying) AB and BC, plus the (square) on CE, is less than the square on BD. I say, therefore, that the square (number created) from (multiplying) AB and BC, plus the (square) on CE, is not square. For if it is square, it is either equal to the (square) on BE, or less than the (square) on BE, but cannot be greater (than the square on BE) any more, lest the unit be divided. First of all, if possible, let the (number created) from (multiplying) AB and BC, plus the (square) on CE, be equal to the (square) on BE. And let GA be double the unit DE. Therefore, since the whole of AC is double the whole of CD, of which AG is double DE, the remainder GC is thus double the remainder EC. Thus, GC has been cut in half at E. Thus, the (number created) from (multiplying) GB and BC, plus the (square) on CE, is equal to the square on BE [Prop. 2.6]. But, the (number created) from (multiplying) AB and BC, plus the (square) on CE, was also assumed (to be) equal to the square on BE. Thus, the (number created) from (multiplying) GB and BC, plus the (square) on CE, is equal to the (number created) from (multiplying) AB and BC, plus the (square) on CE. And subtracting the (square) on CE from both, AB is inferred (to be) equal to GB. The very thing is absurd. Thus, the (number created) from (multiplying) AB and BC, plus the (square) on CE, is not equal to the (square) on BE. So I say that (it is) not less than the (square) on BE either. For, if possible, let it be equal to the (square) on BF . And (let) HA (be) double DF . And it can again be inferred that HC (is) double CF . Hence, CH has also been cut in half at F . And, on account of this, the (number created) from (multiplying) HB and BC, plus the (square) on F C, becomes equal to the (square) on BF [Prop. 2.6]. And the (number created) from (multiplying) AB and BC, plus the (square) on CE, was also assumed (to be) equal to the (square) on BF . Hence, the (number created) from

311

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 (multiplying) HB and BC, plus the (square) on CF , will also be equal to the (number created) from (multiplying) AB and BC, plus the (square) on CE. The very thing is absurd. Thus, the (number created) from (multiplying) AB and BC, plus the (square) on CE, is not equal to less than the (square) on BE. And it was shown that (is it) not equal to the (square) on BE either. Thus, the (number created) from (multiplying) AB and BC, plus the square on CE, is not square. (Which is) the very thing it was required to show.

κθ΄.

Proposition 29

ΕØρε‹ν δύο ·ητ¦ς δυνάµει µόνον συµµέτρους, éστε To find two rational (straight-lines which are) comτ¾ν µείζονα τÁς ™λάσσονος µε‹ζον δύνασθαι τù ¢πÕ mensurable in square only, such that the square on the συµµέτρου ˜αυτÍ µήκει. greater is larger than the (square on the) lesser by the (square) on (some straight-line which is) commensurable in length with the greater.

Ζ

Α Γ

F

Β Ε

A ∆

C

'Εκκείσθω γάρ τις ·ητ¾ ¹ ΑΒ κሠδύο τετράγωνοι ¢ριθµοˆ οƒ Γ∆, ∆Ε, éστε τ¾ν Øπεροχ¾ν αÙτîν τÕν ΓΕ µ¾ εναι τετράγωνον, κሠγεγράφθω ™πˆ τÁς ΑΒ ¹µικύκλιον τÕ ΑΖΒ, κሠπεποιήσθω æς Ð ∆Γ πρÕς τÕν ΓΕ, οÛτως τÕ ¢πÕ τÁς ΒΑ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΑΖ τετράγωνον, κሠ™πεζεύχθω ¹ ΖΒ. 'Επεˆ [οâν] ™στιν æς τÕ ¢πÕ τÁς ΒΑ πρÕς τÕ ¢πÕ τÁς ΑΖ, οÛτως Ð ∆Γ πρÕς τÕν ΓΕ, τÕ ¢πÕ τÁς ΒΑ ¥ρα πρÕς τÕ ¢πÕ τÁς ΑΖ λόγον œχει, Óν ¢ριθµÕς Ð ∆Γ πρÕς ¢ριθµÕν τÕν ΓΕ· σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΒΑ τù ¢πÕ τÁς ΑΖ. ·ητÕν δ τÕ ¢πÕ τÁς ΑΒ· ·ητÕν ¥ρα κሠτÕ ¢πÕ τÁς ΑΖ· ·ητ¾ ¥ρα κሠ¹ ΑΖ. κሠ™πεˆ Ð ∆Γ πρÕς τÕν ΓΕ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ τÕ ¢πÕ τÁς ΒΑ ¥ρα πρÕς τÕ ¢πÕ τÁς ΑΖ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΒ τÍ ΑΖ µήκει· αƒ ΒΑ, ΑΖ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. κሠ™πεί [™στιν] æς Ð ∆Γ πρÕς τÕν ΓΕ, οÛτως τÕ ¢πÕ τÁς ΒΑ πρÕς τÕ ¢πÕ τÁς ΑΖ, ¢ναστρέψαντι ¥ρα æς Ð Γ∆ πρÕς τÕν ∆Ε, οÛτως τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ¢πÕ τÁς ΒΖ. Ð δ Γ∆ πρÕς τÕν ∆Ε λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν·

B E

D

For let some rational (straight-line) AB be laid down, and two square numbers, CD and DE, such that the difference between them, CE, is not square [Prop. 10.28 lem. I]. And let the semi-circle AF B have been drawn on AB. And let it be contrived that as DC (is) to CE, so the square on BA (is) to the square on AF [Prop. 10.6 corr.]. And let F B have been joined. [Therefore,] since as the (square) on BA is to the (square) on AF , so DC (is) to CE, the (square) on BA thus has to the (square) on AF the ratio which the number DC (has) to the number CE. Thus, the (square) on BA is commensurable with the (square) on AF [Prop. 10.6]. And the (square) on AB (is) rational [Def. 10.4]. Thus, the (square) on AF (is) also rational. And since DC does not have to CE the ratio which (some) square number (has) to (some) square number, the (square) on BA thus does not have to the (square) on AF the ratio which (some) square number has to (some) square number either. Thus, AB is incommensurable in length with AF [Prop. 10.9]. Thus, the rational (straightlines) BA and AF are commensurable in square only. And since as DC [is] to CE, so the (square) on BA (is)

312

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ELEMENTS BOOK 10

κሠτÕ ¢πÕ τÁς ΑΒ ¥ρα πρÕς τÕ ¢πÕ τÁς ΒΖ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· σύµµετρος ¥ρα ™στˆν ¹ ΑΒ τÍ ΒΖ µήκει. καί ™στι τÕ ¢πÕ τÁς ΑΒ ‡σον το‹ς ¢πÕ τîν ΑΖ, ΖΒ· ¹ ΑΒ ¥ρα τÁς ΑΖ µε‹ζον δύναται τÍ ΒΖ συµµέτρJ ˜αυτÍ. ΕÛρηνται ¥ρα δύο ·ητሠδυνάµει µόνον σύµµετροι αƒ ΒΑ, ΑΖ, éστε τ¾ν µε‹ζονα τ¾ν ΑΒ τÁς ™λάσσονος τÁς ΑΖ µε‹ζον δύνασθαι τù ¢πÕ τÁς ΒΖ συµµέτρου ˜αυτÍ µήκει· Óπερ œδει δε‹ξαι.

†

√

BA and AF have lengths 1 and

to the (square) on AF , thus, via conversion, as CD (is) to DE, so the (square) on AB (is) to the (square) on BF [Props. 5.19 corr., 3.31, 1.47]. And CD has to DE the ratio which (some) square number (has) to (some) square number. Thus, the (square) on AB also has to the (square) on BF the ratio which (some) square number has to (some) square number. AB is thus commensurable in length with BF [Prop. 10.9]. And the (square) on AB is equal to the (sum of the squares) on AF and F B [Prop. 1.47]. Thus, the square on AB is greater than (the square on) AF by (the square on) BF , (which is) commensurable (in length) with (AB). Thus, two rational (straight-lines), BA and AF , commensurable in square only, have been found such that the square on the greater, AB, is larger than (the square on) the lesser, AF , by the (square) on BF , (which is) commensurable in length with (AB).† (Which is) the very thing it was required to show.

1 − k 2 times that of AB, respectively, where k =

p

DE/CD.

λ΄.

Proposition 30

ΕØρε‹ν δύο ·ητ¦ς δυνάµει µόνον συµµέτρους, éστε To find two rational (straight-lines which are) comτ¾ν µείζονα τÁς ™λάσσονος µε‹ζον δύνασθαι τù ¢πÕ mensurable in square only, such that the square on the ¢συµµέτρου ˜αυτÍ µήκει. greater is larger than the (the square on) lesser by the (square) on (some straight-line which is) incommensurable in length with the greater.

Ζ

F

Α Γ

Β Ε

A ∆

C

'Εκκείσθω ·ητ¾ ¹ ΑΒ κሠδύο τετράγωνοι ¢ριθµοˆ οƒ ΓΕ, Ε∆, éστε τÕν συγκείµενον ™ξ αÙτîν τÕν Γ∆ µ¾ εναι τετράγωνον, κሠγεγράφθω ™πˆ τÁς ΑΒ ¹µικύκλιον τÕ ΑΖΒ, κሠπεποιήσθω æς Ð ∆Γ πρÕς τÕν ΓΕ, οÛτως τÕ ¢πÕ τÁς ΒΑ πρÕς τÕ ¢πÕ τÁς ΑΖ, κሠ™πεζεύχθω ¹ ΖΒ. `Οµοίως δ¾ δείξοµεν τù πρÕ τούτου, Óτι αƒ ΒΑ, ΑΖ ·ηταί ε„σι δυνάµει µόνον σύµµετροι. κሠ™πεί ™στιν æς Ð ∆Γ πρÕς τÕν ΓΕ, οÛτως τÕ ¢πÕ τÁς ΒΑ πρÕς τÕ ¢πÕ τÁς ΑΖ, ¢ναστρέψαντι ¥ρα æς Ð Γ∆ πρÕς τÕν ∆Ε, οÛτως τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ¢πÕ τÁς ΒΖ. Ð δ Γ∆ πρÕς τÕν ∆Ε λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ' ¥ρα τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ

B E

D

Let the rational (straight-line) AB be laid out, and the two square numbers, CE and ED, such that the sum of them, CD, is not square [Prop. 10.28 lem. II]. And let the semi-circle AF B have been drawn on AB. And let it be contrived that as DC (is) to CE, so the (square) on BA (is) to the (square) on AF [Prop. 10.6 corr]. And let F B have been joined. So, similarly to the (proposition) before this, we can show that BA and AF are rational (straight-lines which are) commensurable in square only. And since as DC is to CE, so the (square) on BA (is) to the (square) on AF , thus, via conversion, as CD (is) to DE, so the (square) on AB (is) to the (square) on BF [Props. 5.19 corr., 3.31,

313

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ELEMENTS BOOK 10

¢πÕ τÁς ΒΖ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΒ τÍ ΒΖ µήκει. κሠδύναται ¹ ΑΒ τÁς ΑΖ µε‹ζον τù ¢πÕ τÁς ΖΒ ¢συµµέτρου ˜αυτÍ. Αƒ ΑΒ, ΑΖ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ ΑΒ τÁς ΑΖ µε‹ζον δύναται τù ¢πÕ τÁς ΖΒ ¢συµµέτρου ˜αυτÍ µήκει· Óπερ œδει δε‹ξαι.

†

1.47]. And CD does not have to DE the ratio which (some) square number (has) to (some) square number. Thus, the (square) on AB does not have to the (square) on BF the ratio which (some) square number has to (some) square number either. Thus, AB is incommensurable in length with BF [Prop. 10.9]. And the square on AB is greater than the (square on) AF by the (square) on F B [Prop. 1.47], (which is) incommensurable (in length) with (AB). Thus, AB and AF are rational (straight-lines which are) commensurable in square only, and the square on AB is greater than (the square on) AF by the (square) on F B, (which is) incommensurable (in length) with (AB).† (Which is) the very thing it was required to show.

p √ AB and AF have lengths 1 and 1/ 1 + k 2 times that of AB, respectively, where k = DE/CE.

λα΄.

Proposition 31

ΕØρε‹ν δύο µέσας δυνάµει µόνον συµµέτρους ·ητÕν περιεχούσας, éστε τ¾ν µείζονα τÁς ™λάσσονος µε‹ζον δύνασθαι τù ¢πÕ συµµέτρου ˜αυτÍ µήκει.

To find two medial (straight-lines), commensurable in square only, (and) containing a rational (area), such that the square on the greater is larger than the (square on the) lesser by the (square) on (some straight-line) commensurable in length with the greater.

Α

Β

Γ

∆

A

'Εκκείσθωσαν δύο ·ητሠδυνάµει µόνον σύµµετροι αƒ Α, Β, éστε τ¾ν Α µείζονα οâσαν τÁς ™λάσσονος τÁς Β µε‹ζον δύνασθαι τù ¢πÕ συµµέτρου ˜αυτÍ µήκει. κሠτù ØπÕ τîν Α, Β ‡σον œστω τÕ ¢πÕ τÁς Γ. µέσον δ τÕ ØπÕ τîν Α, Β· µέσον ¥ρα κሠτÕ ¢πÕ τÁς Γ· µέση ¥ρα κሠ¹ Γ. τù δ ¢πÕ τÁς Β ‡σον œστω τÕ ØπÕ τîν Γ, ∆· ·ητÕν δ τÕ ¢πÕ τÁς Β· ·ητÕν ¥ρα κሠτÕ ØπÕ τîν Γ, ∆. κሠ™πεί ™στιν æς ¹ Α πρÕς τ¾ν Β, οÛτως τÕ ØπÕ τîν Α, Β πρÕς τÕ ¢πÕ τÁς Β, ¢λλ¦ τù µν ØπÕ τîν Α, Β ‡σον ™στˆ τÕ ¢πÕ τÁς Γ, τù δ ¢πÕ τÁς Β ‡σον τÕ ØπÕ τîν Γ, ∆, æς ¥ρα ¹ Α πρÕς τ¾ν Β, οÛτως τÕ ¢πÕ τÁς Γ πρÕς τÕ ØπÕ τîν Γ, ∆. æς δ τÕ ¢πÕ τÁς Γ πρÕς τÕ ØπÕ τîν Γ, ∆, οÛτως ¹ Γ πρÕς τ¾ν ∆· κሠæς ¥ρα ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Γ πρÕς τ¾ν ∆. σύµµετρος δ ¹ Α τÍ Β δυνάµει µόνον· σύµµετρος ¥ρα κሠ¹ Γ τÍ ∆ δυνάµει µόνον. καί ™στι µέση ¹ Γ· µέση ¥ρα κሠ¹ ∆. κሠ™πεί ™στιν æς ¹ Α πρÕς τ¾ν Β, ¹ Γ πρÕς τ¾ν ∆, ¹ δ Α τÁς Β µε‹ζον

B

C

D

Let two rational (straight-lines), A and B, commensurable in square only, be laid out, such that the square on the greater A is larger than the (square on the) lesser B by the (square) on (some straight-line) commensurable in length with (A) [Prop. 10.29]. And let the (square) on C be equal to the (rectangle contained) by A and B. And the (rectangle contained by) A and B (is) medial [Prop. 10.21]. Thus, the (square) on C (is) also medial. Thus, C (is) also medial [Prop. 10.21]. And let the (rectangle contained) by C and D be equal to the (square) on B. And the (square) on B (is) rational. Thus, the (rectangle contained) by C and D (is) also rational. And since as A is to B, so the (rectangle contained) by A and B (is) to the (square) on B [Prop. 10.21 lem.], but the (square) on C is equal to the (rectangle contained) by A and B, and the (rectangle contained) by C and D to the (square) on B, thus as A (is) to B, so the (square)

314

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ELEMENTS BOOK 10

δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠ¹ Γ ¥ρα τÁς ∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. ΕÛρηνται ¥ρα δύο µέσαι δυνάµει µόνον σύµµετροι αƒ Γ, ∆ ·ητÕν περιέχουσαι, κሠ¹ Γ τÁς ∆ µε‹ζον δυνάται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει. `Οµοίως δ¾ δειχθήσεται κሠτù ¢πÕ ¢συµµέτρου, Óταν ¹ Α τÁς Β µε‹ζον δύνηται τù ¢πÕ ¢συµµέτρου ˜αυτÍ.

† ‡

on C (is) to the (rectangle contained) by C and D. And as the (square) on C (is) to the (rectangle contained) by C and D, so C (is) to D [Prop. 10.21 lem.]. And thus as A (is) to B, so C (is) to D. And A is commensurable in square only with B. Thus, C (is) also commensurable in square only with D [Prop. 10.11]. And C is medial. Thus, D (is) also medial [Prop. 10.23]. And since as A is to B, (so) C (is) to D, and the square on A is greater than (the square on) B by the (square) on (some straight-line) commensurable (in length) with (A), the square on C is thus also greater than (the square on) D by the (square) on (some straight-line) commensurable (in length) with (C) [Prop. 10.14]. Thus, two medial (straight-lines), C and D, commensurable in square only, (and) containing a rational (area), have been found. And the square on C is greater than (the square on) D by the (square) on (some straight-line) commensurable in length with (C).† So, similarly, (the proposition) can also be demonstrated for (some straight-line) incommensurable (in length with C), provided that the square on A is greater than (the square on B) by the (square) on (some straight-line) incommensurable (in length) with (A) [Prop. 10.30].‡

C and D have lengths (1 − k 2 )1/4 and (1 − k 2 )3/4 times that of A, respectively, where k is defined in the footnote to Prop. 10.29.

C and D would have lengths 1/(1 + k 2 )1/4 and 1/(1 + k 2 )3/4 times that of A, respectively, where k is defined in the footnote to Prop. 10.30.

λβ΄.

Proposition 32

ΕØρε‹ν δύο µέσας δυνάµει µόνον συµµέτρους µέσον περιεχούσας, éστε τ¾ν µείζονα τÁς ™λάσσονος µε‹ζον δύνασθαι τù ¢πÕ συµµέτρου ˜αυτÍ.

To find two medial (straight-lines), commensurable in square only, (and) containing a medial (area), such that the square on the greater is larger than the (square on the) lesser by the (square) on (some straight-line) commensurable (in length) with the greater.

Α Β Γ

∆ Ε

A B C

'Εκκείσθωσαν τρε‹ς ·ητሠδυνάµει µόνον σύµµετροι αƒ Α, Β, Γ, éστε τ¾ν Α τÁς Γ µε‹ζον δύνασθαι τù ¢πÕ συµµέτρου ˜αυτÍ, κሠτù µν ØπÕ τîν Α, Β ‡σον œστω τÕ ¢πÕ τ¾ς ∆. µέσον ¥ρα τÕ ¢πÕ τÁς ∆· κሠ¹ ∆ ¥ρα µέση ™στίν. τù δ ØπÕ τîν Β, Γ ‡σον œστω τÕ ØπÕ τîν ∆, Ε. κሠ™πεί ™στιν æς τÕ ØπÕ τîν Α, Β πρÕς τÕ ØπÕ τîν Β, Γ, οÛτως ¹ Α πρÕς τ¾ν Γ, ¢λλ¦ τù µν ØπÕ τîν Α, Β ‡σον ™στˆ τÕ ¢πÕ τÁς ∆, τù δ ØπÕ τîν Β, Γ ‡σον τÕ ØπÕ τîν ∆, Ε, œστιν ¥ρα æς ¹ Α πρÕς τ¾ν Γ, οÛτως τÕ ¢πÕ τÁς ∆ πρÕς τÕ ØπÕ τîν ∆, Ε. æς δ τÕ ¢πÕ τÁς ∆ πρÕς τÕ ØπÕ τîν ∆, Ε, οÛτως ¹ ∆ πρÕς τ¾ν Ε· κሠæς ¥ρα

D E

Let three rational (straight-lines), A, B and C, commensurable in square only, be laid out such that the square on A is greater than (the square on C) by the (square) on (some straight-line) commensurable (in length) with (A) [Prop. 10.29]. And let the (square) on D be equal to the (rectangle contained) by A and B. Thus, the (square) on D (is) medial. Thus, D is also medial [Prop. 10.21]. And let the (rectangle contained) by D and E be equal to the (rectangle contained) by B and C. And since as the (rectangle contained) by A and B is to the (rectangle contained) by B and C, so A (is) to

315

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ELEMENTS BOOK 10

¹ Α πρÕς τ¾ν Γ, οÛτως ¹ ∆ πρÕς τ¾ν Ε. σύµµετρος δ ¹ Α τÍ Γ δυνάµει [µόνον]. σύµµετρος ¥ρα κሠ¹ ∆ τÍ Ε δυνάµει µόνον. µέση δ ¹ ∆· µέση ¥ρα κሠ¹ Ε. κሠ™πεί ™στιν æς ¹ Α πρÕς τ¾ν Γ, ¹ ∆ πρÕς τ¾ν Ε, ¹ δ Α τÁς Γ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠ¹ ∆ ¥ρα τÁς Ε µε‹ζον δυνήσεται τù ¢πÕ συµµέτρου ˜αυτÍ. λέγω δή, Óτι κሠµέσον ™στˆ τÕ ØπÕ τîν ∆, Ε. ™πεˆ γ¦ρ ‡σον ™στˆ τÕ ØπÕ τîν Β, Γ τù ØπÕ τîν ∆, Ε, µέσον δ τÕ ØπÕ τîν Β, Γ [αƒ γ¦ρ Β, Γ ·ηταί ε„σι δυνάµει µόνον σύµµετροι], µέσον ¥ρα κሠτÕ ØπÕ τîν ∆, Ε. ΕÛρηνται ¥ρα δύο µέσαι δυνάµει µόνον σύµµετροι αƒ ∆, Ε µέσον περιέχουσαι, éστε τ¾ν µείζονα τÁς ™λάσσονος µε‹ζον δύνασθαι τù ¢πÕ συµµέτρου ˜αυτÍ. `Οµοίως δ¾ πάλιν διεχθήσεται κሠτù ¢πÕ ¢συµµέτρου, Óταν ¹ Α τÁς Γ µε‹ζον δύνηται τù ¢πÕ ¢συµµέτρου ˜αυτV.

‡

C [Prop. 10.21 lem.], but the (square) on D is equal to the (rectangle contained) by A and B, and the (rectangle contained) by D and E to the (rectangle contained) by B and C, thus as A is to C, so the (square) on D (is) to the (rectangle contained) by D and E. And as the (square) on D (is) to the (rectangle contained) by D and E, so D (is) to E [Prop. 10.21 lem.]. And thus as A (is) to C, so D (is) to E. And A (is) commensurable in square [only] with C. Thus, D (is) also commensurable in square only with E [Prop. 10.11]. And D (is) medial. Thus, E (is) also medial [Prop. 10.23]. And since as A is to C, (so) D (is) to E, and the square on A is greater than (the square on) C by the (square) on (some straight-line) commensurable (in length) with (A), the square on D is thus also greater than (the square on) E by the (square) on (some straight-line) commensurable (in length) with (D) [Prop. 10.14]. So, I also say that the (rectangle contained) by D and E is medial. For since the (rectangle contained) by B and C is equal to the (rectangle contained) by D and E, and the (rectangle contained) by B and C (is) medial [for B and C are rational (straight-lines which are) commensurable in square only] [Prop. 10.21], the (rectangle contained) by D and E (is) thus also medial. Thus, two medial (straight-lines), D and E, commensurable in square only, (and) containing a medial (area), have been found, such that the square on the greater is larger than the (square on the) lesser by the (square) on (some straight-line) commensurable (in length) with the greater.† . So, similarly, (the proposition) can again also be demonstrated for (some straight-line) incommensurable (in length with the greater), provided that the square on A is greater than (the square on) C by the (square) on (some straight-line) incommensurable (in length) with (A) [Prop. 10.30].‡

√ D and E have lengths k ′1/4 and k ′1/4 1 − k 2 times that of A, respectively, where the length of B is k ′1/2 times that of A, and k is defined in

the footnote to Prop. 10.29. †

√ D and E would have lengths k ′1/4 and k ′1/4 / 1 + k 2 times that of A, respectively, where the length of B is k ′1/2 times that of A, and k is

defined in the footnote to Prop. 10.30.

ΛÁµµα.

Lemma

”Εστω τρίγωνον Ñρθογώνιον τÕ ΑΒΓ Ñρθ¾ν œχον τ¾ν Α, κሠ½χθω κάθετος ¹ Α∆· λέγω, Óτι τÕ µν ØπÕ τîν ΓΒΑ ‡σον ™στˆ τù ¢πÕ τÁς ΒΑ, τÕ δ ØπÕ τîν ΒΓΑ ‡σον τù ¢πÕ τÁς ΓΑ, κሠτÕ ØπÕ τîν Β∆, ∆Γ ‡σον τù ¢πÕ τÁς Α∆, κሠœτι τÕ ØπÕ τîν ΒΓ, Α∆ ‡σον [™στˆ] τù ØπÕ τîν ΒΑ, ΑΓ. Κሠπρîτον, Óτι τÕ ØπÕ τîν ΓΒΑ ‡σον [™στˆ] τù ¢πÕ

Let ABC be a right-angled triangle having the (angle) A a right-angle. And let the perpendicular AD have been drawn. I say that the (rectangle contained) by CBD is equal to the (square) on BA, and the (rectangle contained) by BCD (is) equal to the (square) on CA, and the (rectangle contained) by BD and DC (is) equal to the (square) on AD, and, further, the (rectangle contained)

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ELEMENTS BOOK 10

τÁς ΒΑ.

by BC and AD [is] equal to the (rectangle contained) by BA and AC. And, first of all, (let us prove) that the (rectangle contained) by CBD [is] equal to the (square) on BA.

Α

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'Επεˆ γ¦ρ ™ν ÑρθογωνίJ τριγώνJ ¢πÕ τÁς ÑρθÁς γωνίας ™πˆ τ¾ν βάσιν κάθετος Ãκται ¹ Α∆, τ¦ ΑΒ∆, Α∆Γ ¥ρα τρίγωνα Óµοιά ™στι τù τε ÓλJ τù ΑΒΓ κሠ¢λλήλοις. κሠ™πεˆ Óµοιόν ™στι τÕ ΑΒΓ τρίγωνον τù ΑΒ∆ τριγώνJ, œστιν ¥ρα æς ¹ ΓΒ πρÕς τ¾ν ΒΑ, οÛτως ¹ ΒΑ πρÕς τ¾ν Β∆· τÕ ¥ρα ØπÕ τîν ΓΒ∆ ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ. ∆ι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ØπÕ τîν ΒΓ∆ ‡σον ™στˆ τù ¢πÕ τÁς ΑΓ. Κሠ™πεί, ™¦ν ™ν ÑρθογωνίJ τριγώνJ ¢πÕ τÁς ÑρθÁς γωνίας ™πˆ τ¾ν βάσιν κάθετος ¢χθÍ, ¹ ¢χθε‹σα τîν τÁς βάσεως τµηµάτων µέση ¢νάλογόν ™στιν, œστιν ¥ρα æς ¹ ΒΑ πρÕς τ¾ν ∆Α, οÛτως ¹ Α∆ πρÕς τ¾ν ∆Γ· τÕ ¥ρα ØπÕ τîν Β∆, ∆Γ ‡σον ™στˆ τù ¢πÕ τÁς ∆Α. Λέγω, Óτι κሠτÕ ØπÕ τîν ΒΓ, Α∆ ‡σον ™στˆ τù ØπÕ τîν ΒΑ, ΑΓ. ™πεˆ γ¦ρ, æς œφαµεν, Óµοιόν ™στι τÕ ΑΒΓ τù ΑΒ∆, œστιν ¥ρα æς ¹ ΒΓ πρÕς τ¾ν ΓΑ, οÛτως ¹ ΒΑ πρÕς τ¾ν Α∆. τÕ ¥ρα ØπÕ τîν ΒΓ, Α∆ ‡σον ™στˆ τù ØπÕ τîν ΒΑ, ΑΓ· Óπερ œδει δε‹ξαι.

For since AD has been drawn from the right-angle in a right-angled triangle, perpendicular to the base, ABD and ADC are thus triangles (which are) similar to the whole, ABC, and to one another [Prop. 6.8]. And since triangle ABC is similar to triangle ABD, thus as CB is to BA, so BA (is) to BD [Prop. 6.4]. Thus, the (rectangle contained) by CBD is equal to the (square) on AB [Prop. 6.17]. So, for the same (reasons), the (rectangle contained) by BCD is also equal to the (square) on AC. And since, if a (straight-line) is drawn from the rightangle in a right-angled triangle, perpendicular to the base, the (straight-line so) drawn is the mean proportional to the pieces of the base [Prop. 6.8 corr.], thus as BD is to DA, so AD (is) to DC. Thus, the (rectangle contained) by BD and DC is equal to the (square) on DA [Prop. 6.17]. I also say that the (rectangle contained) by BC and AD is equal to the (rectangle contained) by BA and AC. For since, as we said, ABC is similar to ABD, thus as BC is to CA, so BA (is) to AD [Prop. 6.4]. Thus, the (rectangle contained) by BC and AD is equal to the (rectangle contained) by BA and AC [Prop. 6.16]. (Which is) the very thing it was required to show.

λγ΄.

Proposition 33

ΕØρε‹ν δύο εÙθείας δυνάµει ¢συµµέτρους ποιούσας τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων ·ητόν, τÕ δ' Øπ' αÙτîν µέσον. 'Εκκείσθωσαν δύο ·ητሠδυνάµει µόνον σύµµετροι αƒ ΑΒ, ΒΓ, éστε τ¾ν µείζονα τ¾ν ΑΒ τÁς ™λάσσονος τÁς ΒΓ µείζον δύνασθαι τù ¢πÕ ¢συµµέτρου ˜αυτÍ, κሠτετµήσθω ¹ ΒΓ δίχα κατ¦ τÕ ∆, κሠτù ¢φ' Ðποτέρας τîν Β∆, ∆Γ ‡σον παρ¦ τ¾ν ΑΒ παραβεβλήσθω παραλ-

To find two straight-lines (which are) incommensurable in square, making the sum of the squares on them rational, and the (rectangle contained) by them medial. Let the two rational (straight-lines) AB and BC, (which are) commensurable in square only, be laid out such that the square on the greater, AB, is larger than (the square on) the lesser, BC, by the (square) on (some straight-line which is) incommensurable (in length) with

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ληλόγραµµον ™λλε‹πον ε‡δει τετραγώνJ, κሠœστω τÕ ØπÕ τîν ΑΕΒ, κሠγεγράφθω ™πˆ τÁς ΑΒ ηµικύκλιον τÕ ΑΖΒ, κሠ½χθω τÍ ΑΒ πρÕς Ñρθ¦ς ¹ ΕΖ, κሠ™πεζεύχθωσαν αƒ ΑΖ, ΖΒ.

Z A

E

(AB) [Prop. 10.30]. And let BC have been cut in half at D. And let a parallelogram equal to the (square) on either of BD or DC, (and) falling short by a square figure, have been applied to AB [Prop. 6.28], and let it be the (rectangle contained) by AEB. And let the semi-circle AF B have been drawn on AB. And let EF have been drawn at right-angles to AB. And let AF and F B have been joined.

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A

Κሠ™πεˆ [δύο] εÙθε‹αι ¥νισοί ε„σιν αƒ ΑΒ, ΒΓ, κሠ¹ ΑΒ τÁς ΒΓ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ, τù δ τετάρτJ τοà ¢πÕ τÁς ΒΓ, τουτέστι τù ¢πÕ τÁς ¹µισείας αÙτÁς, ‡σον παρ¦ τ¾ν ΑΒ παραβέβληται παραλληλόγραµµον ™λλε‹πον ε‡δει τετραγώνJ κሠποιε‹ τÕ ØπÕ τîν ΑΕΒ, ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΕ τÍ ΕΒ. καί ™στιν æς ¹ ΑΕ πρÕς ΕΒ, οÛτως τÕ ØπÕ τîν ΒΑ, ΑΕ πρÕς τÕ ØπÕ τîν ΑΒ, ΒΕ, ‡σον δ τÕ µν ØπÕ τîν ΒΑ, ΑΕ τù ¢πÕ τÁς ΑΖ, τÕ δ ØπÕ τîν ΑΒ, ΒΕ τù ¡πÕ τÁς ΒΖ· ¢σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΖ τù ¢πÕ τÁς ΖΒ· αƒ ΑΖ, ΖΒ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι. κሠ™πεˆ ¹ ΑΒ ·ητή ™στιν, ·ητÕν ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς ΑΒ· éστε κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΖ, ΖΒ ·ητόν ™στιν. κሠ™πεˆ πάλιν τÕ ØπÕ τîν ΑΕ, ΕΒ ‡σον ™στˆ τù ¢πÕ τÁς ΕΖ, Øπόκειται δ τÕ ØπÕ τîν ΑΕ, ΕΒ κሠτù ¢πÕ τÁς Β∆ ‡σον, ‡ση ¥ρα ™στˆν ¹ ΖΕ τÍ Β∆· διπλÁ ¥ρα ¹ ΒΓ τ¾ς ΖΕ· éστε κሠτÕ ØπÕ τîν ΑΒ, ΒΓ σύµµετρόν ™στι τù ØπÕ τîν ΑΒ, ΕΖ. µέσον δ τÕ ØπÕ τîν ΑΒ, ΒΓ· µέσον ¥ρα κሠτÕ ØπÕ τîν ΑΒ, ΕΖ. ‡σον δ τÕ ØπÕ τîν ΑΒ, ΕΖ τù ØπÕ τîν ΑΖ, ΖΒ· µέσον ¥ρα κሠτÕ ØπÕ τîν ΑΖ, ΖΒ. ™δείχθη δ κሠ·ητÕν τÕ συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων. ΕÛρηνται ¥ρα δύο εÙθε‹αι δυνάµει ¢σύµµετροι αƒ ΑΖ, ΖΒ ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων ·ητόν, τÕ δ Øπ' αÙτîν µέσον· Óπερ œδει δε‹ξαι.

E

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C

And since AB and BC are [two] unequal straightlines, and the square on AB is greater than (the square on) BC by the (square) on (some straight-line which is) incommensurable (in length) with (AB). And a parallelogram, equal to one quarter of the (square) on BC— that is to say, (equal) to the (square) on half of it—(and) falling short by a square figure, has been applied to AB, and makes the (rectangle contained) by AEB. AE is thus incommensurable (in length) with EB [Prop. 10.18]. And as AE is to EB, so the (rectangle contained) by BA and AE (is) to the (rectangle contained) by AB and EB. And the (rectangle contained) by BA and AE (is) equal to the (square) on AF , and the (rectangle contained) by AB and BE to the (square) on BF [Prop. 10.32 lem.]. The (square) on AF is thus incommensurable with the (square) on F B [Prop. 10.11]. Thus, AF and F B are incommensurable in square. And since AB is rational, the (square) on AB is also rational. Hence, the sum of the (squares) on AF and F B is also rational [Prop. 1.47]. And, again, since the (rectangle contained) by AE and EB is equal to the (square) on EF , and the (rectangle contained) by AE and EB was assumed (to be) equal to the (square) on BD, F E is thus equal to BD. Thus, BC is double F E. And hence the (rectangle contained) by AB and BC is commensurable with the (rectangle contained) by AB and EF [Prop. 10.6]. And the (rectangle contained) by AB and BC (is) medial [Prop. 10.21]. Thus, the (rectangle contained) by AB and EF (is) also medial [Prop. 10.23 corr.]. And the (rectangle contained) by AB and EF (is) equal to the (rectangle contained) by AF and F B [Prop. 10.32 lem.]. Thus, the (rectangle contained) by AF and F B (is) also medial. And the sum of the squares on them was also shown (to be) rational. Thus, the two straight-lines, AF and F B, (which are) incommensurable in square, have been found, making the sum of the squares on them rational, and the (rectan-

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ELEMENTS BOOK 10 gle contained) by them medial. (Which is) the very thing it was required to show.

†

AF and F B have lengths

footnote to Prop. 10.30.

q

[1 + k/(1 + k 2 )1/2 ]/2 and

q

[1 − k/(1 + k 2 )1/2 ]/2 times that of AB, respectively, where k is defined in the

λδ΄.

Proposition 34

ΕØρε‹ν δύο εÙθείας δυνάµει ¢συµµέτρους ποιούσας τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον, τÕ δ' Øπ' αÙτîν ·ητόν.

To find two straight-lines (which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them rational.

∆

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D

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A

'Εκκείσθωσαν δύο µέσαι δυνάµει µόνον σύµµετροι αƒ ΑΒ, ΒΓ ·ητÕν περιέχουσαι τÕ Øπ' αÙτîν, éστε τ¾ν ΑΒ τÁς ΒΓ µε‹ζον δύνασθαι τù ¢πÕ ¢συµµέτρου ˜αυτÍ, κሠγεγράφθω ™πˆ τÁς ΑΒ τÕ Α∆Β ¹µικύκλιον, κሠτετµήσθω ¹ ΒΓ δίχα κατ¦ τÕ Ε, κሠπαραβεβλήσθω παρ¦ τ¾ν ΑΒ τù ¢πÕ τÁς ΒΕ ‡σον παραλληλόγραµµον ™λλε‹πον ε‡δει τετραγώνJ τÕ ØπÕ τîν ΑΖΒ· ¢σύµµετρος ¥ρα [™στˆν] ¹ ΑΖ τÍ ΖΒ µήκει. κሠ½χθω ¢πÕ τοà Ζ τÍ ΑΒ πρÕς Ñρθ¦ς ¹ Ζ∆, κሠ™πεζεύχθωσαν αƒ Α∆, ∆Β. 'Επεˆ ¢σύµµετρός ™στιν ¹ ΑΖ τÍ ΖΒ, ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ØπÕ τîν ΒΑ, ΑΖ τù ØπÕ τîν ΑΒ, ΒΖ. ‡σον δ τÕ µν ØπÕ τîν ΒΑ, ΑΖ τù ¢πÕ τÁς Α∆, τÕ δ ØπÕ τîν ΑΒ, ΒΖ τù ¢πÕ τÁς ∆Β· ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς Α∆ τù ¢πÕ τÁς ∆Β. κሠ™πεˆ µέσον ™στˆ τÕ ¢πÕ τÁς ΑΒ, µέσον ¥ρα κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν Α∆, ∆Β. κሠ™πεˆ διπλÁ ™στιν ¹ ΒΓ τÁς ∆Ζ, διπλάσιον ¥ρα κሠτÕ ØπÕ τîν ΑΒ, ΒΓ τοà ØπÕ τîν ΑΒ, Ζ∆. ·ητÕν δ τÕ ØπÕ τîν ΑΒ, ΒΓ· ·ητÕν ¥ρα κሠτÕ ØπÕ τîν ΑΒ, Ζ∆. τÕ δ ØπÕ τîν ΑΒ, Ζ∆ ‡σον τù ØπÕ τîν Α∆, ∆Β· éστε κሠτÕ ØπÕ τîν Α∆, ∆Β ·ητόν ™στιν. ΕÛρηνται ¥ρα δύο εÙθε‹αι δυνάµει ¢σύµµετροι αƒ Α∆, ∆Β ποιοàσαι τÕ [µν] συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον, τÕ δ' Øπ' αÙτîν ·ητόν· Óπερ œδει δε‹ξαι.

F

B

E

C

Let the two medial (straight-lines) AB and BC, (which are) commensurable in square only, be laid out having the (rectangle contained) by them rational, (and) such that the square on AB is greater than (the square on) BC by the (square) on (some straight-line) incommensurable (in length) with (AB) [Prop. 10.31]. And let the semi-circle ADB have been drawn on AB. And let BC have been cut in half at E. And let a (rectangular) parallelogram equal to the (square) on BE, (and) falling short by a square figure, have been applied to AB, (and let it be) the (rectangle contained by) AF B [Prop. 6.28]. Thus, AF [is] incommensurable in length with F B [Prop. 10.18]. And let F D have been drawn from F at right-angles to AB. And let AD and DB have been joined. Since AF is incommensurable (in length) with F B, the (rectangle contained) by BA and AF is thus also incommensurable with the (rectangle contained) by AB and BF [Prop. 10.11]. And the (rectangle contained) by BA and AF (is) equal to the (square) on AD, and the (rectangle contained) by AB and BF to the (square) on DB [Prop. 10.32 lem.]. Thus, the (square) on AD is also incommensurable with the (square) on DB. And since the (square) on AB is medial, the sum of the (squares) on AD and DB (is) thus also medial [Props. 3.31, 1.47]. And since BC is double DF [see previous proposition], the (rectangle contained) by AB and BC (is) thus also double the (rectangle contained) by AB and F D. And the (rectangle contained) by AB and BC (is) rational. Thus, the (rectangle contained) by AB and F D (is) also rational [Prop. 10.6, Def. 10.4]. And the (rectangle contained) by AB and F D (is) equal to the (rectangle contained) by AD and DB [Prop. 10.32 lem.]. And hence the (rectangle contained) by AD and DB is rational.

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ELEMENTS BOOK 10 Thus, two straight-lines, AD and DB, (which are) incommensurable in square, have been found, making the sum of the squares on them medial, and the (rectangle contained) by them rational.† (Which is) the very thing it was required to show.

q q AD and DB have lengths [(1 + k 2 )1/2 + k]/[2 (1 + k 2 )] and [(1 + k 2 )1/2 − k]/[2 (1 + k 2 )] times that of AB, respectively, where k is defined in the footnote to Prop. 10.29. †

λε΄.

Proposition 35

ΕØρε‹ν δύο εÙθείας δυνάµει ¢συµµέτρους ποιούσας To find two straight-lines (which are) incommensuτό τε συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον rable in square, making the sum of the squares on them κሠτÕ Øπ' αÙτîν µέσον κሠœτι ¢σύµµετρον τù συγ- medial, and the (rectangle contained) by them medial, κειµένJ ™κ τîν ¢π' αÙτîν τετραγώνJ. and, moreover, incommensurable with the sum of the squares on them.

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Γ

A

'Εκκείσθωσαν δύο µέσαι δυνάµει µόνον σύµµετροι αƒ ΑΒ, ΒΓ µέσον περιέχουσαι, éστε τ¾ν ΑΒ τÁς ΒΓ µε‹ζον δύνασθαι τù ¢πÕ ¢συµµέτρου ˜αυτÍ, κሠγεγράφθω ™πˆ τÁς ΑΒ ¹µικύκλιον τÕ Α∆Β, κሠτ¦ λοιπ¦ γεγονέτω το‹ς ™πάνω еοίως. Κሠ™πεˆ ¢σύµµετρός ™στιν ¹ ΑΖ τÍ ΖΒ µήκει, ¢σύµµετρός ™στι κሠ¹ Α∆ τÍ ∆Β δυνάµει. κሠ™πεˆ µέσον ™στˆ τÕ ¢πÕ τÁς ΑΒ, µέσον ¥ρα κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν Α∆, ∆Β. κሠ™πεˆ τÕ ØπÕ τîν ΑΖ, ΖΒ ‡σον ™στˆ τù ¢φ' ˜κατέρας τîν ΒΕ, ∆Ζ, ‡ση ¥ρα ™στˆν ¹ ΒΕ τÍ ∆Ζ· διπλÁ ¥ρα ¹ ΒΓ τÁς Ζ∆· éστε κሠτÕ ØπÕ τîν ΑΒ, ΒΓ διπλάσιόν ™στι τοà ØπÕ τîν ΑΒ, Ζ∆. µέσον δ τÕ ØπÕ τîν ΑΒ, ΒΓ· µέσον ¥ρα κሠτÕ ØπÕ τîν ΑΒ, Ζ∆. καί ™στιν ‡σον τù ØπÕ τîν Α∆, ∆Β· µέσον ¥ρα κሠτÕ ØπÕ τîν Α∆, ∆Β. κሠ™πεˆ ¢σύµµετρός ™στιν ¹ ΑΒ τÍ ΒΓ µήκει, σύµµετρος δ ¹ ΓΒ τÍ ΒΕ, ¢σύµµετρος ¥ρα κሠ¹ ΑΒ τÍ ΒΕ µήκει· éστε κሠτÕ ¢πÕ τÁς ΑΒ τù ØπÕ τîν ΑΒ, ΒΕ ¢σύµµετρόν ™στιν. ¢λλ¦ τù µν ¢πÕ τÁς ΑΒ ‡σα ™στˆ τ¦ ¢πÕ τîν Α∆, ∆Β, τù δ ØπÕ τîν ΑΒ, ΒΕ ‡σον ™στˆ τÕ ØπÕ τîν ΑΒ, Ζ∆, τουτέστι τÕ ØπÕ τîν Α∆, ∆Β· ¢σύµµετρον ¥ρα ™στˆ τÕ συγκείµενον ™κ τîν ¢πÕ τîν Α∆, ∆Β τù ØπÕ τîν Α∆, ∆Β. ΕÛρηνται ¥ρα δύο εÙθε‹αι αƒ Α∆, ∆Β δυνάµει ¢σύµµετροι ποιοàσαι τό τε συγκείµενον ™κ τîν ¢π' αÙτîν µέσον κሠτÕ Øπ' αÙτîν µέσον κሠœτι ¢σύµµετρον τù συγκειµένJ ™κ τîν ¢π' αÙτîν τετραγώνων· Óπερ œδει δε‹ξαι.

F

B

E

C

Let the two medial (straight-lines) AB and BC, (which are) commensurable in square only, be laid out containing a medial (area), such that the square on AB is greater than (the square on) BC by the (square) on (some straight-line) incommensurable (in length) with (AB) [Prop. 10.32]. And let the semi-circle ADB have been drawn on AB. And let the remainder (of the figure) be generated similarly to the above (proposition). And since AF is incommensurable in length with F B [Prop. 10.18], AD is also incommensurable in square with DB [Prop. 10.11]. And since the (square) on AB is medial, the sum of the (squares) on AD and DB (is) thus also medial [Props. 3.31, 1.47]. And since the (rectangle contained) by AF and F B is equal to the (square) on each of BE and DF , BE is thus equal to DF . Thus, BC (is) double F D. And hence the (rectangle contained) by AB and BC is double the (rectangle) contained by AB and F D. And the (rectangle contained) by AB and BC (is) medial. Thus, the (rectangle contained) by AB and F D (is) also medial. And it is equal to the (rectangle contained) by AD and DB [Prop. 10.32 lem.]. Thus, the (rectangle contained) by AD and DB (is) also medial. And since AB is incommensurable in length with BC, and CB (is) commensurable (in length) with BE, AB (is) thus also incommensurable in length with BE [Prop. 10.13]. And hence the (square) on AB is also incommensurable with the (rectangle contained) by AB and BE [Prop. 10.11]. But the (sum of the squares) on AD and DB is equal to the (square) on AB [Prop. 1.47].

320

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 And the (rectangle contained) by AB and F D—that is to say, the (rectangle contained) by AD and DB—is equal to the (rectangle contained) by AB and BE. Thus, the sum of the (squares) on AD and DB is incommensurable with the (rectangle contained) by AD and DB. Thus, two straight-lines, AD and DB, (which are) incommensurable in square, have been found, making the sum of the (squares) on them medial, and the (rectangle contained) by them medial, and, moreover, incommensurable with the sum of the squares on them.† (Which is) the very thing it was required to show.

†

AD and DB have lengths k ′1/4

q

[1 + k/(1 + k 2 )1/2 ]/2 and k ′1/4

defined in the footnote to Prop. 10.32.

q

[1 − k/(1 + k 2 )1/2 ]/2 times that of AB, respectively, where k and k ′ are

λ$΄.

Proposition 36

'Ε¦ν δύο ·ητሠδυνάµει µόνον σύµµετροι συντεθîσιν, ¹ Óλη ¥λογός ™στιν, καλείσθω δ ™κ δύο Ñνοµάτων.

If two rational (straight-lines, which are) commensurable in square only, are added together, then the whole (straight-line) is irrational—let it be called a binomial (straight-line).†

Α

Β

Γ

A

Συγκείσθωσαν γ¦ρ δύο ·ητሠδυνάµει µόνον σύµµετροι αƒ ΑΒ, ΒΓ· λέγω, Óτι Óλη ¹ ΑΓ ¥λογός ™στιν. 'Επεˆ γ¦ρ ¢σύµµετρός ™στιν ¹ ΑΒ τÍ ΒΓ µήκει· δυνάµει γ¦ρ µόνον ε„σˆ σύµµετροι· æς δ ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ØπÕ τîν ΑΒΓ πρÕς τÕ ¢πÕ τÁς ΒΓ, ¢σύµµετρον ¥ρα ™στˆ τÕ ØπÕ τîν ΑΒ, ΒΓ τù ¢πÕ τÁς ΒΓ. ¢λλ¦ τù µν ØπÕ τîν ΑΒ, ΒΓ σύµµετρόν ™στι τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ, τù δ ¢πÕ τÁς ΒΓ σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΒ, ΒΓ· αƒ γ¦ρ ΑΒ, ΒΓ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢σύµµετρον ¥ρα ™στˆ τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ το‹ς ¢πÕ τîν ΑΒ, ΒΓ. κሠσυνθέντι τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ µετ¦ τîν ¢πÕ τîν ΑΒ, ΒΓ, τουτέστι τÕ ¢πÕ τÁς ΑΓ, ¢σύµµετρόν ™στι τù συγκειµένJ ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ· ·ητÕν δ τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ· ¥λογον ¥ρα [™στˆ] τÕ ¢πÕ τÁς ΑΓ· éστε κሠ¹ ΑΓ ¥λογός ™στιν, καλείσθω δ ™κ δύο Ñνοµάτων· Óπερ œδει δε‹ξαι.

†

B

C

For let the two rational (straight-lines), AB and BC, (which are) commensurable in square only, be laid down together. I say that the whole (straight-line), AC, is irrational. For since AB is incommensurable in length with BC—for they are commensurable in square only—and as AB (is) to BC, so the (rectangle contained) by ABC (is) to the (square) on BC, the (rectangle contained) by AB and BC is thus incommensurable with the (square) on BC [Prop. 10.11]. But, twice the (rectangle contained) by AB and BC is commensurable with the (rectangle contained) by AB and BC [Prop. 10.6]. And (the sum of) the (squares) on AB and BC is commensurable with the (square) on BC—for the rational (straight-lines) AB and BC are commensurable in square only [Prop. 10.15]. Thus, twice the (rectangle contained) by AB and BC is incommensurable with (the sum of) the (squares) on AB and BC [Prop. 10.13]. And, via composition, twice the (rectangle contained) by AB and BC, plus (the sum of) the (squares) on AB and BC—that is to say, the (square) on AC [Prop. 2.4]—is incommensurable with the sum of the (squares) on AB and BC [Prop. 10.16]. And the sum of the (squares) on AB and BC (is) rational. Thus, the (square) on AC [is] irrational [Def. 10.4]. Hence, AC is also irrational [Def. 10.4]—let it be called a binomial (straight-line).‡ (Which is) the very thing it was required to show.

Literally, “from two names”.

321

ΣΤΟΙΧΕΙΩΝ ι΄. ‡

ELEMENTS BOOK 10

Thus, a binomial straight-line has a length expressible as 1 + k 1/2 [or, more generally, ρ (1 + k 1/2 ), where ρ is rational—the same proviso

applies to the definitions in the following propositions]. The binomial and the corresponding apotome, whose length is expressible as 1 − k 1/2 (see Prop. 10.73), are the positive roots of the quartic x4 − 2 (1 + k) x2 + (1 − k)2 = 0.

λζ΄.

Proposition 37

'Ε¦ν δύο µέσαι δυνάµει µόνον σύµµετροι συντεθîσι If two medial (straight-lines), commensurable in ·ητÕν περιέχουσαι, ¹ Óλη ¥λογός ™στιν, καλείσθω δ ™κ square only, which contain a rational (area), are added δύο µέσων πρώτη. together, then the whole (straight-line) is irrational—let it be called a first bimedial (straight-line).†

Α

Β

Γ

A

Συγκείσθωσαν γ¦ρ δύο µέσαι δυνάµει µόνον σύµµετροι αƒ ΑΒ, ΒΓ ·ητÕν περιέχουσαι· λέγω, Óτι Óλη ¹ ΑΓ ¥λογός ™στιν. 'Επεˆ γ¦ρ ¢σύµµετρός ™στιν ¹ ΑΒ τÍ ΒΓ µήκει, κሠτ¦ ¢πÕ τîν ΑΒ, ΒΓ ¥ρα ¢σύµµετρά ™στι τù δˆς ØπÕ τîν ΑΒ, ΒΓ· κሠσυνθέντι τ¦ ¢πÕ τîν ΑΒ, ΒΓ µετ¦ τοà δˆς ØπÕ τîν ΑΒ, ΒΓ, Óπερ ™στˆ τÕ ¢πÕ τÁς ΑΓ, ¢σύµµετρόν ™στι τù ØπÕ τîν ΑΒ, ΒΓ. ·ητÕν δ τÕ ØπÕ τîν ΑΒ, ΒΓ· Øπόκεινται γ¦ρ αƒ ΑΒ, ΒΓ ·ητÕν περιέχουσαι· ¥λογον ¥ρα τÕ ¢πÕ τÁς ΑΓ· ¥λογος ¥ρα ¹ ΑΓ, καλείσθω δ ™κ δύο µέσων πρώτη· Óπερ œδει δε‹ξαι.

†

B

C

For let the two medial (straight-lines), AB and BC, commensurable in square only, (and) containing a rational (area), be laid down together. I say that the whole (straight-line), AC, is irrational. For since AB is incommensurable in length with BC, (the sum of) the (squares) on AB and BC is thus also incommensurable with twice the (rectangle contained) by AB and BC [see previous proposition]. And, via composition, (the sum of) the (squares) on AB and BC, plus twice the (rectangle contained) by AB and BC— that is, the (square) on AC [Prop. 2.4]—is incommensurable with the (rectangle contained) by AB and BC [Prop. 10.16]. And the (rectangle contained) by AB and BC (is) rational—for AB and BC were assumed to enclose a rational (area). Thus, the (square) on AC (is) irrational. Thus, AC (is) irrational [Def. 10.4]—let it be called a first bimedial (straight-line).‡ (Which is) the very thing it was required to show.

Literally, “first from two medials”.

‡

Thus, a first bimedial straight-line has a length expressible as k 1/4 + k 3/4 . The first bimedial and the corresponding first apotome of a medial, p whose length is expressible as k 1/4 −k 3/4 (see Prop. 10.74), are the positive roots of the quartic x4 −2 book10eps/k (1+k) x2 +k (1−k)2 = 0.

λη΄.

Proposition 38

'Ε¦ν δύο µέσαι δυνάµει µόνον σύµµετροι συντεθîσι If two medial (straight-lines), commensurable in square µέσον περιέχουσαι, ¹ Óλη ¥λογός ™στιν, καλείσθω δ ™κ only, which contain a medial (area), are added together, δύο µέσων δυετέρα. then the whole (straight-line) is irrational—let it be called a second bimedial (straight-line).

Α ∆

Ε

Β

Γ Θ

A Η

D

Ζ

E 322

B

C H

G

F

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

Συγκείσθωσαν γ¦ρ δύο µέσαι δυνάµει µόνον σύµµετροι αƒ ΑΒ, ΒΓ µέσον περιέχουσαι· λέγω, Óτι ¥λογός ™στιν ¹ ΑΓ. 'Εκκείσθω γ¦ρ ·ητ¾ ¹ ∆Ε, κሠτù ¢πÕ τÁς ΑΓ ‡σον παρ¦ τ¾ν ∆Ε παραβεβλήσθω τÕ ∆Ζ πλάτος ποιοàν τ¾ν ∆Η. κሠ™πεˆ τÕ ¢πÕ τÁς ΑΓ ‡σον ™στˆ το‹ς τε ¢πÕ τîν ΑΒ, ΒΓ κሠτù δˆς ØπÕ τîν ΑΒ, ΒΓ, παραβεβλήσθω δ¾ το‹ς ¢πÕ τîν ΑΒ, ΒΓ παρ¦ τ¾ν ∆Ε ‡σον τÕ ΕΘ· λοιπÕν ¥ρα τÕ ΘΖ ‡σον ™στˆ τù δˆς ØπÕ τîν ΑΒ, ΒΓ. κሠ™πεˆ µέση ™στˆν ˜κατέρα τîν ΑΒ, ΒΓ, µέσα ¥ρα ™στˆ κሠτ¦ ¢πÕ τîν ΑΒ, ΒΓ. µέσον δ Øπόκειται κሠτÕ δˆς ØπÕ τîν ΑΒ, ΒΓ. καί ™στι το‹ς µν ¢πÕ τîν ΑΒ, ΒΓ ‡σον τÕ ΕΘ, τù δ δˆς ØπÕ τîν ΑΒ, ΒΓ ‡σον τÕ ΖΘ· µέσον ¥ρα ˜κάτερον τîν ΕΘ, ΘΖ. κሠπαρ¦ ·ητ¾ν τ¾ν ∆Ε παράκειται· ·ητ¾ ¥ρα ™στˆν ˜κατέρα τîν ∆Θ, ΘΗ κሠ¢σύµµετρος τÍ ∆Ε µήκει. ™πεˆ οâν ¢σύµµετρός ™στιν ¹ ΑΒ τÍ ΒΓ µήκει, καί ™στιν æς ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ØπÕ τîν ΑΒ, ΒΓ, ¢σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΒ τù ØπÕ τîν ΑΒ, ΒΓ. ¢λλ¦ τù µν ¢πÕ τÁς ΑΒ σύµµετρόν ™στι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ τετραγώνων, τù δ ØπÕ τîν ΑΒ, ΒΓ σύµµετρόν ™στι τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ. ¢σύµµετρον ¥ρα ™στˆ τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ τù δˆς ØπÕ τîν ΑΒ, ΒΓ. ¢λλ¦ το‹ς µν ¢πÕ τîν ΑΒ, ΒΓ ‡σον ™στˆ τÕ ΕΘ, τù δ δˆς ØπÕ τîν ΑΒ, ΒΓ ‡σον ™στˆ τÕ ΘΖ. ¢σύµµετρον ¥ρα ™στˆ τÕ ΕΘ τù ΘΖ· éστε κሠ¹ ∆Θ τÍ ΘΗ ™στιν ¢σύµµετρος µήκει. αƒ ∆Θ, ΘΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. éστε ¹ ∆Η ¥λογός ™στιν. ·ητ¾ δ ¹ ∆Ε· τÕ δ ØπÕ ¢λόγου κሠ·ητÁς περιεχόµενον Ñρθογώνιον ¥λογόν ™στιν· ¥λογον ¥ρα ™στˆ τÕ ∆Ζ χωρίον, κሠ¹ δυναµένη [αÙτÕ] ¥λογός ™στιν. δύναται δ τÕ ∆Ζ ¹ ΑΓ· ¥λογος ¥ρα ™στˆν ¹ ΑΓ, καλείσθω δ ™κ δύο µέσων δευτέρα. Óπερ œδει δε‹ξαι.

For let the two medial (straight-lines), AB and BC, commensurable in square only, (and) containing a medial (area), be laid down together [Prop. 10.28]. I say that AC is irrational. For let the rational (straight-line) DE be laid down, and let (the rectangle) DF , equal to the (square) on AC, have been applied to DE, making DG as breadth [Prop. 1.44]. And since the (square) on AC is equal to (the sum of) the (squares) on AB and BC, plus twice the (rectangle contained) by AB and BC [Prop. 2.4], so let (the rectangle) EH, equal to (the sum of) the squares on AB and BC, have been applied to DE. The remainder HF is thus equal to twice the (rectangle contained) by AB and BC. And since AB and BC are each medial, (the sum of) the squares on AB and BC is thus also medial.‡ And twice the (rectangle contained) by AB and BC was also assumed (to be) medial. And EH is equal to (the sum of) the squares on AB and BC, and F H (is) equal to twice the (rectangle contained) by AB and BC. Thus, EH and HF (are) each medial. And they were applied to the rational (straight-line) DE. Thus, DH and HG are each rational, and incommensurable in length with DE [Prop. 10.22]. Therefore, since AB is incommensurable in length with BC, and as AB is to BC, so the (square) on AB (is) to the (rectangle contained) by AB and BC [Prop. 10.21 lem.], the (square) on AB is thus incommensurable with the (rectangle contained) by AB and BC [Prop. 10.11]. But, the sum of the squares on AB and BC is commensurable with the (square) on AB [Prop. 10.15], and twice the (rectangle contained) by AB and BC is commensurable with the (rectangle contained) by AB and BC [Prop. 10.6]. Thus, the sum of the (squares) on AB and BC is incommensurable with twice the (rectangle contained) by AB and BC [Prop. 10.13]. But, EH is equal to (the sum of) the squares on AB and BC, and HF is equal to twice the (rectangle) contained by AB and BC. Thus, EH is incommensurable with HF . Hence, DH is also incommensurable in length with HG [Props. 6.1, 10.11]. Thus, DH and HG are rational (straight-lines which are) commensurable in square only. Hence, DG is irrational [Prop. 10.36]. And DE (is) rational. And the rectangle contained by irrational and rational (straight-lines) is irrational [Prop. 10.20]. The area DF is thus irrational, and (so) the square-root [of it] is irrational [Def. 10.4]. And AC is the square-root of DF . AC is thus irrational—let it be called a second bimedial (straight-line).§ (Which is) the very thing it was required to show.

†

Literally, “second from two medials”.

‡

Since, by hypothesis, the squares on AB and BC are commensurable—see Props. 10.15, 10.23.

323

ΣΤΟΙΧΕΙΩΝ ι΄. §

ELEMENTS BOOK 10

Thus, a second bimedial straight-line has a length expressible as k 1/4 + k ′1/2 /k 1/4 . The second bimedial and the corresponding second

apotome of a medial, whose length is expressible as k 1/4 − k ′1/2 /k 1/4 (see Prop. 10.75), are the positive roots of the quartic x4 − 2 [(k + p k ′ )/ book10eps/k] x2 + [(k − k ′ )2 /k] = 0.

λθ΄.

Proposition 39

'Ε¦ν δύο εÙθε‹αι δυνάµει ¢σύµµετροι συντεθîσι If two straight-lines (which are) incommensurable in ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τε- square, making the sum of the squares on them rational, τραγώνων ·ητόν, τÕ δ' Øπ' αÙτîν µέσον, ¹ Óλη εÙθε‹α and the (rectangle contained) by them medial, are added ¥λογός ™στιν, καλείσθω δ µείζων. together, then the whole straight-line is irrational—let it be called a major (straight-line).

Α

Β

Γ

A

Συγκείσθωσαν γ¦ρ δύο εÙθε‹αι δυνάµει ¢σύµµετροι αƒ ΑΒ, ΒΓ ποιοàσαι τ¦ προκείµενα· λέγω, Óτι ¥λογός ™στιν ¹ ΑΓ. 'Επεˆ γ¦ρ τÕ ØπÕ τîν ΑΒ, ΒΓ µέσον ™στίν, κሠτÕ δˆς [¥ρα] ØπÕ τîν ΑΒ, ΒΓ µέσον ™στίν. τÕ δ συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ ·ητόν· ¢σύµµετρον ¥ρα ™στˆ τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ τù συγκειµένJ ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ· éστε κሠτ¦ ¢πÕ τîν ΑΒ, ΒΓ µετ¦ τοà δˆς ØπÕ τîν ΑΒ, ΒΓ, Óπερ ™στˆ τÕ ¢πÕ τÁς ΑΓ, ¢σύµµετρόν ™στι τù συγκειµένJ ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ [·ητÕν δ τÕ συγµείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ]· ¥λογον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΓ. éστε κሠ¹ ΑΓ ¥λογός ™στιν, καλείσθω δ µείζων. Óπερ œδει δε‹ξαι.

B

C

For let the two straight-lines, AB and BC, incommensurable in square, and fulfilling the prescribed (conditions), be laid down together [Prop. 10.33]. I say that AC is irrational. For since the (rectangle contained) by AB and BC is medial, twice the (rectangle contained) by AB and BC is [thus] also medial [Props. 10.6, 10.23 corr.]. And the sum of the (squares) on AB and BC (is) rational. Thus, twice the (rectangle contained) by AB and BC is incommensurable with the sum of the (squares) on AB and BC [Def. 10.4]. Hence, (the sum of) the squares on AB and BC, plus twice the (rectangle contained) by AB and BC—that is, the (square) on AC [Prop. 2.4]—is also incommensurable with the sum of the (squares) on AB and BC [Prop. 10.16] [and the sum of the (squares) on AB and BC (is) rational]. Thus, the (square) on AC is irrational. Hence, AC is also irrational [Def. 10.4]—let it be called a major (straight-line).† (Which is) the very thing it was required to show.

q q Thus, a major straight-line has a length expressible as [1 + k/(1 + k 2 )1/2 ]/2 + [1 − k/(1 + k 2 )1/2 ]/2. The major and the corresponding q q minor, whose length is expressible as [1 + k/(1 + k 2 )1/2 ]/2 − [1 − k/(1 + k 2 )1/2 ]/2 (see Prop. 10.76), are the positive roots of the quartic x4 − 2 x2 + k 2 /(1 + k 2 ) = 0.

†

µ΄.

Proposition 40

'Ε¦ν δύο εÙθε‹αι δυνάµει ¢σύµµετροι συντεθîσι If two straight-lines (which are) incommensurable ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τε- in square, making the sum of the squares on them τραγώνων µέσον, τÕ δ' Øπ' αÙτîν ·ητόν, ¹ Óλη εÙθε‹α medial, and the (rectangle contained) by them ratio¥λογός ™στιν, καλείσθω δ ·ητÕν κሠµέσον δυναµένη. nal, are added together, then the whole straight-line is irrational—let it be called the square-root of a rational plus a medial (area).

Α

Β

Γ

A

Συγκείσθωσαν γ¦ρ δύο εÙθε‹αι δυνάµει ¢σύµµετροι αƒ ΑΒ, ΒΓ ποιοàσαι τ¦ προκείµενα· λέγω, Óτι ¥λογός ™στιν ¹ ΑΓ. 'Επεˆ γ¦ρ τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ,

B

C

For let the two straight-lines, AB and BC, incommensurable in square, (and) fulfilling the prescribed (conditions), be laid down together [Prop. 10.34]. I say that AC is irrational.

324

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

ΒΓ µέσον ™στίν, τÕ δ δˆς ØπÕ τîν ΑΒ, ΒΓ ·ητόν, ¢σύµµετρον ¥ρα ™στˆ τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ τù δˆς ØπÕ τîν ΑΒ, ΒΓ· éστε κሠτÕ ¡πÕ τÁς ΑΓ ¢σύµµετρόν ™στι τù δˆς ØπÕ τîν ΑΒ, ΒΓ. ·ητÕν δ τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ· ¥λογον ¥ρα τÕ ¢πÕ τÁς ΑΓ. ¥λογος ¥ρα ¹ ΑΓ, καλείσθω δ ·ητÕν κሠµέσον δυναµένη. Óπερ œδει δε‹ξαι.

For since the sum of the (squares) on AB and BC is medial, and twice the (rectangle contained) by AB and BC (is) rational, the sum of the (squares) on AB and BC is thus incommensurable with twice the (rectangle contained) by AB and BC. Hence, the (square) on AC is also incommensurable with twice the (rectangle contained) by AB and BC [Prop. 10.16]. And twice the (rectangle contained) by AB and BC (is) rational. The (square) on AC (is) thus irrational. Thus, AC (is) irrational [Def. 10.4]—let it be called the square-root of a rational plus a medial (area).† (Which is) the very thing it was required to show. q

q [(1 + k 2 )1/2 + k]/[2 (1 + k 2 )]+ [(1 + k 2 )1/2 − k]/[2 (1 + k 2 )]. q q This and the corresponding irrational with a minus sign, whose length is expressible as [(1 + k 2 )1/2 + k]/[2 (1 + k 2 )]− [(1 + k 2 )1/2 − k]/[2 (1 + k 2 )] √ (see Prop. 10.77), are the positive roots of the quartic x4 − (2/ 1 + k 2 ) x2 + k 2 /(1 + k 2 )2 = 0.

†

Thus, the square-root of a rational plus a medial (area) has a length expressible as

µα΄.

Proposition 41

'Ε¦ν δύο εÙθε‹αι δυνάµει ¢σύµµετροι συντεθîσι ποιοàσαι τό τε συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον κሠτÕ Øπ' αÙτîν µέσον κሠœτι ¢σύµµετρον τù συγκειµένJ ™κ τîν ¢π' αÙτîν τετραγώνων, ¹ Óλη εÙθε‹α ¥λογός ™στιν, καλείσθω δ δύο µέσα δυναµένη.

If two straight-lines (which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them medial, and, moreover, incommensurable with the sum of the squares on them, are added together, then the whole straight-line is irrational—let it be called the square-root of (the sum of) two medial (areas).

Κ

Θ

Α Β

Γ

K

H

G

F

D

E

A Η

Ζ

∆

Ε

B

C

Συγκείσθωσαν γ¦ρ δύο εÙθε‹αι δυνάµει ¢σύµµετροι αƒ ΑΒ, ΒΓ ποιοàσαι τ¦ προκείµενα· λέγω, Óτι ¹ ΑΓ ¥λογός ™στιν. 'Εκκείσθω ·ητ¾ ¹ ∆Ε, κሠπαραβεβλήσθω παρ¦ τ¾ν ∆Ε το‹ς µν ¢πÕ τîν ΑΒ, ΒΓ ‡σον τÕ ∆Ζ, τù δ δˆς ØπÕ τîν ΑΒ, ΒΓ ‡σον τÕ ΗΘ· Óλον ¥ρα τÕ ∆Θ ‡σον ™στˆ τù ¢πÕ τÁς ΑΓ τετραγώνJ. κሠ™πεˆ µέσον ™στˆ τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ, καί ™στιν ‡σον τù ∆Ζ, µέσον ¥ρα ™στˆ κሠτÕ ∆Ζ. κሠπαρ¦ ·ητ¾ν τ¾ν ∆Ε παράκειται· ·ητ¾ ¥ρα ™στˆν ¹ ∆Η κሠ¢σύµµετρος τÍ ∆Ε µήκει. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΗΚ ·ητή ™στι καˆ

For let the two straight-lines, AB and BC, incommensurable in square, (and) fulfilling the prescribed (conditions), be laid down together [Prop. 10.35]. I say that AC is irrational. Let the rational (straight-line) DE be laid out, and let (the rectangle) DF , equal to (the sum of) the (squares) on AB and BC, and (the rectangle) GH, equal to twice the (rectangle contained) by AB and BC, have been applied to DE. Thus, the whole of DH is equal to the square on AC [Prop. 2.4]. And since the sum of the (squares) on AB and BC is medial, and is equal to DF ,

325

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ELEMENTS BOOK 10

¢σύµµετρος τÍ ΗΖ, τουτέστι τÍ ∆Ε, µήκει. κሠ™πεˆ ¢σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΒ, ΒΓ τù δˆς ØπÕ τîν ΑΒ, ΒΓ, ¢σύµµετρόν ™στι τÕ ∆Ζ τù ΗΘ· éστε κሠ¹ ∆Η τÍ ΗΚ ¢σύµµετρός ™στιν. κሠε„σι ·ηταί· αƒ ∆Η, ΗΚ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¥λογος ¥ρα ™στˆν ¹ ∆Κ ¹ καλουµένη ™κ δύο Ñνοµάτων. ·ητ¾ δ ¹ ∆Ε· ¥λογον ¥ρα ™στˆ τÕ ∆Θ κሠ¹ δυναµένη αÙτÕ ¥λογός ™στιν. δύναται δ τÕ Θ∆ ¹ ΑΓ· ¥λογος ¥ρα ™στˆν ¹ ΑΓ, καλείσθω δ δύο µέσα δυναµένη. Óπερ œδει δε‹ξαι.

DF is thus also medial. And it is applied to the rational (straight-line) DE. Thus, DG is rational, and incommensurable in length with DE [Prop. 10.22]. So, for the same (reasons), GK is also rational, and incommensurable in length with GF —that is to say, DE. And since (the sum of) the (squares) on AB and BC is incommensurable with twice the (rectangle contained) by AB and BC, DF is incommensurable with GH. Hence, DG is also incommensurable (in length) with GK [Props. 6.1, 10.11]. And they are rational. Thus, DG and GK are rational (straight-lines which are) commensurable in square only. Thus, DK is irrational, and that (straight-line which is) called binomial [Prop. 10.36]. And DE (is) rational. Thus, DH is irrational, and its square-root is irrational [Def. 10.4]. And AC (is) the square-root of HD. Thus, AC is irrational—let it be called the square-root of (the sum of) two medial (areas).† (Which is) the very thing it was required to show.

« „q q [1 + k/(1 + k 2 )1/2 ]/2 + [1 − k/(1 + k 2 )1/2 ]/2 . « „q q [1 + k/(1 + k 2 )1/2 ]/2 − [1 − k/(1 + k 2 )1/2 ]/2 This and the corresponding irrational with a minus sign, whose length is expressible as k ′1/4

†

Thus, the square-root of (the sum of) two medial (areas) has a length expressible as k ′1/4

(see Prop. 10.78), are the positive roots of the quartic x4 − 2 k ′1/2 x2 + k ′ k 2 /(1 + k 2 ) = 0.

ΛÁµµα.

Lemma

“Οτι δ αƒ ε„ρηµέναι ¥λογοι µοναχîς διαιροàνται ε„ς τ¦ς εÙθείας, ™ξ ïν σύγκεινται ποιουσîν τ¦ προκείµενα ε‡δη, δείξοµεν ½δη προεκθέµενοι ληµµάτιον τοιοàτον·

We will now demonstrate that the aforementioned irrational (straight-lines) are uniquely divided into the straight-lines of which they are the sum, and which produce the prescribed types, (after) setting forth the following lemma.

Α

∆ Ε

Γ

Β

A

'Εκκείσθω εÙθε‹α ¹ ΑΒ κሠτετµήσθω ¹ Óλη ε„ς ¥νισα καθ' ˜κάτερον τîν Γ, ∆, Øποκείσθω δ µείζων ¹ ΑΓ τÁς ∆Β· λέγω, Óτι τ¦ ¢πÕ τîν ΑΓ, ΓΒ µείζονά ™στι τîν ¢πÕ τîν Α∆, ∆Β. Τετµήσθω γ¦ρ ¹ ΑΒ δίχα κατ¦ τÕ Ε. κሠ™πεˆ µείζων ™στˆν ¹ ΑΓ τÁς ∆Β, κοιν¾ ¢φVρήσθω ¹ ∆Γ· λοιπ¾ ¥ρα ¹ Α∆ λοιπÁς τÁς ΓΒ µείζων ™στίν. ‡ση δ ¹ ΑΕ τÍ ΕΒ· ™λάττων ¥ρα ¹ ∆Ε τÁς ΕΓ· τ¦ Γ, ∆ ¥ρα σηµε‹α οÙκ ‡σον ¢πέχουσι τÁς διχοτοµίας. κሠ™πεˆ τÕ ØπÕ τîν ΑΓ, ΓΒ µετ¦ τοà ¢πÕ τÁς ΕΓ ‡σον ™στˆ τù ¢πÕ τÁς ΕΒ, ¢λλ¦ µ¾ν κሠτÕ ØπÕ τîν Α∆, ∆Β µετ¦ τοà ¢πÕ ∆Ε ‡σον ™στˆ τù ¢πÕ τÁς ΕΒ, τÕ ¥ρα ØπÕ τîν ΑΓ, ΓΒ µετ¦ τοà ¢πÕ τÁς ΕΓ ‡σον ™στˆ τù ØπÕ τîν Α∆, ∆Β µετ¦ τοà ¢πÕ τÁς ∆Ε· ïν τÕ ¢πÕ τÁς ∆Ε œλασσόν ™στι τοà ¢πÕ τ¾ς ΕΓ· κሠλοιπÕν ¥ρα τÕ ØπÕ τîν ΑΓ, ΓΒ œλασσόν ™στι τοà ØπÕ τîν Α∆, ∆Β. éστε κሠτÕ δˆς ØπÕ τîν ΑΓ, ΓΒ œλασσόν ™στι τοà δˆς ØπÕ τîν Α∆, ∆Β. κሠλοιπÕν ¥ρα τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ µε‹ζόν ™στι τοà

D E

C

B

Let the straight-line AB be laid out, and let the whole (straight-line) have been cut into unequal parts at each of the (points) C and D. And let AC be assumed (to be) greater than DB. I say that (the sum of) the (squares) on AC and CB is greater than (the sum of) the (squares) on AD and DB. For let AB have been cut in half at E. And since AC is greater than DB, let DC have been subtracted from both. Thus, the remainder AD is greater than the remainder CB. And AE (is) equal to EB. Thus, DE (is) less than EC. Thus, points D and C are not equally far from the point of bisection. And since the (rectangle contained) by AC and CB, plus the (square) on EC, is equal to the (square) on EB [Prop. 2.5], but, moreover, the (rectangle contained) by AD and DB, plus the (square) on DE, is also equal to the (square) on EB [Prop. 2.5], the (rectangle contained) by AC and CB, plus the (square) on EC, is thus equal to the (rectangle contained) by AD and

326

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ELEMENTS BOOK 10

συγκειµένου ™κ τîν ¢πÕ τîν Α∆, ∆Β. Óπερ œδει δε‹ξαι.

†

DB, plus the (square) on DE. And, of these, the (square) on DE is less than the (square) on EC. And, thus, the remaining (rectangle contained) by AC and CB is less than the (rectangle contained) by AD and DB. And, hence, twice the (rectangle contained) by AC and CB is less than twice the (rectangle contained) by AD and DB. And thus the remaining sum of the (squares) on AC and CB is greater than the sum of the (squares) on AD and DB.† (Which is) the very thing it was required to show.

Since, AC 2 + CB 2 + 2 AC CB = AD 2 + DB 2 + 2 AD DB = AB 2 .

µβ΄.

Proposition 42

`Η ™κ δύο Ñνοµάτων κατ¦ žν µόνον σηµε‹ον διαιρε‹ται ε„ς τ¦ Ñνόµατα.

A binomial (straight-line) can be divided into its (component) terms at one point only.†

Α

∆

Γ

Β

A

”Εστω ™κ δύο Ñνοµάτων ¹ ΑΒ διVρηµένη ε„ς τ¦ Ñνόµατα κατ¦ τÕ Γ· αƒ ΑΓ, ΓΒ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. λέγω, Óτι ¹ ΑΒ κατ' ¥λλο σηµε‹ον οÙ διαιρε‹ται ε„ς δύο ·ητ¦ς δυνάµει µόνον συµµέτρους. Ε„ γ¦ρ δυνατόν, διVρήσθω κሠκατ¦ τÕ ∆, éστε κሠτ¦ς Α∆, ∆Β ·ητ¦ς εναι δυνάµει µόνον συµµέτρους. φανερÕν δή, Óτι ¹ ΑΓ τÍ ∆Β οÙκ œστιν ¹ αÙτή. ε„ γ¦ρ δυνατόν, œστω. œσται δ¾ κሠ¹ Α∆ τÍ ΓΒ ¹ αÙτή· κሠœσται æς ¹ ΑΓ πρÕς τ¾ν ΓΒ, οÛτως ¹ Β∆ πρÕς τ¾ν ∆Α, κሠœσται ¹ ΑΒ κατ¦ τÕ αÙτÕ τÍ κατ¦ τÕ Γ διαιρέσει διαιρεθε‹σα κሠκατ¦ τÕ ∆· Óπερ οÙχ Øπόκειται. οÙκ ¥ρα ¹ ΑΓ τÍ ∆Β ™στιν ¹ αÙτή. δι¦ δ¾ τοàτο κሠτ¦ Γ, ∆ σηµε‹α οÙκ ‡σον ¢πέχουσι τÁς διχοτοµίας. ú ¥ρα διαφέρει τ¦ ¢πÕ τîν ΑΓ, ΓΒ τîν ¢πÕ τîν Α∆, ∆Β, τούτJ διαφέρει κሠτÕ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ δι¦ τÕ κሠτ¦ ¢πÕ τîν ΑΓ, ΓΒ µετ¦ τοà δˆς ØπÕ τîν ΑΓ, ΓΒ κሠτ¦ ¢πÕ τîν Α∆, ∆Β µετ¦ τοà δˆς ØπÕ τîν Α∆, ∆Β ‡σα εναι τù ¢πÕ τÁς ΑΒ. ¢λλ¦ τ¦ ¢πÕ τîν ΑΓ, ΓΒ τîν ¢πÕ τîν Α∆, ∆Β διαφέρει ·ητù· ·ητ¦ γ¦ρ ¢µφότερα· κሠτÕ δˆς ¥ρα ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ διαφέρει ·ητù µέσα Ôντα· Óπερ ¥τοπον· µέσον γ¦ρ µέσου οÙχ Øπερέχει ·ητù. ΟÙχ ¥ρα ¹ ™κ δύο Ñνοµάτων κατ' ¥λλο κሠ¥λλο σηµε‹ον διαιρε‹ται· καθ' žν ¥ρα µόνον· Óπερ œδει δε‹ξαι.

D

C

B

Let AB be a binomial (straight-line) which has been divided into its (component) terms at C. AC and CB are thus rational (straight-lines which are) commensurable in square only [Prop. 10.36]. I say that AB cannot be divided at another point into two rational (straight-lines which are) commensurable in square only. For, if possible, let it also have been divided at D, such that AD and DB are also rational (straight-lines which are) commensurable in square only. So, (it is) clear that AC is not the same as DB. For, if possible, let it be (the same). So, AD will also be the same as CB. And as AC will be to CB, so BD (will be) to DA. And AB will (thus) also be divided at D in the same (manner) as the division at C. The very opposite was assumed. Thus, AC is not the same as DB. So, on account of this, points C and D are not equally far from the point of bisection. Thus, by whatever (amount the sum of) the (squares) on AC and CB differs from (the sum of) the (squares) on AD and DB, twice the (rectangle contained) by AD and DB also differs from twice the (rectangle contained) by AC and CB by this (same amount)—on account of both (the sum of) the (squares) on AC and CB, plus twice the (rectangle contained) by AC and CB, and (the sum of) the (squares) on AD and DB, plus twice the (rectangle contained) by AD and DB, being equal to the (square) on AB [Prop. 2.4]. But, (the sum of) the (squares) on AC and CB differs from (the sum of) the (squares) on AD and DB by a rational (area). For (they are) both rational (areas). Thus, twice the (rectangle contained) by AD and DB also differs from twice the (rectangle contained) by AC and CB by a rational (area, despite both) being medial (areas) [Prop. 10.21]. The very thing is absurd.

327

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 For a medial (area) cannot exceed a medial (area) by a rational (area) [Prop. 10.26]. Thus, a binomial (straight-line) cannot be divided (into its component terms) at different points. Thus, (it can be so divided) at one point only. (Which is) the very thing it was required to show.

†

In other words, k + k ′1/2 = k ′′ + k ′′′1/2 has only one solution: i.e., k ′′ = k and k ′′′ = k ′ . Likewise, k 1/2 + k ′1/2 = k ′′1/2 + k ′′′1/2 has only

one solution: i.e., k ′′ = k and k ′′′ = k ′ (or, equivalently, k ′′ = k ′ and k ′′′ = k).

µγ΄.

Proposition 43

`Η ™κ δύο µέσων πρώτη καθ' žν µόνον σηµε‹ον διαιρε‹ται.

A first bimedial (straight-line) can be divided (into its component terms) at one point only.†

Α

∆

Γ

Β

A

”Εστω ™κ δύο µέσων πρώτη ¹ ΑΒ διVρηµένη κατ¦ τÕ Γ, éστε τ¦ς ΑΓ, ΓΒ µέσας εναι δυνάµει µόνον συµµέτρους ·ητÕν περιεχούσας· λέγω, Óτι ¹ ΑΒ κατ' ¥λλο σηµε‹ον οÙ διαιρε‹ται. Ε„ γ¦ρ δυνατόν διVρήσθω κሠκατ¦ τÕ ∆, éστε κሠτ¦ς Α∆, ∆Β µέσας εναι δυνάµει µόνον συµµέτρους ·ητÕν περιεχούσας. ™πεˆ οâν, ú διαφέρει τÕ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ, τούτJ διαφέρει τ¦ ¢πÕ τîν ΑΓ, ΓΒ τîν ¢πÕ τîν Α∆, ∆Β, ·ητù δ διαφέρει τÕ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ· ·ητ¦ γ¦ρ ¢µφότερα· ·ητù ¥ρα διαφέρει κሠτ¦ ¢πÕ τîν ΑΓ, ΓΒ τîν ¢πÕ τîν Α∆, ∆Β µέσα Ôντα· Óπερ ¥τοπον. ΟÙκ ¥ρα ¹ ™κ δύο µέσων πρώτη κατ' ¥λλο κሠ¥λλο σηµε‹ον διαιρε‹ται ε„ς τ¦ Ñνόµατα· καθ' žν ¥ρα µόνον· Óπερ œδει δε‹ξαι.

†

D

C

B

Let AB be a first bimedial (straight-line) which has been divided at C, such that AC and CB are medial (straight-lines), commensurable in square only, (and) containing a rational (area) [Prop. 10.37]. I say that AB cannot be (so) divided at another point. For, if possible, let it also have been divided at D, such that AD and DB are also medial (straight-lines), commensurable in square only, (and) containing a rational (area). Since, therefore, by whatever (amount) twice the (rectangle contained) by AD and DB differs from twice the (rectangle contained) by AC and CB, (the sum of) the (squares) on AC and CB differs from (the sum of) the (squares) on AD and DB by this (same amount) [Prop. 10.41 lem.]. And twice the (rectangle contained) by AD and DB differs from twice the (rectangle contained) by AC and CB by a rational (area). For (they are) both rational (areas). (The sum of) the (squares) on AC and CB thus differs from (the sum of) the (squares) on AD and DB by a rational (area, despite both) being medial (areas). The very thing is absurd [Prop. 10.26]. Thus, a first bimedial (straight-line) cannot be divided into its (component) terms at different points. Thus, (it can be so divided) at one point only. (Which is) the very thing it was required to show.

In other words, k 1/4 + k 3/4 = k ′1/4 + k ′3/4 has only one solution: i.e., k ′ = k.

µδ΄.

Proposition 44

`Η ™κ δύο µέσων δευτέρα καθ' žν µόνον σηµε‹ον διαιρε‹ται. ”Εστω ™κ δύο µέσων δευτέρα ¹ ΑΒ διVρηµένη κατ¦ τÕ Γ, éστε τ¦ς ΑΓ, ΓΒ µέσας εναι δυνάµει µόνον συµµέτρους µέσον περιεχούσας· φανερÕν δή, Óτι τÕ Γ οÙκ œστι κατ¦ τÁς διχοτοµίας, Óτι οÙκ ε„σˆ µήκει σύµµετροι. λέγω, Óτι ¹ ΑΒ κατ' ¥λλο σηµε‹ον οÙ

A second bimedial (straight-line) can be divided (into its component terms) at one point only.† Let AB be a second bimedial (straight-line) which has been divided at C, so that AC and BC are medial (straight-lines), commensurable in square only, (and) containing a medial (area) [Prop. 10.38]. So, (it is) clear that C is not (located) at the point of bisection, since (AC

328

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

διαιρε‹ται.

Α

and BC) are not commensurable in length. I say that AB cannot be (so) divided at another point.

∆

Γ

Β

A

C

D

B

Ε

Μ

Θ

Ν

E

M

H

N

Ζ

Λ

Η

Κ

F

L

G

K

Ε„ γ¦ρ δυνατόν, διVρήσθω κሠκατ¦ τÕ ∆, éστε τ¾ν ΑΓ τÍ ∆Β µ¾ εναι τ¾ν αÙτήν, ¢λλ¦ µείζονα καθ' Øπόθεσιν τ¾ν ΑΓ· δÁλον δή, Óτι κሠτ¦ ¢πÕ τîν Α∆, ∆Β, æς ™πάνω ™δείξαµεν, ™λάσσονα τîν ¢πÕ τîν ΑΓ, ΓΒ· κሠτ¦ς Α∆, ∆Β µέσας εναι δυνάµει µόνον συµµέτρους µέσον περιεχούσας. κሠ™κκείσθω ·ητ¾ ¹ ΕΖ, κሠτù µν ¢πÕ τÁς ΑΒ ‡σον παρ¦ τ¾ν ΕΖ παραλληλόγραµµον Ñρθογώνιον παραβεβλήσθω τÕ ΕΚ, το‹ς δ ¢πÕ τîν ΑΓ, ΓΒ ‡σον ¢φVρήσθω τÕ ΕΗ· λοιπÕν ¥ρα τÕ ΘΚ ‡σον ™στˆ τù δˆς ØπÕ τîν ΑΓ, ΓΒ. πάλιν δ¾ το‹ς ¢πÕ τîν Α∆, ∆Β, ¤περ ™λάσσονα ™δείχθη τîν ¢πÕ τîν ΑΓ, ΓΒ, ‡σον ¢φVρήσθω τÕ ΕΛ· κሠλοιπÕν ¥ρα τÕ ΜΚ ‡σον τù δˆς ØπÕ τîν Α∆, ∆Β. κሠ™πεˆ µέσα ™στˆ τ¦ ¢πÕ τîν ΑΓ, ΓΒ, µέσον ¥ρα [καˆ] τÕ ΕΗ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται· ·ητ¾ ¥ρα ™στˆν ¹ ΕΘ κሠ¢σύµµετρος τÍ ΕΖ µήκει. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΘΝ ·ητή ™στι κሠ¢σύµµετρος τÍ ΕΖ µήκει. κሠ™πεˆ αƒ ΑΓ, ΓΒ µέσαι ε„σˆ δυνάµει µόνον σύµµετροι, ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΓ τÍ ΓΒ µήκει. æς δ ¹ ΑΓ πρÕς τ¾ν ΓΒ, οÛτως τÕ ¢πÕ τÁς ΑΓ πρÕς τÕ ØπÕ τîν ΑΓ, ΓΒ· ¢σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΓ τù ØπÕ τîν ΑΓ, ΓΒ. ¢λλ¦ τù µν ¢πÕ τÁς ΑΓ σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΓ, ΓΒ· δυνάµει γάρ ε„σι σύµµετροι αƒ ΑΓ, ΓΒ. τù δ ØπÕ τîν ΑΓ, ΓΒ σύµµετρόν ™στι τÕ δˆς ØπÕ τîν ΑΓ, ΓΒ. κሠτ¦ ¢πÕ τîν ΑΓ, ΓΒ ¥ρα ¢σύµµετρά ™στι τù δˆς ØπÕ τîν ΑΓ, ΓΒ. ¢λλ¦ το‹ς µν ¢πÕ τîν ΑΓ, ΓΒ ‡σον ™στˆ τÕ ΕΗ, τù δ δˆς ØπÕ τîν ΑΓ, ΓΒ ‡σον τÕ ΘΚ· ¢σύµµετρον ¥ρα ™στˆ τÕ ΕΗ τù ΘΚ· éστε κሠ¹ ΕΘ τÍ ΘΝ ¢σύµµετρός ™στι µήκει. καί ε„σι ·ηταί· αƒ ΕΘ, ΘΝ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. ™¦ν δ δύο ·ητሠδυνάµει µόνον σύµµετροι συντεθîσιν, ¹ Óλη ¥λογός ™στιν ¹ καλουµένη ™κ δύο Ñνοµάτων· ¹ ΕΝ ¥ρα ™κ δύο Ñνοµάτων ™στˆ διVρηµένη κατ¦ τÕ Θ. κατ¦ τ¦ αÙτ¦ δ¾ δειχθήσονται καˆ αƒ ΕΜ, ΜΝ ·ητሠδυνάµει µόνον σύµµετροι· κሠœσται ¹ ΕΝ ™κ δύο Ñνοµάτων κατ' ¥λλο κሠ¥λλο διVρηµένη τό τε Θ κሠτÕ Μ, κሠοÙκ œστιν ¹ ΕΘ τÍ ΜΝ ¹ αÙτή, Óτι τ¦ ¢πÕ τîν ΑΓ, ΓΒ µείζονά ™στι τîν ¢πÕ τîν Α∆, ∆Β. ¢λλ¦ τ¦ ¢πÕ τîν Α∆, ∆Β µείζονά ™στι τοà δˆς ØπÕ Α∆, ∆Β· πολλù ¥ρα κሠτ¦ ¢πÕ τîν ΑΓ, ΓΒ, τουτέστι τÕ

For, if possible, let it also have been (so) divided at D, so that AC is not the same as DB, but AC (is), by hypothesis, greater. So, (it is) clear that (the sum of) the (squares) on AD and DB is also less than (the sum of) the (squares) on AC and CB, as we showed above [Prop. 10.41 lem.]. And AD and DB are medial (straight-lines), commensurable in square only, (and) containing a medial (area). And let the rational (straightline) EF be laid down. And let the rectangular parallelogram EK, equal to the (square) on AB, have been applied to EF . And let EG, equal to (the sum of) the (squares) on AC and CB, have been cut off (from EK). Thus, the remainder, HK, is equal to twice the (rectangle contained) by AC and CB [Prop. 2.4]. So, again, let EL, equal to (the sum of) the (squares) on AD and DB—which was shown (to be) less than (the sum of) the (squares) on AC and CB—have been cut off (from EK). And, thus, the remainder, M K, (is) equal to twice the (rectangle contained) by AD and DB. And since (the sum of) the (squares) on AC and CB is medial, EG (is) thus [also] medial. And it is applied to the rational (straight-line) EF . Thus, EH is rational, and incommensurable in length with EF [Prop. 10.22]. So, for the same (reasons), HN is also rational, and incommensurable in length with EF . And since AC and CB are medial (straight-lines which are) commensurable in square only, AC is thus incommensurable in length with CB. And as AC (is) to CB, so the (square) on AC (is) to the (rectangle contained) by AC and CB [Prop. 10.21 lem.]. Thus, the (square) on AC is incommensurable with the (rectangle contained) by AC and CB [Prop. 10.11]. But, (the sum of) the (squares) on AC and CB is commensurable with the (square) on AC. For, AC and CB are commensurable in square [Prop. 10.15]. And twice the (rectangle contained) by AC and CB is commensurable with the (rectangle contained) by AC and CB [Prop. 10.6]. And thus (the sum of) the squares on AC and CB is incommensurable with twice the (rectangle contained) by AC and CB [Prop. 10.13]. But, EG is equal to (the

329

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

ΕΗ, µε‹ζόν ™στι τοà δˆς ØπÕ τîν Α∆, ∆Β, τουτέστι τοà sum of) the (squares) on AC and CB, and HK to twice ΜΚ· éστε κሠ¹ ΕΘ τÁς ΜΝ µείζων ™στίν. ¹ ¥ρα ΕΘ the (rectangle contained) by AC and CB. Thus, EG is τÍ ΜΝ οÙκ œστιν ¹ αÙτή· Óπερ œδει δε‹ξαι. incommensurable with HK. Hence, EH is also incommensurable in length with HN [Props. 6.1, 10.11]. And (they are) rational (straight-lines). Thus, EH and HN are rational (straight-lines which are) commensurable in square only. And if two rational (straight-lines which are) commensurable in square only are added together, then the whole (straight-line) is that irrational called binomial [Prop. 10.36]. Thus, EN is a binomial (straight-line) which has been divided (into its component terms) at H. So, according to the same (reasoning), EM and M N can be shown (to be) rational (straight-lines which are) commensurable in square only. And EN will (thus) be a binomial (straight-line) which has been divided (into its component terms) at the different (points) H and M (which is absurd [Prop. 10.42]). And EH is not the same as M N , since (the sum of) the (squares) on AC and CB is greater than (the sum of) the (squares) on AD and DB. But, (the sum of) the (squares) on AD and DB is greater than twice the (rectangle contained) by AD and DB [Prop. 10.59 lem.]. Thus, (the sum of) the (squares) on AC and CB—that is to say, EG—is also much greater than twice the (rectangle contained) by AD and DB— that is to say, M K. Hence, EH is also greater than M N [Prop. 6.1]. Thus, EH is not the same as M N . (Which is) the very thing it was required to show. †

In other words, k 1/4 + k ′1/2 /k 1/4 = k ′′1/4 + k ′′′1/2 /k ′′1/4 has only one solution: i.e., k ′′ = k and k ′′′ = k ′ .

µε΄.

Proposition 45

`Η µείζων κατ¦ τÕ αÙτÕ µόνον σηµε‹ον διαιρε‹ται.

A major (straight-line) can only be divided (into its component terms) at the same point.†

Α

∆

Γ

Β

A

”Εστω µείζων ¹ ΑΒ διVρηµένη κατ¦ τÕ Γ, éστε τ¦ς ΑΓ, ΓΒ δυνάµει ¢συµµέτρους εναι ποιούσας τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ τετραγώνων ·ητόν, τÕ δ' ØπÕ τîν ΑΓ, ΓΒ µέσον· λέγω, Óτι ¹ ΑΒ κατ' ¥λλο σηµε‹ον οÙ διαιρε‹ται. Ε„ γ¦ρ δυνατόν, διVρήσθω κሠκατ¦ τÕ ∆, éστε κሠτ¦ς Α∆, ∆Β δυνάµει ¢συµµέτρους εναι ποιούσας τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν Α∆, ∆Β ·ητόν, τÕ δ' Øπ' αÙτîν µέσον. κሠ™πεί, ú διαφέρει τ¦ ¢πÕ τîν ΑΓ, ΓΒ τîν ¢πÕ τîν Α∆, ∆Β, τούτJ διαφέρει κሠτÕ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ, ¢λλ¦ τ¦ ¢πÕ τîν ΑΓ, ΓΒ τîν ¢πÕ τîν Α∆, ∆Β Øπερέχει ·ητù· ·ητ¦ γ¦ρ ¢µφότερα· κሠτÕ δˆς ØπÕ τîν Α∆, ∆Β ¥ρα τοà δˆς ØπÕ τîν ΑΓ, ΓΒ Øπερέχει ·ητù µέσα Ôντα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ¹ µείζων κατ' ¥λλο κሠ¥λλο σηµε‹ον

D

C

B

Let AB be a major (straight-line) which has been divided at C, so that AC and CB are incommensurable in square, making the sum of the squares on AC and CB rational, and the (rectangle contained) by AC and CD medial [Prop. 10.39]. I say that AB cannot be (so) divided at another point. For, if possible, let it also have been divided at D, such that AD and DB are also incommensurable in square, making the sum of the (squares) on AD and DB rational, and the (rectangle contained) by them medial. And since, by whatever (amount the sum of) the (squares) on AC and CB differs from (the sum of) the (squares) on AD and DB, twice the (rectangle contained) by AD and DB also differs from twice the (rectangle contained) by AC and CB by this (same amount). But, (the sum of)

330

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

διαιρε‹ται· κατ¦ τÕ αÙτÕ ¥ρα µόνον διαιρε‹ται· Óπερ œδει the (squares) on AC and CB exceeds (the sum of) the δε‹ξαι. (squares) on AD and DB by a rational (area). For (they are) both rational (areas). Thus, twice the (rectangle contained) by AD and DB also exceeds twice the (rectangle contained) by AC and CB by a rational (area), (despite both) being medial (areas). The very thing is impossible [Prop. 10.26]. Thus, a major (straight-line) cannot be divided (into its component terms) at different points. Thus, it can only be (so) divided at the same (point). (Which is) the very thing it was required to show. †

In other words,

solution: i.e.,

k′

q

[1 + k/(1 + k 2 )1/2 ]/2 +

= k.

q

[1 − k/(1 + k 2 )1/2 ]/2 =

q

[1 + k ′ /(1 + k ′2 )1/2 ]/2 +

q

[1 − k ′ /(1 + k ′2 )1/2 ]/2 has only one

µ$΄.

Proposition 46

`Η ·ητÕν κሠµέσον δυναµένη καθ' žν µόνον σηµε‹ον διαιρε‹ται.

The square-root of a rational plus a medial (area) can be divided (into its component terms) at one point only.†

Α

∆

Γ

Β

A

”Εστω ·ητÕν κሠµέσον δυναµένη ¹ ΑΒ διVρηµένη κατ¦ τÕ Γ, éστε τ¦ς ΑΓ, ΓΒ δυνάµει ¢συµµέτρους εναι ποιούσας τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ µέσον, τÕ δ δˆς ØπÕ τîν ΑΓ, ΓΒ ·ητόν· λέγω, Óτι ¹ ΑΒ κατ' ¥λλο σηµε‹ον οÙ διαιρε‹ται. Ε„ γ¦ρ δυνατόν, διVρήσθω κሠκατ¦ τÕ ∆, éστε κሠτ¦ς Α∆, ∆Β δυνάµει ¢συµµέτρους εναι ποιούσας τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν Α∆, ∆Β µέσον, τÕ δ δˆς ØπÕ τîν Α∆, ∆Β ·ητόν. ™πεˆ οâν, ú διαφέρει τÕ δˆς ØπÕ τîν ΑΓ, ΓΒ τοà δˆς ØπÕ τîν Α∆, ∆Β, τούτJ διαφέρει κሠτ¦ ¢πÕ τîν Α∆, ∆Β τîν ¢πÕ τîν ΑΓ, ΓΒ, τÕ δ δˆς ØπÕ τîν ΑΓ, ΓΒ τοà δˆς ØπÕ τîν Α∆, ∆Β Øπερέχει ·ητù, κሠτ¦ ¢πÕ τîν Α∆, ∆Β ¥ρα τîν ¢πÕ τîν ΑΓ, ΓΒ Øπερέχει ·ητù µέσα Ôντα· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ¹ ·ητÕν κሠµέσον δυναµένη κατ' ¥λλο κሠ¥λλο σηµε‹ον διαιρε‹ται. κατ¦ žν ¥ρα σηµε‹ον διαιρε‹ται· Óπερ œδει δε‹ξαι.

D

C

B

Let AB be the square-root of a rational plus a medial (area) which has been divided at C, so that AC and CB are incommensurable in square, making the sum of the (squares) on AC and CB medial, and twice the (rectangle contained) by AC and CB rational [Prop. 10.40]. I say that AB cannot be (so) divided at another point. For, if possible, let it also have been divided at D, so that AD and DB are also incommensurable in square, making the sum of the (squares) on AD and DB medial, and twice the (rectangle contained) by AD and DB rational. Therefore, since by whatever (amount) twice the (rectangle contained) by AC and CB differs from twice the (rectangle contained) by AD and DB, (the sum of) the (squares) on AD and DB also differs from (the sum of) the (squares) on AC and CB by this (same amount). And twice the (rectangle contained) by AC and CB exceeds twice the (rectangle contained) by AD and DB by a rational (area). (The sum of) the (squares) on AD and DB thus also exceeds (the sum of) the (squares) on AC and CB by a rational (area), (despite both) being medial (areas). The very thing is impossible [Prop. 10.26]. Thus, the square-root of a rational plus a medial (area) cannot be divided (into its component terms) at different points. Thus, it can be (so) divided at one point (only). (Which is) the very thing it was required to show.

q q q In other words, [(1 + k 2 )1/2 + k]/[2 (1 + k 2 )] + [(1 + k 2 )1/2 − k]/[2 (1 + k 2 )] = [(1 + k ′2 )1/2 + k ′ ]/[2 (1 + k ′2 )] q + [(1 + k ′2 )1/2 − k ′ ]/[2 (1 + k ′2 )] has only one solution: i.e., k ′ = k.

†

331

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 µζ΄.

Proposition 47

`Η δύο µέσα δυναµένη καθ' žν µόνον σηµε‹ον διαιρε‹ται.

The square-root of (the sum of) two medial (areas) can be divided (into its component terms) at one point only.†

Α

∆

Γ

Β

A

D

C

B

Ε

Ζ

E

F

Μ Θ

Λ Η

M H

L G

Ν

Κ

N

K

”Εστω [δύο µέσα δυναµένη] ¹ ΑΒ διVρηµένη κατ¦ τÕ Γ, éστε τ¦ς ΑΓ, ΓΒ δυνάµει ¢συµµέτρους εναι ποιούσας τό τε συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ µέσον κሠτÕ ØπÕ τîν ΑΓ, ΓΒ µέσον κሠœτι ¢σύµµετρον τù συγκειµένJ ™κ τîν ¢π' αÙτîν. λέγω, Óτι ¹ ΑΒ κατ' ¥λλο σηµε‹ον οÙ διαιρε‹ται ποιοàσα τ¦ προκείµενα. Ε„ γ¦ρ δυνατόν, διVρήσθω κατ¦ τÕ ∆, éστε πάλιν δηλονότι τ¾ν ΑΓ τÍ ∆Β µ¾ εναι τ¾ν αÙτήν, ¢λλ¦ µείζονα καθ' Øπόθεσιν τ¾ν ΑΓ, κሠ™κκείσθω ·ητ¾ ¹ ΕΖ, κሠπαραβεβλήσθω παρ¦ τ¾ν ΕΖ το‹ς µν ¢πÕ τîν ΑΓ, ΓΒ ‡σον τÕ ΕΗ, τù δ δˆς ØπÕ τîν ΑΓ, ΓΒ ‡σον τÕ ΘΚ· Óλον ¥ρα τÕ ΕΚ ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ τετραγώνJ. πάλιν δ¾ παραβεβλήσθω παρ¦ τ¾ν ΕΖ το‹ς ¢πÕ τîν Α∆, ∆Β ‡σον τÕ ΕΛ· λοιπÕν ¥ρα τÕ δˆς ØπÕ τîν Α∆, ∆Β λοιπù τù ΜΚ ‡σον ™στίν. κሠ™πεˆ µέσον Øπόκειται τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ, µέσον ¥ρα ™στˆ κሠτÕ ΕΗ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται· ·ητ¾ ¥ρα ™στˆν ¹ ΘΕ κሠ¢σύµµετρος τÍ ΕΖ µήκει. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΘΝ ·ητή ™στι κሠ¢σύµµετρος τÍ ΕΖ µήκει. κሠ™πεˆ ¢σύµµετρόν ™στι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ τù δˆς ØπÕ τîν ΑΓ, ΓΒ, κሠτÕ ΕΗ ¥ρα τù ΗΝ ¢σύµµετρόν ™στιν· éστε κሠ¹ ΕΘ τÍ ΘΝ ¢σύµµετρός ™στιν. καί ε„σι ·ηταί· αƒ ΕΘ, ΘΝ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¹ ΕΝ ¥ρα ™κ δύο Ñνοµάτων ™στˆ διVρηµένη κατ¦ τÕ Θ. еοίως δ¾ δε‹ξοµεν, Óτι κሠκατ¦ τÕ Μ διÇρηται. κሠοÙκ œστιν ¹ ΕΘ τÍ ΜΝ ¹ αÙτή· ¹ ¥ρα ™κ δύο Ñνοµάτων κατ' ¥λλο κሠ¥λλο σηµε‹ον διÇρηται· Óπερ ™στίν ¥τοπον. οÙκ ¥ρα ¹ δύο µέσα δυναµένη κατ' ¢λλο κሠ¥λλο σηµε‹ον διαιρε‹ται· καθ' žν ¥ρα µόνον [σηµε‹ον] διαιρε‹ται.

Let AB be [the square-root of (the sum of) two medial (areas)] which has been divided at C, such that AC and CB are incommensurable in square, making the sum of the (squares) on AC and CB medial, and the (rectangle contained) by AC and CB medial, and, moreover, incommensurable with the sum of the (squares) on (AC and CB) [Prop. 10.41]. I say that AB cannot be divided at another point fulfilling the prescribed (conditions). For, if possible, let it have been divided at D, such that AC is again manifestly not the same as DB, but AC (is), by hypothesis, greater. And let the rational (straight-line) EF be laid down. And let EG, equal to (the sum of) the (squares) on AC and CB, and HK, equal to twice the (rectangle contained) by AC and CB, have been applied to EF . Thus, the whole of EK is equal to the square on AB [Prop. 2.4]. So, again, let EL, equal to (the sum of) the (squares) on AD and DB, have been applied to EF . Thus, the remainder—twice the (rectangle contained) by AD and DB—is equal to the remainder, M K. And since the sum of the (squares) on AC and CB was assumed (to be) medial, EG is also medial. And it is applied to the rational (straight-line) EF . HE is thus rational, and incommensurable in length with EF [Prop. 10.22]. So, for the same (reasons), HN is also rational, and incommensurable in length with EF . And since the sum of the (squares) on AC and CB is incommensurable with twice the (rectangle contained) by AC and CB, EG is thus also incommensurable with GN . Hence, EH is also incommensurable with HN [Props. 6.1, 10.11]. And they are (both) rational (straight-lines). Thus, EH and HN are rational (straight-lines which are) commensurable in square only. Thus, EN is a binomial (straightline) which has been divided (into its component terms) at H [Prop. 10.36]. So, similarly, we can show that it has

332

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 also been (so) divided at M . And EH is not the same as M N . Thus, a binomial (straight-line) has been divided (into its component terms) at different points. The very thing is absurd [Prop. 10.42]. Thus, the square-root of (the sum of) two medial (areas) cannot be divided (into its component terms) at different points. Thus, it can be (so) divided at one [point] only.

q q q In other words, k ′1/4 [1 + k/(1 + k 2 )1/2 ]/2 + k ′1/4 [1 − k/(1 + k 2 )1/2 ]/2 = k ′′′1/4 [1 + k ′′ /(1 + k ′′2 )1/2 ]/2 q +k ′′′1/4 [1 − k ′′ /(1 + k ′′2 )1/2 ]/2 has only one solution: i.e., k ′′ = k and k ′′′ = k ′ .

†

“Οροι δεύτεροι.

Definitions II

ε΄. `Υποκειµένης ·ητÁς κሠτÁς ™κ δύο Ñνοµάτων διVρηµένης ε„ς τ¦ Ñνόµατα, Âς τÕ µε‹ζον Ôνοµα τοà ™λάσσονος µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει, ™¦ν µν τÕ µε‹ζον Ôνοµα σύµµετρον Ï µήκει τÍ ™κκειµένV ·ητÍ, καλείσθω [¹ Óλη] ™κ δύο Ñνοµάτων πρώτη. $΄. 'Ε¦ν δ τÕ ™λάσσον Ôνοµα σύµµετρον Ï µήκει τÍ ™κκειµένV ·ητÍ, καλείσθω ™κ δύο Ñνοµάτων δευτέρα. ζ΄. 'Ε¦ν δ µηδέτερον τîν Ñνοµάτων σύµµετρον Ï µήκει τÍ ™κκειµένV ·ητÍ, καλείσθω ™κ δύο Ñνοµάτων τρίτη. η΄. Πάλιν δ¾ ™¦ν τÕ µε‹ζον Ôνοµα [τοà ™λάσσονος] µε‹ζον δύνηται τù ¢πÕ ¢συµµέτρου ˜αυτÍ µήκει, ™¦ν µν τÕ µε‹ζον Ôνοµα σύµµετρον Ï µήκει τÍ ™κκειµένV ·ητÍ, καλείσθω ™κ δύο Ñνοµάτων τετάρτη. θ΄. 'Ε¦ν δ τÕ œλασσον, πέµπτη. ι΄. 'Ε¦ν δ µηδέτερον, ›κτη.

5. Given a rational (straight-line), and a binomial (straight-line) which has been divided into its (component) terms, of which the square on the greater term is larger than (the square on) the lesser by the (square) on (some straight-line) commensurable in length with (the greater), then, if the greater term is commensurable in length with the rational (straight-line previously) laid out, let [the whole] (straight-line) be called a first binomial (straight-line). 6. And if the lesser term is commensurable in length with the rational (straight-line previously) laid out, then let (the whole straight-line) be called a second binomial (straight-line). 7. And if neither of the terms is commensurable in length with the rational (straight-line previously) laid out, then let (the whole straight-line) be called a third binomial (straight-line). 8. So, again, if the square on the greater term is larger than (the square on) [the lesser] by the (square) on (some straight-line) incommensurable in length with (the greater), then, if the greater term is commensurable in length with the rational (straight-line previously) laid out, let (the whole straight-line) be called a fourth binomial (straight-line). 9. And if the lesser (term is commensurable), a fifth (binomial straight-line). 10. And if neither (term is commensurable), a sixth (binomial straight-line).

µη΄.

Proposition 48

ΕØρε‹ν τ¾ν ™κ δύο Ñνοµάτων πρώτην. 'Εκκείσθωσαν δύο ¢ριθµοˆ οƒ ΑΓ, ΓΒ, éστε τÕν συγκείµενον ™ξ αÙτîν τÕν ΑΒ πρÕς µν τÕν ΒΓ λόγον œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, πρÕς δ τÕν ΓΑ λόγον µ¾ œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, κሠ™κκείσθω τις ·ητ¾ ¹ ∆, κሠτÍ ∆ σύµµετρος œστω µήκει ¹ ΕΖ. ·ητ¾

To find a first binomial (straight-line). Let the two numbers AC and CB be laid down such that their sum AB has to BC the ratio which (some) square number (has) to (some) square number, and does not have to CA the ratio which (some) square number (has) to (some) square number [Prop. 10.28 lem. I]. And let some rational (straight-line) D be laid down. And

333

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ELEMENTS BOOK 10

¥ρα ™στˆ κሠ¹ ΕΖ. κሠγεγονέτω æς Ð ΒΑ ¢ριθµÕς πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς ΖΗ. Ð δ ΑΒ πρÕς τÕν ΑΓ λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν· κሠτÕ ¢πÕ τÁς ΕΖ ¥ρα πρÕς τÕ ¢πÕ τÁς ΖΗ λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν· éστε σύµµετρόν ™στι τÕ ¢πÕ τÁς ΕΖ τù ¢πÕ τÁς ΖΗ. κሠ™στι ·ητ¾ ¹ ΕΖ· ·ητ¾ ¥ρα κሠ¹ ΖΗ. κሠ™πεˆ Ð ΒΑ πρÕς τÕν ΑΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ τÕ ¢πÕ τÁς ΕΖ ¥ρα πρÕς τÕ ¢πÕ τÁς ΖΗ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΕΖ τÍ ΖΗ µήκει. αƒ ΕΖ, ΖΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ΕΗ. λέγω, Óτι κሠπρώτη.

∆ Ε Α

Θ Ζ Γ

let EF be commensurable in length with D. EF is thus also rational [Def. 10.3]. And let it have been contrived that as the number BA (is) to AC, so the (square) on EF (is) to the (square) on F G [Prop. 10.6 corr.]. And AB has to AC the ratio which (some) number (has) to (some) number. Thus, the (square) on EF also has to the (square) on F G the ratio which (some) number (has) to (some) number. Hence, the (square) on EF is commensurable with the (square) on F G [Prop. 10.6]. And EF is rational. Thus, F G (is) also rational. And since BA does not have to AC the ratio which (some) square number (has) to (some) square number, thus the (square) on EF does not have to the (square) on F G the ratio which (some) square number (has) to (some) square number either. Thus, EF is incommensurable in length with F G [Prop 10.9]. EF and F G are thus rational (straight-lines which are) commensurable in square only. Thus, EG is a binomial (straight-line) [Prop. 10.36]. I say that (it is) also a first (binomial straight-line).

H F

D E

Η Β

A

'Επεˆ γάρ ™στιν æς Ð ΒΑ ¢ριθµÕς πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς ΖΗ, µείζων δ Ð ΒΑ τοà ΑΓ, µε‹ζον ¥ρα κሠτÕ ¢πÕ τÁς ΕΖ τοà ¢πÕ τÁς ΖΗ. œστω οâν τù ¢πÕ τÁς ΕΖ ‡σα τ¦ ¢πÕ τîν ΖΗ, Θ. κሠ™πεί ™στιν æς Ð ΒΑ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς ΖΗ, ¢ναστρέψαντι ¥ρα ™στˆν æς Ð ΑΒ πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς Θ. Ð δ ΑΒ πρÕς τÕν ΒΓ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. κሠτÕ ¢πÕ τÁς ΕΖ ¥ρα πρÕς τÕ ¢πÕ τÁς Θ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. σύµµετρος ¥ρα ™στˆν ¹ ΕΖ τÍ Θ µήκει· ¹ ΕΖ ¥ρα τÁς ΖΗ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. καί ε„σι ·ηταˆ αƒ ΕΖ, ΖΗ, κሠσύµµετρος ¹ ΕΖ τÍ ∆ µήκει. 'Η ΕΗ ¥ρα ™κ δύο Ñνοµάτων ™στˆ πρώτη· Óπερ œδει δε‹ξαι.

C

G B

For since as the number BA is to AC, so the (square) on EF (is) to the (square) on F G, and BA (is) greater than AC, the (square) on EF (is) thus also greater than the (square) on F G [Prop. 5.14]. Therefore, let (the sum of) the (squares) on F G and H be equal to the (square) on EF . And since as BA is to AC, so the (square) on EF (is) to the (square) on F G, thus, via conversion, as AB is to BC, so the (square) on EF (is) to the (square) on H [Prop. 5.19 corr.]. And AB has to BC the ratio which (some) square number (has) to (some) square number. Thus, the (square) on EF also has to the (square) on H the ratio which (some) square number (has) to (some) square number. Thus, EF is commensurable in length with H [Prop. 10.9]. Thus, the square on EF is greater than (the square on) F G by the (square) on (some straight-line) commensurable (in length) with (EF ). And EF and F G are rational (straight-lines). And EF (is) commensurable in length with D. Thus, EG is a first binomial (straight-line) [Def. 10.5].† (Which is) the very thing it was required to show.

√ the rational straight-line has unit length, then the length of a first binomial straight-line is k + k 1 − k ′ 2 . This, and the first apotome, whose √ length is k − k 1 − k ′ 2 [Prop. 10.85], are the roots of x2 − 2 k x + k 2 k ′ 2 = 0.

† If

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ELEMENTS BOOK 10 µθ΄.

Proposition 49

ΕØρε‹ν τ¾ν ™κ δύο Ñνοµάτων δευτέραν.

To find a second binomial (straight-line).

Θ Ε

H E

Α

∆ Ζ Γ

Η Β

A

'Εκκείσθωσαν δύο ¢ριθµοˆ οƒ ΑΓ, ΓΒ, éστε τÕν συγκείµενον ™ξ αÙτîν τÕν ΑΒ πρÕς µν τÕν ΒΓ λόγον œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, πρÕς δ τÕν ΑΓ λόγον µ¾ œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, κሠ™κκείσθω ·ητ¾ ¹ ∆, κሠτÍ ∆ σύµµετρος œστω ¹ ΕΖ µήκει· ·ητ¾ ¥ρα ™στˆν ¹ ΕΖ. γεγονέτω δ¾ κሠæς Ð ΓΑ ¢ριθµÕς πρÕς τÕν ΑΒ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς ΖΗ· σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΕΖ τù ¢πÕ τÁς ΖΗ. ·ητ¾ ¥ρα ™στˆ κሠ¹ ΖΗ. κሠ™πεˆ Ð ΓΑ ¢ριθµÕς πρÕς τÕν ΑΒ λÕγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς ΖΗ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. ¢σύµµετρος ¥ρα ™στˆν ¹ ΕΖ τÍ ΖΗ µήκει· αƒ ΕΖ, ΖΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ΕΗ. δεικτέον δή, Óτι κሠδευτέρα. 'Επεˆ γ¦ρ ¢νάπαλίν ™στιν æς Ð ΒΑ ¢ριθµÕς πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΗΖ πρÕς τÕ ¢πÕ τÁς ΖΕ, µείζων δ Ð ΒΑ τοà ΑΓ, µε‹ζον ¥ρα [καˆ] τÕ ¢πÕ τÁς ΗΖ τοà ¢πÕ τÁς ΖΕ. œστω τù ¢πÕ τÁς ΗΖ ‡σα τ¦ ¢πÕ τîν ΕΖ, Θ· ¢ναστρέψαντι ¥ρα ™στˆν æς Ð ΑΒ πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς Θ. ¢λλ' Ð ΑΒ πρÕς τÕν ΒΓ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· κሠτÕ ¢πÕ τÁς ΖΗ ¥ρα πρÕς τÕ ¢πÕ τÁς Θ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. σύµµετρος ¥ρα ™στˆν ¹ ΖΗ τÍ Θ µήκει· éστε ¹ ΖΗ τÁς ΖΕ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. καί ε„σι ·ηταˆ αƒ ΖΗ, ΖΕ δυνάµει µόνον σύµµετροι, κሠτÕ ΕΖ œλασσον Ôνοµα τÍ ™κκειµένV ·ητÍ σύµµετρόν ™στι τÍ ∆ µήκει. `Η ΕΗ ¥ρα ™κ δύο Ñνοµάτων ™στˆ δευτέρα· Óπερ œδει δε‹ξαι.

D F C

G B

Let the two numbers AC and CB be laid down such that their sum AB has to BC the ratio which (some) square number (has) to (some) square number, and does not have to AC the ratio which (some) square number (has) to (some) square number [Prop. 10.28 lem. I]. And let the rational (straight-line) D be laid down. And let EF be commensurable in length with D. EF is thus a rational (straight-line). So, let it also have been contrived that as the number CA (is) to AB, so the (square) on EF (is) to the (square) on F G [Prop. 10.6 corr.]. Thus, the (square) on EF is commensurable with the (square) on F G [Prop. 10.6]. Thus, F G is also a rational (straightline). And since the number CA does not have to AB the ratio which (some) square number (has) to (some) square number, the (square) on EF does not have to the (square) on F G the ratio which (some) square number (has) to (some) square number either. Thus, EF is incommensurable in length with F G [Prop. 10.9]. EF and F G are thus rational (straight-lines which are) commensurable in square only. Thus, EG is a binomial (straightline) [Prop. 10.36]. So, we must show that (it is) also a second (binomial straight-line). For since, inversely, as the number BA is to AC, so the (square) on GF (is) to the (square) on F E [Prop. 5.7 corr.], and BA (is) greater than AC, the (square) on GF (is) thus [also] greater than the (square) on F E [Prop. 5.14]. Let (the sum of) the (squares) on EF and H be equal to the (square) on GF . Thus, via conversion, as AB is to BC, so the (square) on F G (is) to the (square) on H [Prop. 5.19 corr.]. But, AB has to BC the ratio which (some) square number (has) to (some) square number. Thus, the (square) on F G also has to the (square) on H the ratio which (some) square number (has) to (some) square number. Thus, F G is commensurable in length with H [Prop. 10.9]. Hence, the square on F G is greater than (the square on) F E by the (square) on (some straight-line) commensurable in length with (F G). And F G and F E are rational (straight-lines which are) commensurable in square only. And the lesser term EF is commensurable in length with the rational (straightline) D (previously) laid down. Thus, EG is a second binomial (straight-line) [Def.

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ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 10.6].† (Which is) the very thing it was required to show.

√ If the rational straight-line has unit length, then the length of a second binomial straight-line is k/ 1 − k ′ 2 + k. This, and the second apotome, √ √ whose length is k/ 1 − k ′ 2 − k [Prop. 10.86], are the roots of x2 − (2 k/ 1 − k ′ 2 ) x + k 2 [k ′ 2 /(1 − k ′ 2 )] = 0. †

ν΄.

Proposition 50

ΕØρε‹ν τ¾ν ™κ δύο Ñνοµάτων τρίτην.

Α Ε Ζ

Γ

To find a third binomial (straight-line).

Β

Κ

A ∆

Η

E F

Θ

'Εκκείσθωσαν δύο ¢ριθµοˆ οƒ ΑΓ, ΓΒ, éστε τÕν συγκείµενον ™ξ αÙτîν τÕν ΑΒ πρÕς µν τÕν ΒΓ λόγον œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, πρÕς δ τÕν ΑΓ λόγον µ¾ œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. ™κκείσθω δέ τις κሠ¥λλος µ¾ τετράγωνος ¢ριθµÕς Ð ∆, κሠπρÕς ˜κάτερον τîν ΒΑ, ΑΓ λόγον µ¾ ™χέτω, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· κሠ™κκείσθω τις ·ητ¾ εÙθε‹α ¹ Ε, κሠγεγονέτω æς Ð ∆ πρÕς τÕν ΑΒ, οÛτως τÕ ¢πÕ τÁς Ε πρÕς τÕ ¢πÕ τÁς ΖΗ· σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς Ε τù ¢πÕ τÁς ΖΗ. καί ™στι ·ητ¾ ¹ Ε· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΖΗ. κሠ™πεˆ Ð ∆ πρÕς τÕν ΑΒ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ τÕ ¢πÕ τÁς Ε πρÕς τÕ ¢πÕ τÁς ΖΗ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ Ε τÍ ΖΗ µήκει. γεγονέτω δ¾ πάλιν æς ¹ ΒΑ ¢ριθµÕς πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ· σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΖΗ τù ¢πÕ τÁς ΗΘ. ·ητ¾ δ ¹ ΖΗ· ·ητ¾ ¥ρα κሠ¹ ΗΘ. κሠ™πεˆ Ð ΒΑ πρÕς τÕν ΑΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΘΗ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΖΗ τÍ ΗΘ µήκει. αƒ ΖΗ, ΗΘ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¹ ΖΘ ¥ρα ™κ δύο Ñνοµάτων ™στίν. λέγω δή, Óτι κሠτρίτη. 'Επεˆ γάρ ™στιν æς Ð ∆ πρÕς τÕν ΑΒ, οÛτως τÕ ¢πÕ τÁς Ε πρÕς τÕ ¢πÕ τÁς ΖΗ, æς δ Ð ΒΑ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ, δι' ‡σου ¥ρα ™στˆν æς Ð ∆ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς Ε πρÕς τÕ ¢πÕ τÁς ΗΘ. Ð δ ∆ πρÕς τÕν ΑΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ τÕ ¢πÕ τÁς Ε ¥ρα πρÕς τÕ ¢πÕ τÁς ΗΘ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ Ε τÍ ΗΘ µήκει. κሠ™πεί ™στιν æς Ð ΒΑ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ

C

B

K

D G

H

Let the two numbers AC and CB be laid down such that their sum AB has to BC the ratio which (some) square number (has) to (some) square number, and does not have to AC the ratio which (some) square number (has) to (some) square number. And let some other nonsquare number D also be laid down, and let it not have to each of BA and AC the ratio which (some) square number (has) to (some) square number. And let some rational straight-line E be laid down, and let it have been contrived that as D (is) to AB, so the (square) on E (is) to the (square) on F G [Prop. 10.6 corr.]. Thus, the (square) on E is commensurable with the (square) on F G [Prop. 10.6]. And E is a rational (straight-line). Thus, F G is also a rational (straight-line). And since D does not have to AB the ratio which (some) square number has to (some) square number, the (square) on E does not have to the (square) on F G the ratio which (some) square number (has) to (some) square number either. E is thus incommensurable in length with F G [Prop. 10.9]. So, again, let it have been contrived that as the number BA (is) to AC, so the (square) on F G (is) to the (square) on GH [Prop. 10.6 corr.]. Thus, the (square) on F G is commensurable with the (square) on GH [Prop. 10.6]. And F G (is) a rational (straight-line). Thus, GH (is) also a rational (straight-line). And since BA does not have to AC the ratio which (some) square number (has) to (some) square number, the (square) on F G does not have to the (square) on HG the ratio which (some) square number (has) to (some) square number either. Thus, F G is incommensurable in length with GH [Prop. 10.9]. F G and GH are thus rational (straightlines which are) commensurable in square only. Thus, F H is a binomial (straight-line) [Prop. 10.36]. So, I say that (it is) also a third (binomial straight-line). For since as D is to AB, so the (square) on E (is) to the (square) on F G, and as BA (is) to AC, so the

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ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

¢πÕ τÁς ΗΘ, µε‹ζον ¥ρα τÕ ¢πÕ τÁς ΖΗ τοà ¢πÕ τÁς ΗΘ. œστω οâν τù ¢πÕ τÁς ΖΗ ‡σα τ¦ ¢πÕ τîν ΗΘ, Κ· ¢ναστρέψαντι ¥ρα [™στˆν] æς Ð ΑΒ πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς Κ. Ð δ ΑΒ πρÕς τÕν ΒΓ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· κሠτÕ ¢πÕ τÁς ΖΗ ¥ρα πρÕς τÕ ¢πÕ τÁς Κ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· σύµµετρος ¥ρα [™στˆν] ¹ ΖΗ τÍ Κ µήκει. ¹ ΖΗ ¥ρα τÁς ΗΘ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. καί ε„σιν αƒ ΖΗ, ΗΘ ·ητሠδυνάµει µόνον σύµµετροι, κሠοÙδετέρα αÙτîν σύµµετρός ™στι τÍ Ε µήκει. `Η ΖΘ ¥ρα ™κ δύο Ñνοµάτων ™στˆ τρίτη· Óπερ œδει δε‹ξαι.

(square) on F G (is) to the (square) on GH, thus, via equality, as D (is) to AC, so the (square) on E (is) to the (square) on GH [Prop. 5.22]. And D does not have to AC the ratio which (some) square number (has) to (some) square number. Thus, the (square) on E does not have to the (square) on GH the ratio which (some) square number (has) to (some) square number either. Thus, E is incommensurable in length with GH [Prop. 10.9]. And since as BA is to AC, so the (square) on F G (is) to the (square) on GH, the (square) on F G (is) thus greater than the (square) on GH [Prop. 5.14]. Therefore, let (the sum of) the (squares) on GH and K be equal to the (square) on F G. Thus, via conversion, as AB [is] to BC, so the (square) on F G (is) to the (square) on K [Prop. 5.19 corr.]. And AB has to BC the ratio which (some) square number (has) to (some) square number. Thus, the (square) on F G also has to the (square) on K the ratio which (some) square number (has) to (some) square number. Thus, F G [is] commensurable in length with K [Prop. 10.9]. Thus, the square on F G is greater than (the square on) GH by the (square) on (some straight-line) commensurable (in length) with (F G). And F G and GH are rational (straight-lines which are) commensurable in square only, and neither of them is commensurable in length with E. Thus, F H is a third binomial (straight-line) [Def. 10.7].† (Which is) the very thing it was required to show.

†

If the rational straight-line has unit length, then the length of a third binomial straight-line is k 1/2 (1 + √ whose length is k 1/2 (1 − 1 − k ′ 2 ) [Prop. 10.87], are the roots of x2 − 2 k 1/2 x + k k ′ 2 = 0.

να΄.

Proposition 51

ΕØρε‹ν τ¾ν ™κ δύο Ñνοµάτων τετάρτην.

Ε

Ζ

∆ Α

To find a fourth binomial (straight-line).

Η

E

Θ Γ

√ 1 − k ′ 2 ). This, and the third apotome,

F

D

Β

A

'Εκκείσθωσαν δύο ¢ριθµοˆ οƒ ΑΓ, ΓΒ, éστε τÕν ΑΒ πρÕς τÕν ΒΓ λόγον µ¾ œχειν µήτε µ¾ν πρÕς τÕν ΑΓ, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. κሠ™κκείσθω ·ητ¾ ¹ ∆, κሠτÍ ∆ σύµµετρος œστω µήκει ¹ ΕΖ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΕΖ. κሠγεγονέτω æς Ð ΒΑ ¢ριθµÕς πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς ΖΗ· σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΕΖ τù ¢πÕ τÁς ΖΗ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΖΗ. κሠ™πεˆ Ð ΒΑ πρÕς τÕν ΑΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ τÕ ¢πÕ τÁς ΕΖ πρÕς

G

H C

B

Let the two numbers AC and CB be laid down such that AB does not have to BC, or to AC either, the ratio which (some) square number (has) to (some) square number [Prop. 10.28 lem. I]. And let the rational (straight-line) D be laid down. And let EF be commensurable in length with D. Thus, EF is also a rational (straight-line). And let it have been contrived that as the number BA (is) to AC, so the (square) on EF (is) to the (square) on F G [Prop. 10.6 corr.]. Thus, the (square) on EF is commensurable with the (square) on

337

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

τÕ ¢πÕ τÁς ΖΗ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΕΖ τÍ ΖΗ µήκει. αƒ ΕΖ, ΖΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· éστε ¹ ΕΗ ™κ δύο Ñνοµάτων ™στίν. λέγω δή, Óτι κሠτετάρτη. 'Επεˆ γάρ ™στιν æς Ð ΒΑ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς ΖΗ [µείζων δ Ð ΒΑ τοà ΑΓ], µε‹ζον ¥ρα τÕ ¢πÕ τÁς ΕΖ τοà ¢πÕ τÁς ΖΗ. œστω οâν τù ¢πÕ τÁς ΕΖ ‡σα τ¦ ¢πÕ τîν ΖΗ, Θ· ¢ναστρέψαντι ¥ρα æς Ð ΑΒ ¢ριθµÕς πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς Θ. Ð δ ΑΒ πρÕς τÕν ΒΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ' ¥ρα τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς Θ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. ¢σύµµετρος ¥ρα ™στˆν ¹ ΕΖ τÍ Θ µήκει· ¹ ΕΖ ¥ρα τÁς ΗΖ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καί ε„σιν αƒ ΕΖ, ΖΗ ·ητሠδυνάµει µόνον σύµµετροι, κሠ¹ ΕΖ τÍ ∆ σύµµετρός ™στι µήκει. `Η ΕΗ ¥ρα ™κ δύο Ñνοµάτων ™στˆ τετάρτη· Óπερ œδει δε‹ξαι.

F G [Prop. 10.6]. Thus, F G is also a rational (straightline). And since BA does not have to AC the ratio which (some) square number (has) to (some) square number, the (square) on EF does not have to the (square) on F G the ratio which (some) square number (has) to (some) square number either. Thus, EF is incommensurable in length with F G [Prop. 10.9]. Thus, EF and F G are rational (straight-lines which are) commensurable in square only. Hence, EG is a binomial (straight-line) [Prop. 10.36]. So, I say that (it is) also a fourth (binomial straight-line). For since as BA is to AC, so the (square) on EF (is) to the (square) on F G [and BA (is) greater than AC], the (square) on EF (is) thus greater than the (square) on F G [Prop. 5.14]. Therefore, let (the sum of) the squares on F G and H be equal to the (square) on EF . Thus, via conversion, as the number AB (is) to BC, so the (square) on EF (is) to the (square) on H [Prop. 5.19 corr.]. And AB does not have to BC the ratio which (some) square number (has) to (some) square number. Thus, the (square) on EF does not have to the (square) on H the ratio which (some) square number (has) to (some) square number either. Thus, EF is incommensurable in length with H [Prop. 10.9]. Thus, the square on EF is greater than (the square on) GF by the (square) on (some straight-line) incommensurable (in length) with (EF ). And EF and F G are rational (straight-lines which are) commensurable in square only. And EF is commensurable in length with D. Thus, EG is a fourth binomial (straight-line) [Def. 10.8].† (Which is) the very thing it was required to show.

√ If the rational straight-line has unit length, then the length of a fourth binomial straight-line is k (1 + 1/ 1 + k ′ ). This, and the fourth apotome, √ whose length is k (1 − 1/ 1 + k ′ ) [Prop. 10.88], are the roots of x2 − 2 k x + k 2 k ′ /(1 + k ′ ) = 0. †

νβ΄.

Proposition 52

ΕØρε‹ν τ¾ν ™κ δύο Ñνοµάτων πέµπτην.

Ε

Ζ

∆ Α

To find a fifth binomial straight-line.

Η

E

Θ Γ

F

D

Β

A

'Εκκείσθωσαν δύο ¢ριθµοˆ οƒ ΑΓ, ΓΒ, éστε τÕν ΑΒ πρÕς ˜κάτερον αÙτîν λόγον µ¾ œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, κሠ™κκείσθω ·ητή τις εÙθε‹α ¹ ∆, κሠτÍ ∆ σύµµετρος œστω [µήκει] ¹ ΕΖ· ·ητ¾ ¥ρα ¹ ΕΖ. κሠγεγονέτω æς Ð ΓΑ πρÕς τÕν ΑΒ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς ΖΗ. Ð δ ΓΑ

G H

C

B

Let the two numbers AC and CB be laid down such that AB does not have to either of them the ratio which (some) square number (has) to (some) square number [Prop. 10.38 lem.]. And let some rational straight-line D be laid down. And let EF be commensurable [in length] with D. Thus, EF (is) a rational (straight-

338

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

πρÕς τÕν ΑΒ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οØδ τÕ ¢πÕ τÁς ΕΖ ¥ρα πρÕς τÕ ¢πÕ τÁς ΖΗ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. αƒ ΕΖ, ΖΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ΕΗ. λέγω δή, Óτι κሠπέµπτη. 'Επεˆ γάρ ™στιν æς Ð ΓΑ πρÕς τÕν ΑΒ, οÛτως τÕ ¢πÕ τÁς ΕΖ πρÕς τÕ ¢πÕ τÁς ΖΗ, ¢νάπαλιν æς Ð ΒΑ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΖΕ· µε‹ζον ¥ρα τÕ ¢πÕ τÁς ΗΖ τοà ¢πÕ τÁς ΖΕ. œστω οâν τù ¢πÕ τÁς ΗΖ ‡σα τ¦ ¢πÕ τîν ΕΖ, Θ· ¢ναστρέψαντι ¥ρα ™στˆν æς Ð ΑΒ ¢ριθµÕς πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΗΖ πρÕς τÕ ¢πÕ τÁς Θ. Ð δ ΑΒ πρÕς τÕν ΒΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ' ¥ρα τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς Θ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. ¢σύµµετρος ¥ρα ™στˆν ¹ ΖΗ τÍ Θ µήκει· éστε ¹ ΖΗ τÁς ΖΕ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καί ε„σιν αƒ ΗΖ, ΖΕ ·ητሠδυνάµει µόνον σύµµετροι, κሠτÕ ΕΖ œλαττον Ôνοµα σύµµετρόν ™στι τÍ ™κκειµένV ·ητÍ τÍ ∆ µήκει. `Η ΕΗ ¥ρα ™κ δύο Ñνοµάτων ™στˆ πέµπτη· Óπερ œδει δε‹ξαι.

line). And let it have been contrived that as CA (is) to AB, so the (square) on EF (is) to the (square) on F G [Prop. 10.6 corr.]. And CA does not have to AB the ratio which (some) square number (has) to (some) square number. Thus, the (square) on EF does not have to the (square) on F G the ratio which (some) square number (has) to (some) square number either. Thus, EF and F G are rational (straight-lines which are) commensurable in square only [Prop. 10.9]. Thus, EG is a binomial (straight-line) [Prop. 10.36]. So, I say that (it is) also a fifth (binomial straight-line). For since as CA is to AB, so the (square) on EF (is) to the (square) on F G, inversely, as BA (is) to AC, so the (square) on F G (is) to the (square) on F E [Prop. 5.7 corr.]. Thus, the (square) on GF (is) greater than the (square) on F E [Prop. 5.14]. Therefore, let (the sum of) the (squares) on EF and H be equal to the (square) on GF . Thus, via conversion, as the number AB is to BC, so the (square) on GF (is) to the (square) on H [Prop. 5.19 corr.]. And AB does not have to BC the ratio which (some) square number (has) to (some) square number. Thus, the (square) on F G does not have to the (square) on H the ratio which (some) square number (has) to (some) square number either. Thus, F G is incommensurable in length with H [Prop. 10.9]. Hence, the square on F G is greater than (the square on) F E by the (square) on (some straight-line) incommensurable (in length) with (F G). And GF and F E are rational (straight-lines which are) commensurable in square only. And the lesser term EF is commensurable in length with the rational (straight-line previously) laid down, D. Thus, EG is a fifth binomial (straight-line).† (Which is) the very thing it was required to show.

√ If the rational straight-line has unit length, then the length of a fifth binomial straight-line is k ( 1 + k ′ + 1). This, and the fifth apotome, whose √ √ length is k ( 1 + k ′ − 1) [Prop. 10.89], are the roots of x2 − 2 k 1 + k ′ x + k 2 k ′ = 0.

†

νγ΄.

Proposition 53

ΕØρε‹ν τ¾ν ™κ δύο Ñνοµάτων ›κτην.

Ε ∆ Α

Κ Ζ Γ

To find a sixth binomial (straight-line).

E Η

Θ

D A

Β

'Εκκείσθωσαν δύο ¢ριθµοˆ οƒ ΑΓ, ΓΒ, éστε τÕν ΑΒ πρÕς ˜κάτερον αÙτîν λόγον µ¾ œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· œστω δ κሠ›τερος ¢ριθµÕς Ð ∆ µ¾ τετράγωνος íν µηδ πρÕς ˜κάτερον τîν ΒΑ, ΑΓ λόγον œχων, Öν τετράγωνος ¢ριθµÕς πρÕς

K F C

G

H

B

Let the two numbers AC and CB be laid down such that AB does not have to each of them the ratio which (some) square number (has) to (some) square number. And let D also be another number, which is not square, and does not have to each of BA and AC the ratio which

339

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

τετράγωνον ¢ριθµόν· κሠ™κκείσθω τις ·ητ¾ εÙθε‹α ¹ Ε, κሠγεγονέτω æς Ð ∆ πρÕς τÕν ΑΒ, οÛτως τÕ ¢πÕ τÁς Ε πρÕς τÕ ¢πÕ τÁς ΖΗ· σύµµετρον ¥ρα τÕ ¢πÕ τÁς Ε τù ¢πÕ τÁς ΖΗ. καί ™στι ·ητ¾ ¹ Ε· ·ητ¾ ¥ρα κሠ¹ ΖΗ. κሠ™πεˆ οÙκ œχει Ð ∆ πρÕς τÕν ΑΒ λόγον, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ τÕ ¢πÕ τÁς Ε ¥ρα πρÕς τÕ ¢πÕ τÁς ΖΗ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ¹ Ε τÍ ΖΗ µήκει. γεγονέτω δ¾ πάλιν æς Ð ΒΑ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ. σύµµετρον ¥ρα τÕ ¢πÕ τÁς ΖΗ τù ¢πÕ τÁς ΘΗ. ·ητÕν ¥ρα τÕ ¢πÕ τÁς ΘΗ· ·ητ¾ ¥ρα ¹ ΘΗ. κሠ™πεˆ Ð ΒΑ πρÕς τÕν ΑΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΖΗ τÍ ΗΘ µήκει. αƒ ΖΗ, ΗΘ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ΖΘ. δεικτέον δή, Óτι κሠ›κτη. 'Επεˆ γάρ ™στιν æς Ð ∆ πρÕς τÕν ΑΒ, οÛτως τÕ ¢πÕ τÁς Ε πρÕς τÕ ¢πÕ τÁς ΖΗ, œστι δ κሠæς Ð ΒΑ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ, δι' ‡σου ¥ρα ™στˆν æς Ð ∆ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τÁς Ε πρÕς τÕ ¢πÕ τÁς ΗΘ. Ð δ ∆ πρÕς τÕν ΑΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ τÕ ¢πÕ τÁς Ε ¥ρα πρÕς τÕ ¢πÕ τÁς ΗΘ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ Ε τÍ ΗΘ µήκει. ™δείχθη δ κሠτÍ ΖΗ ¢σύµµετρος· ˜κατέρα ¥ρα τîν ΖΗ, ΗΘ ¢σύµµετρός ™στι τÍ Ε µήκει. κሠ™πεί ™στιν æς Ð ΒΑ πρÕς τÕν ΑΓ, οÛτως τÕ ¢πÕ τ¾ς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ, µε‹ζον ¥ρα τÕ ¢πÕ τÁς ΖΗ τοà ¢πÕ τÁς ΗΘ. œστω οâν τù ¢πÕ [τÁς] ΖΗ ‡σα τ¦ ¢πÕ τîν ΗΘ, Κ· ¢ναστρέψαντι ¥ρα æς Ð ΑΒ πρÕς ΒΓ, οÛτως τÕ ¢πÕ ΖΗ πρÕς τÕ ¢πÕ τÁς Κ. Ð δ ΑΒ πρÕς τÕν ΒΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· éστε οÙδ τÕ ¢πÕ ΖΗ πρÕς τÕ ¢πÕ τÁς Κ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. ¢σύµµετρος ¥ρα ™στˆν ¹ ΖΗ τÍ Κ µήκει· ¹ ΖΗ ¥ρα τÁς ΗΘ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καί ε„σιν αƒ ΖΗ, ΗΘ ·ητሠδυνάµει µόνον σύµµετροι, κሠοÙδετέρα αÙτîν σύµµετρός ™στι µήκει τÍ ™κκειµένV ·ητV τÍ Ε. `Η ΖΘ ¥ρα ™κ δύο Ñνοµάτων ™στˆν ›κτη· Óπερ œδει δε‹ξαι.

(some) square number (has) to (some) square number either [Prop. 10.28 lem. I]. And let some rational straightline E be laid down. And let it have been contrived that as D (is) to AB, so the (square) on E (is) to the (square) on F G [Prop. 10.6 corr.]. Thus, the (square) on E (is) commensurable with the (square) on F G [Prop. 10.6]. And E is rational. Thus, F G (is) also rational. And since D does not have to AB the ratio which (some) square number (has) to (some) square number, the (square) on E thus does not have to the (square) on F G the ratio which (some) square number (has) to (some) square number either. Thus, E (is) incommensurable in length with F G [Prop. 10.9]. So, again, let it have be contrived that as BA (is) to AC, so the (square) on F G (is) to the (square) on GH [Prop. 10.6 corr.]. The (square) on F G (is) thus commensurable with the (square) on HG [Prop. 10.6]. The (square) on HG (is) thus rational. Thus, HG (is) rational. And since BA does not have to AC the ratio which (some) square number (has) to (some) square number, the (square) on F G does not have to the (square) on GH the ratio which (some) square number (has) to (some) square number either. Thus, F G is incommensurable in length with GH [Prop. 10.9]. Thus, F G and GH are rational (straight-lines which are) commensurable in square only. Thus, F H is a binomial (straight-line) [Prop. 10.36]. So, we must show that (it is) also a sixth (binomial straight-line). For since as D is to AB, so the (square) on E (is) to the (square) on F G, and also as BA is to AC, so the (square) on F G (is) to the (square) on GH, thus, via equality, as D is to AC, so the (square) on E (is) to the (square) on GH [Prop. 5.22]. And D does not have to AC the ratio which (some) square number (has) to (some) square number. Thus, the (square) on E does not have to the (square) on GH the ratio which (some) square number (has) to (some) square number either. E is thus incommensurable in length with GH [Prop. 10.9]. And (E) was also shown (to be) incommensurable (in length) with F G. Thus, F G and GH are each incommensurable in length with E. And since as BA is to AC, so the (square) on F G (is) to the (square) on GH, the (square) on F G (is) thus greater than the (square) on GH [Prop. 5.14]. Therefore, let (the sum of) the (squares) on GH and K be equal to the (square) on F G. Thus, via conversion, as AB (is) to BC, so the (square) on F G (is) to the (square) on K [Prop. 5.19 corr.]. And AB does not have to BC the ratio which (some) square number (has) to (some) square number. Hence, the (square) on F G does not have to the (square) on K the ratio which (some) square number (has) to (some) square number either. Thus, F G is

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ELEMENTS BOOK 10 incommensurable in length with K [Prop. 10.9]. The square on F G is thus greater than (the square on) GH by the (square) on (some straight-line which is) incommensurable (in length) with (F G). And F G and GH are rational (straight-lines which are) commensurable in square only, and neither of them is commensurable in length with the rational (straight-line) E (previously) laid down. Thus, F H is a sixth binomial (straight-line) [Def. 10.10].† (Which is) the very thing it was required to show.

†

p p If the rational straight-line has unit length, then the length of a sixth binomial straight-line is book10eps/k + book10eps/k ′ . This, and the p p p sixth apotome, whose length is book10eps/k − book10eps/k ′ [Prop. 10.90], are the roots of x2 − 2 book10eps/k x + (k − k ′ ) = 0.

ΛÁµµα.

Lemma

”Εστω δύο τετράγωνα τ¦ ΑΒ, ΒΓ κሠκείσθωσαν éστε ™π' εÙθείας εναι τ¾ν ∆Β τÍ ΒΕ· ™π' εÙθείας ¥ρα ™στˆ κሠ¹ ΖΒ τÍ ΒΗ. κሠσυµπεπληρώσθω τÕ ΑΓ παραλληλόγραµµον· λέγω, Óτι τετράγωνόν ™στι τÕ ΑΓ, κሠÓτι τîν ΑΒ, ΒΓ µέσον ¢νάλογόν ™στι τÕ ∆Η, κሠœτι τîν ΑΓ, ΓΒ µέσον ¢νάλογόν ™στι τÕ ∆Γ.

Let AB and BC be two squares, and let them be laid down such that DB is straight-on to BE. F B is, thus, also straight-on to BG. And let the parallelogram AC have been completed. I say that AC is a square, and that DG is the mean proportional to AB and BC, and, moreover, DC is the mean proportional to AC and CB.

Κ ∆

Α

Η Β

Ζ

Γ

K

Ε

D

Θ

A

'Επεˆ γ¦ρ ‡ση ™στˆν ¹ µν ∆Β τÍ ΒΖ, ¹ δ ΒΕ τÍ ΒΗ, Óλη ¥ρα ¹ ∆Ε ÓλV τÍ ΖΗ ™στιν ‡ση. ¢λλ' ¹ µν ∆Ε ˜κατέρv τîν ΑΘ, ΚΓ ™στιν ‡ση, ¹ δ ΖΗ ˜κατέρv τîν ΑΚ, ΘΓ ™στιν ‡ση· κሠ˜κατέρα ¥ρα τîν ΑΘ, ΚΓ ˜κατέρv τîν ΑΚ, ΘΓ ™στιν ‡ση. „σόπλευρον ¥ρα ™στˆ τÕ ΑΓ παραλληλόγραµµον· œστι δ κሠÑρθογώνιον· τετράγωνον ¥ρα ™στˆ τÕ ΑΓ. Κሠ™πεˆ ™στιν æς ¹ ΖΒ πρÕς τ¾ν ΒΗ, οÛτως ¹ ∆Β πρÕς τ¾ν ΒΕ, ¢λλ' æς µν ¹ ΖΒ πρÕς τ¾ν ΒΗ, οÛτως τÕ ΑΒ πρÕς τÕ ∆Η, æς δ ¹ ∆Β πρÕς τ¾ν ΒΕ, οÛτως τÕ ∆Η πρÕς τÕ ΒΓ, κሠæς ¥ρα τÕ ΑΒ πρÕς τÕ ∆Η, οÛτως τÕ ∆Η πρÕς τÕ ΒΓ. τîν ΑΒ, ΒΓ ¥ρα µέσον ¢νάλογόν ™στι τÕ ∆Η. Λέγω δή, Óτι κሠτîν ΑΓ, ΓΒ µέσον ¢νάλογόν [™στι] τÕ ∆Γ. 'Επεˆ γάρ ™στιν æς ¹ Α∆ πρÕς τ¾ν ∆Κ, οÛτως ¹ ΚΗ πρÕς τ¾ν ΗΓ· ‡ση γάρ [™στιν] ˜κατέρα ˜κατέρv· κሠσυνθέντι æς ¹ ΑΚ πρÕς Κ∆, οÛτως ¹ ΚΓ πρÕς ΓΗ, ¢λλ'

G

B

F

C E

H

For since DB is equal to BF , and BE to BG, the whole of DE is thus equal to the whole of F G. But DE is equal to each of AH and KC, and F G is equal to each of AK and HC [Prop. 1.34]. Thus, AH and KC are also equal to AK and HC, respectively. Thus, the parallelogram AC is equilateral. And (it is) also right-angled. Thus, AC is a square. And since as F B is to BG, so DB (is) to BE, but as F B (is) to BG, so AB (is) to DG, and as DB (is) to BE, so DG (is) to BC [Prop. 6.1], thus also as AB (is) to DG, so DG (is) to BC [Prop. 5.11]. Thus, DG is the mean proportional to AB and BC. So I also say that DC [is] the mean proportional to AC and CB. For since as AD is to DK, so KG (is) to GC. For [they are] respectively equal. And, via composition, as AK (is) to KD, so KC (is) to CG [Prop. 5.18]. But as AK (is) to KD, so AC (is) to CD, and as KC (is) to CG, so DC

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æς µν ¹ ΑΚ πρÕς Κ∆, οÛτως τÕ ΑΓ πρÕς τÕ Γ∆, æς δ ¹ ΚΓ πρÕς ΓΗ, οÛτως τÕ ∆Γ πρÕς ΓΒ, κሠæς ¥ρα τÕ ΑΓ πρÕς ∆Γ, οÛτως τÕ ∆Γ πρÕς τÕ ΒΓ. τîν ΑΓ, ΓΒ ¥ρα µέσον ¢νάλογόν ™στι τÕ ∆Γ· § προέκειτο δε‹ξαι.

(is) to CB [Prop. 6.1]. Thus, also, as AC (is) to DC, so DC (is) to BC [Prop. 5.11]. Thus, DC is the mean proportional to AC and CB. Which (is the very thing) it was prescribed to show.

νδ΄.

Proposition 54

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο If an area is contained by a rational (straight-line) Ñνοµάτων πρώτης, ¹ τÕ χωρίον δυναµένη ¥λογός ™στιν and a first binomial (straight-line) then the square-root ¹ καλουµένη ™κ δύο Ñνοµάτων. of the area is the irrational (straight-line which is) called binomial.† Α Η Ε Ζ ∆ Ρ Π A G E F D R Q Μ Β

Θ Κ

Λ

Ν

Ξ

M

Γ

B

Ο Σ Χωρίον γ¦ρ τÕ ΑΓ περιεχέσθω ØπÕ ·ητÁς τÁς ΑΒ κሠτÁς ™κ δύο Ñνοµάτων πρώτης τÁς Α∆· λέγω, Óτι ¹ τÕ ΑΓ χωρίον δυναµένη ¥λογός ™στιν ¹ καλουµένη ™κ δύο Ñνοµάτων. 'Επεˆ γ¦ρ ™κ δύο Ñνοµάτων ™στˆ πρώτη ¹ Α∆, διVρήσθω ε„ς τ¦ Ñνόµατα κατ¦ τÕ Ε, κሠœστω τÕ µε‹ζον Ôνοµα τÕ ΑΕ. φανερÕν δή, Óτι αƒ ΑΕ, Ε∆ ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ ΑΕ τÁς Ε∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτV, κሠ¹ ΑΕ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ΑΒ µήκει. τετµήσθω δ¾ ¹ Ε∆ δίχα κατ¦ τÕ Ζ σηµε‹ον. κሠ™πεˆ ¹ ΑΕ τÁς Ε∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, ™¦ν ¥ρα τù τετάρτJ µέρει τοà ¢πÕ τÁς ™λάσσονος, τουτέστι τù ¢πÕ τ¾ς ΕΖ, ‡σον παρ¦ τ¾ν µείζονα τ¾ν ΑΕ παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, ε„ς σύµµετρα αÙτÁν διαιρε‹. παραβεβλήσθω οâν παρ¦ τ¾ν ΑΕ τù ¢πÕ τÁς ΕΖ ‡σον τÕ ØπÕ ΑΗ, ΗΕ· σύµµετρος ¥ρα ™στˆν ¹ ΑΗ τÍ ΕΗ µήκει. κሠ½χθωσαν ¢πÕ τîν Η, Ε, Ζ Ðποτέρv τîν ΑΒ, Γ∆ παράλληλοι αƒ ΗΘ, ΕΚ, ΖΛ· κሠτù µν ΑΘ παραλληλογράµµJ ‡σον τετράγωνον συνεστάτω τÕ ΣΝ, τù δ ΗΚ ‡σον τÕ ΝΠ, κሠκείσθω éστε ™π' εÙθείας εναι τ¾ν ΜΝ τÍ ΝΞ· ™π' εÙθείας ¥ρα ™στˆ κሠ¹ ΡΝ τÍ ΝΟ. κሠσυµπεπληρώσθω τÕ ΣΠ παραλληλόγραµµον· τετράγωνον ¥ρα ™στˆ τÕ ΣΠ. κሠ™πεˆ τÕ ØπÕ τîν ΑΗ, ΗΕ ‡σον ™στˆ τù ¢πÕ τÁς ΕΖ, œστιν ¥ρα æς ¹ ΑΗ πρÕς ΕΖ, οÛτως ¹ ΖΕ πρÕς ΕΗ· κሠæς ¥ρα τÕ ΑΘ πρÕς ΕΛ, τÕ ΕΛ πρÕς ΚΗ· τîν ΑΘ, ΗΚ ¥ρα µέσον ¢νάλογόν ™στι τÕ ΕΛ. ¢λλ¦ τÕ µν ΑΘ ‡σον ™στˆ τù ΣΝ, τÕ δ ΗΚ ‡σον τù ΝΠ· τîν ΣΝ, ΝΠ ¥ρα µέσον ¢νάλογόν ™στι τÕ ΕΛ. œστι δ τîν αÙτîν τîν ΣΝ, ΝΠ µέσον ¢νάλογον κሠτÕ ΜΡ· ‡σον ¥ρα ™στˆ τÕ ΕΛ τù ΜΡ· éστε κሠτù ΟΞ ‡σον ™στίν. œστι δ κሠτ¦ ΑΘ,

H K

L

N

O

C

S

P

For let the area AC be contained by the rational (straight-line) AB and by the first binomial (straightline) AD. I say that square-root of area AC is the irrational (straight-line which is) called binomial. For since AD is a first binomial (straight-line), let it have been divided into its (component) terms at E, and let AE be the greater term. So, (it is) clear that AE and ED are rational (straight-lines which are) commensurable in square only, and that the square on AE is greater than (the square on) ED by the (square) on (some straight-line) commensurable (in length) with (AE), and that AE is commensurable (in length) with the rational (straight-line) AB (first) laid out [Def. 10.5]. So, let ED have been cut in half at point F . And since the square on AE is greater than (the square on) ED by the (square) on (some straight-line) commensurable (in length) with (AE), thus if a (rectangle) equal to the fourth part of the (square) on the lesser (term)—that is to say, the (square) on EF —falling short by a square figure, is applied to the greater (term) AE, then it divides it into (terms which are) commensurable (in length) [Prop 10.17]. Therefore, let the (rectangle contained) by AG and GE, equal to the (square) on EF , have been applied to AE. AG is thus commensurable in length with EG. And let GH, EK, and F L have been drawn from (points) G, E, and F (respectively), parallel to either of AB or CD. And let the square SN , equal to the parallelogram AH, have been constructed, and (the square) N Q, equal to (the parallelogram) GK [Prop. 2.14]. And let M N be laid down so as to be straight-on to N O. RN is thus also straight-on to N P . And let the parallelogram SQ have been completed. SQ is thus a square [Prop. 10.53 lem.]. And since

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ELEMENTS BOOK 10

ΗΚ το‹ς ΣΝ, ΝΠ ‡σα· Óλον ¥ρα τÕ ΑΓ ‡σον ™στˆν ÓλJ τù ΣΠ, τουτέστι τù ¢πÕ τÁς ΜΞ τετραγώνJ· τÕ ΑΓ ¥ρα δύναται ¹ ΜΞ. λέγω, Óτι ¹ ΜΞ ™κ δύο Ñνοµάτων ™στίν. 'Επεˆ γ¦ρ σύµµετρός ™στιν ¹ ΑΗ τÍ ΗΕ, σύµµετρός ™στι κሠ¹ ΑΕ ˜κατέρv τîν ΑΗ, ΗΕ. Øπόκειται δ κሠ¹ ΑΕ τÍ ΑΒ σύµµετρος· καˆ αƒ ΑΗ, ΗΕ ¥ρα τÍ ΑΒ σύµµετροί ε„σιν. καί ™στι ·ητ¾ ¹ ΑΒ· ·ητ¾ ¥ρα ™στˆ κሠ˜κατέρα τîν ΑΗ, ΗΕ· ·ητÕν ¥ρα ™στˆν ˜κάτερον τîν ΑΘ, ΗΚ, καί ™στι σύµµετρον τÕ ΑΘ τù ΗΚ. ¢λλ¦ τÕ µν ΑΘ τù ΣΝ ‡σον ™στίν, τÕ δ ΗΚ τù ΝΠ· κሠτ¦ ΣΝ, ΝΠ ¥ρα, τουτέστι τ¦ ¢πÕ τîν ΜΝ, ΝΞ, ·ητά ™στι κሠσύµµετρα. κሠ™πεˆ ¢σύµµετρός ™στιν ¹ ΑΕ τÍ Ε∆ µήκει, ¢λλ' ¹ µν ΑΕ τÍ ΑΗ ™στι σύµµετρος, ¹ δ ∆Ε τÍ ΕΖ σύµµετρος, ¢σύµµετρος ¥ρα κሠ¹ ΑΗ τÍ ΕΖ· éστε κሠτÕ ΑΘ τù ΕΛ ¢σύµµετρόν ™στιν. ¢λλ¦ τÕ µν ΑΘ τù ΣΝ ™στιν ‡σον, τÕ δ ΕΛ τù ΜΡ· κሠτÕ ΣΝ ¥ρα τù ΜΡ ¢σύµµετρόν ™στιν. ¢λλ' æς τÕ ΣΝ πρÕς ΜΡ, ¹ ΟΝ πρÕς τ¾ν ΝΡ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΟΝ τÍ ΝΡ. ‡ση δ ¹ µν ΟΝ τÍ ΜΝ, ¹ δ ΝΡ τÍ ΝΞ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΜΝ τÍ ΝΞ. καί ™στι τÕ ¢πÕ τÁς ΜΝ σύµµετρον τù ¢πÕ τÁς ΝΞ, κሠ·ητÕν ˜κάτερον· αƒ ΜΝ, ΝΞ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. `Η ΜΞ ¥ρα ™κ δύο Ñνοµάτων ™στˆ κሠδύναται τÕ ΑΓ· Óπερ œδει δε‹ξαι.

the (rectangle contained) by AG and GE is equal to the (square) on EF , thus as AG is to EF , so F E (is) to EG [Prop. 6.17]. And thus as AH (is) to EL, (so) EL (is) to KG [Prop. 6.1]. Thus, EL is the mean proportional to AH and GK. But, AH is equal to SN , and GK (is) equal to N Q. EL is thus the mean proportional to SN and N Q. And M R is also the mean proportional to the same— (namely), SN and N Q [Prop. 10.53 lem.]. EL is thus equal to M R. Hence, it is also equal to P O [Prop. 1.43]. And AH plus GK is equal to SN plus N Q. Thus, the whole of AC is equal to the whole of SQ—that is to say, to the square on M O. Thus, M O (is) the square-root of (area) AC. I say that M O is a binomial (straight-line). For since AG is commensurable (in length) with GE, AE is also commensurable (in length) with each of AG and GE [Prop. 10.15]. And AE was also assumed (to be) commensurable (in length) with AB. Thus, AG and GE are also commensurable (in length) with AB [Prop. 10.12]. And AB is rational. AG and GE are thus each also rational. Thus, AH and GK are each rational (areas), and AH is commensurable with GK [Prop. 10.19]. But, AH is equal to SN , and GK to N Q. SN and N Q—that is to say, the (squares) on M N and N O (respectively)—are thus also rational and commensurable. And since AE is incommensurable in length with ED, but AE is commensurable (in length) with AG, and DE (is) commensurable (in length) with EF , AG (is) thus also incommensurable (in length) with EF [Prop. 10.13]. Hence, AH is also incommensurable with EL [Props. 6.1, 10.11]. But, AH is equal to SN , and EL to M R. Thus, SN is also incommensurable with M R. But, as SN (is) to M R, (so) P N (is) to N R [Prop. 6.1]. P N is thus incommensurable (in length) with N R [Prop. 10.11]. And P N (is) equal to M N , and N R to N O. Thus, M N is incommensurable (in length) with N O. And the (square) on M N is commensurable with the (square) on N O, and each (is) rational. M N and N O are thus rational (straight-lines which are) commensurable in square only. Thus, M O is (both) a binomial (straight-line) [Prop. 10.36], and the square-root of AC. (Which is) the very thing it was required to show.

† If the rational straight-line has unit length, then this proposition states that the square-root of a first binomial straight-line is a binomial p √ straight-line: i.e., a first binomial straight-line has a length k + k 1 − k ′ 2 whose square-root can be written ρ (1 + book10eps/k ′′ ), where p ρ = book10eps/k (1 + k ′ )/2 and k ′′ = (1 − k ′ )/(1 + k ′ ). This is the length of a binomial straight-line (see Prop. 10.36), since ρ is rational.

νε΄.

Proposition 55

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο If an area is contained by a rational (straight-line) and Ðνοµάτων δευτέρας, ¹ τÕ χωρίον δυναµένη ¥λογός ™στιν a second binomial (straight-line) then the square-root of

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ELEMENTS BOOK 10

¹ καλουµένη ™κ δύο µέσων πρώτη. Α

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the area is the irrational (straight-line which is) called first bimedial.†

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Ο Σ Περιεχέσθω γ¦ρ χωρίον τÕ ΑΒΓ∆ ØπÕ ·ητÁς τÁς ΑΒ κሠτÁς ™κ δύο Ñνοµάτων δυετέρας τÁς Α∆· λέγω, Óτι ¹ τÕ ΑΓ χωρίον δυναµένη ™κ δύο µέσων πρώτη ™στίν. 'Επεˆ γ¦ρ ™κ δύο Ñνοµάτων δευτέρα ™στˆν ¹ Α∆, διVρήσθω ε„ς τ¦ Ñνόµατα κατ¦ τÕ Ε, éστε τÕ µε‹ζον Ôνοµα εναι τÕ ΑΕ· αƒ ΑΕ, Ε∆ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ ΑΕ τÁς Ε∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠτÕ œλαττον Ôνοµα ¹ Ε∆ σύµµετρόν ™στι τÍ ΑΒ µήκει. τετµήσθω ¹ Ε∆ δίχα κατ¦ τÕ Ζ, κሠτù ¢πÕ τÁς ΕΖ ‡σον παρ¦ τ¾ν ΑΕ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ τÕ ØπÕ τîν ΑΗΕ· σύµµετρος ¥ρα ¹ ΑΗ τÍ ΗΕ µήκει. κሠδι¦ τîν Η, Ε, Ζ παράλληλοι ½χθωσαν τα‹ς ΑΒ, Γ∆ αƒ ΗΘ, ΕΚ, ΖΛ, κሠτù µν ΑΘ παραλληλογράµµJ ‡σον τετράγωνον συνεστάτω τÕ ΣΝ, τù δ ΗΚ ‡σον τετράγωνον τÕ ΝΠ, κሠκείσθω éστε ™π' εÙθείας εναι τ¾ν ΜΝ τÍ ΝΞ· ™π' εÙθείας ¥ρα [™στˆ] κሠ¹ ΡΝ τÁ ΝΟ. κሠσυµπεπληρώσθω τÕ ΣΠ τετράγωνον· φανερÕν δ¾ ™κ τοà προδεδειγµένου, Óτι τÕ ΜΡ µέσον ¢νάλογόν ™στι τîν ΣΝ, ΝΠ, κሠ‡σον τù ΕΛ, κሠÓτι τÕ ΑΓ χωρίον δύναται ¹ ΜΞ. δεικτέον δή, Óτι ¹ ΜΞ ™κ δύο µέσων ™στˆ πρώτη. 'Επεˆ ¢σύµµετρός ™στιν ¹ ΑΕ τÍ Ε∆ µήκει, σύµµετρος δ ¹ Ε∆ τÍ ΑΒ, ¢σύµµετρος ¥ρα ¹ ΑΕ τÍ ΑΒ. κሠ™πεˆ σύµµετρός ™στιν ¹ ΑΗ τÍ ΕΗ, σύµµετρός ™στι κሠ¹ ΑΕ ˜κατέρv τîν ΑΗ, ΗΕ. ¢λλ¦ ¹ ΑΕ ¢σύµµετρος τÍ ΑΒ µήκει· καˆ αƒ ΑΗ, ΗΕ ¥ρα ¢σύµµετροί ε„σι τÍ ΑΒ. αƒ ΒΑ, ΑΗ, ΗΕ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· éστε µέσον ™στˆν ˜κάτερον τîν ΑΘ, ΗΚ. éστε κሠ˜κάτερον τîν ΣΝ, ΝΠ µέσον ™στίν. καˆ αƒ ΜΝ, ΝΞ ¥ρα µέσαι ε„σίν. κሠ™πεˆ σύµµετρος ¹ ΑΗ τÍ ΗΕ µήκει, σύµµετρόν ™στι κሠτÕ ΑΘ τù ΗΚ, τουτέστι τÕ ΣΝ τù ΝΠ, τουτέστι τÕ ¢πÕ τÁς ΜΝ τù ¢πÕ τÁς ΝΞ [éστε δυνάµει ε„σˆ σύµµετροι αƒ ΜΝ, ΝΞ]. κሠ™πεˆ ¢σύµµετρός ™στιν ¹ ΑΕ τÍ Ε∆ µήκει, ¢λλ' ¹ µν ΑΕ σύµµετρός ™στι τÍ ΑΗ, ¹ δ Ε∆ τÍ ΕΖ σύµµετρος, ¢σύµµετρος ¥ρα ¹ ΑΗ τÍ ΕΖ· éστε κሠτÕ ΑΘ τù ΕΛ ¢σύµµετρόν ™στιν, τουτέστι τÕ ΣΝ τù ΜΡ, τουτέστιν Ð ΟΝ τÍ ΝΡ, τουτέστιν ¹ ΜΝ τÍ ΝΞ ¢σύµµετρός ™στι µήκει. ™δείχθησαν δ αƒ ΜΝ, ΝΞ κሠµέσαι οâσαι κሠδυνάµει σύµµετροι· αƒ ΜΝ,

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For let the area ABCD be contained by the rational (straight-line) AB and by the second binomial (straightline) AD. I say that the square-root of area AC is a first bimedial (straight-line). For since AD is a second binomial (straight-line), let it have been divided into its (component) terms at E, such that AE is the greater term. Thus, AE and ED are rational (straight-lines which are) commensurable in square only, and the square on AE is greater than (the square on) ED by the (square) on (some straight-line) commensurable (in length) with (AE), and the lesser term ED is commensurable in length with AB [Def. 10.6]. Let ED have been cut in half at F . And let the (rectangle contained) by AGE, equal to the (square) on EF , have been applied to AE, falling short by a square figure. AG (is) thus commensurable in length with GE [Prop. 10.17]. And let GH, EK, and F L have been drawn through (points) G, E, and F (respectively), parallel to AB and CD. And let the square SN , equal to the parallelogram AH, have been constructed, and the square N Q, equal to GK. And let M N be laid down so as to be straight-on to N O. Thus, RN [is] also straight-on to N P . And let the square SQ have been completed. So, (it is) clear from what has been previously demonstrated [Prop. 10.53 lem.] that M R is the mean proportional to SN and N Q, and (is) equal to EL, and that M O is the square-root of the area AC. So, we must show that M O is a first bimedial (straight-line). Since AE is incommensurable in length with ED, and ED (is) commensurable (in length) with AB, AE (is) thus incommensurable (in length) with AB [Prop. 10.13]. And since AG is commensurable (in length) with EG, AE is also commensurable (in length) with each of AG and GE [Prop. 10.15]. But, AE is incommensurable in length with AB. Thus, AG and GE are also (both) incommensurable (in length) with AB [Prop. 10.13]. Thus, BA, AG, and (BA, and) GE are (pairs of) rational (straight-lines which are) commensurable in square only. And, hence, each of AH and GK is a medial (area) [Prop. 10.21]. Hence, each of SN and N Q is also a medial (area). Thus, M N and N O

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ELEMENTS BOOK 10

ΝΞ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι. λέγω δή, Óτι κሠ·ητÕν περιέχουσιν. ™πεˆ γ¦ρ ¹ ∆Ε Øπόκειται ˜κατέρv τîν ΑΒ, ΕΖ σύµµετρος, σύµµετρος ¥ρα κሠ¹ ΕΖ τÍ ΕΚ. κሠ·ητ¾ ˜κατέρα αÙτîν· ·ητÕν ¥ρα τÕ ΕΛ, τουτέστι τÕ ΜΡ· τÕ δ ΜΡ ™στι τÕ ØπÕ τîν ΜΝΞ. ™¦ν δ δύο µέσαι δυνάµει µόνον σύµµετροι συντεθîσι ·ητÕν περιέχουσαι, ¹ Óλη ¥λογός ™στιν, καλε‹ται δ ™κ δύο µέσων πρώτη. `Η ¥ρα ΜΞ ™κ δύο µέσων ™στˆ πρώτη· Óπερ œδει δε‹ξαι.

are medial (straight-lines). And since AG (is) commensurable in length with GE, AH is also commensurable with GK—that is to say, SN with N Q—that is to say, the (square) on M N with the (square) on N O [hence, M N and N O are commensurable in square] [Props. 6.1, 10.11]. And since AE is incommensurable in length with ED, but AE is commensurable (in length) with AG, and ED commensurable (in length) with EF , AG (is) thus incommensurable (in length) with EF [Prop. 10.13]. Hence, AH is also incommensurable with EL—that is to say, SN with M R—that is to say, P N with N R—that is to say, M N is incommensurable in length with N O [Props. 6.1, 10.11]. But M N and N O have also been shown to be medial (straight-lines) which are commensurable in square. Thus, M N and N O are medial (straightlines which are) commensurable in square only. So, I say that they also contain a rational (area). For since DE was assumed (to be) commensurable (in length) with each of AB and EF , EF (is) thus also commensurable with EK [Prop. 10.12]. And they (are) each rational. Thus, EL— that is to say, M R—(is) rational [Prop. 10.19]. And M R is the (rectangle contained) by M N O. And if two medial (straight-lines), commensurable in square only, which contain a rational (area), are added together, then the whole is (that) irrational (straight-line which is) called first bimedial [Prop. 10.37]. Thus, M O is a first bimedial (straight-line). (Which is) the very thing it was required to show.

†

If the rational straight-line has unit length, then this proposition states that the square-root of a second binomial straight-line is a first bimedial √ straight-line: i.e., a second binomial straight-line has a length k/ 1 − k ′ 2 + k whose square-root can be written ρ (k ′′1/4 + k ′′3/4 ), where p ρ = (k/2) (1 + k ′ )/(1 − k ′ ) and k ′′ = (1 − k ′ )/(1 + k ′ ). This is the length of a first bimedial straight-line (see Prop. 10.37), since ρ is rational.

ν$΄.

Proposition 56

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο If an area is contained by a rational (straight-line) and Ñνοµάτων τρίτης, ¹ τÕ χωρίον δυναµένη ¥λογός ™στιν ¹ a third binomial (straight-line) then the square-root of καλουµένη ™κ δύο µέσων δευτέρα. the area is the irrational (straight-line which is) called second bimedial.† Α Η Ε Ζ ∆ Ρ Π A G E F D R Q Μ Β

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S P Ο Σ Χωρίον γ¦ρ τÕ ΑΒΓ∆ περιεχέσθω ØπÕ ·ητÁς τÁς For let the area ABCD be contained by the rational ΑΒ κሠτÁς ™κ δύο Ñνοµάτων τρίτης τÁς Α∆ διVρηµένης (straight-line) AB and by the third binomial (straightε„ς τ¦ Ñνόµατα κατ¦ τÕ Ε, ïν µε‹ζόν ™στι τÕ ΑΕ· λέγω, line) AD, which has been divided into its (component) Óτι ¹ τÕ ΑΓ χωρίον δυναµένη ¥λογός ™στιν ¹ καλουµένη terms at E, of which AE is the greater. I say that the

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™κ δύο µέσων δευτέρα. Κατεσκευάσθω γ¦ρ τ¦ αÙτ¦ το‹ς πρότερον. κሠ™πεˆ ™κ δύο Ñνοµάτων ™στˆ τρίτη ¹ Α∆, αƒ ΑΕ, Ε∆ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ ΑΕ τÁς Ε∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠοÙδετέρα τîν ΑΕ, Ε∆ σύµµετρός [™στι] τÍ ΑΒ µήκει. еοίως δ¾ το‹ς προδεδειγµένοις δείξοµεν, Óτι ¹ ΜΞ ™στιν ¹ τÕ ΑΓ χωρίον δυναµένη, καˆ αƒ ΜΝ, ΝΞ µέσαι ε„σˆ δυνάµει µόνον σύµµετροι· éστε ¹ ΜΞ ™κ δύο µέσων ™στίν. δεικτέον δή, Óτι κሠδευτέρα. [Καˆ] ™πεˆ ¢σύµµετρός ™στιν ¹ ∆Ε τÍ ΑΒ µήκει, τουτέστι τÍ ΕΚ, σύµµετρος δ ¹ ∆Ε τÍ ΕΖ, ¢σύµµετρος ¥ρα ™στˆν ¹ ΕΖ τÍ ΕΚ µήκει. καί ε„σι ·ηταί· αƒ ΖΕ, ΕΚ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. µέσον ¥ρα [™στˆ] τÕ ΕΛ, τουτέστι τÕ ΜΡ· κሠπεριέχεται ØπÕ τîν ΜΝΞ· µέσον ¥ρα ™στˆ τÕ ØπÕ τîν ΜΝΞ. `Η ΜΞ ¥ρα ™κ δύο µέσων ™στˆ δευτέρα· Óπερ œδει δε‹ξαι.

square-root of area AC is the irrational (straight-line which is) called second bimedial. For let the same construction be made as previously. And since AD is a third binomial (straight-line), AE and ED are thus rational (straight-lines which are) commensurable in square only, and the square on AE is greater than (the square on) ED by the (square) on (some straight-line) commensurable (in length) with (AE), and neither of AE and ED [is] commensurable in length with AB [Def. 10.7]. So, similarly to that which has been previously demonstrated, we can show that M O is the square-root of area AC, and M N and N O are medial (straight-lines which are) commensurable in square only. Hence, M O is bimedial. So, we must show that (it is) also second (bimedial). [And] since DE is incommensurable in length with AB—that is to say, with EK—and DE (is) commensurable (in length) with EF , EF is thus incommensurable in length with EK [Prop. 10.13]. And they are (both) rational (straight-lines). Thus, F E and EK are rational (straight-lines which are) commensurable in square only. EL—that is to say, M R—[is] thus medial [Prop. 10.21]. And it is contained by M N O. Thus, the (rectangle contained) by M N O is medial. Thus, M O is a second bimedial (straight-line) [Prop. 10.38]. (Which is) the very thing it was required to show.

†

If the rational straight-line has unit length, then this proposition states that the square-root of a third binomial straight-line is a second bimedial √ straight-line: i.e., a third binomial straight-line has a length k 1/2 (1 + 1 − k ′ 2 ) whose square-root can be written ρ (k 1/4 + k ′′1/2 /k 1/4 ), where p ρ = (1 + k ′ )/2 and k ′′ = k (1 − k ′ )/(1 + k ′ ). This is the length of a second bimedial straight-line (see Prop. 10.38), since ρ is rational.

νζ΄.

Proposition 57

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο If an area is contained by a rational (straight-line) and Ñνοµάτων τετάρτης, ¹ τÕ χωρίον δυναµένη ¥λογός ™στιν a fourth binomial (straight-line) then the square-root of ¹ καλουµένη µείζων. the area is the irrational (straight-line which is) called major.† Α Η Ε Ζ ∆ Ρ Π A G E F D R Q Μ Β

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Ο Σ Χωρίον γ¦ρ τÕ ΑΓ περιεχέσθω ØπÕ ·ητÁς τÁς ΑΒ κሠτÁς ™κ δύο Ñνοµάτων τετάρτης τÁς Α∆ διVρηµένης ε„ς τ¦ Ñνόµατα κατ¦ τÕ Ε, ïν µε‹ζον œστω τÕ ΑΕ· λέγω, Óτι ¹ τÕ ΑΓ χωρίον δυναµένη ¥λογός ™στιν ¹ καλουµένη µείζων. 'Επεˆ γ¦ρ ¹ Α∆ ™κ δύο Ñνοµάτων ™στˆ τετάρτη, αƒ

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For let the area AC be contained by the rational (straight-line) AB and the fourth binomial (straight-line) AD, which has been divided into its (component) terms at E, of which let AE be the greater. I say that the squareroot of AC is the irrational (straight-line which is) called major.

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ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

ΑΕ, Ε∆ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ ΑΕ τÁς Ε∆ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ, κሠ¹ ΑΕ τÍ ΑΒ σύµµετρός [™στι] µήκει. τετµήσθω ¹ ∆Ε δίχα κατ¦ τÕ Ζ, κሠτù ¢πÕ τÁς ΕΖ ‡σον παρ¦ τ¾ν ΑΕ παραβεβλήσθω παραλληλόγραµµον τÕ ØπÕ ΑΗ, ΗΕ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΗ τÍ ΗΕ µήκει. ½χθωσαν παράλληλοι τÍ ΑΒ αƒ ΗΘ, ΕΚ, ΖΛ, κሠτ¦ λοιπ¦ τ¦ αÙτ¦ το‹ς πρÕ τούτου γεγονέτω· φανερÕν δή, Óτι ¹ τÕ ΑΓ χωρίον δυναµένη ™στˆν ¹ ΜΞ. δεικτέον δή, Óτι ¹ ΜΞ ¥λογός ™στιν ¹ καλουµένη µείζων. 'Επεˆ ¢σύµµετρός ™στιν ¹ ΑΗ τÍ ΕΗ µήκει, ¢σύµµετρόν ™στι κሠτÕ ΑΘ τù ΗΚ, τουτέστι τÕ ΣΝ τù ΝΠ· αƒ ΜΝ, ΝΞ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι. κሠ™πεˆ σύµµετρός ™στιν ¹ ΑΕ τÍ ΑΒ µήκει, ·ητόν ™στι τÕ ΑΚ· καί ™στιν ‡σον το‹ς ¢πÕ τîν ΜΝ, ΝΞ· ·ητÕν ¥ρα [™στˆ] κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν ΜΝ, ΝΞ. κሠ™πεˆ ¢σύµµετρός [™στιν] ¹ ∆Ε τÍ ΑΒ µήκει, τουτέστι τÍ ΕΚ, ¢λλ¦ ¹ ∆Ε σύµµετρός ™στι τÍ ΕΖ, ¢σύµµετρος ¥ρα ¹ ΕΖ τÍ ΕΚ µήκει. αƒ ΕΚ, ΕΖ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· µέσον ¥ρα τÕ ΛΕ, τουτέστι τÕ ΜΡ. κሠπεριέχεται ØπÕ τîν ΜΝ, ΝΞ· µέσον ¥ρα ™στˆ τÕ ØπÕ τîν ΜΝ, ΝΞ. κሠ·ητÕν τÕ [συγκείµενον] ™κ τîν ¢πÕ τîν ΜΝ, ΝΞ, καί ε„σιν ¢σύµµετροι αƒ ΜΝ, ΝΞ δυνάµει. ™¦ν δ δύο εÙθε‹αι δυνάµει ¢σύµµετροι συντεθîσι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων ·ητόν, τÕ δ' Øπ' αÙτîν µέσον, ¹ Óλη ¥λογός ™στιν, καλε‹ται δ µείζων. `Η ΜΞ ¥ρα ¥λογός ™στιν ¹ καλουµένη µείζων, κሠδύναται τÕ ΑΓ χωρίον· Óπερ œδει δε‹ξαι.

For since AD is a fourth binomial (straight-line), AE and ED are thus rational (straight-lines which are) commensurable in square only, and the square on AE is greater than (the square on) ED by the (square) on (some straight-line) incommensurable (in length) with (AE), and AE [is] commensurable in length with AB [Def. 10.8]. Let DE have been cut in half at F , and let the parallelogram (contained by) AG and GE, equal to the (square) on EF , (and falling short by a square figure) have been applied to AE. AG is thus incommensurable in length with GE [Prop. 10.18]. Let GH, EK, and F L have been drawn parallel to AB, and let the rest (of the construction) have been made the same as the (proposition) before this. So, it is clear that M O is the square-root of area AC. So, we must show that M O is the irrational (straight-line which is) called major. Since AG is incommensurable in length with EG, AH is also incommensurable with GK—that is to say, SN with N Q [Props. 6.1, 10.11]. Thus, M N and N O are incommensurable in square. And since AE is commensurable in length with AB, AK is rational [Prop. 10.19]. And it is equal to the (sum of the squares) on M N and N O. Thus, the sum of the (squares) on M N and N O [is] also rational. And since DE [is] incommensurable in length with AB [Prop. 10.13]—that is to say, with EK—but DE is commensurable (in length) with EF , EF (is) thus incommensurable in length with EK [Prop. 10.13]. Thus, EK and EF are rational (straightlines which are) commensurable in square only. LE— that is to say, M R—(is) thus medial [Prop. 10.21]. And it is contained by M N and N O. The (rectangle contained) by M N and N O is thus medial. And the [sum] of the (squares) on M N and N O (is) rational, and M N and N O are incommensurable in square. And if two straightlines (which are) incommensurable in square, making the sum of the squares on them rational, and the (rectangle contained) by them medial, are added together, then the whole is the irrational (straight-line which is) called major [Prop. 10.39]. Thus, M O is the irrational (straight-line which is) called major. And (it is) the square-root of area AC. (Which is) the very thing it was required to show.

†

If the rational straight-line has unit length, then this proposition states that the square-root of a fourth binomial straight-line is a major straightq √ ′ ′′ line: i.e., a fourth binomial straight-line has a length k (1 + 1/ 1 + k ) whose square-root can be written ρ [1 + k /(1 + k ′′ 2 )1/2 ]/2 + q p ρ [1 − k ′′ /(1 + k ′′ 2 )1/2 ]/2, where ρ = book10eps/k and k ′′ 2 = k ′ . This is the length of a major straight-line (see Prop. 10.39), since ρ is rational.

νη΄.

Proposition 58

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο

347

If an area is contained by a rational (straight-line) and

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

Ñνοµάτων πέµπτης, ¹ τÕ χωρίον δυναµένη ¥λογός ™στιν ¹ καλουµένη ·ητÕν κሠµέσον δυναµένη. Χωρίον γ¦ρ τÕ ΑΓ περιεχέσθω ØπÕ ·ητÁς τÁς ΑΒ κሠτÁς ™κ δύο Ñνοµάτων πέµπτης τÁς Α∆ διVρηµένης ε„ς τ¦ Ñνόµατα κατ¦ τÕ Ε, éστε τÕ µε‹ζον Ôνοµα εναι τÕ ΑΕ· λέγω [δή], Óτι ¹ τÕ ΑΓ χωρίον δυναµένη ¥λογός ™στιν ¹ καλουµένη ·ητÕν κሠµέσον δυναµένη.

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a fifth binomial (straight-line) then the square-root of the area is the irrational (straight-line which is) called the square-root of a rational plus a medial (area).† For let the area AC be contained by the rational (straight-line) AB and the fifth binomial (straight-line) AD, which has been divided into its (component) terms at E, such that AE is the greater term. [So] I say that the square-root of area AC is the irrational (straight-line which is) called the square-root of a rational plus a medial (area). G E F D A R Q

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Σ Ο Κατεσκευάσθω γ¦ρ τ¦ αÙτ¦ το‹ς πρότερον δεδειγµένοις· φανερÕν δή, Óτι ¹ τÕ ΑΓ χωρίον δυναµένη ™στˆν ¹ ΜΞ. δεικτέον δή, Óτι ¹ ΜΞ ™στιν ¹ ·ητÕν κሠµέσον δυναµένη. 'Επεˆ γ¦ρ ¢σύµµετρός ™στιν ¹ ΑΗ τÍ ΗΕ, ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ΑΘ τù ΘΕ, τουτέστι τÕ ¢πÕ τÁς ΜΝ τù ¢πÕ τÁς ΝΞ· αƒ ΜΝ, ΝΞ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι. κሠ™πεˆ ¹ Α∆ ™κ δύο Ñνοµάτων ™στˆ πέµπτη, καί [™στιν] œλασσον αÙτÁς τµÁµα τÕ Ε∆, σύµµετρος ¥ρα ¹ Ε∆ τÍ ΑΒ µήκει. ¢λλ¦ ¹ ΑΕ τÍ Ε∆ ™στιν ¢σύµµετρος· κሠ¹ ΑΒ ¥ρα τÍ ΑΕ ™στιν ¢σύµµετρος µήκει [αƒ ΒΑ, ΑΕ ·ηταί ε„σι δυνάµει µόνον σύµµετροι]· µέσον ¥ρα ™στˆ τÕ ΑΚ, τουτέστι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΜΝ, ΝΞ. κሠ™πεˆ σύµµετρός ™στιν ¹ ∆Ε τÍ ΑΒ µήκει, τουτέστι τÍ ΕΚ, ¢λλ¦ ¹ ∆Ε τÍ ΕΖ σύµµετρός ™στιν, κሠ¹ ΕΖ ¥ρα τÍ ΕΚ σύµµετρός ™στιν. κሠ·ητ¾ ¹ ΕΚ· ·ητÕν ¥ρα κሠτÕ ΕΛ, τουτέστι τÕ ΜΡ, τουτέστι τÕ ØπÕ ΜΝΞ· αƒ ΜΝ, ΝΞ ¥ρα δυνάµει ¢σύµµετροί ε„σι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον, τÕ δ' Øπ' αÙτîν ·ητόν. `Η ΜΞ ¥ρα ·ητÕν κሠµέσον δυναµένη ™στˆ κሠδύναται τÕ ΑΓ χωρίον· Óπερ œδει δε‹ξαι.

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S P For let the same construction be made as that shown previously. So, (it is) clear that M O is the square-root of area AC. So, we must show that M O is the square-root of a rational plus a medial (area). For since AG is incommensurable (in length) with GE [Prop. 10.18], AH is thus also incommensurable with HE—that is to say, the (square) on M N with the (square) on N O [Props. 6.1, 10.11]. Thus, M N and N O are incommensurable in square. And since AD is a fifth binomial (straight-line), and ED [is] its lesser segment, ED (is) thus commensurable in length with AB [Def. 10.9]. But, AE is incommensurable (in length) with ED. Thus, AB is also incommensurable in length with AE [BA and AE are rational (straight-lines which are) commensurable in square only] [Prop. 10.13]. Thus, AK—that is to say, the sum of the (squares) on M N and N O—is medial [Prop. 10.21]. And since DE is commensurable in length with AB—that is to say, with EK—but, DE is commensurable (in length) with EF , EF is thus also commensurable (in length) with EK [Prop. 10.12]. And EK (is) rational. Thus, EL—that is to say, M R—that is to say, the (rectangle contained) by M N O—(is) also rational [Prop. 10.19]. M N and N O are thus (straight-lines which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them rational. Thus, M O is the square-root of a rational plus a medial (area) [Prop. 10.40]. And (it is) the square-root of area AC. (Which is) the very thing it was required to show.

†

If the rational straight-line has unit length, then this proposition states that the square-root of a fifth binomial straight-line is the square root of √ a rational plus a medial area: i.e., a fifth binomial straight-line has a length k ( 1 + k ′ + 1) whose square-root can be written

348

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

q q p ρ [(1 + k ′′ 2 )1/2 + k ′′ ]/[2 (1 + k ′′ 2 )] + ρ [(1 + k ′′ 2 )1/2 − k ′′ ]/[2 (1 + k ′′ 2 )], where ρ = book10eps/k (1 + k ′′ 2 ) and k ′′ 2 = k ′ . This is the length of the square root of a rational plus a medial area (see Prop. 10.40), since ρ is rational.

νθ΄.

Proposition 59

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο If an area is contained by a rational (straight-line) and Ñνοµάτων ›κτης, ¹ τÕ χωρίον δυναµένη ¥λογός ™στιν ¹ a sixth binomial (straight-line) then the square-root of καλουµένη δύο µέσα δυναµένη. the area is the irrational (straight-line which is) called the square-root of (the sum of) two medial (areas).† Π Α Η Ε Ζ ∆ Ρ G E F D A R Q Μ Β

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Σ Ο Χωρίον γ¦ρ τÕ ΑΒΓ∆ περιεχέσθω ØπÕ ·ητÁς τÁς ΑΒ κሠτÁς ™κ δύο Ñνοµάτων ›κτης τÁς Α∆ διVρηµένης ε„ς τ¦ Ñνόµατα κατ¦ τÕ Ε, éστε τÕ µε‹ζον Ôνοµα εναι τÕ ΑΕ· λέγω, Óτι ¹ τÕ ΑΓ δυναµένη ¹ δύο µέσα δυναµένη ™στίν. Κατεσκευάσθω [γ¦ρ] τ¦ αÙτ¦ το‹ς προδεδειγµένοις. φανερÕν δή, Óτι [¹] τÕ ΑΓ δυναµένη ™στˆν ¹ ΜΞ, κሠÓτι ¢σύµµετρός ™στιν ¹ ΜΝ τÍ ΝΞ δυνάµει. κሠ™πεˆ ¢σύµµετρός ™στιν ¹ ΕΑ τÍ ΑΒ µήκει, αƒ ΕΑ, ΑΒ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· µέσον ¥ρα ™στˆ τÕ ΑΚ, τουτέστι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΜΝ, ΝΞ. πάλιν, ™πεˆ ¢σύµµετρός ™στιν ¹ Ε∆ τÍ ΑΒ µήκει, ¢σύµµετρος ¥ρα ™στˆ κሠ¹ ΖΕ τÍ ΕΚ· αƒ ΖΕ, ΕΚ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· µέσον ¥ρα ™στˆ τÕ ΕΛ, τουτέστι τÕ ΜΡ, τουτέστι τÕ ØπÕ τîν ΜΝΞ. κሠ™πεˆ ¢σύµµετρος ¹ ΑΕ τÍ ΕΖ, κሠτÕ ΑΚ τù ΕΛ ¢σύµµετρόν ™στιν. ¢λλ¦ τÕ µν ΑΚ ™στι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΜΝ, ΝΞ, τÕ δ ΕΛ ™στι τÕ ØπÕ τîν ΜΝΞ· ¢σύµµετρον ¥ρα ™στˆ τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΜΝΞ τù ØπÕ τîν ΜΝΞ. καί ™στι µέσον ˜κάτερον αÙτîν, καˆ αƒ ΜΝ, ΝΞ δυνάµει ε„σˆν ¢σύµµετροι. `Η ΜΞ ¥ρα δύο µέσα δυναµένη ™στˆ κሠδύναται τÕ ΑΓ· Óπερ œδει δε‹ξαι.

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S P For let the area ABCD be contained by the rational (straight-line) AB and the sixth binomial (straight-line) AD, which has been divided into its (component) terms at E, such that AE is the greater term. So, I say that the square-root of AC is the square-root of (the sum of) two medial (areas). [For] let the same construction be made as that shown previously. So, (it is) clear that M O is the square-root of AC, and that M N is incommensurable in square with N O. And since EA is incommensurable in length with AB [Def. 10.10], EA and AB are thus rational (straightlines which are) commensurable in square only. Thus, AK—that is to say, the sum of the (squares) on M N and N O—is medial [Prop. 10.21]. Again, since ED is incommensurable in length with AB [Def. 10.10], F E is thus also incommensurable (in length) with EK [Prop. 10.13]. Thus, F E and EK are rational (straightlines which are) commensurable in square only. Thus, EL—that is to say, M R—that is to say, the (rectangle contained) by M N O—is medial [Prop. 10.21]. And since AE is incommensurable (in length) with EF , AK is also incommensurable with EL [Props. 6.1, 10.11]. But, AK is the sum of the (squares) on M N and N O, and EL is the (rectangle contained) by M N O. Thus, the sum of the (squares) on M N O is incommensurable with the (rectangle contained) by M N O. And each of them is medial. And M N and N O are incommensurable in square. Thus, M O is the square-root of (the sum of) two medial (areas) [Prop. 10.41]. And (it is) the square-root of AC. (Which is) the very thing it was required to show.

†

If the rational straight-line has unit length, then this proposition states that the square-root of a sixth binomial straight-line is the square root of p p the sum of two medial areas: i.e., a sixth binomial straight-line has a length book10eps/k + book10eps/k ′ whose square-root can be written

349

ΣΤΟΙΧΕΙΩΝ ι΄. k 1/4

ELEMENTS BOOK 10

« „q q [1 + k ′′ /(1 + k ′′ 2 )1/2 ]/2 + [1 − k ′′ /(1 + k ′′ 2 )1/2 ]/2 , where k ′′ 2 = (k − k ′ )/k ′ . This is the length of the square-root of the sum of

two medial areas (see Prop. 10.41).

ΛÁµµα.

Lemma

'Ε¦ν εÙθε‹α γραµµ¾ τµηθÍ ε„ς ¥νισα, τ¦ ¢πÕ τîν If a straight-line is cut unequally, then (the sum of) ¢νίσων τετράγωνα µείζονά ™στι τοà δˆς ØπÕ τîν ¢νίσων the squares on the unequal (parts) is greater than twice περιεχοµένου Ñρθογωνίου. the rectangle contained by the unequal (parts).

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”Εστω εÙθε‹α ¹ ΑΒ κሠτετµήσθω ε„ς ¥νισα κατ¦ τÕ Γ, κሠœστω µείζων ¹ ΑΓ· λέγω, Óτι τ¦ ¢πÕ τîν ΑΓ, ΓΒ µείζονά ™στι τοà δˆς ØπÕ τîν ΑΓ, ΓΒ. Τετµήσθω γ¦ρ ¹ ΑΒ δίχα κατ¦ τÕ ∆. ™πεˆ οâν εÙθε‹α γραµµ¾ τέτµηται ε„ς µν ‡σα κατ¦ τÕ ∆, ε„ς δ ¥νισα κατ¦ τÕ Γ, τÕ ¥ρα ØπÕ τîν ΑΓ, ΓΒ µετ¦ τοà ¢πÕ Γ∆ ‡σον ™στˆ τù ¢πÕ Α∆· éστε τÕ ØπÕ τîν ΑΓ, ΓΒ œλαττόν ™στι τοà ¢πÕ Α∆· τÕ ¥ρα δˆς ØπÕ τîν ΑΓ, ΓΒ œλαττον À διπλάσιόν ™στι τοà ¢πÕ Α∆. ¢λλ¦ τ¦ ¢πÕ τîν ΑΓ, ΓΒ διπλάσιά [™στι] τîν ¢πÕ τîν Α∆, ∆Γ· τ¦ ¥ρα ¢πÕ τîν ΑΓ, ΓΒ µείζονά ™στι τοà δˆς ØπÕ τîν ΑΓ, ΓΒ· Óπερ œδει δε‹ξαι.

Let AB be a straight-line, and let it have been cut unequally at C, and let AC be greater (than CB). I say that (the sum of) the (squares) on AC and CB is greater than twice the (rectangle contained) by AC and CB. For let AB have been cut in half at D. Therefore, since a straight-line has been cut into equal (parts) at D, and into unequal (parts) at C, the (rectangle contained) by AC and CB, plus the (square) on CD, is thus equal to the (square) on AD [Prop. 2.5]. Hence, the (rectangle contained) by AC and CB is less than the (square) on AD. Thus, twice the (rectangle contained) by AC and CB is less than double the (square) on AD. But, (the sum of) the (squares) on AC and CB [is] double (the sum of) the (squares) on AD and DC [Prop. 2.9]. Thus, (the sum of) the (squares) on AC and CB is greater than twice the (rectangle contained) by AC and CB. (Which is) the very thing it was required to show.

ξ΄.

Proposition 60

ΤÕ ¢πÕ τÁς ™κ δύο Ñνοµάτων παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων πρώτην.

The square on a binomial (straight-line) applied to a rational (straight-line) produces as breadth a first binomial (straight-line).†

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”Εστω ™κ δύο Ñνοµάτων ¹ ΑΒ διVρηµένη ε„ς τ¦ Ñνόµατα κατ¦ τÕ Γ, éστε τÕ µε‹ζον Ôνοµα εναι τÕ ΑΓ, κሠ™κκείσθω ·ητ¾ ¹ ∆Ε, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ τ¾ν ∆Ε παραβεβλήσθω τÕ ∆ΕΖΗ πλάτος ποιοàν τ¾ν ∆Η· λέγω, Óτι ¹ ∆Η ™κ δύο Ñνοµάτων ™στˆ πρώτη. Παραβεβλήσθω γ¦ρ παρ¦ τ¾ν ∆Ε τù µν ¢πÕ τÁς ΑΓ ‡σον τÕ ∆Θ, τù δ ¢πÕ τÁς ΒΓ ‡σον τÕ ΚΛ· λοιπÕν ¥ρα τÕ δˆς ØπÕ τîν ΑΓ, ΓΒ ‡σον ™στˆ τù ΜΖ. τετµήσθω

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Let AB be a binomial (straight-line), having been divided into its (component) terms at C, such that AC is the greater term. And let the rational (straight-line) DE be laid down. And let the (rectangle) DEF G, equal to the (square) on AB, have been applied to DE, producing DG as breadth. I say that DG is a first binomial (straightline). For let DH, equal to the (square) on AC, and KL,

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ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

¹ ΜΗ δίχα κατ¦ τÕ Ν, κሠπαράλληλος ½χθω ¹ ΝΞ [˜κατέρv τîν ΜΛ, ΗΖ]. ˜κάτερον ¥ρα τîν ΜΞ, ΝΖ ‡σον ™στˆ τù ¤παξ ØπÕ τîν ΑΓΒ. κሠ™πεˆ ™κ δύο Ñνοµάτων ™στˆν ¹ ΑΒ διVρηµένη ε„ς τ¦ Ñνόµατα κατ¦ τÕ Γ, αƒ ΑΓ, ΓΒ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· τ¦ ¥ρα ¢πÕ τîν ΑΓ, ΓΒ ·ητά ™στι κሠσύµµετρα ¢λλήλοις· éστε κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ. καί ™στιν ‡σον τù ∆Λ· ·ητÕν ¥ρα ™στˆ τÕ ∆Λ. κሠπαρ¦ ·ητ¾ν τ¾ν ∆Ε παράκειται· ·ητ¾ ¥ρα ™στˆν ¹ ∆Μ κሠσύµµετρος τÍ ∆Ε µήκει. πάλιν, ™πεˆ αƒ ΑΓ, ΓΒ ·ηταί ε„σι δυνάµει µόνον σύµµετροι, µέσον ¥ρα ™στˆ τÕ δˆς ØπÕ τîν ΑΓ, ΓΒ, τουτέστι τÕ ΜΖ. κሠπαρ¦ ·ητ¾ν τ¾ν ΜΛ παράκειται· ·ητ¾ ¥ρα κሠ¹ ΜΗ κሠ¢σύµµετρος τÍ ΜΛ, τουτέστι τÍ ∆Ε, µήκει. œστι δ κሠ¹ Μ∆ ·ητ¾ κሠτÍ ∆Ε µήκει σύµµετρος· ¢σύµµετρος ¥ρα ™στˆν ¹ ∆Μ τÍ ΜΗ µήκει. καί ε„σι ·ηταί· αƒ ∆Μ, ΜΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ∆Η. δεικτέον δή, Óτι κሠπρώτη. 'Επεˆ τîν ¢πÕ τîν ΑΓ, ΓΒ µέσον ¢νάλογόν ™στι τÕ ØπÕ τîν ΑΓΒ, κሠτîν ∆Θ, ΚΛ ¥ρα µέσον ¢νάλογόν ™στι τÕ ΜΞ. œστιν ¥ρα æς τÕ ∆Θ πρÕς τÕ ΜΞ, οÛτως τÕ ΜΞ πρÕς τÕ ΚΛ, τουτέστιν æς ¹ ∆Κ πρÕς τ¾ν ΜΝ, ¹ ΜΝ πρÕς τ¾ν ΜΚ· τÕ ¥ρα ØπÕ τîν ∆Κ, ΚΜ ‡σον ™στˆ τù ¢πÕ τÁς ΜΝ. κሠ™πεˆ σύµµετρόν ™στι τÕ ¢πÕ τÁς ΑΓ τù ¢πÕ τÁς ΓΒ, σύµµετρόν ™στι κሠτÕ ∆Θ τù ΚΛ· éστε κሠ¹ ∆Κ τÍ ΚΜ σύµµετρός ™στιν. κሠ™πεˆ µείζονά ™στι τ¦ ¢πÕ τîν ΑΓ, ΓΒ τοà δˆς ØπÕ τîν ΑΓ, ΓΒ, µε‹ζον ¥ρα κሠτÕ ∆Λ τοà ΜΖ· éστε κሠ¹ ∆Μ τÁς ΜΗ µείζων ™στίν. καί ™στιν ‡σον τÕ ØπÕ τîν ∆Κ, ΚΜ τù ¢πÕ τÁς ΜΝ, τουτέστι τù τετάρτJ τοà ¢πÕ τÁς ΜΗ, κሠσύµµετρος ¹ ∆Κ τÍ ΚΜ. ™¦ν δ ðσι δύο εÙθε‹αι ¥νισοι, τù δ τετάρτJ µέρει τοà ¢πÕ τÁς ™λάσσονος ‡σον παρ¦ τ¾ν µείζονα παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ κሠε„ς σύµµετρα αÙτ¾ν διαιρÍ, ¹ µείζων τÁς ™λάσσονος µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ· ¹ ∆Μ ¥ρα τÁς ΜΗ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. καί ε„σι ·ηταˆ αƒ ∆Μ, ΜΗ, κሠ¹ ∆Μ µε‹ζον Ôνοµα οâσα σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ∆Ε µήκει. `Η ∆Η ¥ρα ™κ δύο Ñνοµάτων ™στˆ πρώτη· Óπερ œδει δε‹ξαι.

equal to the (square) on BC, have been applied to DE. Thus, the remaining twice the (rectangle contained) by AC and CB is equal to M F [Prop. 2.4]. Let M G have been cut in half at N , and let N O have been drawn parallel [to each of M L and GF ]. M O and N F are thus each equal to once the (rectangle contained) by ACB. And since AB is a binomial (straight-line), having been divided into its (component) terms at C, AC and CB are thus rational (straight-lines which are) commensurable in square only [Prop. 10.36]. Thus, the (squares) on AC and CB are rational, and commensurable with one another. And hence the sum of the (squares) on AC and CB (is rational) [Prop. 10.15], and is equal to DL. Thus, DL is rational. And it is applied to the rational (straightline) DE. DM is thus rational, and commensurable in length with DE [Prop. 10.20]. Again, since AC and CB are rational (straight-lines which are) commensurable in square only, twice the (rectangle contained) by AC and CB—that is to say, M F —is thus medial [Prop. 10.21]. And it is applied to the rational (straight-line) M L. M G is thus also rational, and incommensurable in length with M L—that is to say, with DE [Prop. 10.22]. And M D is also rational, and commensurable in length with DE. Thus, DM is incommensurable in length with M G [Prop. 10.13]. And they are rational. DM and M G are thus rational (straight-lines which are) commensurable in square only. Thus, DG is a binomial (straight-line) [Prop. 10.36]. So, we must show that (it is) also a first (binomial straight-line). Since the (rectangle contained) by ACB is the mean proportional to the squares on AC and CB [Prop. 10.53 lem.], M O is thus also the mean proportional to DH and KL. Thus, as DH is to M O, so M O (is) to KL—that is to say, as DK (is) to M N , (so) M N (is) to M K [Prop. 6.1]. Thus, the (rectangle contained) by DK and KM is equal to the (square) on M N [Prop. 6.17]. And since the (square) on AC is commensurable with the (square) on CB, DH is also commensurable with KL. Hence, DK is also commensurable with KM [Props. 6.1, 10.11]. And since (the sum of) the squares on AC and CB is greater than twice the (rectangle contained) by AC and CB [Prop. 10.59 lem.], DL (is) thus also greater than M F . Hence, DM is also greater than M G [Props. 6.1, 5.14]. And the (rectangle contained) by DK and KM is equal to the (square) on M N —that is to say, to one quarter the (square) on M G. And DK (is) commensurable (in length) with KM . And if there are two unequal straight-lines, and a (rectangle) equal to the fourth part of the (square) on the lesser, falling short by a square figure, is applied to the greater, and divides it into (parts which are) commensu-

351

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 rable (in length), then the square on the greater is larger than (the square on) the lesser by the (square) on (some straight-line) commensurable (in length) with the greater [Prop. 10.17]. Thus, the square on DM is greater than (the square on) M G by the (square) on (some straightline) commensurable (in length) with (DM ). And DM and M G are rational. And DM , which is the greater term, is commensurable in length with the (previously) laid down rational (straight-line) DE. Thus, DG is a first binomial (straight-line) [Def. 10.5]. (Which is) the very thing it was required to show.

†

In other words, the square of a binomial is a first binomial. See Prop. 10.54.

ξα΄.

Proposition 61

ΤÕ ¢πÕ τÁς ™κ δύο µέσων πρώτης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων δευτέραν.

The square on a first bimedial (straight-line) applied to a rational (straight-line) produces as breadth a second binomial (straight-line).†

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Β

”Εστω ™κ δύο µέσων πρώτη ¹ ΑΒ διVρηµένη ε„ς τ¦ς µέσας κατ¦ τÕ Γ, ïν µείζων ¹ ΑΓ, κሠ™κκείσθω ·ητ¾ ¹ ∆Ε, κሠπαραβεβλήσθω παρ¦ τ¾ν ∆Ε τù ¢πÕ τÁς ΑΒ ‡σον παραλληλόγραµµον τÕ ∆Ζ πλάτος ποιοàν τ¾ν ∆Η· λέγω, Óτι ¹ ∆Η ™κ δύο Ñνοµάτων ™στˆ δευτέρα. Κατεσκευάσθω γ¦ρ τ¦ αÙτ¦ το‹ς πρÕ τούτου. κሠ™πεˆ ¹ ΑΒ ™κ δύο µέσων ™στˆ πρώτη διVρηµένη κατ¦ τÕ Γ, αƒ ΑΓ, ΓΒ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι ·ητÕν περιέχουσαι· éστε κሠτ¦ ¢πÕ τîν ΑΓ, ΓΒ µέσα ™στίν. µέσον ¥ρα ™στˆ τÕ ∆Λ. κሠπαρ¦ ·ητ¾ν τ¾ν ∆Ε παραβέβληται· ·ητ¾ ¥ρα ™στίν ¹ Μ∆ κሠ¢σύµµετρος τÍ ∆Ε µήκει. πάλιν, ™πεˆ ·ητόν ™στι τÕ δˆς ØπÕ τîν ΑΓ, ΓΒ, ·ητόν ™στι κሠτÕ ΜΖ. κሠπαρ¦ ·ητ¾ν τ¾ν ΜΛ παράκειται· ·ητ¾ ¥ρα [™στˆ] κሠ¹ ΜΗ κሠµήκει σύµµετρος τÍ ΜΛ, τουτέστι τÍ ∆Ε· ¢σύµµετρος ¥ρα ™στˆν ¹ ∆Μ τÍ ΜΗ µήκει. καί ε„σι ·ηταί· αƒ ∆Μ, ΜΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ∆Η. δεικτέον δή, Óτι κሠδευτέρα. 'Επεˆ γ¦ρ τ¦ ¢πÕ τîν ΑΓ, ΓΒ µείζονά ™στι τοà δˆς ØπÕ τîν ΑΓ, ΓΒ, µε‹ζον ¥ρα κሠτÕ ∆Λ τοà ΜΖ· éστε κሠ¹ ∆Μ τÁς ΜΗ. κሠ™πεˆ σύµµετρόν ™στι τÕ ¢πÕ τÁς ΑΓ τù ¢πÕ τÁς ΓΒ, σύµµετρόν ™στι κሠτÕ ∆Θ τù ΚΛ· éστε κሠ¹ ∆Κ τÍ ΚΜ σύµµετρός ™στιν. καί ™στι τÕ

B

Let AB be a first bimedial (straight-line) having been divided into its (component) medial (straight-lines) at C, of which AC (is) the greater. And let the rational (straight-line) DE be laid down. And let the parallelogram DF , equal to the (square) on AB, have been applied to DE, producing DG as breadth. I say that DG is a second binomial (straight-line). For let the same construction have been made as in the (proposition) before this. And since AB is a first bimedial (straight-line), having been divided at C, AC and CB are thus medial (straight-lines) commensurable in square only, and containing a rational (area) [Prop. 10.37]. Hence, the (squares) on AC and CB are also medial [Prop. 10.21]. Thus, DL is medial [Props. 10.15, 10.23 corr.]. And it has been applied to the rational (straight-line) DE. M D is thus rational, and incommensurable in length with DE [Prop. 10.22]. Again, since twice the (rectangle contained) by AC and CB is rational, M F is also rational. And it is applied to the rational (straight-line) M L. Thus, M G [is] also rational, and commensurable in length with M L—that is to say, with DE [Prop. 10.20]. DM is thus incommensurable in length with M G [Prop. 10.13]. And they are

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ELEMENTS BOOK 10

ØπÕ τîν ∆ΚΜ ‡σον τù ¢πÕ τÁς ΜΝ· ¹ ∆Μ ¥ρα τÁς ΜΗ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. καί ™στιν ¹ ΜΗ σύµµετρος τÍ ∆Ε µήκει. `Η ∆Η ¥ρα ™κ δύο Ñνοµάτων ™στˆ δευτέρα.

† In

rational. DM and M G are thus rational, and commensurable in square only. DG is thus a binomial (straight-line) [Prop. 10.36]. So, we must show that (it is) also a second (binomial straight-line). For since (the sum of) the squares on AC and CB is greater than twice the (rectangle contained) by AC and CB [Prop. 10.59], DL (is) thus also greater than M F . Hence, DM (is) also (greater) than M G [Prop. 6.1]. And since the (square) on AC is commensurable with the (square) on CB, DH is also commensurable with KL. Hence, DK is also commensurable (in length) with KM [Props. 6.1, 10.11]. And the (rectangle contained) by DKM is equal to the (square) on M N . Thus, the square on DM is greater than (the square on) M G by the (square) on (some straight-line) commensurable (in length) with (DM ) [Prop. 10.17]. And M G is commensurable in length with DE. Thus, DG is a second binomial (straight-line) [Def. 10.6].

other words, the square of a first bimedial is a second binomial. See Prop. 10.55.

ξβ΄.

Proposition 62

ΤÕ ¢πÕ τÁς ™κ δύο µέσων δευτέρας παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων τρίτην.

The square on a second bimedial (straight-line) applied to a rational (straight-line) produces as breadth a third binomial (straight-line).†

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”Εστω ™κ δύο µέσων δευτέρα ¹ ΑΒ διVρηµένη ε„ς τ¦ς µέσας κατ¦ τÕ Γ, éστε τÕ µε‹ζον τµÁµα εναι τÕ ΑΓ, ·ητ¾ δέ τις œστω ¹ ∆Ε, κሠπαρ¦ τ¾ν ∆Ε τù ¢πÕ τÁς ΑΒ ‡σον παραλληλόγραµµον παραβεβλήσθω τÕ ∆Ζ πλάτος ποιοàν τ¾ν ∆Η· λέγω, Óτι ¹ ∆Η ™κ δύο Ñνοµάτων ™στˆ τρίτη. Κατεσκευάσθω τ¦ αÙτ¦ το‹ς προδεδειγµένοις. κሠ™πεˆ ™κ δύο µέσων δευτέρα ™στˆν ¹ ΑΒ διVρηµένη κατ¦ τÕ Γ, αƒ ΑΓ, ΓΒ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι µέσον περιέχουσαι· éστε κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ µέσον ™στίν. καί ™στιν ‡σον τù ∆Λ· µέσον ¥ρα κሠτÕ ∆Λ. κሠπαράκειται παρ¦ ·ητ¾ν τ¾ν ∆Ε· ·ητ¾ ¥ρα ™στˆ κሠ¹ Μ∆ κሠ¢σύµµετρος τÍ ∆Ε µήκει. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΜΗ ·ητή ™στι κሠ¢σύµµετρος τÍ ΜΛ, τουτέστι τÍ ∆Ε, µήκει· ·ητ¾ ¥ρα

B

Let AB be a second bimedial (straight-line) having been divided into its (component) medial (straight-lines) at C, such that AC is the greater segment. And let DE be some rational (straight-line). And let the parallelogram DF , equal to the (square) on AB, have been applied to DE, producing DG as breadth. I say that DG is a third binomial (straight-line). Let the same construction be made as that shown previously. And since AB is a second bimedial (straightline), having been divided at C, AC and CB are thus medial (straight-lines) commensurable in square only, and containing a medial (area) [Prop. 10.38]. Hence, the sum of the (squares) on AC and CB is also medial [Props. 10.15, 10.23 corr.]. And it is equal to DL. Thus, DL (is) also medial. And it is applied to the rational

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ELEMENTS BOOK 10

™στˆν ˜κατέρα τîν ∆Μ, ΜΗ κሠ¢σύµµετρος τÍ ∆Ε µήκει. κሠ™πεˆ ¢σύµµετρός ™στιν ¹ ΑΓ τÍ ΓΒ µήκει, æς δ ¹ ΑΓ πρÕς τ¾ν ΓΒ, οÛτως τÕ ¢πÕ τÁς ΑΓ πρÕς τÕ ØπÕ τîν ΑΓΒ, ¢σύµµετρον ¥ρα κሠτÕ ¢πÕ τÁς ΑΓ τù ØπÕ τîν ΑΓΒ. éστε κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ τù δˆς ØπÕ τîν ΑΓΒ ¢σύµµετρόν ™στιν, τουτέστι τÕ ∆Λ τù ΜΖ· éστε κሠ¹ ∆Μ τù ΜΗ ¢σύµµετρός ™στιν. καί ε„σι ·ηταί· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ∆Η. δεικτέον [δή], Óτι κሠτρίτη. `Οµοίως δ¾ το‹ς προτέροις ™πιλογιούµεθα, Óτι µείζων ™στˆν ¹ ∆Μ τÁς ΜΗ, κሠσύµµετρος ¹ ∆Κ τÍ ΚΜ. καί ™στι τÕ ØπÕ τîν ∆ΚΜ ‡σον τù ¢πÕ τÁς ΜΝ· ¹ ∆Μ ¥ρα τÁς ΜΗ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. κሠοÙδετέρα τîν ∆Μ, ΜΗ σύµµετρός ™στι τÍ ∆Ε µήκει. `Η ∆Η ¥ρα ™κ δύο Ñνοµάτων ™στˆ τρίτη· Óπερ œδει δε‹ξαι.

†

(straight-line) DE. M D is thus also rational, and incommensurable in length with DE [Prop. 10.22]. So, for the same (reasons), M G is also rational, and incommensurable in length with M L—that is to say, with DE. Thus, DM and M G are each rational, and incommensurable in length with DE. And since AC is incommensurable in length with CB, and as AC (is) to CB, so the (square) on AC (is) to the (rectangle contained) by ACB [Prop. 10.21 lem.], the (square) on AC (is) also incommensurable with the (rectangle contained) by ACB [Prop. 10.11]. And hence the sum of the (squares) on AC and CB is incommensurable with twice the (rectangle contained) by ACB—that is to say, DL with M F [Props. 10.12, 10.13]. Hence, DM is also incommensurable (in length) with M G [Props. 6.1, 10.11]. And they are rational. DG is thus a binomial (straight-line) [Prop. 10.36]. [So] we must show that (it is) also a third (binomial straight-line). So, similarly to the previous (propositions), we can conclude that DM is greater than M G, and DK (is) commensurable (in length) with KM . And the (rectangle contained) by DKM is equal to the (square) on M N . Thus, the square on DM is greater than (the square on) M G by the (square) on (some straight-line) commensurable (in length) with (DM ) [Prop. 10.17]. And neither of DM and M G is commensurable in length with DE. Thus, DG is a third binomial (straight-line) [Def. 10.7]. (Which is) the very thing it was required to show.

In other words, the square of a second bimedial is a third binomial. See Prop. 10.56.

ξγ΄.

Proposition 63

ΤÕ ¢πÕ τÁς µείζονος παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων τετάρτην.

The square on a major (straight-line) applied to a rational (straight-line) produces as breadth a fourth binomial (straight-line).†

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”Εστω µείζων ¹ ΑΒ διVρηµένη κατ¦ τÕ Γ, éστε µείζονα εναι τ¾ν ΑΓ τÁς ΓΒ, ·ητ¾ δ ¹ ∆Ε, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ τ¾ν ∆Ε παραβεβλήσθω τÕ ∆Ζ παραλληλόγραµµον πλάτος ποιοàν τ¾ν ∆Η· λέγω, Óτι ¹ ∆Η ™κ δύο Ñνοµάτων ™στˆ τετάρτη.

B

Let AB be a major (straight-line) having been divided at C, such that AC is greater than CB, and (let) DE (be) a rational (straight-line). And let the parallelogram DF , equal to the (square) on AB, have been applied to DE, producing DG as breadth. I say that DG is a fourth

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ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

Κατεσκευάσθω τ¦ αÙτ¦ το‹ς προδεδειγµένοις. κሠ™πεˆ µείζων ™στˆν ¹ ΑΒ διVρηµένη κατ¦ τÕ Γ, αƒ ΑΓ, ΓΒ δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων ·ητόν, τÕ δ Øπ' αÙτîν µέσον. ™πεˆ οâν ·ητόν ™στι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ, ·ητόν ¥ρα ™στˆ τÕ ∆Λ· ·ητ¾ ¥ρα κሠ¹ ∆Μ κሠσύµµετρος τÍ ∆Ε µήκει. πάλιν, ™πεˆ µέσον ™στˆ τÕ δˆς ØπÕ τîν ΑΓ, ΓΒ, τουτέστι τÕ ΜΖ, κሠπαρ¦ ·ητήν ™στι τ¾ν ΜΛ, ·ητ¾ ¥ρα ™στˆ κሠ¹ ΜΗ κሠ¢σύµµετρος τÍ ∆Ε µήκει· ¢σύµµετρος ¥ρα ™στˆ κሠ¹ ∆Μ τÍ ΜΗ µήκει. αƒ ∆Μ, ΜΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ∆Η. δεικτέον [δή], Óτι κሠτετάρτη. `Οµοίως δ¾ δείξοµεν το‹ς πρότερον, Óτι µείζων ™στˆν ¹ ∆Μ τÁς ΜΗ, κሠÓτι τÕ ØπÕ ∆ΚΜ ‡σον ™στˆ τù ¢πÕ τÁς ΜΝ. ™πεˆ οâν ¢σύµµετρόν ™στι τÕ ¢πÕ τÁς ΑΓ τù ¢πÕ τÁς ΓΒ, ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ∆Θ τù ΚΛ· éστε ¢σύµµετρος κሠ¹ ∆Κ τÍ ΚΜ ™στιν. ™¦ν δ ðσι δύο εÙθε‹αι ¥νισοι, τù δ τετάρτJ µέρει τοà ¢πÕ τ¾ς ™λάσσονος ‡σον παραλληλόγραµµον παρ¦ τ¾ν µείζονα παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ κሠε„ς ¢σύµµετρα αÙτ¾ν διαιρÍ, ¹ µείζων τÁς ™λάσσονος µε‹ζον δυνήσεται τù ¢πÕ ¢σύµµέτρου ˜αυτÍ µήκει· ¹ ∆Μ ¥ρα τ¾ς ΜΗ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καί ε„σιν αƒ ∆Μ, ΜΗ ·ητሠδυνάµει µόνον σύµµετροι, κሠ¹ ∆Μ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ∆Ε. `Η ∆Η ¥ρα ™κ δύο Ñνοµάτων ™στˆ τετάρτη· Óπερ œδει δε‹ξαι.

†

binomial (straight-line). Let the same construction be made as that shown previously. And since AB is a major (straight-line), having been divided at C, AC and CB are incommensurable in square, making the sum of the squares on them rational, and the (rectangle contained) by them medial [Prop. 10.39]. Therefore, since the sum of the (squares) on AC and CB is rational, DL is thus rational. Thus, DM (is) also rational, and commensurable in length with DE [Prop. 10.20]. Again, since twice the (rectangle contained) by AC and CB—that is to say, M F —is medial, and is (applied to) the rational (straight-line) M L, M G is thus also rational, and incommensurable in length with DE [Prop. 10.22]. DM is thus also incommensurable in length with M G [Prop. 10.13]. DM and M G are thus rational (straight-lines which are) commensurable in square only. Thus, DG is a binomial (straight-line) [Prop. 10.36]. [So] we must show that (it is) also a fourth (binomial straight-line). So, similarly to the previous (propositions), we can show that DM is greater than M G, and that the (rectangle contained) by DKM is equal to the (square) on M N . Therefore, since the (square) on AC is incommensurable with the (square) on CB, DH is also incommensurable with KL. Hence, DK is also incommensurable with KM [Props. 6.1, 10.11]. And if there are two unequal straight-lines, and a parallelogram equal to the fourth part of the (square) on the lesser, falling short by a square figure, is applied to the greater, and divides it into (parts which are) incommensurable (in length), then the square on the greater will be larger than (the square on) the lesser by the (square) on (some straight-line) incommensurable in length with the greater [Prop. 10.18]. Thus, the square on DM is greater than (the square on) M G by the (square) on (some straight-line) commensurable (in length) with (DM ). And DM and M G are rational (straight-lines which are) commensurable in square only. And DM is commensurable (in length) with the (previously) laid down rational (straight-line) DE. Thus, DG is a fourth binomial (straight-line) [Def. 10.8]. (Which is) the very thing it was required to show.

In other words, the square of a major is a fourth binomial. See Prop. 10.57.

ξδ΄.

Proposition 64

ΤÕ ¢πÕ τÁς ·ητÕν κሠµέσον δυναµένης παρ¦ ·ητ¾ν The square on the square-root of a rational plus a meπαραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων dial (area) applied to a rational (straight-line) produces πέµπτην. as breadth a fifth binomial (straight-line).† ”Εστω ·ητÕν κሠµέσον δυναµένη ¹ ΑΒ διVρηµένη Let AB be the square-root of a rational plus a medial ε„ς τ¦ς εÙθείας κατ¦ τÕ Γ, éστε µείζονα εναι τ¾ν ΑΓ, (area) having been divided into its (component) straight-

355

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

κሠ™κκείσθω ·ητ¾ ¹ ∆Ε, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ lines at C, such that AC is greater. And let the rational τ¾ν ∆Ε παραβεβλήσθω τÕ ∆Ζ πλάτος ποιοàν τ¾ν ∆Η· (straight-line) DE be laid down. And let the (paralleloλέγω, Óτι ¹ ∆Η ™κ δύο Ñνοµάτων ™στˆ πέµπτη. gram) DF , equal to the (square) on AB, have been applied to DE, producing DG as breadth. I say that DG is a fifth binomial straight-line.

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Κατεσκευάσθω τ¦ αÙτα το‹ς πρÕ τούτου. ™πεˆ οâν ·ητÕν κሠµέσον δυναµένη ™στˆν ¹ ΑΒ διVρηµένη κατ¦ τÕ Γ, αƒ ΑΓ, ΓΒ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον, τÕ δ' Øπ' αÙτîν ·ητόν. ™πεˆ οâν µέσον ™στˆ τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ, µέσον ¥ρα ™στˆ τÕ ∆Λ· éστε ·ητή ™στιν ¹ ∆Μ κሠµήκει ¢σύµµετρος τÍ ∆Ε. πάλιν, ™πεˆ ·ητόν ™στι τÕ δˆς ØπÕ τîν ΑΓΒ, τουτέστι τÕ ΜΖ, ·ητ¾ ¥ρα ¹ ΜΗ κሠσύµµετρος τÍ ∆Ε. ¢σύµµετρος ¥ρα ¹ ∆Μ τÍ ΜΗ· αƒ ∆Μ, ΜΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ∆Η. λέγω δή, Óτι κሠπέµπτη. `Οµοίως γ¦ρ διεχθήσεται, Óτι τÕ ØπÕ τîν ∆ΚΜ ‡σον ™στˆ τù ¢πÕ τÁς ΜΝ, κሠ¢σύµµετρος ¹ ∆Κ τÍ ΚΜ µήκει· ¹ ∆Μ ¥ρα τÁς ΜΗ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καί ε„σιν αƒ ∆Μ, ΜΗ [·ηταˆ] δυνάµει µόνον σύµµετροι, κሠ¹ ™λάσσων ¹ ΜΗ σύµµετρος τÍ ∆Ε µήκει. `Η ∆Η ¥ρα ™κ δύο Ñνοµάτων ™στˆ πέµπτη· Óπερ œδει δε‹ξαι.

B

Let the same construction be made as in the (propositions) before this. Therefore, since AB is the square-root of a rational plus a medial (area), having been divided at C, AC and CB are thus incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them rational [Prop. 10.40]. Therefore, since the sum of the (squares) on AC and CB is medial, DL is thus medial. Hence, DM is rational and incommensurable in length with DE [Prop. 10.22]. Again, since twice the (rectangle contained) by ACB— that is to say, M F —is rational, M G (is) thus rational and commensurable (in length) with DE [Prop. 10.20]. DM (is) thus incommensurable (in length) with M G [Prop. 10.13]. Thus, DM and M G are rational (straightlines which are) commensurable in square only. Thus, DG is a binomial (straight-line) [Prop. 10.36]. So, I say that (it is) also a fifth (binomial straight-line). For, similarly (to the previous propositions), it can be shown that the (rectangle contained) by DKM is equal to the (square) on M N , and DK (is) incommensurable in length with KM . Thus, the square on DM is greater than (the square on) M G by the (square) on (some straight-line) incommensurable (in length) with (DM ) [Prop. 10.18]. And DM and M G are [rational] (straight-lines which are) commensurable in square only, and the lesser M G is commensurable in length with DE. Thus, DG is a fifth binomial (straight-line) [Def. 10.9]. (Which is) the very thing it was required to show.

†

In other words, the square of the square-root of a rational plus medial is a fifth binomial. See Prop. 10.58.

ξε΄.

Proposition 65

ΤÕ ¢πÕ τÁς δύο µέσα δυναµένης παρ¦ ·ητ¾ν παραThe square on the square-root of (the sum of) two meβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων ›κτην. dial (areas) applied to a rational (straight-line) produces ”Εστω δύο µέσα δυναµένη ¹ ΑΒ διVρηµένη κατ¦ τÕ as breadth a sixth binomial (straight-line).† 356

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

Γ, ·ητ¾ δ œστω ¹ ∆Ε, κሠπαρ¦ τ¾ν ∆Ε τù ¢πÕ τÁς ΑΒ ‡σον παραβεβλήσθω τÕ ∆Ζ πλάτος ποιοàν τ¾ν ∆Η· λέγω, Óτι ¹ ∆Η ™κ δύο Ñνοµάτων ™στˆν ›κτη.

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Κατεσκευάσθω γ¦ρ τÕ αÙτ¦ το‹ς πρότερον. κሠ™πεˆ ¹ ΑΒ δύο µέσα δυναµένη ™στˆ διVρηµένη κατ¦ τÕ Γ, αƒ ΑΓ, ΓΒ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τό τε συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον κሠτÕ Øπ' αÙτîν µέσον κሠœτι ¢σύµµετρον τÕ ™κ τîν ¢π' αÙτîν τετραγώνων συγκείµενον τù Øπ' αÙτîν· éστε κατ¦ τ¦ προδεδειγµένα µέσον ™στˆν ˜κάτερον τîν ∆Λ, ΜΖ. κሠπαρ¦ ·ητ¾ν τ¾ν ∆Ε παράκειται· ·ητ¾ ¥ρα ™στˆν ˜κατέρα τîν ∆Μ, ΜΗ κሠ¢σύµµετρος τÍ ∆Ε µήκει. κሠ™πεˆ ¢σύµµετρόν ™στι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ τù δˆς ØπÕ τîν ΑΓ, ΓΒ, ¢σύµµετρον ¥ρα ™στˆ τÕ ∆Λ τù ΜΖ. ¢σύµµετρος ¥ρα κሠ¹ ∆Μ τÍ ΜΗ· αƒ ∆Μ, ΜΗ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ∆Η. λέγω δή, Óτι κሠ›κτη. `Οµοίως δ¾ πάλιν δε‹ξοµεν, Óτι τÕ ØπÕ τîν ∆ΚΜ ‡σον ™στˆ τù ¢πÕ τÁς ΜΝ, κሠÓτι ¹ ∆Κ τÍ ΚΜ µήκει ™στˆν ¢σύµµετρος· κሠδι¦ τ¦ αÙτ¦ δ¾ ¹ ∆Μ τÁς ΜΗ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ µήκει. κሠοÙδετέρα τîν ∆Μ, ΜΗ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ∆Ε µήκει. `Η ∆Η ¥ρα ™κ δύο Ñνοµάτων ™στˆν ›κτη· Óπερ œδει δε‹ξαι.

†

Let AB be the square-root of (the sum of) two medial (areas), having been divided at C. And let DE be a rational (straight-line). And let the (parallelogram) DF , equal to the (square) on AB, have been applied to DE, producing DG as breadth. I say that DG is a sixth binomial (straight-line).

B

For let the same construction be made as in the previous (propositions). And since AB is the square-root of (the sum of) two medial (areas), having been divided at C, AC and CB are thus incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them medial, and, moreover, the sum of the squares on them incommensurable with the (rectangle contained) by them [Prop. 10.41]. Hence, according to what has been previously demonstrated, DL and M F are each medial. And they are applied to the rational (straight-line) DE. Thus, DM and M G are each rational, and incommensurable in length with DE [Prop. 10.22]. And since the sum of the (squares) on AC and CB is incommensurable with twice the (rectangle contained) by AC and CB, DL is thus incommensurable with M F . Thus, DM (is) also incommensurable (in length) with M G [Props. 6.1, 10.11]. DM and M G are thus rational (straight-lines which are) commensurable in square only. Thus, DG is a binomial (straight-line) [Prop. 10.36]. So, I say that (it is) also a sixth (binomial straight-line). So, similarly (to the previous propositions), we can again show that the (rectangle contained) by DKM is equal to the (square) on M N , and that DK is incommensurable in length with KM . And so, for the same (reasons), the square on DM is greater than (the square on) M G by the (square) on (some straight-line) incommensurable with (DM ) [Prop. 10.18]. And neither of DM and M G is commensurable in length with the (previously) laid down rational (straight-line) DE. Thus, DG is a sixth binomial (straight-line) [Def. 10.10]. (Which is) the very thing it was required to show.

In other words, the square of the square-root of two medials is a sixth binomial. See Prop. 10.59.

357

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 ξ$΄.

Proposition 66

`Η τÍ ™κ δύο Ñνοµάτων µήκει σύµµετρος κሠαÙτ¾ ™κ δύο Ñνοµάτων ™στˆ κሠτÍ τάξει ¹ αÙτή. ”Εστω ™κ δύο Ñνοµάτων ¹ ΑΒ, κሠτÍ ΑΒ µήκει σύµµετρος œστω ¹ Γ∆· λέγω, Óτι ¹ Γ∆ ™κ δύο Ñνοµάτων ™στˆ κሠτÍ τάξει ¹ αÙτ¾ τÍ ΑΒ.

A (straight-line) commensurable in length with a binomial (straight-line) is itself also binomial, and the same in order. Let AB be a binomial (straight-line), and let CD be commensurable in length with AB. I say that CD is a binomial (straight-line), and (is) the same in order as AB.

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'Επεˆ γ¦ρ ™κ δύο Ñνοµάτων ™στˆν ¹ ΑΒ, διVρήσθω ε„ς τ¦ Ñνόµατα κατ¦ τÕ Ε, κሠœστω µε‹ζον Ôνοµα τÕ ΑΕ· αƒ ΑΕ, ΕΒ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. γεγονέτω æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΑΕ πρÕς τ¾ν ΓΖ· κሠλοιπ¾ ¥ρα ¹ ΕΒ πρÕς λοιπ¾ν τ¾ν Ζ∆ ™στιν, æς ¹ ΑΒ πρÕς τ¾ν Γ∆. σύµµετρος δ ¹ ΑΒ τÍ Γ∆ µήκει· σύµµετρος ¥ρα ™στˆ κሠ¹ µν ΑΕ τÍ ΓΖ, ¹ δ ΕΒ τÍ Ζ∆. καί ε„σι ·ηταˆ αƒ ΑΕ, ΕΒ· ·ητሠ¥ρα ε„σˆ καˆ αƒ ΓΖ, Ζ∆. κሠ™στιν æς ¹ ΑΕ πρÕς ΓΖ, ¹ ΕΒ πρÕς Ζ∆. ™ναλλ¦ξ ¥ρα ™στˆν æς ¹ ΑΕ πρÕς ΕΒ, ¹ ΓΖ πρÕς Ζ∆. αƒ δ ΑΕ, ΕΒ δυνάµει µόνον [ε„σˆ] σύµµετροι· καˆ αƒ ΓΖ, Ζ∆ ¥ρα δυνάµει µόνον ε„σˆ σύµµετροι. καί ε„σι ·ηταί· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ Γ∆. λέγω δή, Óτι τÍ τάξει ™στˆν ¹ αÙτ¾ τÍ ΑΒ. `Η γ¦ρ ΑΕ τÁς ΕΒ µε‹ζον δύναται ½τοι τù ¢πÕ συµµέτρου ˜αυτÍ À τù ¢πÕ ¢συµµέτρου. ε„ µν οâν ¹ ΑΕ τÁς ΕΒ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠ¹ ΓΖ τÁς Ζ∆ µε‹ζον δυνήσεται τù ¢πÕ συµµέτρου ˜αυτÍ. καˆ ε„ µν σύµµετρός ™στιν ¹ ΑΕ τÍ ™κκειµένV ·ητÍ, κሠ¹ ΓΖ σύµµετρος αÙτÍ œσται, κሠδι¦ τοàτο ˜κατέρα τîν ΑΒ, Γ∆ ™κ δύο Ñνοµάτων ™στˆ πρώτη, τουτέστι τÍ τάξει ¹ αÙτή. ε„ δ ¹ ΕΒ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ, κሠ¹ Ζ∆ σύµµετρός ™στιν αÙτÍ, κሠδι¦ τοàτο πάλιν τÍ τάξει ¹ αÙτ¾ œσται τÍ ΑΒ· ˜κατέρα γ¦ρ αÙτîν œσται ™κ δύο Ñνοµάτων δευτέρα. ε„ δ οÙδετέρα τîν ΑΕ, ΕΒ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ, οÙδετέρα τîν ΓΖ, Ζ∆ σύµµετρος αÙτÍ œσται, καί ™στιν ˜κατέρα τρίτη. ε„ δ ¹ ΑΕ τÁς ΕΒ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ, κሠ¹ ΓΖ τ¾ς Ζ∆ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καˆ ε„ µν ¹ ΑΕ σύµµετρός ™στι τÍ ™κκειµένV ·ητV, κሠ¹ ΓΖ σύµµετρός ™στιν αÙτÍ, και ™στιν ˜κατέρα τετάρτη. ε„ δ ¹ ΕΒ, κሠ¹ Ζ∆, κሠœσται ˜κατέρα πέµπτη. ε„ δ οÙδετέρα τîν ΑΕ, ΕΒ, κሠτîν ΓΖ, Ζ∆ οÙδετέρα σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ, κሠœσται ˜κατέρα ›κτη. “Ωστε ¹ τÍ ™κ δύο Ñνοµάτων µήκει σύµµετρος ™κ δύο Ñνοµάτων ™στˆ κሠτÍ τάξει ¹ αÙτή· Óπερ œδει δε‹ξαι.

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D

For since AB is a binomial (straight-line), let it have been divided into its (component) terms at E, and let AE be the greater term. AE and EB are thus rational (straight-lines which are) commensurable in square only [Prop. 10.36]. Let it have been contrived that as AB (is) to CD, so AE (is) to CF [Prop. 6.12]. Thus, the remainder EB is also to the remainder F D, as AB (is) to CD [Props. 6.16, 5.19 corr.]. And AB (is) commensurable in length with CD. Thus, AE is also commensurable (in length) with CF , and EB with F D [Prop. 10.11]. And AE and EB are rational. Thus, CF and F D are also rational. And as AE is to CF , (so) EB (is) to F D [Prop. 5.11]. Thus, alternately, as AE is to EB, (so) CF (is) to F D [Prop. 5.16]. And AE and EB [are] commensurable in square only. Thus, CF and F D are also commensurable in square only [Prop. 10.11]. And they are rational. CD is thus a binomial (straight-line) [Prop. 10.36]. So, I say that it is the same in order as AB. For the square on AE is greater than (the square on) EB by the (square) on (some straight-line) either commensurable or incommensurable (in length) with (AE). Therefore, if the square on AE is greater than (the square on) EB by the (square) on (some straight-line) commensurable (in length) with (AE), then the square on CF will also be greater than (the square on) F D by the (square) on (some straight-line) commensurable (in length) with (CF ) [Prop. 10.14]. And if AE is commensurable (in length) with (some previously) laid down rational (straight-line), then CF will also be commensurable (in length) with it [Prop. 10.12]. And, on account of this, AB and CD are each first binomial (straightlines) [Def. 10.5]—that is to say, the same in order. And if EB is commensurable (in length) with the (previously) laid down rational (straight-line), then F D is also commensurable (in length) with it [Prop. 10.12], and, again, on account of this, (CD) will be the same in order as AB. For each of them will be second binomial (straightlines) [Def. 10.6]. And if neither of AE and EB is com-

358

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 mensurable (in length) with the (previously) laid down rational (straight-line), then neither of CF and F D will be commensurable (in length) with it [Prop. 10.13], and each (of AB and CD) is a third (binomial straight-line) [Def. 10.7]. And if the square on AE is greater than (the square on) EB by the (square) on (some straightline) incommensurable (in length) with (AE), then the square on CF is also greater than (the square on) F D by the (square) on (some straight-line) incommensurable (in length) with (CF ) [Prop. 10.14]. And if AE is commensurable (in length) with the (previously) laid down rational (straight-line), then CF is also commensurable (in length) with it [Prop. 10.12], and each (of AB and CD) is a fourth (binomial straight-line) [Def. 10.8]. And if EB (is commensurable in length with the previously laid down rational straight-line), then F D (is) also (commensurable in length with it), and each (of AB and CD) will be a fifth (binomial straight-line) [Def. 10.9]. And if neither of AE and EB (is commensurable in length with the previously laid down rational straight-line), then also neither of CF and F D is commensurable (in length) with the laid down rational (straight-line), and each (of AB and CD) will be a sixth (binomial straight-line) [Def. 10.10]. Hence, a (straight-line) commensurable in length with a binomial (straight-line) is a binomial (straightline), and the same in order. (Which is) the very thing it was required to show.

ξζ΄.

Proposition 67

`Η τÍ ™κ δύο µέσων µήκει σύµµετρος κሠαÙτ¾ ™κ A (straight-line) commensurable in length with a biδύο µέσων ™στˆ κሠτÍ τάξει ¹ αÙτή. medial (straight-line) is itself also bimedial, and the same in order.

Α Γ

Ε

Β Ζ

A ∆

C

”Εστω ™κ δύο µέσων ¹ ΑΒ, κሠτÍ ΑΒ σύµµετρος œστω µήκει ¹ Γ∆· λέγω, Óτι ¹ Γ∆ ™κ δύο µέσων ™στˆ κሠτÍ τάξει ¹ αÙτ¾ τÍ ΑΒ. 'Επεˆ γ¦ρ ™κ δύο µέσων ™στˆν ¹ ΑΒ, διVρήσθω ε„ς τ¦ς µέσας κατ¦ τÕ Ε· αƒ ΑΕ, ΕΒ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι. κሠγεγονέτω æς ¹ ΑΒ πρÕς Γ∆, ¹ ΑΕ πρÕς ΓΖ· κሠλοιπ¾ ¥ρα ¹ ΕΒ πρÕς λοιπ¾ν τ¾ν Ζ∆ ™στιν, æς ¹ ΑΒ πρÕς Γ∆. σύµµετρος δ ¹ ΑΒ τÍ Γ∆ µήκει· σύµµετρος ¥ρα κሠ˜κατέρα τîν ΑΕ, ΕΒ ˜κατέρv τîν ΓΖ, Ζ∆. µέσαι δ αƒ ΑΕ, ΕΒ· µέσαι ¥ρα καˆ αƒ ΓΖ, Ζ∆. κሠ™πεί ™στιν æς ¹ ΑΕ πρÕς ΕΒ, ¹ ΓΖ πρÕς Ζ∆, αƒ δ ΑΕ, ΕΒ δυνάµει µόνον σύµµετροί ε„σιν, καˆ αƒ ΓΖ, Ζ∆ [¥ρα] δυνάµει µόνον σύµµετροί ε„σιν, ™δείχθησαν δ

E

B F

D

Let AB be a bimedial (straight-line), and let CD be commensurable in length with AB. I say that CD is bimedial, and the same in order as AB. For since AB is a bimedial (straight-line), let it have been divided into its (component) medial (straight-lines) at E. Thus, AE and EB are medial (straight-lines which are) commensurable in square only [Props. 10.37, 10.38]. And let it have been contrived that as AB (is) to CD, (so) AE (is) to CF [Prop. 6.12]. And thus as the remainder EB is to the remainder F D, so AB (is) to CD [Props. 5.19 corr., 6.16]. And AB (is) commensurable in length with CD. Thus, AE and EB are also commensurable (in length) with CF and F D, respectively

359

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

κሠµέσαι· ¹ Γ∆ ¥ρα ™κ δύο µέσων ™στίν. λέγω δή, Óτι κሠτÍ τάξει ¹ αÙτή ™στι τÍ ΑΒ. 'Επεˆ γάρ ™στιν æς ¹ ΑΕ πρÕς ΕΒ, ¹ ΓΖ πρÕς Ζ∆, κሠæς ¥ρα τÕ ¢πÕ τÁς ΑΕ πρÕς τÕ ØπÕ τîν ΑΕΒ, οÛτως τÕ ¢πÕ τÁς ΓΖ πρÕς τÕ ØπÕ τîν ΓΖ∆· ™ναλλ¦ξ æς τÕ ¢πÕ τÁς ΑΕ πρÕς τÕ ¢πÕ τÁς ΓΖ, οÛτως τÕ ØπÕ τîν ΑΕΒ πρÕς τÕ ØπÕ τîν ΓΖ∆. σύµµετρον δ τÕ ¢πÕ τÁς ΑΕ τù ¢πÕ τÁς ΓΖ· σύµµετρον ¥ρα κሠτÕ ØπÕ τîν ΑΕΒ τù ØπÕ τîν ΓΖ∆. ε‡τε οâν ·ητόν ™στι τÕ ØπÕ τîν ΑΕΒ, κሠτÕ ØπÕ τîν ΓΖ∆ ·ητόν ™στιν [κሠδι¦ τοàτό ™στιν ™κ δύο µέσων πρώτη]. ε‡τε µέσον, µέσον, καί ™στιν ˜κατέρα δευτέρα. Κሠδι¦ τοàτο œσται ¹ Γ∆ τÍ ΑΒ τÍ τάξει ¹ αÙτή· Óπερ œδει δε‹ξαι.

[Prop. 10.11]. And AE and EB (are) medial. Thus, CF and F D (are) also medial [Prop. 10.23]. And since as AE is to EB, (so) CF (is) to F D, and AE and EB are commensurable in square only, CF and F D are [thus] also commensurable in square only [Prop. 10.11]. And they were also shown (to be) medial. Thus, CD is a bimedial (straight-line). So, I say that it is also the same in order as AB. For since as AE is to EB, (so) CF (is) to F D, thus also as the (square) on AE (is) to the (rectangle contained) by AEB, so the (square) on CF (is) to the (rectangle contained) by CF D [Prop. 10.21 lem.]. Alternately, as the (square) on AE (is) to the (square) on CF , so the (rectangle contained) by AEB (is) to the (rectangle contained) by CF D [Prop. 5.16]. And the (square) on AE (is) commensurable with the (square) on CF . Thus, the (rectangle contained) by AEB (is) also commensurable with the (rectangle contained) by CF D [Prop. 10.11]. Therefore, either the (rectangle contained) by AEB is rational, and the (rectangle contained) by CF D is rational [and, on account of this, (AE and CD) are first bimedial (straight-lines)], or (the rectangle contained by AEB is) medial, and (the rectangle contained by CF D is) medial, and (AB and CD) are each second (bimedial straight-lines) [Props. 10.23, 10.37, 10.38]. And, on account of this, CD will be the same in order as AB. (Which is) the very thing it was required to show.

ξη΄.

Proposition 68

`Η τÍ µείζονι σύµµετρος κሠαÙτ¾ µείζων ™στίν.

Α Γ

Ε

A (straight-line) commensurable (in length) with a major (straight-line) is itself also major.

Β Ζ

A ∆

C

”Εστω µείζων ¹ ΑΒ, κሠτÍ ΑΒ σύµµετρος œστω ¹ Γ∆· λέγω, Óτι ¹ Γ∆ µείζων ™στίν. ∆ιVρήσθω ¹ ΑΒ κατ¦ τÕ Ε· αƒ ΑΕ, ΕΒ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων ·ητόν, τÕ δ' Øπ' αÙτîν µέσον· κሠγεγονέτω τ¦ αÙτ¦ το‹ς πρότερον. κሠ™πεί ™στιν æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¼ τε ΑΕ πρÕς τ¾ν ΓΖ κሠ¹ ΕΒ πρÕς τ¾ν Ζ∆, κሠæς ¥ρα ¹ ΑΕ πρÕς τ¾ν ΓΖ, οÛτως ¹ ΕΒ πρÕς τ¾ν Ζ∆. σύµµετρος δ ¹ ΑΒ τÍ Γ∆· σύµµετρος ¥ρα κሠ˜κατέρα τîν ΑΕ, ΕΒ ˜κατέρv τîν ΓΖ, Ζ∆. κሠ™πεί ™στιν æς ¹ ΑΕ πρÕς τ¾ν ΓΖ, οÛτως ¹ ΕΒ πρÕς τ¾ν Ζ∆, κሠ™ναλλ¦ξ æς ¹ ΑΕ πρÕς ΕΒ, οÛτως ¹ ΓΖ πρÕς Ζ∆, κሠσυνθέντι ¥ρα ˜στˆν æς ¹ ΑΒ πρÕς τ¾ν ΒΕ, οÛτως ¹ Γ∆ πρÕς τ¾ν ∆Ζ· κሠæς ¥ρα

E

B F

D

Let AB be a major (straight-line), and let CD be commensurable (in length) with AB. I say that CD is a major (straight-line). Let AB have been divided (into its component terms) at E. AE and EB are thus incommensurable in square, making the sum of the squares on them rational, and the (rectangle contained) by them medial [Prop. 10.39]. And let (the) same (things) have been contrived as in the previous (propositions). And since as AB is to CD, so AE (is) to CF and EB to F D, thus also as AE (is) to CF , so EB (is) to F D [Prop. 5.11]. And AB (is) commensurable (in length) with CD. Thus, AE and EB (are) also commensurable (in length) with CF and F D, respectively [Prop. 10.11]. And since as AE is to CF , so

360

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ¢πÕ τÁς ΒΕ, οÛτως τÕ ¢πÕ τÁς Γ∆ πρÕς τÕ ¢πÕ τÁς ∆Ζ. еοίως δ¾ δείξοµεν, Óτι κሠæς τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ¢πÕ τÁς ΑΕ, οÛτως τÕ ¢πÕ τÁς Γ∆ πρÕς τÕ ¢πÕ τÁς ΓΖ. κሠæς ¥ρα τÕ ¢πÕ τÁς ΑΒ πρÕς τ¦ ¢πÕ τîν ΑΕ, ΕΒ, οÛτως τÕ ¢πÕ τÁς Γ∆ πρÕς τ¦ ¢πÕ τîν ΓΖ, Ζ∆· κሠ™ναλλ¦ξ ¥ρα ™στˆν æς τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ¢πÕ τÁς Γ∆, οÛτως τ¦ ¢πÕ τîν ΑΕ, ΕΒ πρÕς τ¦ ¢πÕ τîν ΓΖ, Ζ∆. σύµµετρον δ τÕ ¢πÕ τÁς ΑΒ τù ¢πÕ τÁς Γ∆· σύµµετρα ¥ρα κሠτ¦ ¢πÕ τîν ΑΕ, ΕΒ το‹ς ¢πÕ τîν ΓΖ, Ζ∆. καί ™στι τ¦ ¢πÕ τîν ΑΕ, ΕΒ ¤µα ·ητόν, κሠτ¦ ¢πÕ τîν ΓΖ, Ζ∆ ¤µα ·ητόν ™στιν. еοίως δ κሠτÕ δˆς ØπÕ τîν ΑΕ, ΕΒ σύµµετρόν ™στι τù δˆς ØπÕ τîν ΓΖ, Ζ∆. καί ™στι µέσον τÕ δˆς ØπÕ τîν ΑΕ, ΕΒ· µέσον ¥ρα κሠτÕ δˆς ØπÕ τîν ΓΖ, Ζ∆. αƒ ΓΖ, Ζ∆ ¥ρα δυνάµει ¢σύµµετροί ε„σι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων ¤µα ·ητόν, τÕ δ δˆς Øπ' αÙτîν µέσον· Óλη ¥ρα ¹ Γ∆ ¥λογός ™στιν ¹ καλουµένη µείζων. `Η ¥ρα τÍ µείζονι σύµµετρος µείζων ™στίν· Óπερ œδει δε‹ξαι.

EB (is) to F D, also, alternately, as AE (is) to EB, so CF (is) to F D [Prop. 5.16], and thus, via composition, as AB is to BE, so CD (is) to DF [Prop. 5.18]. And thus as the (square) on AB (is) to the (square) on BE, so the (square) on CD (is) to the (square) on DF [Prop. 6.20]. So, similarly, we can also show that as the (square) on AB (is) to the (square) on AE, so the (square) on CD (is) to the (square) on CF . And thus as the (square) on AB (is) to (the sum of) the (squares) on AE and EB, so the (square) on CD (is) to (the sum of) the (squares) on CF and F D. And thus, alternately, as the (square) on AB is to the (square) on CD, so (the sum of) the (squares) on AE and EB (is) to (the sum of) the (squares) on CF and F D [Prop. 5.16]. And the (square) on AB (is) commensurable with the (square) on CD. Thus, (the sum of) the (squares) on AE and EB (is) also commensurable with (the sum of) the (squares) on CF and F D [Prop. 10.11]. And the (squares) on AE and EB (added) together are rational. The (squares) on CF and F D (added) together (are) thus also rational. So, similarly, twice the (rectangle contained) by AE and EB is also commensurable with twice the (rectangle contained) by CF and F D. And twice the (rectangle contained) by AE and EB is medial. Therefore, twice the (rectangle contained) by CF and F D (is) also medial [Prop. 10.23 corr.]. CF and F D are thus (straight-lines which are) incommensurable in square [Prop 10.13], simultaneously making the sum of the squares on them rational, and twice the (rectangle contained) by them medial. The whole, CD, is thus that irrational (straight-line) called major [Prop. 10.39]. Thus, a (straight-line) commensurable (in length) with a major (straight-line) is major. (Which is) the very thing it was required to show.

ξθ΄.

Proposition 69

`Η τÍ ·ητÕν κሠµέσον δυναµένV σύµµετρος [κሠαÙτ¾] ·ητÕν κሠµέσον δυναµένη ™στίν.

A (straight-line) commensurable (in length) with the square-root of a rational plus a medial (area) is [itself also] the square-root of a rational plus a medial (area).

Α Γ

Ε

Β Ζ

A ∆

C

”Εστω ·ητÕν κሠµέσον δυναµένη ¹ ΑΒ, κሠτÍ ΑΒ σύµµετρος œστω ¹ Γ∆· δεικτέον, Óτι κሠ¹ Γ∆ ·ητÕν κሠµέσον δυναµένη ™στίν. ∆ιVρήσθω ¹ ΑΒ ε„ς τ¦ς εÙθείας κατ¦ τÕ Ε· αƒ ΑΕ, ΕΒ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον, τÕ δ' Øπ' αÙτîν ·ητόν· κሠτ¦ αÙτ¦ κατεσκευάσθω το‹ς πρότερον. еοίως δ¾ δείξοµεν, Óτι καˆ αƒ ΓΖ,

E

B F

D

Let AB be the square-root of a rational plus a medial (area), and let CD be commensurable (in length) with AB. We must show that CD is also the square-root of a rational plus a medial (area). Let AB have been divided into its (component) straight-lines at E. AE and EB are thus incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them rational

361

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

Ζ∆ δυνάµει ε„σˆν ¢σύµµετροι, κሠσύµµετρον τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΑΕ, ΕΒ τù συγκειµένJ ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆, τÕ δ ØπÕ ΑΕ, ΕΒ τù ØπÕ ΓΖ, Ζ∆· éστε κሠτÕ [µν] συγκείµενον ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆ τετραγώνων ™στˆ µέσον, τÕ δ' ØπÕ τîν ΓΖ, Ζ∆ ·ητόν. `ΡητÕν ¥ρα κሠµέσον δυναµένη ™στˆν ¹ Γ∆· Óπερ œδει δε‹ξαι.

[Prop. 10.40]. And let the same construction have been made as in the previous (propositions). So, similarly, we can show that CF and F D are also incommensurable in square, and that the sum of the (squares) on AE and EB (is) commensurable with the sum of the (squares) on CF and F D, and the (rectangle contained) by AE and EB with the (rectangle contained) by CF and F D. And hence the sum of the squares on CF and F D is medial, and the (rectangle contained) by CF and F D (is) rational. Thus, CD is the square-root of a rational plus a medial (area) [Prop. 10.40]. (Which is) the very thing it was required to show.

ο΄.

Proposition 70

`Η τÍ δύο µέσα δυναµένV σύµµετρος δύο µέσα δυναµένη ™στίν.

A (straight-line) commensurable (in length) with the square-root of (the sum of) two medial (areas) is (itself also) the square-root of (the sum of) two medial (areas).

Α Γ

Ε

Β Ζ

A ∆

C

”Εστω δύο µέσα δυναµένη ¹ ΑΒ, κሠτÍ ΑΒ σύµµετρος ¹ Γ∆· δεικτέον, Óτι κሠ¹ Γ∆ δύο µέσα δυναµένη ™στίν. 'Επεˆ γ¦ρ δύο µέσα δυναµένη ™στˆν ¹ ΑΒ, διVρήσθω ε„ς τ¦ς εÙθείας κατ¦ τÕ Ε· αƒ ΑΕ, ΕΒ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τό τε συγκείµενον ™κ τîν ¢π' αÙτîν [τετραγώνων] µέσον κሠτÕ Øπ' αÙτîν µέσον κሠœτι ¢σύµµετρον τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΕ, ΕΒ τετραγώνων τù ØπÕ τîν ΑΕ, ΕΒ· κሠκατεσκευάσθω τ¦ αÙτ¦ το‹ς πρότερον. еοίως δ¾ δείξοµεν, Óτι καˆ αƒ ΓΖ, Ζ∆ δυνάµει ε„σˆν ¢σύµµετροι κሠσύµµετρον τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΑΕ, ΕΒ τù συγκειµένJ ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆, τÕ δ ØπÕ τîν ΑΕ, ΕΒ τù ØπÕ τîν ΓΖ, Ζ∆· éστε κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆ τετραγώνων µέσον ™στˆ κሠτÕ ØπÕ τîν ΓΖ, Ζ∆ µέσον κሠœτι ¢σύµµετρον τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆ τετραγώνων τù ØπÕ τîν ΓΖ, Ζ∆. `Η ¥ρα Γ∆ δύο µέσα δυναµένη ™στίν· Óπερ œδει δε‹ξαι.

E

B F

D

Let AB be the square-root of (the sum of) two medial (areas), and (let) CD (be) commensurable (in length) with AB. We must show that CD is also the square-root of (the sum of) two medial (areas). For since AB is the square-root of (the sum of) two medial (areas), let it have been divided into its (component) straight-lines at E. Thus, AE and EB are incommensurable in square, making the sum of the [squares] on them medial, and the (rectangle contained) by them medial, and, moreover, the sum of the (squares) on AE and EB incommensurable with the (rectangle) contained by AE and EB [Prop. 10.41]. And let the same construction have been made as in the previous (propositions). So, similarly, we can show that CF and F D are also incommensurable in square, and (that) the sum of the (squares) on AE and EB (is) commensurable with the sum of the (squares) on CF and F D, and the (rectangle contained) by AE and EB with the (rectangle contained) by CF and F D. Hence, the sum of the squares on CF and F D is also medial, and the (rectangle contained) by CF and F D (is) medial, and, moreover, the sum of the squares on CF and F D (is) incommensurable with the (rectangle contained) by CF and F D. Thus, CD is the square-root of (the sum of) two medial (areas) [Prop. 10.41]. (Which is) the very thing it

362

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 was required to show.

οα΄.

Proposition 71

`Ρητοà κሠµέσου συντιθεµένου τέσσαρες ¥λογοι γίγνονται ½τοι ™κ δύο Ñνοµάτων À ™κ δύο µέσων πρώτη À µείζων À ·ητÕν κሠµέσον δυναµένη. ”Εστω ·ητÕν µν τÕ ΑΒ, µέσον δ τÕ Γ∆· λέγω, Óτι ¹ τÕ Α∆ χωρίον δυναµένη ½τοι ™κ δύο Ñνοµάτων ™στˆν À ™κ δύο µέσων πρώτη À µείζων À ·ητÕν κሠµέσον δυναµένη.

When a rational and a medial (area) are added together, four irrational (straight-lines) arise (as the squareroots of the total area)—either a binomial, or a first bimedial, or a major, or the square-root of a rational plus a medial (area). Let AB be a rational (area), and CD a medial (area). I say that the square-root of area AD is either binomial, or first bimedial, or major, or the square-root of a rational plus a medial (area).

Α

Γ

Β

Ε

Θ

Κ

Ζ

Η

Ι

A

∆

C

B

ΤÕ γ¦ρ ΑΒ τοà Γ∆ ½τοι µε‹ζόν ™στιν À œλασσον. œστω πρότερον µε‹ζον· κሠ™κκείσθω ·ητ¾ ¹ ΕΖ, κሠπαραβεβλήσθω παρ¦ τ¾ν ΕΖ τù ΑΒ ‡σον τÕ ΕΗ πλάτος ποιοàν τ¾ν ΕΘ· τù δ ∆Γ ‡σον παρ¦ τ¾ν ΕΖ παραβεβλήσθω τÕ ΘΙ πλάτος ποιοàν τ¾ν ΘΚ. κሠ™πεˆ ·ητόν ™στι τÕ ΑΒ καί ™στιν ‡σον τù ΕΗ, ·ητÕν ¥ρα κሠτÕ ΕΗ. κሠπαρ¦ [·ητ¾ν] τ¾ν ΕΖ παραβέβληται πλάτος ποιοàν τ¾ν ΕΘ· ¹ ΕΘ ¥ρα ·ητή ™στι κሠσύµµετρος τÍ ΕΖ µήκει. πάλιν, ™πεˆ µέσον ™στˆ τÕ Γ∆ καί ™στιν ‡σον τù ΘΙ, µέσον ¥ρα ™στˆ κሠτÕ ΘΙ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται πλάτος ποιοàν τ¾ν ΘΚ· ·ητ¾ ¥ρα ™στˆν ¹ ΘΚ κሠ¢σύµµετρος τÍ ΕΖ µήκει. κሠ™πεˆ µέσον ™στˆ τÕ Γ∆, ·ητÕν δ τÕ ΑΒ, ¢σύµµετρον ¥ρα ™στˆ τÕ ΑΒ τù Γ∆· éστε κሠτÕ ΕΗ ¢σύµµετρόν ™στι τù ΘΙ. æς δ τÕ ΕΗ πρÕς τÕ ΘΙ, οÛτως ™στˆν ¹ ΕΘ πρÕς τ¾ν ΘΚ· ¢σύµµετρος ¥ρα ™στˆ κሠ¹ ΕΘ τÍ ΘΚ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ΕΘ, ΘΚ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ΕΚ διVρηµένη κατ¦ τÕ Θ. κሠ™πεˆ µε‹ζόν ™στι τÕ ΑΒ τοà Γ∆, ‡σον δ τÕ µν ΑΒ τù ΕΗ, τÕ δ Γ∆ τù ΘΙ, µε‹ζον ¥ρα κሠτÕ ΕΗ τοà ΘΙ· κሠ¹ ΕΘ ¥ρα µείζων ™στˆ τÁς ΘΚ. ½τοι οâν ¹ ΕΘ τÁς ΘΚ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει À τù ¢πÕ ¢συµµέτρου. δυνάσθω πρότερον τù ¢πÕ συµµέτρου ˜αυτÍ· καί ™στιν ¹ µείζων ¹ ΘΕ σύµµετρος τÍ ™κκειµένV ·ητV τÍ ΕΖ· ¹ ¥ρα ΕΚ ™κ δύο Ñνοµάτων ™στˆ πρώτη. ·ητ¾ δ ¹ ΕΖ· ™¦ν δ

E

H

K

F

G

I

D

For AB is either greater or less than CD. Let it, first of all, be greater. And let the rational (straight-line) EF be laid down. And let (the rectangle) EG, equal to AB, have been applied to EF , producing EH as breadth. And let (the recatangle) HI, equal to DC, have been applied to EF , producing HK as breadth. And since AB is rational, and is equal to EG, EG is thus also rational. And it has been applied to the [rational] (straight-line) EF , producing EH as breadth. EH is thus rational, and commensurable in length with EF [Prop. 10.20]. Again, since CD is medial, and is equal to HI, HI is thus also medial. And it is applied to the rational (straight-line) EF , producing HK as breadth. HK is thus rational, and incommensurable in length with EF [Prop. 10.22]. And since CD is medial, and AB rational, AB is thus incommensurable with CD. Hence, EG is also incommensurable with HI. And as EG (is) to HI, so EH is to HK [Prop. 6.1]. Thus, EH is also incommensurable in length with HK [Prop. 10.11]. And they are both rational. Thus, EH and HK are rational (straight-lines which are) commensurable in square only. EK is thus a binomial (straight-line), having been divided (into its component terms) at H [Prop. 10.36]. And since AB is greater than CD, and AB (is) equal to EG, and CD to HI, EG (is) thus also greater than HI. Thus, EH is also greater than HK [Prop. 5.14]. Therefore, the square

363

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο Ñνοµάτων πρώτης, ¹ τÕ χωρίον δυναµένη ™κ δύο Ñνοµάτων ™στίν. ¹ ¥ρα τÕ ΕΙ δυναµένη ™κ δύο Ñνοµάτων ™στίν· éστε κሠ¹ τÕ Α∆ δυναµένη ™κ δύο Ñνοµάτων ™στίν. ¢λλ¦ δ¾ δυνάσθω ¹ ΕΘ τÁς ΘΚ µε‹ζον τù ¢πÕ ¢συµµέτρου ˜αυτÍ· καί ™στιν ¹ µείζων ¹ ΕΘ σύµµετρος τÍ ™κκειµένV ·ητÍ τÍ ΕΖ µήκει· ¹ ¥ρα ΕΚ ™κ δύο Ñνοµάτων ™στˆ τετάρτη. ·ητ¾ δ ¹ ΕΖ· ™¦ν δ χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο Ñνοµάτων τετάρτης, ¹ τÕ χωρίον δυναµένη ¥λογός ™στιν ¹ καλουµένη µείζων. ¹ ¥ρα τÕ ΕΙ χωρίον δυναµένη µείζων ™στίν· éστε κሠ¹ τÕ Α∆ δυναµένη µείζων ™στίν. 'Αλλ¦ δ¾ œστω œλασσον τÕ ΑΒ τοà Γ∆· κሠτÕ ΕΗ ¥ρα œλασσόν ™στι τοà ΘΙ· éστε κሠ¹ ΕΘ ™λάσσων ™στˆ τÁς ΘΚ. ½τοι δ ¹ ΘΚ τÁς ΕΘ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ À τù ¢πÕ ¢συµµέτρου. δυνάσθω πρότερον τù ¢πÕ συµµέτρου ˜αυτÍ µήκει· καί ™στιν ¹ ™λάσσων ¹ ΕΘ σύµµετρος τÍ ™κκειµένV ·ητÍ τÍ ΕΖ µήκει· ¹ ¥ρα ΕΚ ™κ δύο Ñνοµάτων ™στˆ δευτέρα. ·ητ¾ δ ¹ ΕΖ· ™¦ν δ χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο Ñνοµάτων δευτέρας, ¹ τÕ χωρίον δυναµένη ™κ δύο µέσων ™στˆ πρώτη. ¹ ¥ρα τÕ ΕΙ χωρίον δυναµένη ™κ δύο µέσων ™στˆ πρώτη· éστε κሠ¹ τÕ Α∆ δυναµένη ™κ δύο µέσων ™στˆ πρώτη. ¢λλ¦ δ¾ ¹ ΘΚ τÁς ΘΕ µε‹ζον δυνάσθω τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καί ™στιν ¹ ™λάσσων ¹ ΕΘ σύµµετρος τÍ ™κκειµένV ·ητÍ τÍ ΕΖ· ¹ ¥ρα ΕΚ ™κ δύο Ñνοµάτων ™στˆ πέµπτη. ·ητ¾ δ ¹ ΕΖ· ™¦ν δ χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο Ñνοµάτων πέµπτÁς, ¹ τÕ χωρίον δυναµένη ·ητÕν κሠµέσον δυναµένη ™στίν. ¹ ¥ρα τÕ ΕΙ χωρίον δυναµένη ·ητÕν κሠµέσον δυναµένη ™στίν· éστε κሠ¹ τÕ Α∆ χωρίον δυναµένη ·ητÕν κሠµέσον δυναµένη ™στίν. `Ρητοà ¥ρα κሠµέσου συντιθεµένου τέσσαρες ¥λογοι γίγνονται ½τοι ™κ δύο Ñνοµάτων À ™κ δύο µέσων πρώτη À µείζων À ·ητÕν κሠµέσον δυναµένη· Óπερ œδει δε‹ξαι.

on EH is greater than (the square on) HK either by the (square) on (some straight-line) commensurable in length with (EH), or by the (square) on (some straightline) incommensurable (in length with EH). Let it, first of all, be greater by the (square) on (some straight-line) commensurable (in length with EH). And the greater (of the two components of EK) HE is commensurable (in length) with the (previously) laid down (straightline) EF . EK is thus a first binomial (straight-line) [Def. 10.5]. And EF (is) rational. And if an area is contained by a rational (straight-line) and a first binomial (straight-line), then the square-root of the area is a binomial (straight-line) [Prop. 10.54]. Thus, the square-root of EI is a binomial (straight-line). Hence the squareroot of AD is also a binomial (straight-line). And, so, let the square on EH be greater than (the square on) HK by the (square) on (some straight-line) incommensurable (in length) with (EH). And the greater (of the two components of EK) EH is commensurable in length with the (previously) laid down rational (straight-line) EF . Thus, EK is a fourth binomial (straight-line) [Def. 10.8]. And EF (is) rational. And if an area is contained by a rational (straight-line) and a fourth binomial (straight-line), then the square-root of the area is the irrational (straight-line) called major [Prop. 10.57]. Thus, the square-root of area EI is a major (straight-line). Hence, the square-root of AD is also major. And so, let AB be less than CD. Thus, EG is also less than HI. Hence, EH is also less than HK [Props. 6.1, 5.14]. And the square on HK is greater than (the square on) EH either by the (square) on (some straightline) commensurable (in length) with (HK), or by the (square) on (some straight-line) incommensurable (in length) with (HK). Let it, first of all, be greater by the square on (some straight-line) commensurable in length with (HK). And the lesser (of the two components of EK) EH is commensurable in length with the (previously) laid down rational (straight-line) EF . Thus, EK is a second binomial (straight-line) [Def. 10.6]. And EF (is) rational. And if an area is contained by a rational (straight-line) and a second binomial (straight-line), then the square-root of the area is a first bimedial (straightline) [Prop. 10.55]. Thus, the square-root of area EI is a first bimedial (straight-line). Hence, the square-root of AD is also a first bimedial (straight-line). And so, let the square on HK be greater than (the square on) HE by the (square) on (some straight-line) incommensurable (in length) with (HK). And the lesser (of the two components of EK) EH is commensurable (in length) with the (previously) laid down rational (straight-line) EF . Thus, EK is a fifth binomial (straight-line) [Def. 10.9].

364

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 And EF (is) rational. And if an area is contained by a rational (straight-line) and a fifth binomial (straight-line), then the square-root of the area is the square-root of a rational plus a medial (area) [Prop. 10.58]. Thus, the square-root of area EI is the square-root of a rational plus a medial (area). Hence, the square-root of area AD is also the square-root of a rational plus a medial (area). Thus, when a rational and a medial area are added together, four irrational (straight-lines) arise (as the squareroots of the total area)—either a binomial, or a first bimedial, or a major, or the square-root of a rational plus a medial (area). (Which is) the very thing it was required to show.

ξβ΄.

Proposition 72

∆ύο µέσων ¢συµµέτρων ¢λλήλοις συντιθεµένων αƒ When two medial (areas which are) incommensuλοιπሠδύο ¥λογοι γίγνονται ½τοι ™κ δύο µέσων δευτέρα rable with one another are added together, the remaining À [¹] δύο µέσα δυναµένη. two irrational (straight-lines) arise (as the square-roots of the total area)—either a second bimedial, or the squareroot of (the sum of) two medial (areas).

Α

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Ζ

Η

Ι

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∆

C

B

Συγκείσθω γ¦ρ δύο µέσα ¢σύµµετρα ¢λλήλοις τ¦ ΑΒ, Γ∆· λέγω, Óτι ¹ τÕ Α∆ χωρίον δυναµένη ½τοι ™κ δύο µέσων ™στˆ δευτέρα À δύο µέσα δυναµένη. ΤÕ γ¦ρ ΑΒ τοà Γ∆ ½τοι µε‹ζόν ™στιν À œλασσον. œστω, ε„ τύχον, πρότερον µε‹ζον τÕ ΑΒ τοà Γ∆· κሠ™κκείσθω ·ητ¾ ¹ ΕΖ, κሠτù µν ΑΒ ‡σον παρ¦ τ¾ν ΕΖ παραβεβλήσθω τÕ ΕΗ πλάτος ποιοàν τ¾ν ΕΘ, τù δ Γ∆ ‡σον τÕ ΘΙ πλάτος ποιοàν τ¾ν ΘΚ. κሠ™πεˆ µέσον ™στˆν ˜κάτερον τîν ΑΒ, Γ∆, µέσον ¥ρα κሠ™κάτερον τîν ΕΗ, ΘΙ. κሠπαρ¦ ·ητ¾ν τ¾ν ΖΕ παράκειται πλάτος ποιοàν τ¦ς ΕΘ, ΘΚ· ˜κατέρα ¥ρα τîν ΕΘ, ΘΚ ·ητή ™στι κሠ¢σύµµετρος τÍ ΕΖ µήκει. κሠ™πεˆ ¢σύµµετρόν ™στι τÕ ΑΒ τù Γ∆, καί ™στιν ‡σον τÕ µν ΑΒ τù ΕΗ, τÕ δ Γ∆ τù ΘΙ, ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ΕΗ τù ΘΙ. æς δ τÕ ΕΗ πρÕς τÕ ΘΙ, οÛτως ™στˆν ¹ ΕΘ πρÕς ΘΚ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΕΘ τÍ ΘΚ µήκει. αƒ ΕΘ, ΘΚ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ΕΚ. ½τοι δ ¹ ΕΘ τÁς

E

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For let the two medial (areas) AB and CD, (which are) incommensurable with one another, have been added together. I say that the square-root of area AD is either a second bimedial, or the square-root of (the sum of) two medial (areas). For AB is either greater than or less than CD. By chance, let AB, first of all, be greater than CD. And let the rational (straight-line) EF be laid down. And let EG, equal to AB, have been applied to EF , producing EH as breadth, and HI, equal to CD, producing HK as breadth. And since AB and CD are each medial, EG and HI (are) thus also each medial. And they are applied to the rational straight-line F E, producing EH and HK (respectively) as breadth. Thus, EH and HK are each rational (straight-lines which are) incommensurable in length with EF [Prop. 10.22]. And since AB is incommensurable with CD, and AB is equal to EG, and CD to HI, EG is thus also incommensurable with HI. And

365

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

ΘΚ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ À τù ¢πÕ ¢συµµέτρου. δυνάσθω πρότερον τù ¢πÕ συµµέτρου ˜αυτÍ µήκει· κሠοÙδετέρα τîν ΕΘ, ΘΚ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ΕΖ µήκει· ¹ ΕΚ ¥ρα ™κ δύο Ñνοµάτων ™στˆ τρίτη. ·ητ¾ δ ¹ ΕΖ· ™¦ν δ χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο Ñνοµάτων τρίτης, ¹ τÕ χωρίον δυναµένη ™κ δύο µέσων ™στˆ δευτέρα· ¹ ¥ρα τÕ ΕΙ, τουτέστι τÕ Α∆, δυναµένη ™κ δύο µέσων ™στˆ δευτέρα. ¢λλα δ¾ ¹ ΕΘ τÁς ΘΚ µε‹ζον δυνάσθω τù ¢πÕ ¢συµµέτρου ˜αυτÍ µήκει· κሠ¢σύµµετρός ™στιν ˜κατέρα τîν ΕΘ, ΘΚ τÍ ΕΖ µήκει· ¹ ¥ρα ΕΚ ™κ δύο Ñνοµάτων ™στˆν ›κτη. ™¦ν δ χωρίον περιέχηται ØπÕ ·ητÁς κሠτÁς ™κ δύο Ñνοµάτων ›κτης, ¹ τÕ χωρίον δυναµένη ¹ δύο µέσα δυναµένη ™στίν· éστε κሠ¹ τÕ Α∆ χωρίον δυναµένη ¹ δύο µέσα δυναµένη ™στίν. [`Οµοίως δ¾ δείξοµεν, Óτι κ¨ν œλαττον Ï τÕ ΑΒ τοà Γ∆, ¹ τÕ Α∆ χωρίον δυναµένη À ™κ δύο µέσων δευτέρα ™στˆν ½τοι δύο µέσα δυναµένη]. ∆ύο ¥ρα µέσων ¢συµµέτρων ¢λλήλοις συντιθεµένων αƒ λοιπሠδύο ¥λογοι γίγνονται ½τοι ™κ δύο µέσων δευτέρα À δύο µέσα δυναµένη.

as EG (is) to HI, so EH is to HK [Prop. 6.1]. EH is thus incommensurable in length with HK [Prop. 10.11]. Thus, EH and HK are rational (straight-lines which are) commensurable in square only. EK is thus a binomial (straight-line) [Prop. 10.36]. And the square on EH is greater than (the square on) HK either by the (square) on (some straight-line) commensurable (in length) with (EH), or by the (square) on (some straight-line) incommensurable (in length with EH). Let it, first of all, be greater by the square on (some straight-line) commensurable in length with (EH). And neither of EH or HK is commensurable in length with the (previously) laid down rational (straight-line) EF . Thus, EK is a third binomial (straight-line) [Def. 10.7]. And EF (is) rational. And if an area is contained by a rational (straight-line) and a third binomial (straight-line), then the square-root of the area is a second bimedial (straight-line) [Prop. 10.56]. Thus, the square-root of EI—that is to say, of AD— is a second bimedial. And so, let the square on EH be greater than (the square) on HK by the (square) on (some straight-line) incommensurable in length with (EH). And EH and HK are each incommensurable in length with EF . Thus, EK is a sixth binomial (straightline) [Def. 10.10]. And if an area is contained by a rational (straight-line) and a sixth binomial (straight-line), then the square-root of the area is the square-root of (the sum of) two medial (areas) [Prop. 10.59]. Hence, the square-root of area AD is also the square-root of (the sum of) two medial (areas). [So, similarly, we can show that, even if AB is less than CD, the square-root of area AD is either a second bimedial or the square-root of (the sum of) two medial (areas).] Thus, when two medial (areas which are) incommensurable with one another are added together, the remaining two irrational (straight-lines) arise (as the squareroots of the total area)—either a second bimedial, or the square-root of (the sum of) two medial (areas).

'Η ™κ δύο Ñνοµάτων καˆ αƒ µετ' αÙτ¾ν ¥λογοι οÜτε τÍ µέσV οÜτε ¢λλήλαις ε„σˆν αƒ αÙταί. τÕ µν γ¦ρ ¢πÕ µέσης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ·ητ¾ν κሠ¢σύµµετρον τÍ παρ' ¿ν παράκειται µήκει. τÕ δ ¢πÕ τÁς ™κ δύο Ñνοµάτων παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων πρώτην. τÕ δ ¢πÕ τÁς ™κ δύο µέσων πρώτης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων δευτέραν. τÕ δ ¢πÕ τÁς ™κ δύο µέσων δευτέρας παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων τρίτην. τÕ δ ¢πÕ τÁς µείζονος παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων τετάρτην. τÕ δ ¢πÕ τÁς ·ητÕν κሠµέσον δυναµένης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων πέµπτην. τÕ δ ¢πÕ τÁς δύο µέσα δυναµένης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων ›κτην. τ¦ δ' ε„ρηµένα πλάτη διαφέρει τοà τε πρώτου κሠ¢λλήλων, τοà µν πρώτου, Óτι ·ητή ™στιν, ¢λλήλων δέ, Óτι τÍ τάξει οÙκ ε„σˆν αƒ A binomial (straight-line), and the (other) irrational αÙταί· éστε κሠαÙταˆ αƒ ¥λογοι διαφέρουσιν ¢λλήλων. (straight-lines) after it, are neither the same as a medial (straight-line) nor (the same) as one another. For the (square) on a medial (straight-line), applied to a rational (straight-line), produces as breadth a rational (straightline which is) also incommensurable in length with (the straight-line) to which it is applied [Prop. 10.22]. And the (square) on a binomial (straight-line), applied to a rational (straight-line), produces as breadth a first binomial [Prop. 10.60]. And the (square) on a first bimedial (straight-line), applied to a rational (straight-line), produces as breadth a second binomial [Prop. 10.61]. And 366

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 the (square) on a second bimedial (straight-line), applied to a rational (straight-line), produces as breadth a third binomial [Prop. 10.62]. And the (square) on a major (straight-line), applied to a rational (straight-line), produces as breadth a fourth binomial [Prop. 10.63]. And the (square) on the square-root of a rational plus a medial (area), applied to a rational (straight-line), produces as breadth a fifth binomial [Prop. 10.64]. And the (square) on the square-root of (the sum of) two medial (areas), applied to a rational (straight-line), produces as breadth a sixth binomial [Prop. 10.65]. And the aforementioned breadths differ from the first (breadth), and from one another—from the first, because it is rational—and from one another, because they are not the same in order. Hence, the (previously mentioned) irrational (straightlines) themselves also differ from one another.

ογ΄.

Proposition 73

'Ε¦ν ¢πÕ ·ητÁς ·ητ¾ ¢φαιρεθÍ δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV, ¹ λοιπ¾ ¥λογός ™στιν· καλείσθω δ ¢ποτοµή.

If a rational (straight-line), which is commensurable in square only with the whole, is subtracted from a(nother) rational (straight-line), then the remainder is an irrational (straight-line). Let it be called an apotome.

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A

'ΑπÕ γ¦ρ ·ητÁς τÁς ΑΒ ·ητ¾ ¢φVρήσθω ¹ ΒΓ δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV· λέγω, Óτι ¹ λοιπ¾ ¹ ΑΓ ¥λογός ™στιν ¹ καλουµένη ¢ποτοµή. 'Επεˆ γ¦ρ ¢σύµµετρός ™στιν ¹ ΑΒ τÍ ΒΓ µήκει, καί ™στιν æς ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ØπÕ τîν ΑΒ, ΒΓ, ¢σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΒ τù ØπÕ τîν ΑΒ, ΒΓ. ¢λλ¦ τù µν ¢πÕ τÁς ΑΒ σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΒ, ΒΓ τετράγωνα, τù δ ØπÕ τîν ΑΒ, ΒΓ σύµµετρόν ™στι τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ. κሠ™πειδήπερ τ¦ ¢πÕ τîν ΑΒ, ΒΓ ‡σα ™στˆ τù δˆς ØπÕ τîν ΑΒ, ΒΓ µετ¦ τοà ¢πÕ ΓΑ, κሠλοιπù ¥ρα τù ¢πÕ τÁς ΑΓ ¢σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΒ, ΒΓ. ·ητ¦ δ τ¦ ¢πÕ τîν ΑΒ, ΒΓ· ¥λογος ¥ρα ™στˆν ¹ ΑΓ· καλείσθω δ ¢ποτοµή. Óπερ œδει δε‹ξαι.

C

B

For let the rational (straight-line) BC, which commensurable in square only with the whole, have been subtracted from the rational (straight-line) AB. I say that the remainder AC is that irrational (straight-line) called an apotome. For since AB is incommensurable in length with BC, and as AB is to BC, so the (square) on AB (is) to the (rectangle contained) by AB and BC [Prop. 10.21 lem.], the (square) on AB is thus incommensurable with the (rectangle contained) by AB and BC [Prop. 10.11]. But, the (sum of the) squares on AB and BC is commensurable with the (square) on AB [Prop. 10.15], and twice the (rectangle contained) by AB and BC is commensurable with the (rectangle contained) by AB and BC [Prop. 10.6]. And, inasmuch as the (sum of the squares) on AB and BC is equal to twice the (rectangle contained) by AB and BC plus the (square) on AC [Prop. 2.7], the (sum of the squares) on AB and BC is thus also incommensurable with the remaining (square) on AC [Props. 10.13, 10.16]. And the (sum of the squares) on AB and BC is rational. AC is thus an irrational (straight-line) [Def. 10.4]. And let it be called an apotome.† (Which is) the very thing it was required to show.

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ELEMENTS BOOK 10

See footnote to Prop. 10.36.

οδ΄.

Proposition 74

'Ε¦ν ¢πÕ µέσης µέση ¢φαιρεθÍ δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης ·ητÕν περιέχουσα, ¹ λοιπ¾ ¥λογός ™στιν· καλείσθω δ µέσης ¢ποτοµ¾ πρώτη.

If a medial (straight-line), which is commensurable in square only with the whole, and which contains a rational (area) with the whole, is subtracted from a(nother) medial (straight-line), then the remainder is an irrational (straight-line). Let it be called a first apotome of a medial (straight-line).

Α

Γ

Β

A

'ΑπÕ γ¦ρ µέσης τÁς ΑΒ µέση ¢φVρήσθω ¹ ΒΓ δυνάµει µόνον σύµµετρος οâσα τÍ ΑΒ, µετ¦ δ τÁς ΑΒ ·ητÕν ποιοàσα τÕ ØπÕ τîν ΑΒ, ΒΓ· λέγω, Óτι ¹ λοιπ¾ ¹ ΑΓ ¥λογός ™στιν· καλείσθω δ µέσης ¢ποτοµ¾ πρώτη. 'Επεˆ γ¦ρ αƒ ΑΒ, ΒΓ µέσαι ε„σίν, µέσα ™στˆ κሠτ¦ ¢πÕ τîν ΑΒ, ΒΓ. ·ητÕν δ τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ· ¢σύµµετρα ¥ρα τ¦ ¢πÕ τîν ΑΒ, ΒΓ τù δˆς ØπÕ τîν ΑΒ, ΒΓ· κሠλοιπù ¥ρα τù ¢πÕ τÁς ΑΓ ¢σύµµετρόν ™στι τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ, ™πεˆ κ¨ν τÕ Óλον ˜νˆ αÙτîν ¢σύµµετρον Ï, κሠτ¦ ™ξ ¢ρχÁς µεγέθη ¢σύµµετρα œσται. ·ητÕν δ τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ· ¥λογον ¥ρα τÕ ¢πÕ τÁς ΑΓ· ¥λογος ¥ρα ™στˆν ¹ ΑΓ· καλείσθω δ µέσης ¢ποτοµ¾ πρώτη.

†

C

B

For let the medial (straight-line) BC, which is commensurable in square only with AB, and which makes with AB the rational (rectangle contained) by AB and BC, have been subtracted from the medial (straight-line) AB [Prop. 10.27]. I say that the remainder AC is an irrational (straight-line). Let it be called the first apotome of a medial (straight-line). For since AB and BC are medial (straight-lines), the (sum of the squares) on AB and BC is also medial. And twice the (rectangle contained) by AB and BC (is) rational. The (sum of the squares) on AB and BC (is) thus incommensurable with twice the (rectangle contained) by AB and BC. Thus, twice the (rectangle contained) by AB and BC is also incommensurable with the remaining (square) on AC [Prop. 2.7], since if the whole is incommensurable with one of the (constituent magnitudes), then the original magnitudes will also be incommensurable (with one another) [Prop. 10.16]. And twice the (rectangle contained) by AB and BC (is) rational. Thus, the (square) on AC is irrational. Thus, AC is an irrational (straight-line) [Def. 10.4]. Let it be called a first apotome of a medial (straight-line).†

See footnote to Prop. 10.37.

οε΄.

Proposition 75

'Ε¦ν ¢πÕ µέσης µέση ¢φαιρεθÍ δυνάµει µόνον σύµµετρος οâσα τÍ Óλη, µετ¦ δ τÁς Óλης µέσον περιέχουσα, ¹ λοιπ¾ ¥λογός ™στιν· καλείσθω δ µέσης ¢ποτοµ¾ δευτέρα. 'ΑπÕ γ¦ρ µέσης τÁς ΑΒ µέση ¢φVρήσθω ¹ ΓΒ δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV τÍ ΑΒ, µετ¦ δ τÁς Óλης τÁς ΑΒ µέσον περιέχουσα τÕ ØπÕ τîν ΑΒ, ΒΓ· λέγω, Óτι ¹ λοιπ¾ ¹ ΑΓ ¥λογός ™στιν· καλείσθω δ µέσης ¢ποτοµ¾ δευτέρα.

If a medial (straight-line), which is commensurable in square only with the whole, and which contains a medial (area) with the whole, is subtracted from a(nother) medial (straight-line), then the remainder is an irrational (straight-line). Let it be called a second apotome of a medial (straight-line). For let the medial (straight-line) CB, which is commensurable in square only with the whole, AB, and which contains with the whole, AB, the medial (rectangle contained) by AB and BC, have been subtracted from the medial (straight-line) AB [Prop. 10.28]. I say that the remainder AC is an irrational (straight-line). Let

368

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 it be called a second apotome of a medial (straight-line).

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Β

A D

F

G

I

H

E

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Ι

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Ε

'Εκκείσθω γ¦ρ ·ητ¾ ¹ ∆Ι, κሠτο‹ς µν ¢πÕ τîν ΑΒ, ΒΓ ‡σον παρ¦ τ¾ν ∆Ι παραβεβλήσθω τÕ ∆Ε πλάτος ποιοàν τ¾ν ∆Η, τù δ δˆς ØπÕ τîν ΑΒ, ΒΓ ‡σον παρ¦ τ¾ν ∆Ι παραβεβλήσθω τÕ ∆Θ πλάτος ποιοàν τ¾ν ∆Ζ· λοιπÕν ¥ρα τÕ ΖΕ ‡σον ™στˆ τù ¢πÕ τÁς ΑΓ. κሠ™πεˆ µέσα κሠσύµµετρά ™στι τ¦ ¢πÕ τîν ΑΒ, ΒΓ, µέσον ¥ρα κሠτÕ ∆Ε. κሠπαρ¦ ·ητ¾ν τ¾ν ∆Ι παράκειται πλάτος ποιοàν τ¾ν ∆Η· ·ητ¾ ¥ρα ™στˆν ¹ ∆Η κሠ¢σύµµετρος τÍ ∆Ι µήκει. πάλιν, ™πεˆ µέσον ™στˆ τÕ ØπÕ τîν ΑΒ, ΒΓ, κሠτÕ δˆς ¥ρα ØπÕ τîν ΑΒ, ΒΓ µέσον ™στίν. καί ™στιν ‡σον τù ∆Θ· κሠτÕ ∆Θ ¥ρα µέσον ™στίν. κሠπαρ¦ ·ητ¾ν τ¾ν ∆Ι παραβέβληται πλάτος ποιοàν τ¾ν ∆Ζ· ·ητ¾ ¥ρα ™στˆν ¹ ∆Ζ κሠ¢σύµµετρος τÍ ∆Ι µήκει. κሠ™πεˆ αƒ ΑΒ, ΒΓ δυνάµει µόνον σύµµετροί ε„σιν, ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΒ τÍ ΒΓ µήκει· ¢σύµµετρον ¥ρα κሠτÕ ¢πÕ τÁς ΑΒ τετράγωνον τù ØπÕ τîν ΑΒ, ΒΓ. ¢λλ¦ τù µν ¢πÕ τÁς ΑΒ σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΒ, ΒΓ, τù δ ØπÕ τîν ΑΒ, ΒΓ σύµµετρόν ™στι τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ· ¢σύµµετρον ¥ρα ™στˆ τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ το‹ς ¢πÕ τîν ΑΒ, ΒΓ. ‡σον δ το‹ς µν ¢πÕ τîν ΑΒ, ΒΓ τÕ ∆Ε, τù δ δˆς ØπÕ τîν ΑΒ, ΒΓ τÕ ∆Θ· ¢σύµµετρον ¥ρα [™στˆ] τÕ ∆Ε τù ∆Θ. æς δ τÕ ∆Ε πρÕς τÕ ∆Θ, οÛτως ¹ Η∆ πρÕς τ¾ν ∆Ζ· ¢σύµµετρος ¥ρα ™στˆν ¹ Η∆ τÍ ∆Ζ. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ¥ρα Η∆, ∆Ζ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¹ ΖΗ ¥ρα ¢ποτοµή ™στιν. ·ητ¾ δ ¹ ∆Ι· τÕ δ ØπÕ ·ητÁς κሠ¢λόγου περιεχόµενον ¥λογόν ™στιν, κሠ¹ δυναµένη αÙτÕ ¥λογός ™στιν. κሠδύναται τÕ ΖΕ ¹ ΑΓ· ¹ ΑΓ ¥ρα ¥λογός ™στιν· καλείσθω δ µέσης ¢ποτοµ¾ δευτέρα. Óπερ œδει δε‹ξαι.

C

B

For let the rational (straight-line) DI be laid down. And let DE, equal to the (sum of the squares) on AB and BC, have been applied to DI, producing DG as breadth. And let DH, equal to twice the (rectangle contained) by AB and BC, have been applied to DI, producing DF as breadth. The remainder F E is thus equal to the (square) on AC [Prop. 2.7]. And since the (squares) on AB and BC are medial and commensurable (with one another), DE (is) thus also medial [Props. 10.15, 10.23 corr.]. And it is applied to the rational (straightline) DI, producing DG as breadth. Thus, DG is rational, and incommensurable in length with DI [Prop. 10.22]. Again, since the (rectangle contained) by AB and BC is medial, twice the (rectangle contained) by AB and BC is thus also medial [Prop. 10.23 corr.]. And it is equal to DH. Thus, DH is also medial. And it has been applied to the rational (straight-line) DI, producing DF as breadth. DF is thus rational, and incommensurable in length with DI [Prop. 10.22]. And since AB and BC are commensurable in square only, AB is thus incommensurable in length with BC. Thus, the square on AB (is) also incommensurable with the (rectangle contained) by AB and BC [Props. 10.21 lem., 10.11]. But, the (sum of the squares) on AB and BC is commensurable with the (square) on AB [Prop. 10.15], and twice the (rectangle contained) by AB and BC is commensurable with the (rectangle contained) by AB and BC [Prop. 10.6]. Thus, twice the (rectangle contained) by AB and BC is incommensurable with the (sum of the squares) on AB and BC [Prop. 10.13]. And DE is equal to the (sum of the squares) on AB and BC, and DH to twice the (rectangle contained) by AB and BC. Thus, DE [is] incommensurable with DH. And as DE (is) to DH, so GD (is) to DF [Prop. 6.1]. Thus, GD is incommensurable with DF [Prop. 10.11]. And they are both rational (straight-lines). Thus, GD and DF are rational (straight-lines which are) commensurable in square only. Thus, F G is an apotome [Prop. 10.73]. And DI (is) rational. And the (area) contained by a rational and an irrational (straight-line) is

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ELEMENTS BOOK 10 irrational [Prop. 10.20], and its square-root is irrational. And AC is the square-root of F E. Thus, AC is an irrational (straight-line) [Def. 10.4]. And let it be called the second apotome of a medial (straight-line).† (Which is) the very thing it was required to show.

†

See footnote to Prop. 10.38.

ο$΄.

Proposition 76

'Ε¦ν ¢πÕ εÙθείας εÙθε‹α ¢φαιρεθÊ δυνάµει ¢σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης ποιοàσα τ¦ µν ¢π' αÙτîν ¤µα ·ητόν, τÕ δ' Øπ' αÙτîν µέσον, ¹ λοιπ¾ ¥λογός ™στιν· καλείσθω δ ™λάσσων.

If a straight-line, which is incommensurable in square with the whole, and with the whole makes the (squares) on them (added) together rational, and the (rectangle contained) by them medial, is subtracted from a(nother) straight-line, then the remainder is an irrational (straightline). Let it be called a minor (straight-line).

Α

Γ

Β

A

'ΑπÕ γ¦ρ εÙθείας τÁς ΑΒ εÙθε‹α ¢φVρήσθω ¹ ΒΓ δυνάµει ¢σύµµετρος οâσα τÍ ÓλV ποιοàσα τ¦ προκείµενα. λέγω, Óτι ¹ λοιπ¾ ¹ ΑΓ ¥λογός ™στιν ¹ καλουµένη ™λάσσων. 'Επεˆ γ¦ρ τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ τετραγώνων ·ητόν ™στιν, τÕ δ δˆς ØπÕ τîν ΑΒ, ΒΓ µέσον, ¢σύµµετρα ¥ρα ™στˆ τ¦ ¢πÕ τîν ΑΒ, ΒΓ τù δˆς ØπÕ τîν ΑΒ, ΒΓ· κሠ¢ναστρέψαντι λοιπù τù ¢πÕ τÁς ΑΓ ¢σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΒ, ΒΓ. ·ητ¦ δ τ¦ ¢πÕ τîν ΑΒ, ΒΓ· ¥λογον ¥ρα τÕ ¢πÕ τÁς ΑΓ· ¥λογος ¥ρα ¹ ΑΓ· καλείσθω δ ™λάσσων. Óπερ œδει δε‹ξαι.

C

B

For let the straight-line BC, which is incommensurable in square with the whole, and fulfils the (other) prescribed (conditions), have been subtracted from the straight-line AB [Prop. 10.33]. I say that the remainder AC is that irrational (straight-line) called minor. For since the sum of the squares on AB and BC is rational, and twice the (rectangle contained) by AB and BC (is) medial, the (sum of the squares) on AB and BC is thus incommensurable with twice the (rectangle contained) by AB and BC. And, via conversion, the (sum of the squares) on AB and BC is incommensurable with the remaining (square) on AC [Props. 2.7, 10.16]. And the (sum of the squares) on AB and BC (is) rational. The (square) on AC (is) thus irrational. Thus, AC (is) an irrational (straight-line) [Def. 10.4]. Let it be called a minor (straight-line).† (Which is) the very thing it was required to show.

See footnote to Prop. 10.39.

οζ΄.

Proposition 77

'Ε¦ν ¢πÕ εÙθείας εÙθε‹α ¢φαιρεθÍ δυνάµει ¢σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης ποιοàσα τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον, τÕ δ δˆς Øπ' αÙτîν ·ητόν, ¹ λοιπ¾ ¥λογός ™στιν· καλείσθω δ ¹ µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσα.

If a straight-line, which is incommensurable in square with the whole, and with the whole makes the sum of the squares on them medial, and twice the (rectangle contained) by them rational, is subtracted from a(nother) straight-line, then the remainder is an irrational (straightline). Let it be called that which makes with a rational (area) a medial whole.

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A

C

B

'ΑπÕ γ¦ρ εÙθείας τÁς ΑΒ εÙθε‹α ¢φVρήσθω ¹ ΒΓ For let the straight-line BC, which is incommensuδυνάµει ¢σύµµετος οâσα τÍ ΑΒ ποιοàσα τ¦ προκείµενα· rable in square with AB, and fulfils the (other) prescribed λέγω, Óτι ¹ λοιπ¾ ¹ ΑΓ ¥λογός ™στιν ¹ προειρηµένη. (conditions), have been subtracted from the straight-line 370

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'Επεˆ γ¦ρ τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ τετραγώνων µέσον ™στίν, τÕ δ δˆς ØπÕ τîν ΑΒ, ΒΓ ·ητόν, ¢σύµµετρα ¥ρα ™στˆ τ¦ ¢πÕ τîν ΑΒ, ΒΓ τù δˆς ØπÕ τîν ΑΒ, ΒΓ· κሠλοιπÕν ¥ρα τÕ ¢πÕ τÁς ΑΓ ¢σύµµετρόν ™στι τù δˆς ØπÕ τîν ΑΒ, ΒΓ. καί ™στι τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ ·ητόν· τÕ ¥ρα ¢πÕ τÁς ΑΓ ¥λογόν ™στιν· ¥λογος ¥ρα ™στˆν ¹ ΑΓ· καλείσθω δ ¹ µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσα. Óπερ œδει δε‹ξαι.

AB [Prop. 10.34]. I say that the remainder AC is the aforementioned irrational (straight-line). For since the sum of the squares on AB and BC is medial, and twice the (rectangle contained) by AB and BC rational, the (sum of the squares) on AB and BC is thus incommensurable with twice the (rectangle contained) by AB and BC. Thus, the remaining (square) on AC is also incommensurable with twice the (rectangle contained) by AB and BC [Props. 2.7, 10.16]. And twice the (rectangle contained) by AB and BC is rational. Thus, the (square) on AC is irrational. Thus, AC is an irrational (straight-line) [Def. 10.4]. And let it be called that which makes with a rational (area) a medial whole.† (Which is) the very thing it was required to show.

See footnote to Prop. 10.40.

οη΄.

Proposition 78

'Ε¦ν ¢πÕ εÙθείας εÙθε‹α ¢φαιρεθÍ δυνάµει ¢σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης ποιοàσα τό τε συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον τό τε δˆς Øπ' αÙτîν µέσον κሠœτι τ¦ ¢π' αÙτîν τετράγωνα ¢σύµµετρα τù δˆς Øπ' αÙτîν, ¹ λοιπ¾ ¥λογός ™στιν· καλείσθω δ ¹ µετ¦ µέσου µέσον τÕ Óλον ποιοàσα.

If a straight-line, which is incommensurable in square with the whole, and with the whole makes the sum of the squares on them medial, and twice the (rectangle contained) by them medial, and, moreover, the (sum of the) squares on them incommensurable with twice the (rectangle contained) by them, is subtracted from a(nother) straight-line, then the remainder is an irrational (straightline). Let it be called that which makes with a medial (area) a medial whole.

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I

H

E

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A

'ΑπÕ γ¦ρ εÙθείας τÁς ΑΒ εÙθε‹α ¢φVρήσθω ¹ ΒΓ δυνάµει ¢σύµµετρος οâσα τÍ ΑΒ ποιοàσα τ¦ προκείµενα· λέγω, Óτι ¹ λοιπ¾ ¹ ΑΓ ¥λογός ™στιν ¹ καλουµένη ¹ µετ¦ µέσου µέσον τÕ Óλον ποιοàσα. 'Εκκείσθω γ¦ρ ·ητ¾ ¹ ∆Ι, κሠτο‹ς µν ¢πÕ τîν ΑΒ, ΒΓ ‡σον παρ¦ τ¾ν ∆Ι παραβεβλήσθω τÕ ∆Ε πλάτος ποιοàν τ¾ν ∆Η, τù δ δˆς ØπÕ τîν ΑΒ, ΒΓ ‡σον ¢φVρήσθω τÕ ∆Θ [πλάτος ποιοàν τ¾ν ∆Ζ]. λοιπÕν ¥ρα τÕ ΖΕ ‡σον ™στˆ τù ¢πÕ τÁς ΑΓ· éστε ¹ ΑΓ δύναται τÕ ΖΕ. κሠ™πεˆ τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΒ, ΒΓ τετραγώνων µέσον ™στˆ καί ™στιν ‡σον τù ∆Ε, µέσον ¥ρα [™στˆ] τÕ ∆Ε. κሠπαρ¦ ·ητ¾ν τ¾ν ∆Ι παράκειται πλάτος ποιοàν τ¾ν ∆Η· ·ητ¾ ¥ρα ™στˆν ¹ ∆Η κሠ¢σύµµετρος

C

B

For let the straight-line BC, which is incommensurable in square AB, and fulfils the (other) prescribed (conditions), have been subtracted from the (straightline) AB [Prop. 10.35]. I say that the remainder AC is the irrational (straight-line) called that which makes with a medial (area) a medial whole. For let the rational (straight-line) DI be laid down. And let DE, equal to the (sum of the squares) on AB and BC, have been applied to DI, producing DG as breadth. And let DH, equal to twice the (rectangle contained) by AB and BC, have been subtracted (from DE) [producing DF as breadth]. Thus, the remainder F E is equal to the (square) on AC [Prop. 2.7]. Hence, AC is the

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ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

τÍ ∆Ι µήκει. πάλιν, ™πεˆ τÕ δˆς ØπÕ τîν ΑΒ, ΒΓ µέσον ™στˆ καί ™στιν ‡σον τù ∆Θ, τÕ ¥ρα ∆Θ µέσον ™στίν. κሠπαρ¦ ·ητ¾ν τ¾ν ∆Ι παράκειται πλάτος ποιοàν τ¾ν ∆Ζ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ∆Ζ κሠ¢σύµµετρος τÍ ∆Ι µήκει. κሠ™πεˆ ¢σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΒ, ΒΓ τù δˆς ØπÕ τîν ΑΒ, ΒΓ, ¢σύµµετρον ¥ρα κሠτÕ ∆Ε τù ∆Θ. æς δ τÕ ∆Ε πρÕς τÕ ∆Θ, οÛτως ™στˆ κሠ¹ ∆Η πρÕς τ¾ν ∆Ζ· ¢σύµµετρος ¥ρα ¹ ∆Η τÍ ∆Ζ. καί ε„σιν ¢µφότεραι ·ηταί· αƒ Η∆, ∆Ζ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. ¢ποτοµ¾ ¥ρα ™στίν ¹ ΖΗ· ·ητ¾ δ ¹ ΖΘ. τÕ δ ØπÕ ·ητÁς κሠ¢ποτοµÁς περιεχόµενον [Ñρθογώνιον] ¥λογόν ™στιν, κሠ¹ δυναµένη αÙτÕ ¥λογός ™στιν· κሠδύναται τÕ ΖΕ ¹ ΑΓ· ¹ ΑΓ ¥ρα ¥λογός ™στιν· καλείσθω δ ¹ µετ¦ µέσου µέσον τÕ Óλον ποιοàσα. Óπερ œδει δε‹ξαι.

square-root of F E. And since the sum of the squares on AB and BC is medial, and is equal to DE, DE [is] thus medial. And it is applied to the rational (straight-line) DI, producing DG as breadth. Thus, DG is rational, and incommensurable in length with DI [Prop 10.22]. Again, since twice the (rectangle contained) by AB and BC is medial, and is equal to DH, DH is thus medial. And it is applied to the rational (straight-line) DI, producing DF as breadth. Thus, DF is also rational, and incommensurable in length with DI [Prop. 10.22]. And since the the (sum of the squares) on AB and BC is incommensurable with twice the (rectangle contained) by AB and BC, DE (is) also incommensurable with DH. And as DE (is) to DH, so DG also is to DF [Prop. 6.1]. Thus, DG (is) incommensurable (in length) with DF [Prop. 10.11]. And they are both rational. Thus, GD and DF are rational (straight-lines which are) commensurable in square only. Thus, F G is an apotome [Prop. 10.73]. And F H (is) rational. And the [rectangle] contained by a rational (straight-line) and an apotome is irrational [Prop. 10.20], and its square-root is irrational. And AC is the squareroot of F E. Thus, AC is irrational. Let it be called that which makes with a medial (area) a medial whole.† (Which is) the very thing it was required to show.

See footnote to Prop. 10.41.

οθ΄.

Proposition 79

ΤÍ ¢ποτοµÍ µία [µόνον] προσαρµόζει εÙθε‹α ·ητ¾ [Only] one rational straight-line, which is commensuδυνάµει µόνον σύµµετρος οâσα τÍ ÓλV. rable in square only with the whole, can be attached to an apotome.†

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”Εστω ¢ποτοµ¾ ¹ ΑΒ, προσαρµόζουσα δ αÙτÍ ¹ ΒΓ· αƒ ΑΓ, ΓΒ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· λέγω, Óτι τÍ ΑΒ ˜τέρα οÙ προσαρµόζει ·ητ¾ δυνάµει µόνον σύµµετρος οâσα τÍ ÓλÍ. Ε„ γ¦ρ δυνατόν, προσαρµοζέτω ¹ Β∆· καˆ αƒ Α∆, ∆Β ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. κሠ™πεί, ú Øπερέχει τ¦ ¢πÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν Α∆, ∆Β, τούτJ Øπερέχει κሠτ¦ ¢πÕ τîν ΑΓ, ΓΒ τοà δˆς ØπÕ τîν ΑΓ, ΓΒ· τù γ¦ρ αÙτù τù ¢πÕ τÁς ΑΒ ¢µφότερα Øπερέχει· ™ναλλ¦ξ ¥ρα, ú Øπερέχει τ¦ ¢πÕ τîν Α∆, ∆Β τîν ¢πÕ τîν ΑΓ, ΓΒ, τούτJ Øπερέχει [καˆ] τÕ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ. τ¦ δ ¢πÕ τîν Α∆, ∆Β τîν ¢πÕ τîν ΑΓ, ΓΒ Øπερέχει ·ητù· ·ητ¦ γ¦ρ ¢µφότερα. κሠτÕ δˆς ¥ρα ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ Øπερέχει ·ητù· Óπερ ™στˆν ¢δύνατον· µέσα γ¦ρ ¢µφότερα, µέσον δ µέσου οÙχ Øπερέχει ·ητù. τÍ ¥ρα ΑΒ ˜τέρα οÙ προσαρµόζει ·ητ¾ δυνάµει µόνον

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Let AB be an apotome, with BC (so) attached to it. AC and CB are thus rational (straight-lines which are) commensurable in square only [Prop. 10.73]. I say that another rational (straight-line), which is commensurable in square only with the whole, cannot be attached to AB. For, if possible, let BD be (so) attached (to AB). Thus, AD and DB are also rational (straight-lines which are) commensurable in square only [Prop. 10.73]. And since by whatever (area) the (sum of the squares) on AD and DB exceeds twice the (rectangle contained) by AD and DB, the (sum of the squares) on AC and CB also exceeds twice the (rectangle contained) by AC and CB by this (same area). For both exceed by the same (area)— (namely), the (square) on AB [Prop. 2.7]. Thus, alternately, by whatever (area) the (sum of the squares) on AD and DB exceeds the (sum of the squares) on AC and CB, twice the (rectangle contained) by AD and DB

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σύµµετρος οâσα τÍ ÓλV. [also] exceeds twice the (rectangle contained) by AC and Μία ¥ρα µόνη τÍ ¢ποτοµÍ προσαρµόζει ·ητ¾ CB by this (same area). And the (sum of the squares) on δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV· Óπερ œδει δε‹ξαι. AD and DB exceeds the (sum of the squares) on AC and AC by a rational (area). For both (are) rational (areas). Thus, twice the (rectangle contained) by AD and DB also exceeds twice the (rectangle contained) by AC and CB by a rational (area). The very thing is impossible. For both are medial (areas) [Prop. 10.21], and a medial (area) cannot exceed a(nother) medial (area) by a rational (area) [Prop. 10.26]. Thus, another rational (straight-line), which is commensurable in square only with the whole, cannot be attached to AB. Thus, only one rational (straight-line), which is commensurable in square only with the whole, can be attached to an apotome. (Which is) the very thing it was required to show. †

This proposition is equivalent to Prop. 10.42, with minus signs instead of plus signs.

π΄.

Proposition 80

ΤÍ µέσης ¢ποτοµÍ πρώτV µία µόνον προσαρµόζει Only one medial straight-line, which is commensuεÙθε‹α µέση δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV, rable in square only with the whole, and contains a raµετ¦ δ τÁς Óλης ·ητÕν περιέχουσα. tional (area) with the whole, can be attached to a first apotome of a medial (straight-line).†

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”Εστω γ¦ρ µέσης ¢ποτοµ¾ πρώτη ¹ ΑΒ, κሠτÍ ΑΒ προσαρµοζέτω ¹ ΒΓ· αƒ ΑΓ, ΓΒ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι ·ητÕν περιέχουσαι τÕ ØπÕ τîν ΑΓ, ΓΒ· λέγω, Óτι τÍ ΑΒ ˜τέρα οÙ προσαρµόζει µέση δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης ·ητÕν περιέχουσα. Ε„ γ¦ρ δυνατόν, προσαρµοζέτω κሠ¹ ∆Β· αƒ ¥ρα Α∆, ∆Β µέσαι ε„σˆ δυνάµει µόνον σύµµετροι ·ητÕν περιέχουσαι τÕ ØπÕ τîν Α∆, ∆Β. κሠ™πεί, ú Øπερέχει τ¦ ¢πÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν Α∆, ∆Β, τούτJ Øπερέχει κሠτ¦ ¢πÕ τîν ΑΓ, ΓΒ τοà δˆς ØπÕ τîν ΑΓ, ΓΒ· τù γ¦ρ αÙτù [πάλιν] Øπερέχουσι τù ¢πÕ τÁς ΑΒ· ™ναλλ¦ξ ¥ρα, ú Øπερέχει τ¦ ¢πÕ τîν Α∆, ∆Β τîν ¢πÕ τîν ΑΓ, ΓΒ, τούτJ Øπερέχει κሠτÕ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ. τÕ δ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ Øπερέχει ·ητù· ·ητ¦ γ¦ρ ¢µφότερα. κሠτ¦ ¢πÕ τîν Α∆, ∆Β ¥ρα τîν ¢πÕ τîν ΑΓ, ΓΒ [τετραγώνων] Øπερέχει ·ητù· Óπερ ™στˆν ¢δύνατον· µέσα γάρ ™στιν ¢µφότερα, µέσον δ µέσου οÙχ Øπερέχει ·ητù. ΤÍ ¥ρα µέσης ¢ποτοµÍ πρώτV µία µόνον προσαρµόζει εÙθε‹α µέση δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης ·ητÕν περιέχουσα· Óπερ œδει δε‹ξαι.

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For let AB be a first apotome of a medial (straightline), and let BC be (so) attached to AB. Thus, AC and CB are medial (straight-lines which are) commensurable in square only, containing a rational (area)— (namely, that contained) by AB and CB [Prop. 10.74]. I say that a(nother) medial (straight-line), which is commensurable in square only with the whole, and contains a rational (area) with the whole, cannot be attached to AB. For, if possible, let DB also be (so) attached to AB. Thus, AD and DB are medial (straight-lines which are) commensurable in square only, containing a rational (area)—(namely, that) contained by AD and DB [Prop. 10.74]. And since by whatever (area) the (sum of the squares) on AD and DB exceeds twice the (rectangle contained) by AD and DB, the (sum of the squares) on AC and CB also exceeds twice the (rectangle contained) by AC and CB by this (same area). For [again] both exceed by the same (area)—(namely), the (square) on AB [Prop. 2.7]. Thus, alternately, by whatever (area) the (sum of the squares) on AD and DB exceeds the (sum of the squares) on AC and CB, twice the (rectangle contained) by AD and DB also exceeds twice the (rectangle contained) by AC and CB by this (same area). And twice

373

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ELEMENTS BOOK 10 the (rectangle contained) by AD and DB exceeds twice the (rectangle contained) by AC and AC by a rational (area). For both (are) rational (areas). Thus, the (sum of the squares) on AD and DB also exceeds the (sum of the) [squares] on AC and CB by a rational (area). The very thing is impossible. For both are medial (areas) [Props. 10.15, 10.23 corr.], and a medial (area) cannot exceed a(nother) medial (area) by a rational (area) [Prop. 10.26]. Thus, only one medial (straight-line), which is commensurable in square only with the whole, and contains a rational (area) with the whole, can be attached to a first apotome of a medial (straight-line). (Which is) the very thing it was required to show.

†

This proposition is equivalent to Prop. 10.43, with minus signs instead of plus signs.

πα΄.

Proposition 81

ΤÍ µέσης ¢ποτοµÍ δευτέρv µία µόνον προσαρµόζει Only one medial straight-line, which is commensuεÙθε‹α µέση δυνάµει µόνον σύµµετρος τÍ ÓλV, µετ¦ δ rable in square only with the whole, and contains a meτÁς Óλης µέσον περιέχουσα. dial (area) with the whole, can be attached to a second apotome of a medial (straight-line).†

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”Εστω µέσης ¢ποτοµ¾ δευτέρα ¹ ΑΒ κሠτÍ ΑΒ προσαρµόζουσα ¹ ΒΓ· αƒ ¥ρα ΑΓ, ΓΒ µέσαι ε„σˆ δυνάµει µόνον σύµµετροι µέσον περιέχουσαι τÕ ØπÕ τîν ΑΓ, ΓΒ· λέγω, Óτι τÍ ΑΒ ˜τέρα οÙ προσαρµόσει εÙθε‹α µέση δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης µέσον περιέχουσα. Ε„ γ¦ρ δυνατόν, προσαρµοζέτω ¹ Β∆· καˆ αƒ Α∆, ∆Β ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι µέσον περιέχουσαι τÕ ØπÕ τîν Α∆, ∆Β. κሠ™κκείσθω ·ητ¾ ¹ ΕΖ, κሠτο‹ς µν ¢πÕ τîν ΑΓ, ΓΒ ‡σον παρ¦ τ¾ν ΕΖ παραβεβλήσθω τÕ ΕΗ πλάτος ποιοàν τ¾ν ΕΜ· τù δ δˆς ØπÕ τîν ΑΓ, ΓΒ ‡σον ¢φVρήσθω τÕ ΘΗ πλάτος ποιοàν τ¾ν ΘΜ· λοιπÕν ¥ρα τÕ ΕΛ ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ· éστε ¹ ΑΒ δύναται τÕ ΕΛ. πάλιν δ¾ το‹ς ¢πÕ τîν Α∆, ∆Β ‡σον παρ¦ τ¾ν ΕΖ παραβεβλήσθω τÕ ΕΙ πλάτος ποιοàν τ¾ν ΕΝ· œστι δ κሠτÕ ΕΛ ‡σον τù ¢πÕ τÁς ΑΒ τετραγώνJ· λοιπÕν ¥ρα τÕ ΘΙ ‡σον ™στˆ τù δˆς ØπÕ τîν Α∆, ∆Β. κሠ™πεˆ µέσαι ε„σˆν αƒ ΑΓ, ΓΒ, µέσα ¥ρα ™στˆ

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Let AB be a second apotome of a medial (straightline), with BC (so) attached to AB. Thus, AC and CB are medial (straight-lines which are) commensurable in square only, containing a medial (area)—(namely, that contained) by AC and CB [Prop. 10.75]. I say that a(nother) medial straight-line, which is commensurable in square only with the whole, and contains a medial (area) with the whole, cannot be attached to AB. For, if possible, let BD be (so) attached. Thus, AD and DB are also medial (straight-lines which are) commensurable in square only, containing a medial (area)— (namely, that contained) by AD and DB [Prop. 10.75]. And let the rational (straight-line) EF be laid down. And let EG, equal to the (sum of the squares) on AC and CB, have been applied to EF , producing EM as breadth. And let HG, equal to twice the (rectangle contained) by AC and CB, have been subtracted (from EG), producing HM as breadth. The remainder EL is thus equal

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κሠτ¦ ¢πÕ τîν ΑΓ, ΓΒ. καί ™στιν ‡σα τù ΕΗ· µέσον ¥ρα κሠτÕ ΕΗ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται πλάτος ποιοàν τ¾ν ΕΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΕΜ κሠ¢σύµµετρος τÍ ΕΖ µήκει. πάλιν, ™πεˆ µέσον ™στˆ τÕ ØπÕ τîν ΑΓ, ΓΒ, κሠτÕ δˆς ØπÕ τîν ΑΓ, ΓΒ µέσον ™στίν. καί ™στιν ‡σον τù ΘΗ· κሠτÕ ΘΗ ¥ρα µέσον ™στίν. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται πλάτος ποιοàν τ¾ν ΘΜ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΘΜ κሠ¢σύµµετρος τÍ ΕΖ µήκει. κሠ™πεˆ αƒ ΑΓ, ΓΒ δυνάµει µόνον σύµµετροί ε„σιν, ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΓ τÍ ΓΒ µήκει. æς δ ¹ ΑΓ πρÕς τ¾ν ΓΒ, οÛτως ™στˆ τÕ ¢πÕ τÁς ΑΓ πρÕς τÕ ØπÕ τîν ΑΓ, ΓΒ· ¢σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΓ τù ØπÕ τîν ΑΓ, ΓΒ. ¢λλ¦ τù µν ¢πÕ τÁς ΑΓ σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΓ, ΓΒ, τù δ ØπÕ τîν ΑΓ, ΓΒ σύµµετρόν ™στι τÕ δˆς ØπÕ τîν ΑΓ, ΓΒ· ¢σύµµετρα ¥ρα ™στˆ τ¦ ¢πÕ τîν ΑΓ, ΓΒ τù δˆς ØπÕ τîν ΑΓ, ΓΒ. καί ™στι το‹ς µν ¢πÕ τîν ΑΓ, ΓΒ ‡σον τÕ ΕΗ, τù δ δˆς ØπÕ τîν ΑΓ, ΓΒ ‡σον τÕ ΗΘ· ¢σύµµετρον ¥ρα ™στˆ τÕ ΕΗ τù ΘΗ. æς δ τÕ ΕΗ πρÕς τÕ ΘΗ, οÛτως ™στˆν ¹ ΕΜ πρÕς τ¾ν ΘΜ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΕΜ τÍ ΜΘ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ΕΜ, ΜΘ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΕΘ, προσαρµόζουσα δ αÙτÍ ¹ ΘΜ. еοίως δ¾ δείξοµεν, Óτι κሠ¹ ΘΝ αÙτÍ προσαρµόζει· τÍ ¥ρα ¢ποτοµÍ ¥λλη κሠ¥λλη προσαρµόζει εÙθε‹α δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV· Óπερ ™στˆν ¢δύνατον. ΤÍ ¥ρα µέσης ¢ποτοµÍ δευτέρv µία µόνον προσαρµόζει εÙθε‹α µέση δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης µέσον περιέχουσα· Óπερ œδει δε‹ξαι.

to the (square) on AB [Prop. 2.7]. Hence, AB is the square-root of EL. So, again, let EI, equal to the (sum of the squares) on AD and DB have been applied to EF , producing EN as breadth. And EL is also equal to the square on AB. Thus, the remainder HI is equal to twice the (rectangle contained) by AD and DB [Prop. 2.7]. And since AC and CB are (both) medial (straight-lines), the (sum of the squares) on AC and CB is also medial. And it is equal to EG. Thus, EG is also medial [Props. 10.15, 10.23 corr.]. And it is applied to the rational (straight-line) EF , producing EM as breadth. Thus, EM is rational, and incommensurable in length with EF [Prop. 10.22]. Again, since the (rectangle contained) by AC and CB is medial, twice the (rectangle contained) by AC and CB is also medial [Prop. 10.23 corr.]. And it is equal to HG. Thus, HG is also medial. And it is applied to the rational (straight-line) EF , producing HM as breadth. Thus, HM is also rational, and incommensurable in length with EF [Prop. 10.22]. And since AC and CB are commensurable in square only, AC is thus incommensurable in length with CB. And as AC (is) to CB, so the (square) on AC is to the (rectangle contained) by AC and CB [Prop. 10.21 corr.]. Thus, the (square) on AC is incommensurable with the (rectangle contained) by AC and CB [Prop. 10.11]. But, the (sum of the squares) on AC and CB is commensurable with the (square) on AC, and twice the (rectangle contained) by AC and CB is commensurable with the (rectangle contained) by AC and CB [Prop. 10.6]. Thus, the (sum of the squares) on AC and CB is incommensurable with twice the (rectangle contained) by AC and CB [Prop. 10.13]. And EG is equal to the (sum of the squares) on AC and CB. And GH is equal to twice the (rectangle contained) by AC and CB. Thus, EG is incommensurable with HG. And as EG (is) to HG, so EM is to HM [Prop. 6.1]. Thus, EM is incommensurable in length with M H [Prop. 10.11]. And they are both rational (straight-lines). Thus, EM and M H are rational (straight-lines which are) commensurable in square only. Thus, EH is an apotome [Prop. 10.73], and HM (is) attached to it. So, similarly, we can show that HN (is) also (commensurable in square only with EN and is) attached to (EH). Thus, different straight-lines, which are commensurable in square only with the whole, are attached to an apotome. The very thing is impossible [Prop. 10.79]. Thus, only one medial straight-line, which is commensurable in square only with the whole, and contains a medial (area) with the whole, can be attached to a second apotome of a medial (straight-line). (Which is) the very thing it was required to show.

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This proposition is equivalent to Prop. 10.44, with minus signs instead of plus signs.

πβ΄.

Proposition 82

ΤÍ ™λάσσονι µία µόνον προσαρµόζει εÙθε‹α δυνάµει ¢σύµµετρος οâσα τÍ ÓλV ποιοàσα µετ¦ τÁς Óλης τÕ µν ™κ τîν ¢π' αÙτîν τετραγώνων ·ητόν, τÕ δ δˆς Øπ' αÙτîν µέσον.

Only one straight-line, which is incommensurable in square with the whole, and (together) with the whole makes the (sum of the) squares on them rational, and twice the (rectangle contained) by them medial, can be attached to a minor (straight-line).

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”Εστω ¹ ™λάσσων ¹ ΑΒ, κሠτÍ ΑΒ προσαρµόζουσα œστω ¹ ΒΓ· αƒ ¥ρα ΑΓ, ΓΒ δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων ·ητόν, τÕ δ δˆς Øπ' αÙτîν µέσον· λέγω, Óτι τÍ ΑΒ ˜τέρα εÙθε‹α οÙ προσαρµόσει τ¦ αÙτ¦ ποιοàσα. Ε„ γ¦ρ δυνατόν, προσαρµοζέτω ¹ Β∆· καˆ αƒ Α∆, ∆Β ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τ¦ προειρηµένα. κሠ™πεί, ú Øπερέχει τ¦ ¢πÕ τîν Α∆, ∆Β τîν ¢πÕ τîν ΑΓ, ΓΒ, τούτJ Øπερέχει κሠτÕ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ, τ¦ δ ¢πÕ τîν Α∆, ∆Β τετράγωνα τîν ¢πÕ τîν ΑΓ, ΓΒ τετραγώνων Øπερέχει ·ητù· ·ητ¦ γάρ ™στιν ¢µφότερα· κሠτÕ δˆς ØπÕ τîν Α∆, ∆Β ¥ρα τοà δˆς ØπÕ τîν ΑΓ, ΓΒ Øπερέχει ·ητù· Óπερ ™στˆν ¢δύνατον· µέσα γάρ ™στιν ¢µφότερα. ΤÍ ¥ρα ™λάσσονι µία µόνον προσαρµόζει εÙθε‹α δυνάµει ¢σύµµετρος οâσα τÍ ÓλV κሠποιοàσα τ¦ µν ¢π' αÙτîν τετράγωνα ¤µα ·ητόν, τÕ δ δˆς Øπ' αÙτîν µέσον· Óπερ œδει δε‹ξαι.

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Let AB be a minor (straight-line), and let BC be (so) attached to AB. Thus, AC and CB are (straight-lines which are) incommensurable in square, making the sum of the squares on them rational, and twice the (rectangle contained) by them medial [Prop. 10.76]. I say that another another straight-line fulfilling the same (conditions) cannot be attached to AB. For, if possible, let BD be (so) attached (to AB). Thus, AD and DB are also (straight-lines which are) incommensurable in square, fulfilling the (other) aforementioned (conditions) [Prop. 10.76]. And since by whatever (area) the (sum of the squares) on AD and DB exceeds the (sum of the squares) on AC and CB, twice the (rectangle contained) by AD and DB also exceeds twice the (rectangle contained) by AC and CB by this (same area) [Prop. 2.7]. And the (sum of the) squares on AD and DB exceeds the (sum of the) squares on AC and CB by a rational (area). For both are rational (areas). Thus, twice the (rectangle contained) by AD and DB also exceeds twice the (rectangle contained) by AC and CB by a rational (area). The very thing is impossible. For both are medial (areas) [Prop. 10.26]. Thus, only one straight-line, which is incommensurable in square with the whole, and (with the whole) makes the squares on them (added) together rational, and twice the (rectangle contained) by them medial, can be attached to a minor (straight-line). (Which is) the very thing it was required to show.

This proposition is equivalent to Prop. 10.45, with minus signs instead of plus signs.

πγ΄.

Proposition 83

ΤÍ µετ¦ ·ητοà µέσον τÕ Óλον ποιούσV µία µόνον Only one straight-line, which is incommensurable in προσαρµόζει εÙθε‹α δυνάµει ¢σύµµετρος οâσα τÍ ÓλV, square with the whole, and (together) with the whole µετ¦ δ τÁς Óλης ποιοàσα τÕ µν συγκείµενον ™κ τîν makes the sum of the squares on them medial, and twice ¢π' αÙτîν τετραγώνων µέσον, τÕ δ δˆς Øπ' αÙτîν ·ητόν. the (rectangle contained) by them rational, can be attached to that (straight-line) which with a rational (area) makes a medial whole.†

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ELEMENTS BOOK 10

”Εστω ¹ µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσα ¹ ΑΒ, κሠτÍ ΑΒ προσαρµοζέτω ¹ ΒΓ· αƒ ¥ρα ΑΓ, ΓΒ δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τ¦ προκείµενα· λέγω, Óτι τÍ ΑΒ ˜τέρα οÙ προσαρµόσει τ¦ αÙτ¦ ποιοàσα. Ε„ γ¦ρ δυνατόν, προσαρµοζέτω ¹ Β∆· καˆ αƒ Α∆, ∆Β ¥ρα εÙθε‹αι δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τ¦ προκείµενα. ™πεˆ οÙν, ú Øπερέχει τ¦ ¢πÕ τîν Α∆, ∆Β τîν ¢πÕ τîν ΑΓ, ΓΒ, τούτJ Øπερέχει κሠτÕ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ ¢κολούθως το‹ς πρÕ αÙτοà, τÕ δ δˆς ØπÕ τîν Α∆, ∆Β τοà δˆς ØπÕ τîν ΑΓ, ΓΒ Øπερέχει ·ητù· ·ητ¦ γάρ ™στιν ¢µφότερα· κሠτ¦ ¢πÕ τîν Α∆, ∆Β ¥ρα τîν ¢πÕ τîν ΑΓ, ΓΒ Øπερέχει ·ητù· Óπερ ™στˆν ¢δύνατον· µέσα γάρ ™στιν ¢µφότερα. ΟÙκ ¥ρα τÍ ΑΒ ˜τέρα προσαρµόσει εÙθε‹α δυνάµει ¢σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης ποιοàσα τ¦ προειρηµένα· µία ¥ρα µόνον προσαρµόσει· Óπερ œδει δε‹ξαι.

†

Let AB be a (straight-line) which with a rational (area) makes a medial whole, and let BC be (so) attached to AB. Thus, AC and CB are (straight-lines which are) incommensurable in square, fulfilling the (other) proscribed (conditions) [Prop. 10.77]. I say that another (straight-line) fulfilling the same (conditions) cannot be attached to AB. For, if possible, let BD be (so) attached (to AB). Thus, AD and DB are also straight-lines (which are) incommensurable in square, fulfilling the (other) prescribed (conditions) [Prop. 10.77]. Therefore, analogously to the (propositions) before this, since by whatever (area) the (sum of the squares) on AD and DB exceeds the (sum of the squares) on AC and CB, twice the (rectangle contained) by AD and DB also exceeds twice the (rectangle contained) by AC and CB by this (same area). And twice the (rectangle contained) by AD and DB exceeds twice the (rectangle contained) by AC and CB by a rational (area). For they are (both) rational (areas). Thus, the (sum of the squares) on AD and DB also exceeds the (sum of the squares) on AC and CB by a rational (area). The very thing is impossible. For both are medial (areas) [Prop. 10.26]. Thus, another straight-line cannot be attached to AB, which is incommensurable in square with the whole, and fulfills the (other) aforementioned (conditions) with the whole. Thus, only one (such straight-line) can be (so) attached. (Which is) the very thing it was required to show.

This proposition is equivalent to Prop. 10.46, with minus signs instead of plus signs.

πδ΄.

Proposition 84

ΤÍ µετ¦ µέσου µέσον τÕ Óλον ποιούσV µία µόνη προσαρµόζει εÙθε‹α δυνάµει ¢σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης ποιοàσα τό τε συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον τό τε δˆς Øπ' αÙτîν µέσον κሠœτι ¢σύµµετρον τù συγκειµένJ ™κ τîν ¢π' αÙτîν.

Only one straight-line, which is incommensurable in square with the whole, and (together) with the whole makes the sum of the squares on them medial, and twice the (rectangle contained) by them medial, and, moreover, incommensurable with the sum of the (squares) on them, can be attached to that (straight-line) which with a medial (area) makes a medial whole.†

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”Εστω ¹ µετ¦ µέσου µέσον τÕ Óλον ποιοàσα ¹ ΑΒ, προσαρµόζουσα δ αÙτÍ ¹ ΒΓ· αƒ ¥ρα ΑΓ, ΓΒ δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τ¦ προειρηµένα. λέγω, Óτι τÍ ΑΒ ˜τέρα οÙ προσαρµόσει ποιοàσα προειρηµένα. Ε„ γ¦ρ δυνατόν, προσαρµοζέτω ¹ Β∆, éστε κሠτ¦ς Α∆, ∆Β δυνάµει ¢συµµέτρους εναι ποιούσας τά τε ¢πÕ τîν Α∆, ∆Β τετράγωνα ¤µα µέσον κሠτÕ δˆς ØπÕ τîν Α∆, ∆Β µέσον κሠœτι τ¦ ¢πÕ τîν Α∆, ∆Β ¢σύµµετρα τù δˆς ØπÕ τîν Α∆, ∆Β· κሠ™κκείσθω ·ητ¾ ¹ ΕΖ, κሠτο‹ς µν ¢πÕ τîν ΑΓ, ΓΒ ‡σον παρ¦ τ¾ν ΕΖ παραβεβλήσθω τÕ ΕΗ πλάτος ποιοàν τ¾ν ΕΜ, τù δ δˆς ØπÕ τîν ΑΓ, ΓΒ ‡σον παρ¦ τ¾ν ΕΖ παραβεβλήσθω τÕ ΘΗ πλάτος ποιοàν τ¾ν ΘΜ· λοιπÕν ¥ρα τÕ ¢πÕ τÁς ΑΒ ‡σον ™στˆ τù ΕΛ· ¹ ¥ρα ΑΒ δύναται τÕ ΕΛ. πάλιν το‹ς ¢πÕ τîν Α∆, ∆Β ‡σον παρ¦ τ¾ν ΕΖ παραβεβλήσθω τÕ ΕΙ πλάτος ποιοàν τ¾ν ΕΝ. œστι δ κሠτÕ ¢πÕ τÁς ΑΒ ‡σον τù ΕΛ· λοιπÕν ¥ρα τÕ δˆς ØπÕ τîν Α∆, ∆Β ‡σον [™στˆ] τù ΘΙ. κሠ™πεˆ µέσον ™στˆ τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΓ, ΓΒ καί ™στιν ‡σον τù ΕΗ, µέσον ¥ρα ™στˆ κሠτÕ ΕΗ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται πλάτος ποιοàν τ¾ν ΕΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΕΜ κሠ¢σύµµετρος τÍ ΕΖ µήκει. πάλιν, ™πεˆ µέσον ™στˆ τÕ δˆς ØπÕ τîν ΑΓ, ΓΒ καί ™στιν ‡σον τù ΘΗ, µέσον ¥ρα κሠτÕ ΘΗ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται πλάτος ποιοàν τ¾ν ΘΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΘΜ κሠ¢σύµµετρος τÍ ΕΖ µήκει. κሠ™πεˆ ¢σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΓ, ΓΒ τù δˆς ØπÕ τîν ΑΓ, ΓΒ, ¢σύµµετρόν ™στι κሠτÕ ΕΗ τù ΘΗ· ¢σύµµετρος ¥ρα ™στˆ κሠ¹ ΕΜ τÍ ΜΘ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ¥ρα ΕΜ, ΜΘ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΕΘ, προσαρµόζουσα δ αÙτÍ ¹ ΘΜ. еοίως δ¾ δείξοµεν, Óτι ¹ ΕΘ πάλιν ¢ποτοµή ™στιν, προσαρµόζουσα δ αÙτÍ ¹ ΘΝ. τÍ ¥ρα ¢ποτοµÍ ¥λλη κሠ¥λλη προσαρµόζει ·ητ¾ δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV· Óπερ ™δείχθη ¢δύνατον. οÙκ ¥ρα τÍ ΑΒ ˜τέρα προσαρµόσει εÙθε‹α. ΤÍ ¥ρα ΑΒ µία µόνον προσαρµόζει εÙθε‹α δυνάµει ¢σύµµετρος οâσα τÍ ÓλV, µετ¦ δ τÁς Óλης ποιοàσα τά τε ¢π' αÙτîν τετράγωνα ¤µα µέσον κሠτÕ δˆς Øπ' αÙτîν µέσον κሠœτι τ¦ ¢π' αÙτîν τετράγωνα ¢σύµµετρα τù δˆς Øπ' αÙτîν· Óπερ œδει δε‹ξαι.

Let AB be a (straight-line) which with a medial (area) makes a medial whole, BC being (so) attached to it. Thus, AC and CB are incommensurable in square, fulfilling the (other) aforementioned (conditions) [Prop. 10.78]. I say that a(nother) (straight-line) fulfilling the aforementioned (conditions) cannot be attached to AB. For, if possible, let BD be (so) attached. Hence, AD and DB are also (straight-lines which are) incommensurable in square, making the squares on AD and DB (added) together medial, and twice the (rectangle contained) by AD and DB medial, and, moreover, the (sum of the squares) on AD and DB incommensurable with twice the (rectangle contained) by AD and DB [Prop. 10.78]. And let the rational (straight-line) EF be laid down. And let EG, equal to the (sum of the squares) on AC and CB, have been applied to EF , producing EM as breadth. And let HG, equal to twice the (rectangle contained) by AC and CB, have been applied to EF , producing HM as breadth. Thus, the remaining (square) on AB is equal to EL [Prop. 2.7]. Thus, AB is the square-root of EL. Again, let EI, equal to the (sum of the squares) on AD and DB, have been applied to EF , producing EN as breadth. And the (square) on AB is also equal to EL. Thus, the remaining twice the (rectangle contained) by AD and DB [is] equal to HI [Prop. 2.7]. And since the sum of the (squares) on AC and CB is medial, and is equal to EG, EG is thus medial. And it is applied to the rational (straight-line) EF , producing EM as breadth. EM is thus rational, and incommensurable in length with EF [Prop. 10.22]. Again, since twice the (rectangle contained) by AC and CB is medial, and is equal to HG, HG is thus also medial. And it is applied to the rational (straight-line) EF , producing HM as breadth. HM is thus rational, and incommensurable in length with EF [Prop. 10.22]. And since the (sum of the squares) on AC and CB is incommensurable with twice the (rectangle contained) by AC and CB, EG is also incommensurable with HG. Thus, EM is also incommensurable in length with M H [Props. 6.1, 10.11]. And they are both rational (straight-lines). Thus, EM and M H are rational (straight-lines which are) commensurable in square only. Thus, EH is an apotome [Prop. 10.73], with HM attached to it. So, similarly, we can show that EH is again an apotome, with HN attached to it. Thus, different rational (straight-lines), which are commensurable in square only with the whole, are attached to an apotome. The very thing was shown (to be) impossible [Prop. 10.79]. Thus, another straightline cannot be (so) attached to AB. Thus, only one straight-line, which is incommensu-

378

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ELEMENTS BOOK 10 rable in square with the whole, and (together) with the whole makes the squares on them (added) together medial, and twice the (rectangle contained) by them medial, and, moreover, the (sum of the) squares on them incommensurable with the (rectangle contained) by them, can be attached to AB. (Which is) the very thing it was required to show.

†

This proposition is equivalent to Prop. 10.47, with minus signs instead of plus signs.

“Οροι τρίτοι.

Definitions III

ια΄. `Υποκειµένης ·ητÁς κሠ¢ποτοµÁς, ™¦ν µν ¹ Óλη τÁς προσαρµοζούσης µε‹ζον δύνηται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει, κሠ¹ Óλη σύµµετρος Ï τÍ ™κκειµένV ·ητÍ µήκει, καλείσθω ¢ποτοµ¾ πρώτη. ιβ΄. 'Ε¦ν δ ¹ προσαρµόζουσα σύµµετρος Ï τÍ ™κκειµένV ·ητÍ µήκει, κሠ¹ Óλη τÁς προσαρµοζούσης µε‹ζον δύνηται τù ¢πÕ συµµέτρου ˜αυτÍ, καλείσθω ¢ποτοµ¾ δευτέρα. ιγ΄. 'Ε¦ν δ µηδετέρα σύµµετρος Ï τÍ ™κκειµένV ·ητÍ µήκει, ¹ δ Óλη τÁς προσαρµοζούσης µε‹ζον δύνηται τù ¢πÕ συµµέτρου ˜αυτÍ, καλείσθω ¢ποτοµ¾ τρίτη. ιδ΄. Πάλιν, ™¦ν ¹ Óλη τÁς προσαρµοζούσης µε‹ζον δύνηται τù ¢πÕ ¢συµµέτρου ˜αυτÍ [µήκει], ™¦ν µν ¹ Óλη σύµµετρος Ï τÍ ™κκειµένV ·ητÍ µήκει, καλείσθω ¢ποτοµ¾ τετάρτη. ιε΄. 'Ε¦ν δ ¹ προσαρµόζουσα, πέµπτη. ι$΄. 'Ε¦ν δ µηδετέρα, ›κτη.

11. Given a rational (straight-line) and an apotome, if the square on the whole is greater than the (square on a straight-line) attached (to the apotome) by the (square) on (some straight-line) commensurable in length with (the whole), and the whole is commensurable in length with the (previously) laid down rational (straight-line), then let the (apotome) be called a first apotome. 12. And if the attached (straight-line) is commensurable in length with the (previously) laid down rational (straight-line), and the square on the whole is greater than (the square on) the attached (straight-line) by the (square) on (some straight-line) commensurable (in length) with (the whole), then let the (apotome) be called a second apotome. 13. And if neither of (the whole or the attached straight-line) is commensurable in length with the (previously) laid down rational (straight-line), and the square on the whole is greater than (the square on) the attached (straight-line) by the (square) on (some straight-line) commensurable (in length) with (the whole), then let the (apotome) be called a third apotome. 14. Again, if the square on the whole is greater than (the square on) the attached (straight-line) by the (square) on (some straight-line) incommensurable [in length] with (the whole), and the whole is commensurable in length with the (previously) laid down rational (straight-line), then let the (apotome) be called a fourth apotome. 15. And if the attached (straight-line is commensurable), a fifth (apotome). 16. And if neither (the whole nor the attached straight-line is commensurable), a sixth (apotome).

πε΄.

Proposition 85

ΕØρε‹ν τ¾ν πρώτην ¢ποτοµήν.

Α Θ

To find a first apotome.

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'Εκκείσθω ·ητ¾ ¹ Α, κሠτÍ Α µήκει σύµµετρος œστω ¹ ΒΗ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΒΗ. κሠ™κκείσθωσαν δύο τετράγωνοι ¢ριθµοˆ οƒ ∆Ε, ΕΖ, ïν ¹ Øπεροχ¾ Ð Ζ∆ µ¾ œστω τετράγωνος· οÙδ' ¥ρα Ð Ε∆ πρÕς τÕν ∆Ζ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. κሠπεποιήσθω æς Ð Ε∆ πρÕς τÕν ∆Ζ, οÛτως τÕ ¢πÕ τÁς ΒΗ τετράγωνον πρÕς τÕ ¢πÕ τ¾ς ΗΓ τετράγωνον· σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΒΗ τù ¢πÕ τÁς ΗΓ. ·ητÕν δ τÕ ¢πÕ τÁς ΒΗ· ·ητÕν ¥ρα κሠτÕ ¢πÕ τÁς ΗΓ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΗΓ. κሠ™πεˆ Ð Ε∆ πρÕς τÕν ∆Ζ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ' ¥ρα τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς ΗΓ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΒΗ τÍ ΗΓ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ΒΗ, ΗΓ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¹ ¥ρα ΒΓ ¢ποτοµή ™στιν. λέγω δή, Óτι κሠπρώτη. ‘Ωι γ¦ρ µε‹ζόν ™στι τÕ ¢πÕ τÁς ΒΗ τοà ¢πÕ τÁς ΗΓ, œστω τÕ ¢πÕ τÁς Θ. κሠ™πεί ™στιν æς Ð Ε∆ πρÕς τÕν Ζ∆, οÛτως τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς ΗΓ, κሠ¢ναστρέψαντι ¥ρα ™στˆν æς Ð ∆Ε πρÕς τÕν ΕΖ, οÛτως τÕ ¢πÕ τÁς ΗΒ πρÕς τÕ ¢πÕ τÁς Θ. Ð δ ∆Ε πρÕς τÕν ΕΖ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ˜κάτερος γ¦ρ τετράγωνός ™στιν· κሠτÕ ¢πÕ τÁς ΗΒ ¥ρα πρÕς τÕ ¢πÕ τÁς Θ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· σύµµετρος ¥ρα ™στˆν ¹ ΒΗ τÍ Θ µήκει. κሠδύναται ¹ ΒΗ τÁς ΗΓ µε‹ζον τù ¢πÕ τÁς Θ· ¹ ΒΗ ¥ρα τÁς ΗΓ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει. καί ™στιν ¹ Óλη ¹ ΒΗ σύµµετρος τÍ ™κκειµένV ·ητÍ µήκει τÍ Α. ¹ ΒΓ ¥ρα ¢ποτοµή ™στι πρώτη. ΕÛρηται ¥ρα ¹ πρώτη ¢ποτοµ¾ ¹ ΒΓ· Óπερ œδει εØρε‹ν.

†

Let the rational (straight-line) A be laid down. And let BG be commensurable in length with A. BG is thus also a rational (straight-line). And let two square numbers DE and EF be laid down, and let their difference F D be not square [Prop. 10.28 lem. I]. Thus, ED does not have to DF the ratio which (some) square number (has) to (some) square number. And let it have been contrived that as ED (is) to DF , so the square on BG (is) to the square on GC [Prop. 10.6. corr.]. Thus, the (square) on BG is commensurable with the (square) on GC [Prop. 10.6]. And the (square) on BG (is) rational. Thus, the (square) on GC (is) also rational. Thus, GC is also rational. And since ED does not have to DF the ratio which (some) square number (has) to (some) square number, the (square) on BG thus does not have to the (square) on GC the ratio which (some) square number (has) to (some) square number either. Thus, BG is incommensurable in length with GC [Prop. 10.9]. And they are both rational (straight-lines). Thus, BG and GC are rational (straight-lines which are) commensurable in square only. Thus, BC is an apotome [Prop. 10.73]. So, I say that (it is) also a first (apotome). Let the (square) on H be that (area) by which the (square) on BG is greater than the (square) on GC [Prop. 10.13 lem.]. And since as ED is to F D, so the (square) on BG (is) to the (square) on GC, thus, via conversion, as DE is to EF , so the (square) on GB (is) to the (square) on H [Prop. 5.19 corr.]. And DE has to EF the ratio which (some) square-number (has) to (some) square-number. For each is a square (number). Thus, the (square) on GB also has to the (square) on H the ratio which (some) square number (has) to (some) square number. Thus, BG is commensurable in length with H [Prop. 10.9]. And the square on BG is greater than (the square on) GC by the (square) on H. Thus, the square on BG is greater than (the square on) GC by the (square) on (some straight-line) commensurable in length with (BG). And the whole, BG, is commensurable in length with the (previously) laid down rational (straight-line) A. Thus, BC is a first apotome [Def. 10.11]. Thus, the first apotome BC has been found. (Which is) the very thing it was required to find.

See footnote to Prop. 10.48.

π$΄.

Proposition 86

ΕØρε‹ν τ¾ν δευτέραν ¢ποτοµήν. To find a second apotome. 'Εκκείσθω ·ητ¾ ¹ Α κሠτÍ Α σύµµετρος µήκει Let the rational (straight-line) A, and GC (which is) ¹ ΗΓ. ·ητ¾ ¥ρα ™στˆν ¹ ΗΓ. κሠ™κκείσθωσαν δύο commensurable in length with A, be laid down. Thus, τετράγωνοι ¢ριθµοˆ οƒ ∆Ε, ΕΖ, ïν ¹ Øπεροχ¾ Ð ∆Ζ GC is a rational (straight-line). And let the two square µ¾ œστω τετράγωνος. κሠπεποιήσθω æς Ð Ζ∆ πρÕς τÕν numbers DE and EF be laid down, and let their differ380

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

∆Ε, οÛτως τÕ ¢πÕ τÁς ΓΗ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΗΒ τετράγωνον. σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΓΗ τετράγωνον τù ¢πÕ τÁς ΗΒ τετραγώνJ. ·ητÕν δ τÕ ¢πÕ τÁς ΓΗ. ·ητÕν ¥ρα [™στˆ] κሠτÕ ¢πÕ τÁς ΗΒ· ·ητ¾ ¥ρα ™στˆν ¹ ΒΗ. κሠ™πεˆ τÕ ¢πÕ τÁς ΗΓ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΗΒ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, ¢σύµµετρός ™στιν ¹ ΓΗ τÍ ΗΒ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ΓΗ, ΗΒ ¥ρα ρηταί ε„σι δυνάµει µόνον σύµµετροι· ¹ ΒΓ ¥ρα ¢ποτοµή ™στιν. λέγω δή, Óτι κሠδευτέρα.

Α Θ

Β

Γ

Ε

Ζ

Η

B

C

E

F

G

A

∆

D

H

‘Ωι γ¦ρ µε‹ζόν ™στι τÕ ¢πÕ τÁς ΒΗ τοà ¢πÕ τÁς ΗΓ, œστω τÕ ¢πÕ τÁς Θ. ™πεˆ οâν ™στιν æς τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς ΗΓ, οÛτως Ð Ε∆ ¢ριθµÕς πρÕς τÕν ∆Ζ ¢ριθµόν, ¢ναστρέψαντι ¥ρα ™στˆν æς τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς Θ, οÛτως Ð ∆Ε πρÕς τÕν ΕΖ. καί ™στιν ˜κάτερος τîν ∆Ε, ΕΖ τετράγωνος· τÕ ¥ρα ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς Θ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· σύµµετρος ¥ρα ™στˆν ¹ ΒΗ τÍ Θ µήκει. κሠδύναται ¹ ΒΗ τÁς ΗΓ µε‹ζον τù ¢πÕ τÁς Θ· ¹ ΒΗ ¥ρα τÁς ΗΓ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει. καί ™στιν ¹ προσαρµόζουσα ¹ ΓΗ τÍ ™κκειµένV ·ητÍ σύµµετρος τÍ Α. ¹ ΒΓ ¥ρα ¢ποτοµή ™στι δευτέτα. ΕÛρηται ¥ρα δευτέρα ¢ποτοµ¾ ¹ ΒΓ· Óπερ œδει δε‹ξαι.

†

ence DF be not square [Prop. 10.28 lem. I]. And let it have been contrived that as F D (is) to DE, so the square on CG (is) to the square on GB [Prop. 10.6 corr.]. Thus, the square on CG is commensurable with the square on GB [Prop. 10.6]. And the (square) on CG (is) rational. Thus, the (square) on GB [is] also rational. Thus, BG is a rational (straight-line). And since the square on GC does not have to the (square) on GB the ratio which (some) square number (has) to (some) square number, CG is incommensurable in length with GB [Prop. 10.9]. And they are both rational (straight-lines). Thus, CG and GB are rational (straight-lines which are) commensurable in square only. Thus, BC is an apotome [Prop. 10.73]. So, I say that it is also a second (apotome).

For let the (square) on H be that (area) by which the (square) on BG is greater than the (square) on GC [Prop. 10.13 lem.]. Therefore, since as the (square) on BG is to the (square) on GC, so the number ED (is) to the number DF , thus, also, via conversion, as the (square) on BG is to the (square) on H, so DE (is) to EF [Prop. 5.19 corr.]. And DE and EF are each square (numbers). Thus, the (square) on BG has to the (square) on H the ratio which (some) square number (has) to (some) square number. Thus, BG is commensurable in length with H [Prop. 10.9]. And the square on BG is greater than (the square on) GC by the (square) on H. Thus, the square on BG is greater than (the square on) GC by the (square) on (some straight-line) commensurable in length with (BG). And the attachment CG is commensurable (in length) with the (prevously) laid down rational (straight-line) A. Thus, BC is a second apotome [Def. 10.12].† Thus, the second apotome BC has been found. (Which is) the very thing it was required to show.

See footnote to Prop. 10.49.

πζ΄.

Proposition 87

ΕØρε‹ν τ¾ν τρίτην ¢ποτοµήν.

Α Ε

To find a third apotome.

Β

∆

Ζ

Θ

Γ

B

D

F

H

C

A Η

G

E

Κ

K

'Εκκείσθω ·ητ¾ ¹ Α, κሠ™κκείσθωσαν τρε‹ς ¢ριθµοˆ 381

Let the rational (straight-line) A be laid down. And

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

οƒ Ε, ΒΓ, Γ∆ λόγον µ¾ œχοντες πρÕς ¢λλήλους, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, Ð δ ΓΒ πρÕς τÕν Β∆ λόγον ™χέτω, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, κሠπεποιήσθω æς µν Ð Ε πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΗ τετράγωνον, æς δ Ð ΒΓ πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς ΖΗ τετράγωνον πρÕς τÕ ¢πÕ τ¾ς ΗΘ. ™πεˆ οâν ™στιν æς Ð Ε πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΗ τετράγωνον, σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς Α τετράγωνον τù ¢πÕ τÁς ΖΗ τετραγώνJ. ·ητÕν δ τÕ ¢πÕ τÁς Α τετράγωνον. ·ητÕν ¥ρα κሠτÕ ¢πÕ τÁς ΖΗ· ·ητ¾ ¥ρα ™στˆν ¹ ΖΗ. κሠ™πεˆ Ð Ε πρÕς τÕν ΒΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ' ¥ρα τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΗ [τετράγωνον] λόγον œχει, Óν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ Α τÍ ΖΗ µήκει. πάλιν, ™πεί ™στιν æς Ð ΒΓ πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς ΖΗ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΗΘ, σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΖΗ τù ¢πÕ τÁς ΗΘ. ·ητÕν δ τÕ ¢πÕ τÁς ΖΗ· ·ητÕν ¥ρα κሠτÕ ¢πÕ τÁς ΗΘ· ·ητ¾ ¥ρα ™στˆν ¹ ΗΘ. κሠ™πεˆ Ð ΒΓ πρÕς τÕν Γ∆ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ' ¥ρα τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΖΗ τÍ ΗΘ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ΖΗ, ΗΘ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΖΘ. λέγω δή, Óτι κሠτρίτη. 'Επεˆ γάρ ™στιν æς µν Ð Ε πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς Α τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΗ, æς δ Ð ΒΓ πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΘΗ, δι' ‡σου ¥ρα ™στˆν æς Ð Ε πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς ΘΗ. Ð δ Ε πρÕς τÕν Γ∆ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ' ¥ρα τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς ΗΘ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ¹ Α τÍ ΗΘ µήκει. οÙδετέρα ¥ρα τîν ΖΗ, ΗΘ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ Α µήκει. ú οâν µε‹ζόν ™στι τÕ ¢πÕ τÁς ΖΗ τοà ¢πÕ τÁς ΗΘ, œστω τÕ ¢πÕ τÁς Κ. ™πεˆ οâν ™στιν æς Ð ΒΓ πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ, ¢ναστρέψαντι ¥ρα ™στˆν æς Ð ΒΓ πρÕς τÕν Β∆, οÛτως τÕ ¢πÕ τÁς ΖΗ τετράγωνον πρÕς τÕ ¢πÕ τÁς Κ. Ð δ ΒΓ πρÕς τÕν Β∆ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. κሠτÕ ¡πÕ τÁς ΖΗ ¥ρα πρÕς τÕ ¢πÕ τÁς Κ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. σύµµετρός ¥ρα ™στˆν ¹ ΖΗ τÍ Κ µήκει, κሠδύναται ¹ ΖΗ τÁς ΗΘ µε‹ζον τù ¢πÕ συµµέτρου ˜αυτÍ. κሠοÙδετέρα τîν ΖΗ, ΗΘ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ Α µήκει· ¹

let the three numbers, E, BC, and CD, not having to one another the ratio which (some) square number (has) to (some) square number, be laid down. And let CB have to BD the ratio which (some) square number (has) to (some) square number. And let it have been contrived that as E (is) to BC, so the square on A (is) to the square on F G, and as BC (is) to CD, so the square on F G (is) to the (square) on GH [Prop. 10.6 corr.]. Therefore, since as E is to BC, so the square on A (is) to the square on F G, the square on A is thus commensurable with the square on F G [Prop. 10.6]. And the square on A (is) rational. Thus, the (square) on F G (is) also rational. Thus, F G is a rational (straight-line). And since E does not have to BC the ratio which (some) square number (has) to (some) square number, the square on A thus does not have to the [square] on F G the ratio which (some) square number (has) to (some) square number either. Thus, A is incommensurable in length with F G [Prop. 10.9]. Again, since as BC is to CD, so the square on F G is to the (square) on GH, the square on F G is thus commensurable with the (square) on GH [Prop. 10.6]. And the (square) on F G (is) rational. Thus, the (square) on GH (is) also rational. Thus, GH is a rational (straightline). And since BC does not have to CD the ratio which (some) square number (has) to (some) square number, the (square) on F G thus does not have to the (square) on GH the ratio which (some) square number (has) to (some) square number either. Thus, F G is incommensurable in length with GH [Prop. 10.9]. And both are rational (straight-lines). F G and GH are thus rational (straight-lines which are) commensurable in square only. Thus, F H is an apotome [Prop. 10.73]. So, I say that (it is) also a third (apotome). For since as E is to BC, so the square on A (is) to the (square) on F G, and as BC (is) to CD, so the (square) on F G (is) to the (square) on HG, thus, via equality, as E is to CD, so the (square) on A (is) to the (square) on HG [Prop. 5.22]. And E does not have to CD the ratio which (some) square number (has) to (some) square number. Thus, the (square) on A does not have to the (square) on GH the ratio which (some) square number (has) to (some) square number either. A (is) thus incommensurable in length with GH [Prop. 10.9]. Thus, neither of F G and GH is commensurable in length with the (previously) laid down rational (straight-line) A. Therefore, let the (square) on K be that (area) by which the (square) on F G is greater than the (square) on GH [Prop. 10.13 lem.]. Therefore, since as BC is to CD, so the (square) on F G (is) to the (square) on GH, thus, via conversion, as BC is to BD, so the square on F G (is) to the square on K [Prop. 5.19 corr.]. And BC has to BD

382

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

ΖΘ ¥ρα ¢ποτοµή ™στι τρίτη. the ratio which (some) square number (has) to (some) ΕÛρηται ¥ρα ¹ τρίτη ¢ποτοµ¾ ¹ ΖΘ· Óπερ œδει square number. Thus, the (square) on F G also has to δε‹ξαι. the (square) on K the ratio which (some) square number (has) to (some) square number. F G is thus commensurable in length with K [Prop. 10.9]. And the square on F G is (thus) greater than (the square on) GH by the (square) on (some straight-line) commensurable (in length) with (F G). And neither of F G and GH is commensurable in length with the (previously) laid down rational (straight-line) A. Thus, F H is a third apotome [Def. 10.13]. Thus, the third apotome F H has been found. (Which is) very thing it was required to show. †

See footnote to Prop. 10.50.

πη΄.

Proposition 88

ΕØρε‹ν τ¾ν τετάρτην ¢ποτοµήν.

Α Θ

To find a fourth apotome.

Β

Γ

∆

Ζ

Η

B

C

E

F

G

A

Ε

D

H

'Εκκείσθω ·ητ¾ ¹ Α κሠτÍ Α µήκει σύµµετρος ¹ ΒΗ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΒΗ. κሠ™κκείσθωσαν δύο ¢ριθµοˆ οƒ ∆Ζ, ΖΕ, éστε τÕν ∆Ε Óλον πρÕς ˜κάτερον τîν ∆Ζ, ΕΖ λόγον µ¾ œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν. κሠπεποιήσθω æς Ð ∆Ε πρÕς τÕν ΕΖ, οÛτως τÕ ¢πÕ τÁς ΒΗ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΗΓ· σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΒΗ τù ¢πÕ τÁς ΗΓ. ·ητÕν δ τÕ ¢πÕ τÁς ΒΗ· ·ητÕν ¥ρα κሠτÕ ¢πÕ τÁς ΗΓ· ·ητ¾ ¥ρα ™στˆν ¹ ΗΓ. κሠ™πεˆ Ð ∆Ε πρÕς τÕν ΕΖ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ' ¥ρα τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς ΗΓ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΒΗ τÍ ΗΓ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ΒΗ, ΗΓ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΒΓ. [λέγω δή, Óτι κሠτετάρτη.] ‘Ωι οâν µε‹ζόν ™στι τÕ ¢πÕ τÁς ΒΗ τοà ¢πÕ τÁς ΗΓ, œστω τÕ ¢πÕ τÁς Θ. ™πεˆ οâν ™στιν æς Ð ∆Ε πρÕς τÕν ΕΖ, οÛτως τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς ΗΓ, κሠ¢ναστρέψαντι ¥ρα ™στˆν æς Ð Ε∆ πρÕς τÕν ∆Ζ, οÛτως τÕ ¢πÕ τÁς ΗΒ πρÕς τÕ ¢πÕ τÁς Θ. Ð δ Ε∆ πρÕς τÕν ∆Ζ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ' ¥ρα τÕ ¢πÕ τÁς ΗΒ πρÕς τÕ ¢πÕ τÁς Θ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΒΗ τÍ Θ µήκει. κሠδύναται ¹ ΒΗ τÁς ΗΓ µε‹ζον τù ¢πÕ τÁς Θ· ¹ ¥ρα ΒΗ τÁς ΗΓ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. κሠ™στιν Óλη ¹ ΒΗ σύµµετρος τÍ ™κκειµένV

Let the rational (straight-line) A, and BG (which is) commensurable in length with A, be laid down. Thus, BG is also a rational (straight-line). And let the two numbers DF and F E be laid down such that the whole, DE, does not have to each of DF and EF the ratio which (some) square number (has) to (some) square number. And let it have been contrived that as DE (is) to EF , so the square on BG (is) to the (square) on GC [Prop. 10.6 corr.]. The (square) on BG is thus commensurable with the (square) on GC [Prop. 10.6]. And the (square) on BG (is) rational. Thus, the (square) on GC (is) also rational. Thus, GC (is) a rational (straightline). And since DE does not have to EF the ratio which (some) square number (has) to (some) square number, the (square) on BG thus does not have to the (square) on GC the ratio which (some) square number (has) to (some) square number either. Thus, BG is incommensurable in length with GC [Prop. 10.9]. And they are both rational (straight-lines). Thus, BG and GC are rational (straight-lines which are) commensurable in square only. Thus, BC is an apotome [Prop. 10.73]. [So, I say that (it is) also a fourth (apotome).] Now, let the (square) on H be that (area) by which the (square) on BG is greater than the (square) on GC [Prop. 10.13 lem.]. Therefore, since as DE is to EF , so the (square) on BG (is) to the (square) on GC, thus, also, via conversion, as ED is to DF , so the (square) on GB (is) to the (square) on H [Prop. 5.19 corr.]. And ED

383

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

·ητÍ µήκει τÍ Α. ¹ ¥ρα ΒΓ ¢ποτοµή ™στι τετάρτη. ΕÛρηται ¥ρα ¹ τετάρτη ¢ποτοµή· Óπερ œδει δε‹ξαι.

†

does not have to DF the ratio which (some) square number (has) to (some) square number. Thus, the (square) on GB does not have to the (square) on H the ratio which (some) square number (has) to (some) square number either. Thus, BG is incommensurable in length with H [Prop. 10.9]. And the square on BG is greater than (the square on) GC by the (square) on H. Thus, the square on BG is greater than (the square) on GC by the (square) on (some straight-line) incommensurable (in length) with (BG). And the whole, BG, is commensurable in length with the the (previously) laid down rational (straightline) A. Thus, BC is a fourth apotome [Def. 10.14].† Thus, a fourth apotome has been found. (Which is) the very thing it was required to show.

See footnote to Prop. 10.51.

πθ΄.

Proposition 89

ΕÙρε‹ν τ¾ν πέµπτην ¢ποτοµήν.

Α Θ

Β ∆

To find a fifth apotome.

Γ

Η Ζ

B

C

G

A

Ε

D

F

E

H

'Εκκείσθω ·ητ¾ ¹ Α, κሠτÍ Α µήκει σύµµετρος œστω ¹ ΓΗ· ·ητ¾ ¥ρα [™στˆν] ¹ ΓΗ. κሠ™κκείσθωσαν δύο ¢ριθµοˆ οƒ ∆Ζ, ΖΕ, éστε τÕν ∆Ε πρÕς ˜κάτερον τîν ∆Ζ, ΖΕ λόγον πάλιν µ¾ œχειν, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· κሠπεποιήσθω æς Ð ΖΕ πρÕς τÕν Ε∆, οÛτως τÕ ¢πÕ τÁς ΓΗ πρÕς τÕ ¢πÕ τÁς ΗΒ. ·ητÕν ¥ρα κሠτÕ ¢πÕ τÁς ΗΒ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΒΗ. κሠ™πεί ™στιν æς Ð ∆Ε πρÕς τÕν ΕΖ, οÛτως τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς ΗΓ, Ð δ ∆Ε πρÕς τÕν ΕΖ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ' ¥ρα τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς ΗΓ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΒΗ τÍ ΗΓ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ΒΗ, ΗΓ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¹ ΒΓ ¥ρα ¢ποτοµή ™στιν. λέγω δή, Óτι κሠπέµπτη. ‘Ωι γ¦ρ µε‹ζόν ™στι τÕ ¢πÕ τÁς ΒΗ τοà ¢πÕ τÁς ΗΓ, œστω τÕ ¢πÕ τÁς Θ. ™πεˆ οâν ™στιν æς τÕ ¢πÕ τ¾ς ΒΗ πρÕς τÕ ¢πÕ τÁς ΗΓ, οÛτως Ð ∆Ε πρÕς τÕν ΕΖ, ¢ναστρέψαντι ¥ρα ™στˆν æς Ð Ε∆ πρÕς τÕν ∆Ζ, οÛτως τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς Θ, Ð δ Ε∆ πρÕς τÕν ∆Ζ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ' ¥ρα τÕ ¢πÕ τÁς ΒΗ πρÕς τÕ ¢πÕ τÁς Θ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΒΗ τÍ Θ µήκει. κሠδύναται ¹ ΒΗ τÁς ΗΓ µε‹ζον τù ¢πÕ τÁς Θ· ¹ ΗΒ ¥ρα τÁς ΗΓ µε‹ζον δύναται τù ¢πÕ

Let the rational (straight-line) A be laid down, and let CG be commensurable in length with A. Thus, CG [is] a rational (straight-line). And let the two numbers DF and F E be laid down such that DE again does not have to each of DF and F E the ratio which (some) square number (has) to (some) square number. And let it have been contrived that as F E (is) to ED, so the (square) on CG (is) to the (square) on GB. Thus, the (square) on GB (is) also rational [Prop. 10.6]. Thus, BG is also rational. And since as DE is to EF , so the (square) on BG (is) to the (square) on GC. And DE does not have to EF the ratio which (some) square number (has) to (some) square number. The (square) on BG thus does not have to the (square) on GC the ratio which (some) square number (has) to (some) square number either. Thus, BG is incommensurable in length with GC [Prop. 10.9]. And they are both rational (straight-lines). BG and GC are thus rational (straight-lines which are) commensurable in square only. Thus, BC is an apotome [Prop. 10.73]. So, I say that (it is) also a fifth (apotome). For, let the (square) on H be that (area) by which the (square) on BG is greater than the (square) on GC [Prop. 10.13 lem.]. Therefore, since as the (square) on BG (is) to the (square) on GC, so DE (is) to EF , thus, via conversion, as ED is to DF , so the (square) on BG (is) to the (square) on H [Prop. 5.19 corr.]. And ED does not have to DF the ratio which (some) square number

384

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

¢συµµέτρου ˜αυτÍ µήκει. καί ™στιν ¹ προσαρµόζουσα ¹ ΓΗ σύµµετρος τÍ ™κκειµένV ·ητÍ τÍ Α µήκει· ¹ ¥ρα ΒΓ ¢ποτοµή ™στι πέµπτη. ΕÛρηται ¥ρα ¹ πέµπτη ¢ποτοµ¾ ¹ ΒΓ· Óπερ œδει δε‹ξαι.

†

(has) to (some) square number. Thus, the (square) on BG does not have to the (square) on H the ratio which (some) square number (has) to (some) square number either. Thus, BG is incommensurable in length with H [Prop. 10.9]. And the square on BG is greater than (the square on) GC by the (square) on H. Thus, the square on GB is greater than (the square on) GC by the (square) on (some straight-line) incommensurable in length with (GB). And the attachment CG is commensurable in length with the (previously) laid down rational (straightline) A. Thus, BC is a fifth apotome [Def. 10.15].† Thus, the fifth apotome BC has been found. (Which is) the very thing it was required to show.

See footnote to Prop. 10.52.

&΄.

Proposition 90

ΕØρε‹ν τ¾ν ›κτην ¢ποτοµήν. 'Εκκείσθω ·ητ¾ ¹ Α κሠτρε‹ς ¢ριθµοˆ οƒ Ε, ΒΓ, Γ∆ λόγον µ¾ œχοντες πρÕς ¢λλήλους, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· œτι δ καˆ Ð ΓΒ πρÕς τÕν Β∆ λόγον µ¾ ™χετώ, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· κሠπεποιήσθω æς µν Ð Ε πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς ΖΗ, æς δ Ð ΒΓ πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ.

To find a sixth apotome. Let the rational (straight-line) A, and the three numbers E, BC, and CD, not having to one another the ratio which (some) square number (has) to (some) square number, be laid down. Furthermore, let CB also not have to BD the ratio which (some) square number (has) to (some) square number. And let it have been contrived that as E (is) to BC, so the (square) on A (is) to the (square) on F G, and as BC (is) to CD, so the (square) on F G (is) to the (square) on GH [Prop. 10.6 corr.].

Α Ε

Β Ζ

∆ Θ

Γ

D

B

C

A Η

F

H

G

E

Κ

K

'Επεˆ οâν ™στιν æς Ð Ε πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς ΖΗ, σύµµετρον ¥ρα τÕ ¢πÕ τÁς Α τù ¢πÕ τÁς ΖΗ. ·ητÕν δ τÕ ¢πÕ τÁς Α· ·ητÕν ¥ρα κሠτÕ ¢πÕ τÁς ΖΗ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΖΗ. κሠ™πεˆ Ð Ε πρÕς τÕν ΒΓ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ' ¥ρα τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς ΖΗ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ Α τÁ ΖΗ µήκει. πάλιν, ™πεί ™στιν æς Ð ΒΓ πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ, σύµµετρον ¥ρα τÕ ¢πÕ τÁς ΖΗ τù ¢πÕ τÁς ΗΘ. ·ητÕν δ τÕ ¢πÕ τ¾ς ΖΗ· ·ητÕν ¥ρα κሠτÕ ¢πÕ τÁς ΗΘ· ·ητ¾ ¥ρα κሠ¹ ΗΘ. κሠ™πεˆ Ð ΒΓ πρÕς τÕν Γ∆ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, οÙδ' ¥ρα τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΖΗ τÍ ΗΘ µήκει. καί ε„σιν

Therefore, since as E is to BC, so the (square) on A (is) to the (square) on F G, the (square) on A (is) thus commensurable with the (square) on F G [Prop. 10.6]. And the (square) on A (is) rational. Thus, the (square) on F G (is) also rational. Thus, F G is also a rational (straight-line). And since E does not have to BC the ratio which (some) square number (has) to (some) square number, the (square) on A thus does not have to the (square) on F G the ratio which (some) square number (has) to (some) square number either. Thus, A is incommensurable in length with F G [Prop. 10.9]. Again, since as BC is to CD, so the (square) on F G (is) to the (square) on GH, the (square) on F G (is) thus commensurable with the (square) on GH [Prop. 10.6]. And the (square) on F G (is) rational. Thus, the (square) on GH (is) also rational. Thus, GH (is) also rational. And since BC does not have to CD the ratio which (some) square

385

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

¢µφότεραι ·ηταί· αƒ ΖΗ, ΗΘ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¹ ¥ρα ΖΘ ¢ποτοµή ™στιν. λέγω δή, Óτι κሠ›κτη. 'Επεˆ γάρ ™στιν æς µν Ð Ε πρÕς τÕν ΒΓ, οÛτως τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς ΖΗ, æς δ Ð ΒΓ πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ, δι' ‡σου ¥ρα ™στˆν æς Ð Ε πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς ΗΘ. Ð δ Ε πρÕς τÕν Γ∆ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ' ¥ρα τÕ ¢πÕ τÁς Α πρÕς τÕ ¢πÕ τÁς ΗΘ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ Α τÍ ΗΘ µήκει· οÙδετέρα ¥ρα τîν ΖΗ, ΗΘ σύµµετρός ™στι τÍ Α ·ητÍ µήκει. ú οâν µε‹ζόν ™στι τÕ ¢πÕ τÁς ΖΗ τοà ¢πÕ τÁς ΗΘ, œστω τÕ ¢πÕ τÁς Κ. ™πεˆ οâν ™στιν æς Ð ΒΓ πρÕς τÕν Γ∆, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς ΗΘ, ¢ναστρέψαντι ¥ρα ™στˆν æς Ð ΓΒ πρÕς τÕν Β∆, οÛτως τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς Κ. Ð δ ΓΒ πρÕς τÕν Β∆ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· οÙδ' ¥ρα τÕ ¢πÕ τÁς ΖΗ πρÕς τÕ ¢πÕ τÁς Κ λόγον œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΖΗ τÍ Κ µήκει. κሠδύναται ¹ ΖΗ τÁς ΗΘ µε‹ζον τù ¢πÕ τÁς Κ· ¹ ΖΗ ¥ρα τÁς ΗΘ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ µήκει. κሠοÙδετέρα τîν ΖΗ, ΗΘ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ µήκει τÍ Α. ¹ ¥ρα ΖΘ ¢ποτοµή ™στιν ›κτη. ΕÛρηται ¥ρα ¹ ›κτη ¢ποτοµ¾ ¹ ΖΘ· Óπερ œδει δε‹ξαι.

†

number (has) to (some) square number, the (square) on F G thus does not have to the (square) on GH the ratio which (some) square (number) has to (some) square (number) either. Thus, F G is incommensurable in length with GH [Prop. 10.9]. And both are rational (straightlines). Thus, F G and GH are rational (straight-lines which are) commensurable in square only. Thus, F H is an apotome [Prop. 10.73]. So, I say that (it is) also a sixth (apotome). For since as E is to BC, so the (square) on A (is) to the (square) on F G, and as BC (is) to CD, so the (square) on F G (is) to the (square) on GH, thus, via equality, as E is to CD, so the (square) on A (is) to the (square) on GH [Prop. 5.22]. And E does not have to CD the ratio which (some) square number (has) to (some) square number. Thus, the (square) on A does not have to the (square) GH the ratio which (some) square number (has) to (some) square number either. A is thus incommensurable in length with GH [Prop. 10.9]. Thus, neither of F G and GH is commensurable in length with the rational (straight-line) A. Therefore, let the (square) on K be that (area) by which the (square) on F G is greater than the (square) on GH [Prop. 10.13 lem.]. Therefore, since as BC is to CD, so the (square) on F G (is) to the (square) on GH, thus, via conversion, as CB is to BD, so the (square) on F G (is) to the (square) on K [Prop. 5.19 corr.]. And CB does not have to BD the ratio which (some) square number (has) to (some) square number. Thus, the (square) on F G does not have to the (square) on K the ratio which (some) square number (has) to (some) square number either. F G is thus incommensurable in length with K [Prop. 10.9]. And the square on F G is greater than (the square on) GH by the (square) on K. Thus, the square on F G is greater than (the square on) GH by the (square) on (some straightline) incommensurable in length with (F G). And neither of F G and GH is commensurable in length with the (previously) laid down rational (straight-line) A. Thus, F H is a sixth apotome [Def. 10.16]. Thus, the sixth apotome F H has been found. (Which is) the very thing it was required to show.

See footnote to Prop. 10.53.

&α΄.

Proposition 91

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠ¢ποτοµÁς If an area is contained by a rational (straight-line) and πρώτης, ¹ τÕ χωρίον δυναµένη ¢ποροµή ™στιν. a first apotome then the square-root of the area is an apoΠεριεχέσθω γ¦ρ χωρίον τÕ ΑΒ ØπÕ ·ητÁς τÁς ΑΓ tome. κሠ¢ποτοµÁς πρώτης τÁς Α∆· λέγω, Óτι ¹ τÕ ΑΒ χωρίον For let the area AB have been contained by the ratioδυναµένη ¢ποτοµή ™στιν. nal (straight-line) AC and the first apotome AD. I say

386

ΣΤΟΙΧΕΙΩΝ ι΄.

Α

∆

Ε

ELEMENTS BOOK 10

Ζ Η

Σ Γ

Β

Θ

Ν

Λ

Ι Κ Ρ

Φ Υ

that the square-root of area AB is an apotome. A D E F G N L

Ο

P V

Ξ

S

Χ

O

U W

C

Μ Τ 'Επεˆ γ¦ρ ¢ποτοµή ™στι πρώτη ¹ Α∆, œστω αÙτÍ προσαρµόζουσα ¹ ∆Η· αƒ ΑΗ, Η∆ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. κሠÓλη ¹ ΑΗ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ΑΓ, κሠ¹ ΑΗ τÁς Η∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει· ™¦ν ¥ρα τù τετάρτJ µέρει τοà ¢πÕ τÁς ∆Η ‡σον παρ¦ τ¾ν ΑΗ παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, ε„ς σύµµετρα αÙτ¾ν διαιρε‹. τετµήσθω ¹ ∆Η δίχα κατ¦ τÕ Ε, κሠτù ¢πÕ τÁς ΕΗ ‡σον παρ¦ τ¾ν ΑΗ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ, κሠœστω τÕ ØπÕ τîν ΑΖ, ΖΗ· σύµµετρος ¥ρα ™στˆν ¹ ΑΖ τÍ ΖΗ. κሠδι¦ τîν Ε, Ζ, Η σηµείων τÍ ΑΓ παράλληλοι ½χθωσαν αƒ ΕΘ, ΖΙ, ΗΚ. Κሠ™πεˆ σύµµετρός ™στιν ¹ ΑΖ τÍ ΖΗ µήκει, κሠ¹ ΑΗ ¥ρα ˜κατέρv τîν ΑΖ, ΖΗ σύµµετρός ™στι µήκει. ¢λλ¦ ¹ ΑΗ σύµµετρός ™στι τÍ ΑΓ· κሠ˜κατέρα ¥ρα τîν ΑΖ, ΖΗ σύµµετρός ™στι τÍ ΑΓ µήκει. καί ™στι ·ητ¾ ¹ ΑΓ· ·ητ¾ ¥ρα κሠ˜κατέρα τîν ΑΖ, ΖΗ· éστε κሠ˜κάτερον τîν ΑΙ, ΖΚ ·ητόν ™στιν. κሠ™πεˆ σύµµετρός ™στιν ¹ ∆Ε τÍ ΕΗ µήκει, κሠ¹ ∆Η ¥ρα ˜κατέρv τîν ∆Ε, ΕΗ σύµµετρός ™στι µήκει. ·ητ¾ δ ¹ ∆Η κሠ¢σύµµετρος τÍ ΑΓ µήκει· ·ητ¾ ¥ρα κሠ˜κατέρα τîν ∆Ε, ΕΗ κሠ¢σύµµετρος τÍ ΑΓ µήκει· ˜κάτερον ¥ρα τîν ∆Θ, ΕΚ µέσον ™στίν. Κείσθω δ¾ τù µν ΑΙ ‡σον τετράγωνον τÕ ΛΜ, τù δ ΖΚ ‡σον τετράγωνον ¢φVρήσθω κοιν¾ν γωνίαν œχον αÙτù τ¾ν ØπÕ ΛΟΜ τÕ ΝΞ· περˆ τ¾ν αÙτ¾ν ¥ρα διάµετρόν ™στι τ¦ ΛΜ, ΝΞ τετράγωνα. œστω αÙτîν διάµετρος ¹ ΟΡ, κሠκαταγεγράφθω τÕ σχÁµα. ™πεˆ οâν ‡σον ™στˆ τÕ ØπÕ τîν ΑΖ, ΖΗ περιεχόµενον Ñρθογώνιον τù ¢πÕ τÁς ΕΗ τετραγώνJ, œστιν ¥ρα æς ¹ ΑΖ πρÕς τ¾ν ΕΗ, οÛτως ¹ ΕΗ πρÕς τ¾ν ΖΗ. ¢λλ' æς µν ¹ ΑΖ πρÕς τ¾ν ΕΗ, οÛτως τÕ ΑΙ πρÕς τÕ ΕΚ, æς δ ¹ ΕΗ πρÕς τ¾ν ΖΗ, οÛτως ™στˆ τÕ ΕΚ πρÕς τÕ ΚΖ· τîν ¥ρα ΑΙ, ΚΖ µέσον ¢νάλογόν ™στι τÕ ΕΚ. œστι δ κሠτîν ΛΜ, ΝΞ µέσον ¢νάλογον τÕ ΜΝ, æς ™ν το‹ς œµπροσθεν ™δείχθη, καί ™στι τÕ [µν] ΑΙ τù ΛΜ τετραγώνJ ‡σον, τÕ δ ΚΖ τù ΝΞ· κሠτÕ ΜΝ ¥ρα τù ΕΚ ‡σον ™στίν. ¢λλ¦ τÕ µν ΕΚ τù ∆Θ ™στιν ‡σον, τÕ δ ΜΝ τù ΛΞ· τÕ ¥ρα ∆Κ ‡σον ™στˆ τù ΥΦΧ γνώµονι κሠτù ΝΞ. œστι δ κሠτÕ ΑΚ ‡σον το‹ς ΛΜ, ΝΞ τετραγώνοις· λοιπÕν ¥ρα τÕ ΑΒ ‡σον ™στˆ τù ΣΤ. τÕ δ ΣΤ τÕ ¢πÕ τÁς ΛΝ ™στι τετράγωνον· τÕ ¥ρα ¢πÕ τÁς ΛΝ τετράγωνον ‡σον

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M T For since AD is a first apotome, let DG be its attachment. Thus, AG and DG are rational (straight-lines which are) commensurable in square only [Prop. 10.73]. And the whole, AG, is commensurable (in length) with the (previously) laid down rational (straight-line) AC, and the square on AG is greater than (the square on) GD by the (square) on (some straight-line) commensurable in length with (AG) [Def. 10.11]. Thus, if (an area) equal to the fourth part of the (square) on DG is applied to AG, falling short by a square figure, then it divides (AG) into (parts which are) commensurable (in length) [Prop. 10.17]. Let DG have been cut in half at E. And let (an area) equal to the (square) on EG have been applied to AG, falling short by a square figure. And let it be the (rectangle contained) by AF and F G. AF is thus commensurable (in length) with F G. And let EH, F I, and GK have been drawn through points E, F , and G (respectively), parallel to AC. And since AF is commensurable in length with F G, AG is thus also commensurable in length with each of AF and F G [Prop. 10.15]. But AG is commensurable (in length) with AC. Thus, each of AF and F G is also commensurable in length with AC [Prop. 10.12]. And AC is a rational (straight-line). Thus, AF and F G (are) each also rational (straight-lines). Hence, AI and F K are also each rational (areas) [Prop. 10.19]. And since DE is commensurable in length with EG, DG is thus also commensurable in length with each of DE and EG [Prop. 10.15]. And DG (is) rational, and incommensurable in length with AC. DE and EG (are) thus each rational, and incommensurable in length with AC [Prop. 10.13]. Thus, DH and EK are each medial (areas) [Prop. 10.21]. So let the square LM , equal to AI, be laid down. And let the square N O, equal to F K, have been subtracted (from LM ), having with it the common angle LP M . Thus, the squares LM and N O are about the same diagonal [Prop. 6.26]. Let P R be their (common) diagonal, and let the (rest of the) figure have been drawn. Therefore, since the rectangle contained by AF and F G is equal to the square EG, thus as AF is to EG, so EG (is) to F G [Prop. 6.17]. But, as AF (is)

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ELEMENTS BOOK 10

™στˆ τù ΑΒ· ¹ ΛΝ ¥ρα δύναται τÕ ΑΒ. λέγω δή, Óτι ¹ ΛΝ ¢ποτοµή ™στιν. 'Επεˆ γ¦ρ ·ητόν ™στιν ˜κάτερον τîν ΑΙ, ΖΚ, καί ™στιν ‡σον το‹ς ΛΜ, ΝΞ, κሠ˜κάτερον ¥ρα τîν ΛΜ, ΝΞ ·ητόν ™στιν, τουτέστι τÕ ¢πÕ ˜κατέρας τîν ΛΟ, ΟΝ· κሠ˜κατέρα ¥ρα τîν ΛΟ, ΟΝ ·ητή ™στιν. πάλιν, ™πεˆ µέσον ™στˆ τÕ ∆Θ καί ™στιν ‡σον τù ΛΞ, µέσον ¥ρα ™στˆ κሠτÕ ΛΞ. ™πεˆ οâν τÕ µν ΛΞ µέσον ™στίν, τÕ δ ΝΞ ·ητόν, ¢σύµµετρον ¥ρα ™στˆ τÕ ΛΞ τù ΝΞ· æς δ τÕ ΛΞ πρÕς τÕ ΝΞ, οÛτως ™στˆν ¹ ΛΟ πρÕς τ¾ν ΟΝ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΛΟ τÍ ΟΝ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ΛΟ, ΟΝ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΛΝ. κሠδύναται τÕ ΑΒ χωρίον· ¹ ¥ρα τÕ ΑΒ χωρίον δυναµένη ¢ποτοµή ™στιν. 'Ε¦ν ¥ρα χωρίον περιέχηται ØπÕ ·ητÁς κሠτ¦ ˜ξÁς.

to EG, so AI (is) to EK, and as EG (is) to F G, so EK is to KF [Prop. 6.1]. Thus, EK is the mean proportional to AI and KF [Prop. 5.11]. And M N is also the mean proportional to LM and N O, as shown before [Prop. 10.53 lem.]. And AI is equal to the square LM , and KF to N O. Thus, M N is also equal to EK. But, EK is equal to DH, and M N to LO [Prop. 1.43]. Thus, DK is equal to the gnomon U V W and N O. And AK is also equal to (the sum of) the squares LM and N O. Thus, the remainder AB is equal to ST . And ST is the square on LN . Thus, the square on LN is equal to AB. Thus, LN is the square-root of AB. So, I say that LN is an apotome. For since AI and F K are each rational (areas), and are equal to LM and N O (respectively), thus LM and N O—that is to say, the (squares) on each of LP and P N (respectively)—are also each rational (areas). Thus, LP and P N are also each rational (straight-lines). Again, since DH is a medial (area), and is equal to LO, LO is thus also a medial (area). Therefore, since LO is medial, and N O rational, LO is thus incommensurable with N O. And as LO (is) to N O, so LP is to P N [Prop. 6.1]. LP is thus incommensurable in length with P N [Prop. 10.11]. And they are both rational (straightlines). Thus, LP and P N are rational (straight-lines which are) commensurable in square only. Thus, LN is an apotome [Prop. 10.73]. And it is the square-root of area AB. Thus, the square-root of area AB is an apotome. Thus, if an area is contained by a rational (straightline), and so on . . . .

&β΄.

Proposition 92

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠ¢ποτοµÁς If an area is contained by a rational (straight-line) and δευτέρας, ¹ τÕ χωρίον δυναµένη µέσης ¢ποτοµή ™στι a second apotome then the square-root of the area is a πρώτη. first apotome of a medial (straight-line). E F G N A D Ε Ζ Η Ν Α ∆ L O Λ Ο Φ V

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Μ Τ Χωρίον γ¦ρ τÕ ΑΒ περιεχέσθω ØπÕ ·ητÁς τÁς ΑΓ κሠ¢ποτοµÁς δευτέρας τÁς Α∆· λέγω, Óτι ¹ τÕ ΑΒ χωρίον δυναµένη µέσης ¢ποτοµή ™στι πρώτη. ”Εστω γ¦ρ τÍ Α∆ προσαρµόζουσα ¹ ∆Η· αƒ ¥ρα ΑΗ, Η∆ ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ προσαρµόζουσα ¹ ∆Η σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ΑΓ, ¹ δ Óλη ¹ ΑΗ τÁς προσαρµοζούσης τÁς

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M T For let the area AB have been contained by the rational (straight-line) AC and the second apotome AD. I say that the square-root of area AB is the first apotome of a medial (straight-line). For let DG be an attachment to AD. Thus, AG and GD are rational (straight-lines which are) commensurable in square only [Prop. 10.73], and the attachment

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ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

Η∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει. ™πεˆ οâν ¹ ΑΗ τÁς Η∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, ™¦ν ¥ρα τù τετάρτJ µέρει τοà ¢πÕ τÁς Η∆ ‡σον παρ¦ τ¾ν ΑΗ παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, ε„ς σύµµετρα αÙτ¾ν διαιρε‹. τετµήσθω οâν ¹ ∆Η δίχα κατ¦ τÕ Ε· κሠτù ¢πÕ τÁς ΕΗ ‡σον παρ¦ τ¾ν ΑΗ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ, κሠœστω τÕ ØπÕ τîν ΑΖ, ΖΗ· σύµµετρος ¥ρα ™στˆν ¹ ΑΖ τÍ ΖΗ µήκει. κሠ¹ ΑΗ ¥ρα ˜κατέρv τîν ΑΖ, ΖΗ σύµµετρός ™στι µήκει. ·ητ¾ δ ¹ ΑΗ κሠ¢σύµµετρος τÍ ΑΓ µήκει· κሠ˜κατέρα ¥ρα τîν ΑΖ, ΖΗ ·ητή ™στι κሠ¢σύµµετρος τÍ ΑΓ µήκει· ˜κάτερον ¥ρα τîν ΑΙ, ΖΚ µέσον ™στίν. πάλιν, ™πεˆ σύµµετρός ™στιν ¹ ∆Ε τÍ ΕΗ, κሠ¹ ∆Η ¥ρα ˜κατέρv τîν ∆Ε, ΕΗ σύµµετρός ™στιν. ¢λλ' ¹ ∆Η σύµµετρός ™στι τÍ ΑΓ µήκει [·ητ¾ ¥ρα κሠ˜κατέρα τîν ∆Ε, ΕΗ κሠσύµµετρος τÍ ΑΓ µήκει]. ˜κάτερον ¥ρα τîν ∆Θ, ΕΚ ·ητόν ™στιν. Συνεστάτω οâν τù µν ΑΙ ‡σον τετράγωνον τÕ ΛΜ, τù δ ΖΚ ‡σον ¢φVρήσθω τÕ ΝΞ περˆ τ¾ν αÙτ¾ν γωνίαν ×ν τù ΛΜ τ¾ν ØπÕ τîν ΛΟΜ· περˆ τ¾ν αÙτ¾ν ¥ρα ™στˆ διάµετρον τ¦ ΛΜ, ΝΞ τετράγωνα. œστω αÙτîν διάµετρος ¹ ΟΡ, κሠκαταγεγράφθω τÕ σχÁµα. ™πεˆ οâν τ¦ ΑΙ, ΖΚ µέσα ™στˆ καί ™στιν ‡σα το‹ς ¢πÕ τîν ΛΟ, ΟΝ, κሠτ¦ ¢πÕ τîν ΛΟ, ΟΝ [¥ρα] µέσα ™στίν· καˆ αƒ ΛΟ, ΟΝ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι. κሠ™πεˆ τÕ ØπÕ τîν ΑΖ, ΖΗ ‡σον ™στˆ τù ¢πÕ τÁς ΕΗ, œστιν ¥ρα æς ¹ ΑΖ πρÕς τ¾ν ΕΗ, οÛτως ¹ ΕΗ πρÕς τ¾ν ΖΗ· ¢λλ' æς µν ¹ ΑΖ πρÕς τ¾ν ΕΗ, οÛτως τÕ ΑΙ πρÕς τÕ ΕΚ· æς δ ¹ ΕΗ πρÕς τ¾ν ΖΗ, οÛτως [™στˆ] τÕ ΕΚ πρÕς τÕ ΖΚ· τîν ¥ρα ΑΙ, ΖΚ µέσον ¢νάλογόν ™στι τÕ ΕΚ. œστι δ κሠτîν ΛΜ, ΝΞ τετραγώνων µέσον ¢νάλογον τÕ ΜΝ· καί ™στιν ‡σον τÕ µν ΑΙ τù ΛΜ, τÕ δ ΖΚ τù ΝΞ· κሠτÕ ΜΝ ¥ρα ‡σον ™στˆ τù ΕΚ. ¢λλ¦ τù µν ΕΚ ‡σον [™στˆ] τÕ ∆Θ, τù δ ΜΝ ‡σον τÕ ΛΞ· Óλον ¥ρα τÕ ∆Κ ‡σον ™στˆ τù ΥΦΧ γνώµονι κሠτù ΝΞ. ™πεˆ οâν Óλον τÕ ΑΚ ‡σον ™στˆ το‹ς ΛΜ, ΝΞ, ïν τÕ ∆Κ ‡σον ™στˆ τù ΥΦΧ γνώµονι κሠτù ΝΞ, λοιπÕν ¥ρα τÕ ΑΒ ‡σον ™στˆ τù ΤΣ. τÕ δ ΤΣ ™στι τÕ ¢πÕ τÁς ΛΝ· τÕ ¢πÕ τÁς ΛΝ ¥ρα ‡σον ™στˆ τù ΑΒ χωρίJ· ¹ ΛΝ ¥ρα δύναται τÕ ΑΒ χωρίον. λέγω [δή], Óτι ¹ ΛΝ µέσης ¢ποτοµή ™στι πρώτη.

DG is commensurable (in length) with the (previously) laid down rational (straight-line) AC, and the square on the whole, AG, is greater than (the square on) the attachment, GD, by the (square) on (some straight-line) commensurable in length with (AG) [Def. 10.12]. Therefore, since the square on AG is greater than (the square on) GD by the (square) on (some straight-line) commensurable (in length) with (AG), thus if (an area) equal to the fourth part of the (square) on GD is applied to AG, falling short by a square figure, then it divides (AG) into (parts which are) commensurable (in length) [Prop. 10.17]. Therefore, let DG have been cut in half at E. And let (an area) equal to the (square) on EG have been applied to AG, falling short by a square figure. And let it be the (rectangle contained) by AF and F G. Thus, AF is commensurable in length with F G. AG is thus also commensurable in length with each of AF and F G [Prop. 10.15]. And AG (is) a rational (straight-line), and incommensurable in length with AC. AF and F G are thus also each rational (straight-lines), and incommensurable in length with AC [Prop. 10.13]. Thus, AI and F K are each medial (areas) [Prop. 10.21]. Again, since DE is commensurable (in length) with EG, thus DG is also commensurable (in length) with each of DE and EG [Prop. 10.15]. But, DG is commensurable in length with AC [thus, DE and EG are also each rational, and commensurable in length with AC]. Thus, DH and EK are each rational (areas) [Prop. 10.19]. Therefore, let the square LM , equal to AI, have been constructed. And let N O, equal to F K, which is about the same angle LP M as LM , have been subtracted (from LM ). Thus, the squares LM and N O are about the same diagonal [Prop. 6.26]. Let P R be their (common) diagonal, and let the (rest of the) figure have been drawn. Therefore, since AI and F K are medial (areas), and are equal to the (squares) on LP and P N (respectively), [thus] the (squares) on LP and P N are also medial. Thus, LP and P N are also medial (straight-lines which are) commensurable in square only.† And since the (rectangle contained) by AF and F G is equal to the (square) on EG, thus as AF is to EG, so EG (is) to F G [Prop. 10.17]. But, as AF (is) to EG, so AI 'Επεˆ γ¦ρ ·ητόν ™στι τÕ ΕΚ καί ™στιν ‡σον τù ΛΞ, ·ητÕν (is) to EK. And as EG (is) to F G, so EK [is] to F K ¥ρα ™στˆ τÕ ΛΞ, τουτέστι τÕ ØπÕ τîν ΛΟ, ΟΝ. µέσον δ [Prop. 6.1]. Thus, EK is the mean proportional to AI ™δείχθη τÕ ΝΞ· ¢σύµµετρον ¥ρα ™στˆ τÕ ΛΞ τù ΝΞ· æς and F K [Prop. 5.11]. And M N is also the mean proδ τÕ ΛΞ πρÕς τÕ ΝΞ, οÛτως ™στˆν ¹ ΛΟ πρÕς ΟΝ· αƒ portional to the squares LM and N O [Prop. 10.53 lem.]. ΛΟ, ΟΝ ¥ρα ¢σύµµετροί ε„σι µήκει. αƒ ¥ρα ΛΟ, ΟΝ And AI is equal to LM , and F K to N O. Thus, M N is µέσαι ε„σˆ δυνάµει µόνον σύµµετροι ·ητÕν περιέχουσαι· also equal to EK. But, DH [is] equal to EK, and LO ¹ ΛΝ ¥ρα µέσης ¢ποτοµή ™στι πρώτη· κሠδύναται τÕ equal to M N [Prop. 1.43]. Thus, the whole (of) DK is ΑΒ χωρίον. equal to the gnomon U V W and N O. Therefore, since the whole (of) AK is equal to LM and N O, of which DK is 389

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ELEMENTS BOOK 10

`Η ¥ρα τÕ ΑΒ χωρίον δυναµένη µέσης ¢ποτοµή ™στι equal to the gnomon U V W and N O, the remainder AB πρώτη· Óπερ œδει δε‹ξαι. is thus equal to T S. And T S is the (square) on LN . Thus, the (square) on LN is equal to the area AB. LN is thus the square-root of area AB. [So], I say that LN is the first apotome of a medial (straight-line). For since EK is a rational (area), and is equal to LO, LO—that is to say, the (rectangle contained) by LP and P N —is thus a rational (area). And N O was shown (to be) a medial (area). Thus, LO is incommensurable with N O. And as LO (is) to N O, so LP is to P N [Prop. 6.1]. Thus, LP and P N are incommensurable in length [Prop. 10.11]. LP and P N are thus medial (straight-lines which are) commensurable in square only, and which contain a rational (area). Thus, LN is the first apotome of a medial (straight-line) [Prop. 10.74]. And it is the square-root of area AB. Thus, the square root of area AB is the first apotome of a medial (straight-line). (Which is) the very thing it was required to show. †

There is an error in the argument here. It should just say that LP and P N are commensurable in square, rather than in square only, since LP

and P N are only shown to be incommensurable in length later on.

&γ΄.

Proposition 93

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠ¢ποτοµÁς If an area is contained by a rational (straight-line) and τρίτης, ¹ τÕ χωρίον δυναµένη µέσης ¢ποτοµή ™στι a third apotome then the square-root of the area is a secδευτέρα. ond apotome of a medial (straight-line). E F G N A D Ε Ζ Η Ν Α ∆ L O Ο Λ Φ V

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Μ Τ Χωρίον γ¦ρ τÕ ΑΒ περιεχέσθω ØπÕ ·ητÁς τÁς ΑΓ κሠ¢ποτοµÁς τρίτης τÁς Α∆· λέγω, Óτι ¹ τÕ ΑΒ χωρίον δυναµένη µέσης ¢ποτοµή ™στι δευτέρα. ”Εστω γ¦ρ τÍ Α∆ προσαρµόζουσα ¹ ∆Η· αƒ ΑΗ, Η∆ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠοÙδετέρα τîν ΑΗ, Η∆ σύµµετρός ™στι µήκει τÍ ™κκειµένV ·ητÍ τÍ ΑΓ, ¹ δ Óλη ¹ ΑΗ τ¾ς προσαρµοζούσης τÁς ∆Η µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. ™πεˆ οâν ¹ ΑΗ τÁς Η∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, ™¦ν ¥ρα τù τετάρτJ µέρει τοà ¢πÕ τÁς ∆Η ‡σον παρ¦ τ¾ν ΑΗ παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, ε„ς σύµµετρα αÙτ¾ν διελε‹. τετµήσθω οâν ¹ ∆Η δίχα κατ¦ τÕ Ε, κሠτù ¢πÕ τÁς ΕΗ ‡σον παρ¦ τ¾ν ΑΗ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ, κሠœστω τÕ ØπÕ τîν ΑΖ, ΖΗ. κሠ½χθωσαν δι¦ τîν Ε, Ζ, Η

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M T For let the area AB have been contained by the rational (straight-line) AC and the third apotome AD. I say that the square-root of area AB is the second apotome of a medial (straight-line). For let DG be an attachment to AD. Thus, AG and GD are rational (straight-lines which are) commensurable in square only [Prop. 10.73], and neither of AG and GD is commensurable in length with the (previously) laid down rational (straight-line) AC, and the square on the whole, AG, is greater than (the square on) the attachment, DG, by the (square) on (some straightline) commensurable (in length) with (AG) [Def. 10.13]. Therefore, since the square on AG is greater than (the square on) GD by the (square) on (some straight-line) commensurable (in length) with (AG), thus if (an area)

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σηµείων τÍ ΑΓ παράλληλοι αƒ ΕΘ, ΖΙ, ΗΚ· σύµµετροι ¥ρα ε„σˆν αƒ ΑΖ, ΖΗ· σύµµετρον ¥ρα κሠτÕ ΑΙ τù ΖΚ. κሠ™πεˆ αƒ ΑΖ, ΖΗ σύµµετροί ε„σι µήκει, κሠ¹ ΑΗ ¥ρα ˜κατέρv τîν ΑΖ, ΖΗ σύµµετρός ™στι µήκει. ·ητ¾ δ ¹ ΑΗ κሠ¢σύµµετρος τÍ ΑΓ µήκει· éστε καˆ αƒ ΑΖ, ΖΗ. ˜κάτερον ¥ρα τîν ΑΙ, ΖΚ µέσον ™στίν. πάλιν, ™πεˆ σύµµετρός ™στιν ¹ ∆Ε τÍ ΕΗ µήκει, κሠ¹ ∆Η ¥ρα ˜κατέρv τîν ∆Ε, ΕΗ σύµµετρός ™στι µήκει. ·ητ¾ δ ¹ Η∆ κሠ¢σύµµετρος τÍ ΑΓ µήκει· ·ητ¾ ¥ρα κሠ˜κατέρα τîν ∆Ε, ΕΗ κሠ¢σύµµετρος τÍ ΑΓ µήκει· ˜κάτερον ¥ρα τîν ∆Θ, ΕΚ µέσον ™στίν. κሠ™πεˆ αƒ ΑΗ, Η∆ δυνάµει µόνον σύµµετροί ε„σιν, ¢σύµµετρος ¥ρα ™στˆ µήκει ¹ ΑΗ τÍ Η∆. ¢λλ' ¹ µν ΑΗ τÍ ΑΖ σύµµετρός ™στι µήκει ¹ δ ∆Η τÍ ΕΗ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΖ τÍ ΕΗ µήκει. æς δ ¹ ΑΖ πρÕς τ¾ν ΕΗ, οÛτως ™στˆ τÕ ΑΙ πρÕς τÕ ΕΚ· ¢σύµµετρον ¥ρα ™στˆ τÕ ΑΙ τù ΕΚ. Συνεστάτω οâν τù µν ΑΙ ‡σον τετράγωνον τÕ ΛΜ, τù δ ΖΚ ‡σον ¢φÍρήσθω τÕ ΝΞ περˆ τ¾ν αÙτ¾ν γωνίαν ×ν τù ΛΜ· περˆ τ¾ν αÙτ¾ν ¥ρα διάµετρόν ™στι τ¦ ΛΜ, ΝΞ. œστω αÙτîν διάµετρος ¹ ΟΡ, κሠκαταγεγράφθω τÕ σχÁµα. ™πεˆ οâν τÕ ØπÕ τîν ΑΖ, ΖΗ ‡σον ™στˆ τù ¢πÕ τÁς ΕΗ, œστιν ¥ρα æς ¹ ΑΖ πρÕς τ¾ν ΕΗ, οÛτως ¹ ΕΗ πρÕς τ¾ν ΖΗ. ¢λλ' æς µν ¹ ΑΖ πρÕς τ¾ν ΕΗ, οÛτως ™στˆ τÕ ΑΙ πρÕς τÕ ΕΚ· æς δ ¹ ΕΗ πρÕς τ¾ν ΖΗ, οÛτως ™στˆ τÕ ΕΚ πρÕς τÕ ΖΚ· κሠæς ¥ρα τÕ ΑΙ πρÕς τÕ ΕΚ, οÛτως τÕ ΕΚ πρÕς τÕ ΖΚ· τîν ¥ρα ΑΙ, ΖΚ µέσον ¢νάλογόν ™στι τÕ ΕΚ. œστι δ κሠτîν ΛΜ, ΝΞ τετραγώνων µέσον ¢νάλογον τÕ ΜΝ· καί ™στιν ‡σον τÕ µν ΑΙ τù ΛΜ, τÕ δ ΖΚ τù ΝΞ· κሠτÕ ΕΚ ¥ρα ‡σον ™στˆ τù ΜΝ. ¢λλ¦ τÕ µν ΜΝ ‡σον ™στˆ τù ΛΞ, τÕ δ ΕΚ ‡σον [™στˆ] τù ∆Θ· κሠÓλον ¥ρα τÕ ∆Κ ‡σον ™στˆ τù ΥΦΧ γνώµονι κሠτù ΝΞ. œστι δ κሠτÕ ΑΚ ‡σον το‹ς ΛΜ, ΝΞ· λοιπÕν ¥ρα τÕ ΑΒ ‡σον ™στˆ τù ΣΤ, τουτέστι τù ¢πÕ τÁς ΛΝ τετραγώνJ· ¹ ΛΝ ¥ρα δύναται τÕ ΑΒ χωρίον. λέγω, Óτι ¹ ΛΝ µέσης ¢ποτοµή ™στι δευτέρα. 'Επεˆ γ¦ρ µέσα ™δείχθη τ¦ ΑΙ, ΖΚ καί ™στιν ‡σα το‹ς ¢πÕ τîν ΛΟ, ΟΝ, µέσον ¥ρα κሠ˜κάτερον τîν ¢πÕ τîν ΛΟ, ΟΝ· µέση ¥ρα ˜κατέρα τîν ΛΟ, ΟΝ. κሠ™πεˆ σύµµετρόν ™στι τÕ ΑΙ τù ΖΚ, σύµµετρον ¥ρα κሠτÕ ¢πÕ τÁς ΛΟ τù ¢πÕ τÁς ΟΝ. πάλιν, ™πεˆ ¢σύµµετρον ™δείχθη τÕ ΑΙ τù ΕΚ, ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ΛΜ τù ΜΝ, τουτέστι τÕ ¢πÕ τÁς ΛΟ τù ØπÕ τîν ΛΟ, ΟΝ· éστε κሠ¹ ΛΟ ¢σύµµετρός ™στι µήκει τÍ ΟΝ· αƒ ΛΟ, ΟΝ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι. λέγω δή, Óτι κሠµέσον περιέχουσιν. 'Επεˆ γ¦ρ µέσον ™δείχθη τÕ ΕΚ καί ™στιν ‡σον τù ØπÕ τîν ΛΟ, ΟΝ, µέσον ¥ρα ™στˆ κሠτÕ ØπÕ τîν ΛΟ, ΟΝ· éστε αƒ ΛΟ, ΟΝ µέσαι ε„σˆ δυνάµει µόνον σύµµετροι µέσον περιέχουσαι. ¹ ΛΝ ¥ρα µέσης ¢πο-

equal to the fourth part of the square on DG is applied to AG, falling short by a square figure, then it divides (AG) into (parts which are) commensurable (in length) [Prop. 10.17]. Therefore, let DG have been cut in half at E. And let (an area) equal to the (square) on EG have been applied to AG, falling short by a square figure. And let it be the (rectangle contained) by AF and F G. And let EH, F I, and GK have been drawn through points E, F , and G (respectively), parallel to AC. Thus, AF and F G are commensurable (in length). AI (is) thus also commensurable with F K [Props. 6.1, 10.11]. And since AF and F G are commensurable in length, AG is thus also commensurable in length with each of AF and F G [Prop. 10.15]. And AG (is) rational, and incommensurable in length with AC. Hence, AF and F G (are) also (rational, and incommensurable in length with AC) [Prop. 10.13]. Thus, AI and F K are each medial (areas) [Prop. 10.21]. Again, since DE is commensurable in length with EG, DG is also commensurable in length with each of DE and EG [Prop. 10.15]. And GD (is) rational, and incommensurable in length with AC. Thus, DE and EG (are) each also rational, and incommensurable in length with AC [Prop. 10.13]. DH and EK are thus each medial (areas) [Prop. 10.21]. And since AG and GD are commensurable in square only, AG is thus incommensurable in length with GD. But, AG is commensurable in length with AF , and DG with EG. Thus, AF is incommensurable in length with EG [Prop. 10.13]. And as AF (is) to EG, so AI is to EK [Prop. 6.1]. Thus, AI is incommensurable with EK [Prop. 10.11]. Therefore, let the square LM , equal to AI, have been constructed. And let N O, equal to F K, which is about the same angle as LM , have been subtracted (from LM ). Thus, LM and N O are about the same diagonal [Prop. 6.26]. Let P R be their (common) diagonal, and let the (rest of the) figure have been drawn. Therefore, since the (rectangle contained) by AF and F G is equal to the (square) on EG, thus as AF is to EG, so EG (is) to F G [Prop. 6.17]. But, as AF (is) to EG, so AI is to EK [Prop. 6.1]. And as EG (is) to F G, so EK is to F K [Prop. 6.1]. And thus as AI (is) to EK, so EK (is) to F K [Prop. 5.11]. Thus, EK is the mean proportional to AI and F K. And M N is also the mean proportional to the squares LM and N O [Prop. 10.53 lem.]. And AI is equal to LM , and F K to N O. Thus, EK is also equal to M N . But, M N is equal to LO, and EK [is] equal to DH [Prop. 1.43]. And thus the whole of DK is equal to the gnomon U V W and N O. And AK (is) also equal to LM and N O. Thus, the remainder AB is equal to ST —that is to say, to the square on LN . Thus, LN is the square-root of area AB. I say that LN is the second apotome of a

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τοµή ™στι δευτέρα· κሠδύναται τÕ ΑΒ χωρίον. medial (straight-line). `Η ¥ρα τÕ ΑΒ χωρίον δυναµένη µέσης ¢ποτοµή ™στι For since AI and F K were shown (to be) medial (arδευτέρα· Óπερ œδει δε‹ξαι. eas), and are equal to the (squares) on LP and P N (respectively), the (squares) on each of LP and P N (are) thus also medial. Thus, LP and P N (are) each medial (straight-lines). And since AI is commensurable with F K [Props. 6.1, 10.11], the (square) on LP (is) thus also commensurable with the (square) on P N . Again, since AI was shown (to be) incommensurable with EK, LM is thus also incommensurable with M N —that is to say, the (square) on LP with the (rectangle contained) by LP and P N . Hence, LP is also incommensurable in length with P N [Props. 6.1, 10.11]. Thus, LP and P N are medial (straight-lines which are) commensurable in square only. So, I say that they also contain a medial (area). For since EK was shown (to be) a medial (area), and is equal to the (rectangle contained) by LP and P N , the (rectangle contained) by LP and P N is thus also medial. Hence, LP and P N are medial (straight-lines which are) commensurable in square only, and which contain a medial (area). Thus, LN is the second apotome of a medial (straight-line) [Prop. 10.75]. And it is the square-root of area AB. Thus, the square-root of area AB is the second apotome of a medial (straight-line). (Which is) the very thing it was required to show.

&δ΄.

Proposition 94

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠ¢ποτοµÁς τετάρτης, ¹ τÕ χωρίον δυναµένη ™λάσσων ™στίν.

If an area is contained by a rational (straight-line) and a fourth apotome then the square-root of the area is a minor (straight-line). E F G N A D L O V

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Μ Τ Χωρίον γ¦ρ τÕ ΑΒ περιεχέσθω ØπÕ ·ητÁς τÁς ΑΓ κሠ¢ποτοµÁς τετάρτης τÁς Α∆· λέγω, Óτι ¹ τÕ ΑΒ χωρίον δυναµένη ™λάσσων ™στίν. ”Εστω γ¦ρ τÍ Α∆ προσαρµόζουσα ¹ ∆Η· αƒ ¥ρα ΑΗ, Η∆ ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ ΑΗ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ΑΓ µήκει, ¹ δ Óλη ¹ ΑΗ τÁς προσαρµοζούσης τÁς ∆Η µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ µήκει. ™πεˆ οâν ¹ ΑΗ τÁς Η∆ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ µήκει, ™¦ν ¥ρα τù τετάρτJ µέρει τοà ¢πÕ τÁς ∆Η ‡σον

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M T For let the area AB have been contained by the rational (straight-line) AC and the fourth apotome AD. I say that the square-root of area AB is a minor (straightline). For let DG be an attachment to AD. Thus, AG and DG are rational (straight-lines which are) commensurable in square only [Prop. 10.73], and AG is commensurable in length with the (previously) laid down rational (straight-line) AC, and the square on the whole, AG, is greater than (the square on) the attachment, DG, by the square on (some straight-line) incommensurable

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παρ¦ τ¾ν ΑΗ παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, ε„ς ¢σύµµετρα αÙτ¾ν διελε‹. τετµήσθω οâν ¹ ∆Η δίχα κατ¦ τÕ Ε, κሠτù ¢πÕ τÁς ΕΗ ‡σον παρ¦ τ¾ν ΑΗ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ, κሠœστω τÕ ØπÕ τîν ΑΖ, ΖΗ· ¢σύµµετρος ¥ρα ™στˆ µήκει ¹ ΑΖ τÍ ΖΗ. ½χθωσαν οâν δι¦ τîν Ε, Ζ, Η παράλληλοι τα‹ς ΑΓ, Β∆ αƒ ΕΘ, ΖΙ, ΗΚ. ™πεˆ οâν ·ητή ™στιν ¹ ΑΗ κሠσύµµετρος τÍ ΑΓ µήκει, ·ητÕν ¥ρα ™στˆν Óλον τÕ ΑΚ. πάλιν, ™πεˆ ¢σύµµετρός ™στιν ¹ ∆Η τÍ ΑΓ µήκει, καί ε„σιν ¢µφότεραι ·ηταί, µέσον ¥ρα ™στˆ τÕ ∆Κ. πάλιν, ™πεˆ ¢σύµµετρός ™στˆν ¹ ΑΖ τÍ ΖΗ µήκει, ¢σύµµετρον ¥ρα κሠτÕ ΑΙ τù ΖΚ. Συνεστάτω οâν τù µν ΑΙ ‡σον τετράγωνον τÕ ΛΜ, τù δ ΖΚ ‡σον ¢φVρήσθω περˆ τ¾ν αÙτ¾ν γωνίαν τ¾ν ØπÕ τîν ΛΟΜ τÕ ΝΞ. περˆ τ¾ν αÙτ¾ν ¥ρα διάµετόν ™στι τ¦ ΛΜ, ΝΞ τετράγωνα. œστω αÙτîν διάµετος ¹ ΟΡ, κሠκαταγεγράφθω τÕ σχÁµα. ™πεˆ οâν τÕ ØπÕ τîν ΑΖ, ΖΗ ‡σον ™στˆ τù ¢πÕ τÁς ΕΗ, ¢νάλογον ¥ρα ™στˆν æς ¹ ΑΖ πρÕς τ¾ν ΕΗ, οÛτως ¹ ΕΗ πρÕς τ¾ν ΖΗ. ¢λλ' æς µν ¹ ΑΖ πρÕς τ¾ν ΕΗ, οÛτως ™στˆ τÕ ΑΙ πρÕς τÕ ΕΚ, æς δ ¹ ΕΗ πρÕς τ¾ν ΖΗ, οÛτως ™στˆ τÕ ΕΚ πρÕς τÕ ΖΚ· τîν ¥ρα ΑΙ, ΖΚ µέσον ¢νάλογόν ™στι τÕ ΕΚ. œστι δ κሠτîν ΛΜ, ΝΞ τετραγώνων µέσον ¢νάλογον τÕ ΜΝ, καί ™στιν ‡σον τÕ µν ΑΙ τù ΛΜ, τÕ δ ΖΚ τù ΝΞ· κሠτÕ ΕΚ ¥ρα ‡σον ™στˆ τù ΜΝ. ¢λλ¦ τù µν ΕΚ ‡σον ™στˆ τÕ ∆Θ, τù δ ΜΝ ‡σον ™στˆ τÕ ΛΞ· Óλον ¥ρα τÕ ∆Κ ‡σον ™στˆ τù ΥΦΧ γνώµονι κሠτù ΝΞ. ™πεˆ οÙν Óλον τÕ ΑΚ ‡σον ™στˆ το‹ς ΛΜ, ΝΞ τετραγώνοις, ïν τÕ ∆Κ ‡σον ™στˆ τù ΥΦΧ γνώµονι κሠτù ΝΞ τετραγώνJ, λοιπÕν ¥ρα τÕ ΑΒ ‡σον ™στˆ τù ΣΤ, τουτέστι τù ¢πÕ τÁς ΛΝ τετραγώνJ· ¹ ΛΝ ¥ρα δύναται τÕ ΑΒ χωρίον. λέγω, Óτι ¹ ΛΝ ¥λογός ™στιν ¹ καλουµένη ™λάσσων. 'Επεˆ γ¦ρ ·ητόν ™στι τÕ ΑΚ καί ™στιν ‡σον το‹ς ¢πÕ τîν ΛΟ, ΟΝ τετράγωνοις, τÕ ¥ρα συγκείµενον ™κ τîν ¢πÕ τîν ΛΟ, ΟΝ ·ητόν ™στιν. πάλιν, ™πεˆ τÕ ∆Κ µέσον ™στίν, καί ™στιν ‡σον τÕ ∆Κ τù δˆς ØπÕ τîν ΛΟ, ΟΝ, τÕ ¥ρα δˆς ØπÕ τîν ΛΟ, ΟΝ µέσον ™στίν. κሠ™πεˆ ¢σύµµετρον ™δείχθη τÕ ΑΙ τù ΖΚ, ¢σύµµετρον ¥ρα κሠτÕ ¢πÕ τÁς ΛΟ τετράγωνον τù ¢πÕ τÁς ΟΝ τετραγώνJ. αƒ ΛΟ, ΟΝ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων ·ητόν, τÕ δ δˆς Øπ' αÙτîν µέσον. ¹ ΛΝ ¥ρα ¥λογός ™στιν ¹ καλουµένη ™λάσσων· κሠδύναται τÕ ΑΒ χωρίον. `Η ¥ρα τÕ ΑΒ χωρίον δυναµένη ™λάσσων ™στίν· Óπερ œδει δε‹ξαι.

in length with (AG) [Def. 10.14]. Therefore, since the square on AG is greater than (the square on) GD by the (square) on (some straight-line) incommensurable in length with (AG), thus if (some area), equal to the fourth part of the (square) on DG, is applied to AG, falling short by a square figure, then it divides (AG) into (parts which are) incommensurable (in length) [Prop. 10.18]. Therefore, let DG have been cut in half at E, and let (some area), equal to the (square) on EG, have been applied to AG, falling short by a square figure, and let it be the (rectangle contained) by AF and F G. Thus, AF is incommensurable in length with F G. Therefore, let EH, F I, and GK have been drawn through E, F , and G (respectively), parallel to AC and BD. Therefore, since AG is rational, and commensurable in length with AC, the whole (area) AK is thus rational [Prop. 10.19]. Again, since DG is incommensurable in length with AC, and both are rational (straight-lines), DK is thus a medial (area) [Prop. 10.21]. Again, since AF is incommensurable in length with F G, AI (is) thus also incommensurable with F K [Props. 6.1, 10.11]. Therefore, let the square LM , equal to AI, have been constructed. And let N O, equal to F K, (and) about the same angle, LP M , have been subtracted (from LM ). Thus, the squares LM and N O are about the same diagonal [Prop. 6.26]. Let P R be their (common) diagonal, and let the (rest of the) figure have been drawn. Therefore, since the (rectangle contained) by AF and F G is equal to the (square) on EG, thus, proportionally, as AF is to EG, so EG (is) to F G [Prop. 6.17]. But, as AF (is) to EG, so AI is to EK, and as EG (is) to F G, so EK is to F K [Prop. 6.1]. Thus, EK is the mean proportional to AI and F K [Prop. 5.11]. And M N is also the mean proportional to the squares LM and N O [Prop. 10.13 lem.], and AI is equal to LM , and F K to N O. EK is thus also equal to M N . But, DH is equal to EK, and LO is equal to M N [Prop. 1.43]. Thus, the whole of DK is equal to the gnomon U V W and N O. Therefore, since the whole of AK is equal to the (sum of the) squares LM and N O, of which DK is equal to the gnomon U V W and the square N O, the remainder AB is thus equal to ST —that is to say, to the square on LN . Thus, LN is the square-root of area AB. I say that LN is the irrational (straight-line which is) called minor. For since AK is rational, and is equal to the (sum of the) squares LP and P N , the sum of the (squares) on LP and P N is thus rational. Again, since DK is medial, and DK is equal to twice the (rectangle contained) by LP and P N , thus twice the (rectangle contained) by LP and P N is medial. And since AI was shown (to be) incommensurable with F K, the square on LP (is) thus

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ELEMENTS BOOK 10 also incommensurable with the square on P N . Thus, LP and P N are (straight-lines which are) incommensurable in square, making the sum of the squares on them rational, and twice the (rectangle contained) by them medial. LN is thus the irrational (straight-line) called minor [Prop. 10.76]. And it is the square-root of area AB. Thus, the square-root of area AB is a minor (straightline). (Which is) the very thing it was required to show.

&ε΄.

Proposition 95

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠ¢ποτοµÁς πέµπτης, ¹ τÕ χωρίον δυναµένη [¹] µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσά ™στιν.

If an area is contained by a rational (straight-line) and a fifth apotome then the square-root of the area is that (straight-line) which with a rational (area) makes a medial whole. E F G N A D L O V

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Μ Τ Χωρίον γ¦ρ τÕ ΑΒ περιεχέσθω ØπÕ ·ητÁς τÁς ΑΓ κሠ¢ποτοµÁς πέµπτης τÁς Α∆· λέγω, Óτι ¹ τÕ ΑΒ χωρίον δυναµένη [¹] µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσά ™στιν. ”Εστω γ¦ρ τÍ Α∆ προσαρµόζουσα ¹ ∆Η· αƒ ¥ρα ΑΗ, Η∆ ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ προσαρµόζουσα ¹ Η∆ σύµµετρός ™στι µήκει τÍ ™κκειµένV ·ητÍ τÍ ΑΓ, ¹ δ Óλη ¹ ΑΗ τÁς προσαρµοζούσης τÁς ∆Η µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. ™¦ν ¥ρα τù τετάρτJ µέρει τοà ¢πÕ τÁς ∆Η ‡σον παρ¦ τ¾ν ΑΗ παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, ε„ς ¢σύµµετρα αÙτ¾ν διελε‹. τετµήσθω οâν ¹ ∆Η δίχα κατ¦ τÕ Ε σηµε‹ον, κሠτù ¢πÕ τÁς ΕΗ ‡σον παρ¦ τ¾ν ΑΗ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ κሠœστω τÕ ØπÕ τîν ΑΖ, ΖΗ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΖ τÍ ΖΗ µήκει. κሠ™πεˆ ¢σύµµετρός ™στˆν ¹ ΑΗ τÍ ΓΑ µήκει, καί ε„σιν ¢µφότεραι ·ηταί, µέσον ¥ρα ™στˆ τÕ ΑΚ. πάλιν, ™πεˆ ·ητή ™στιν ¹ ∆Η κሠσύµµετρος τÍ ΑΓ µήκει, ·ητόν ™στι τÕ ∆Κ. Συνεστάτω οâν τù µν ΑΙ ‡σον τετράγωνον τÕ ΛΜ, τù δ ΖΚ ‡σον τετράγωνον ¢φVρήσθω τÕ ΝΞ περˆ τ¾ν αÙτ¾ν γωνίαν τ¾ν ØπÕ ΛΟΜ· περˆ τ¾ν αÙτ¾ν ¥ρα διάµετρόν ™στι τ¦ ΛΜ, ΝΞ τετράγωνα. œστω αÙτîν διάµετρος ¹ ΟΡ, κሠκαταγεγράφθω τÕ σχÁµα. еοίως δ¾ δείξοµεν, Óτι ¹ ΛΝ δύναται τÕ ΑΒ χωρίον. λέγω, Óτι ¹ ΛΝ ¹ µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσά ™στιν. 'Επεˆ γ¦ρ µέσον ™δείχθη τÕ ΑΚ καί ™στιν ‡σον το‹ς ¢πÕ τîν ΛΟ, ΟΝ, τÕ ¥ρα συγκείµενον ™κ τîν ¢πÕ

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M T For let the area AB have been contained by the rational (straight-line) AC and the fifth apotome AD. I say that the square-root of area AB is that (straight-line) which with a rational (area) makes a medial whole. For let DG be an attachment to AD. Thus, AG and DG are rational (straight-lines which are) commensurable in square only [Prop. 10.73], and the attachment GD is commensurable in length the the (previously) laid down rational (straight-line) AC, and the square on the whole, AG, is greater than (the square on) the attachment, DG, by the (square) on (some straight-line) incommensurable (in length) with (AG) [Def. 10.15]. Thus, if (some area), equal to the fourth part of the (square) on DG, is applied to AG, falling short by a square figure, then it divides (AG) into (parts which are) incommensurable (in length) [Prop. 10.18]. Therefore, let DG have been divided in half at point E, and let (some area), equal to the (square) on EG, have been applied to AG, falling short by a square figure, and let it be the (rectangle contained) by AF and F G. Thus, AF is incommensurable in length with F G. And since AG is incommensurable in length with CA, and both are rational (straight-lines), AK is thus a medial (area) [Prop. 10.21]. Again, since DG is rational, and commensurable in length with AC, DK is a rational (area) [Prop. 10.19]. Therefore, let the square LM , equal to AI, have been constructed. And let the square N O, equal to F K, (and) about the same angle, LP M , have been subtracted (from

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τîν ΛΟ, ΟΝ µέσον ™στίν. πάλιν, ™πεˆ ·ητόν ™στι τÕ ∆Κ καί ™στιν ‡σον τù δˆς ØπÕ τîν ΛΟ, ΟΝ, κሠαØτÕ ·ητόν ™στιν. κሠ™πεˆ ¢σύµµετρόν ™στι τÕ ΑΙ τù ΖΚ, ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς ΛΟ τù ¢πÕ τÁς ΟΝ· αƒ ΛΟ, ΟΝ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον, τÕ δ δˆς Øπ' αÙτîν ·ητόν. ¹ λοιπ¾ ¥ρα ¹ ΛΝ ¥λογός ™στιν ¹ καλουµένη µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσα· κሠδύναται τÕ ΑΒ χωρίον. `Η τÕ ΑΒ ¥ρα χωρίον δυναµένη µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσά ™στιν· Óπερ œδει δε‹ξαι.

N O). Thus, the squares LM and N O are about the same diagonal [Prop. 6.26]. Let P R be their (common) diagonal, and let (the rest of) the figure have been drawn. So, similarly (to the previous propositions), we can show that LN is the square-root of area AB. I say that LN is that (straight-line) which with a rational (area) makes a medial whole. For since AK was shown (to be) a medial (area), and is equal to (the sum of) the squares on LP and P N , the sum of the (squares) on LP and P N is thus medial. Again, since DK is rational, and is equal to twice the (rectangle contained) by LP and P N , (the latter) is also rational. And since AI is incommensurable with F K, the (square) on LP is thus also incommensurable with the (square) on P N . Thus, LP and P N are (straight-lines which are) incommensurable in square, making the sum of the squares on them medial, and twice the (rectangle contained) by them rational. Thus, the remainder LN is the irrational (straight-line) called that which with a rational (area) makes a medial whole [Prop. 10.77]. And it is the square-root of area AB. Thus, the square-root of area AB is that (straightline) which with a rational (area) makes a medial whole. (Which is) the very thing it was required to show.

&$΄.

Proposition 96

'Ε¦ν χωρίον περιέχηται ØπÕ ·ητÁς κሠ¢ποτοµÁς ›κτης, ¹ τÕ χωρίον δυναµένη µετ¦ µέσου µέσον τÕ Óλον ποιοàσά ™στιν.

If an area is contained by a rational (straight-line) and a sixth apotome then the square-root of the area is that (straight-line) which with a medial (area) makes a medial whole. E F G N A D L O V

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Μ Τ Χωρίον γ¦ρ τÕ ΑΒ περιεχέσθω ØπÕ ·ητÁς τÁς ΑΓ κሠ¢ποτοµÁς ›κτης τÁς Α∆· λέγω, Óτι ¹ τÕ ΑΒ χωρίον δυναµένη [¹] µετ¦ µέσου µέσον τÕ Óλον ποιοàσά ™στιν. ”Εστω γ¦ρ τÍ Α∆ προσαρµόζουσα ¹ ∆Η· αƒ ¥ρα ΑΗ, Η∆ ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠοÙδετέρα αÙτîν σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ΑΓ µήκει, ¹ δ Óλη ¹ ΑΗ τÁς προσαρµοζούσης τÁς ∆Η µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ µήκει. ™πεˆ οâν ¹ ΑΗ τÁς Η∆ µε‹ζον δύναται τù ¢πÕ ¡συµµέτρου ™αυτÍ µήκει, ˜¦ν ¥ρα τù τετάρτJ µέρει τοà ¢πÕ τÁς ∆Η ‡σον παρ¦ τ¾ν ΑΗ παραβληθÍ ™λλε‹πον ε‡δει τετραγώνJ, ε„ς ¢σύµµετρα αÙτ¾ν διελε‹. τετµήσθω οâν ¹

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M T For let the area AB have been contained by the rational (straight-line) AC and the sixth apotome AD. I say that the square-root of area AB is that (straight-line) which with a medial (area) makes a medial whole. For let DG be an attachment to AD. Thus, AG and DG are rational (straight-lines which are) commensurable in square only [Prop. 10.73], and neither of them is commensurable in length with the (previously) laid down rational (straight-line) AC, and the square on the whole, AG, is greater than (the square on) the attachment, DG, by the (square) on (some straight-line) incommensurable in length with (AG) [Def. 10.16]. Therefore, since the

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∆Η δίχα κατ¦ τÕ Ε [σηµε‹ον], κሠτù ¢πÕ τÁς ΕΗ ‡σον παρ¦ τ¾ν ΑΗ παραβεβλήσθω ™λλε‹πον ε‡δει τετραγώνJ, κሠœστω τÕ ØπÕ τîν ΑΖ, ΖΗ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΖ τÍ ΖΗ µήκει. æς δ ¹ ΑΖ πρÕς τ¾ν ΖΗ, οÛτως ™στˆ τÕ ΑΙ πρÕς τÕ ΖΚ· ¢σύµµετρον ¥ρα ™στˆ τÕ ΑΙ τù ΖΚ. κሠ™πεˆ αƒ ΑΗ, ΑΓ ·ηταί ε„σι δυνάµει µόνον σύµµετροι, µέσον ™στˆ τÕ ΑΚ. πάλιν, ™πεˆ αƒ ΑΓ, ∆Η ·ηταί ε„σι κሠ¢σύµµετροι µήκει, µέσον ™στˆ κሠτÕ ∆Κ. ™πεˆ οâν αƒ ΑΗ, Η∆ δυνάµει µόνον σύµµετροί ε„σιν, ¢σύµµετρος ¥ρα ™στˆν ¹ ΑΗ τÍ Η∆ µήκει. æς δ ¹ ΑΗ πρÕς τ¾ν Η∆, οÛτως ™στˆ τÕ ΑΚ πρÕς τÕ Κ∆· ¢σύµµετρον ¥ρα ™στˆ τÕ ΑΚ τù Κ∆. Συνεστάτω οâν τù µν ΑΙ ‡σον τετράγωνον τÕ ΛΜ, τù δ ΖΚ ‡σον ¢φVρήσθω περˆ τ¾ν αÙτ¾ν γωνίαν τÕ ΝΞ· περˆ τ¾ν αÙτ¾ν ¥ρα διάµετρόν ™στι τ¦ ΛΜ, ΝΞ τετράγωνα. œστω αÙτîν διάµετρος ¹ ΟΡ, κሠκαταγεγράφθω τÕ σχÁµα. еοίως δ¾ το‹ς ™πάνω δε‹ξοµεν, Óτι ¹ ΛΝ δύναται τÕ ΑΒ χωρίον. λέγω, Óτι ¹ ΛΝ [¹] µετ¦ µέσου µέσον τÕ Óλον ποιοàσά ™στιν. 'Επεˆ γ¦ρ µέσον ™δείχθη τÕ ΑΚ καί ™στιν ‡σον το‹ς ¢πÕ τîν ΛΟ, ΟΝ, τÕ ¥ρα συγκείµενον ™κ τîν ¢πÕ τîν ΛΟ, ΟΝ µέσον ™στίν. πάλιν, ™πεˆ µέσον ™δείχθη τÕ ∆Κ καί ™στιν ‡σον τù δˆς ØπÕ τîν ΛΟ, ΟΝ, κሠτÕ δˆς ØπÕ τîν ΛΟ, ΟΝ µέσον ™στίν. κሠ™πεˆ ¢σύµµετρον ™δείχθη τÕ ΑΚ τù ∆Κ, ¢σύµµετρα [¥ρα] ™στˆ κሠτ¦ ¢πÕ τîν ΛΟ, ΟΝ τετράγωνα τù δˆς ØπÕ τîν ΛΟ, ΟΝ. κሠ™πεˆ ¢σύµµετρόν ™στι τÕ ΑΙ τù ΖΚ, ¢σύµµετρον ¥ρα κሠτÕ ¢πÕ τÁς ΛΟ τù ¢πÕ τÁς ΟΝ· αƒ ΛΟ, ΟΝ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τό τε συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον κሠτÕ δˆς Øπ' αÙτîν µέσον œτι τε τ¦ ¢π' αÙτîν τετράγωνα ¢σύµµετρα τù δˆς Øπ' αÙτîν. ¹ ¥ρα ΛΝ ¥λογός ™στιν ¹ καλουµέµη µετ¦ µέσου µέσον τÕ Óλον ποιοàσα· κሠδύναται τÕ ΑΒ χωρίον. `Η ¥ρα τÕ χωρίον δυναµένη µετ¦ µέσου µέσον τÕ Óλον ποιοàσά ™στιν· Óπερ œδει δε‹ξαι.

square on AG is greater than (the square on) GD by the (square) on (some straight-line) incommensurable in length with (AG), thus if (some area), equal to the fourth part of square on DG, is applied to AG, falling short by a square figure, then it divides (AG) into (parts which are) incommensurable (in length) [Prop. 10.18]. Therefore, let DG have been cut in half at [point] E. And let (some area), equal to the (square) on EG, have been applied to AG, falling short by a square figure. And let it be the (rectangle contained) by AF and F G. AF is thus incommensurable in length with F G. And as AF (is) to F G, so AI is to F K [Prop. 6.1]. Thus, AI is incommensurable with F K [Prop. 10.11]. And since AG and AC are rational (straight-lines which are) commensurable in square only, AK is a medial (area) [Prop. 10.21]. Again, since AC and DG are rational (straight-lines which are) incommensurable in length, DK is also a medial (area) [Prop. 10.21]. Therefore, since AG and GD are commensurable in square only, AG is thus incommensurable in length with GD. And as AG (is) to GD, so AK is to KD [Prop. 6.1]. Thus, AK is incommensurable with KD [Prop. 10.11]. Therefore, let the square LM , equal to AI, have been constructed. And let N O, equal to F K, (and) about the same angle, have been subtracted (from LM ). Thus, the squares LM and N O are about the same diagonal [Prop. 6.26]. Let P R be their (common) diagonal, and let (the rest of) the figure have been drawn. So, similarly to the above, we can show that LN is the square-root of area AB. I say that LN is that (straight-line) which with a medial (area) makes a medial whole. For since AK was shown (to be) a medial (area), and is equal to the (sum of the) squares on LP and P N , the sum of the (squares) on LP and P N is medial. Again, since DK was shown (to be) a medial (area), and is equal to twice the (rectangle contained) by LP and P N , twice the (rectangle contained) by LP and P N is also medial. And since AK was shown (to be) incommensurable with DK, [thus] the (sum of the) squares on LP and P N is also incommensurable with twice the (rectangle contained) by LP and P N . And since AI is incommensurable with F K, the (square) on LP (is) thus also incommensurable with the (square) on P N . Thus, LP and P N are (straight-lines which are) incommensurable in square, making the sum of the squares on them medial, and twice the (rectangle contained) by medial, and, furthermore, the (sum of the) squares on them incommensurable with twice the (rectangle contained) by them. Thus, LN is the irrational (straight-line) called that which with a medial (area) makes a medial whole [Prop. 10.78]. And it is the square-root of area AB.

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ELEMENTS BOOK 10 Thus, the square-root of area (AB) is that (straightline) which with a medial (area) makes a medial whole. (Which is) the very thing it was required to show.

&ζ΄.

Proposition 97

ΤÕ ¢πÕ ¢ποτοµÁς παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν πρώτην.

The (square) on an apotome, applied to a rational (straight-line), produces a first apotome as breadth.

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”Εστω ¢ποτοµ¾ ¾ ΑΒ, ·ητ¾ δ ¹ Γ∆, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΕ πλάτος ποιοàν τ¾ν ΓΖ· λέγω, Óτι ¹ ΓΖ ¢ποτοµή ™στι πρώτη. ”Εστω γ¦ρ τÍ ΑΒ προσαρµόζουσα ¹ ΒΗ· αƒ ¥ρα ΑΗ, ΗΒ ·ηταί ε„σι δυνάµει µόνον σύµµετροι. κሠτù µν ¢πÕ τÁς ΑΗ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΘ, τù δ ¢πÕ τÁς ΒΗ τÕ ΚΛ. Óλον ¥ρα τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ· ïν τÕ ΓΕ ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ· λοιπÕν ¥ρα τÕ ΖΛ ‡σον ™στˆ τù δˆς ØπÕ τîν ΑΗ, ΗΒ. τετµήσθω ¹ ΖΜ δίχα κατ¦ τÕ Ν σηµε‹ον, κሠ½χθω δι¦ τοà Ν τÍ Γ∆ παράλληλος ¹ ΝΞ· ˜κάτερον ¥ρα τîν ΖΞ, ΛΝ ‡σον ™στˆ τù ØπÕ τîν ΑΗ, ΗΒ. κሠ™πεˆ τ¦ ¢πÕ τîν ΑΗ, ΗΒ ·ητά ™στιν, καί εστι το‹ς ¢πÕ τîν ΑΗ, ΗΒ ‡σον τÕ ∆Μ, ·ητÕν ¥ρα ™στˆ τÕ ∆Μ. κሠπαρ¦ ·ητ¾ν τ¾ν Γ∆ παραβέβληται πλάτος ποιοàν τ¾ν ΓΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΓΜ κሠσύµµετρος τÍ Γ∆ µήκει. πάλιν, ™πεˆ µέσον ™στˆ τÕ δˆς ØπÕ τîν ΑΗ, ΗΒ, κሠτù δˆς ØπÕ τîν ΑΗ, ΗΒ ‡σον τÕ ΖΛ, µέσον ¥ρα τÕ ΖΛ. κሠπαρ¦ ·ητ¾ν τ¾ν Γ∆ παράκειται πλάτος ποιοàν τ¾ν ΖΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΖΜ κሠ¢σύµµετρος τÍ Γ∆ µήκει. κሠ™πεˆ τ¦ µν ¢πÕ τîν ΑΗ, ΗΒ ·ητά ™στιν, τÕ δ δˆς ØπÕ τîν ΑΗ, ΗΒ µέσον, ¢σύµµετρα ¥ρα ™στˆ τ¦ ¢πÕ τîν ΑΗ, ΗΒ τù δˆς ØπÕ τîν ΑΗ, ΗΒ. κሠτο‹ς µν ¢πÕ τîν ΑΗ, ΗΒ ‡σον ™στˆ τÕ ΓΛ, τù δ δˆς ØπÕ τîν ΑΗ, ΗΒ τÕ ΖΛ· ¢σύµµετρον ¥ρα ™στˆ τÕ ∆Μ τù ΖΛ. æς δ τÕ ∆Μ πρÕς τÕ ΖΛ, οÛτως ™στˆν ¹ ΓΜ πρÕς τ¾ν ΖΜ. ¢σύµµετρος ¥ρα ™στˆν ¹ ΓΜ τÍ ΖΜ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ¥ρα ΓΜ, ΜΖ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¹ ΓΖ ¥ρα ¢ποτοµή ™στιν. λέγω δή, Óτι κሠπρώτη. 'Επεˆ γ¦ρ τîν ¢πÕ τîν ΑΗ, ΗΒ µέσον ¢νάλογόν ™στι τÕ ØπÕ τîν ΑΗ, ΗΒ, καί ™στι τù µν ¢πÕ τÁς ΑΗ ‡σον τÕ ΓΘ, τù δ ¢πÕ τÁς ΒΗ ‡σον τÕ ΚΛ, τù δ

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Let AB be an apotome, and CD a rational (straightline). And let CE, equal to the (square) on AB, have been applied to CD, producing CF as breadth. I say that CF is a first apotome. For let BG be an attachment to AB. Thus, AG and GB are rational (straight-lines which are) commensurable in square only [Prop. 10.73]. And let CH, equal to the (square) on AG, and KL, (equal) to the (square) on BG, have been applied to CD. Thus, the whole of CL is equal to the (sum of the squares) on AG and GB, of which CE is equal to the (square) on AB. The remainder F L is thus equal to twice the (rectangle contained) by AG and GB [Prop. 2.7]. Let F M have been cut in half at point N . And let N O have been drawn through N , parallel to CD. Thus, F O and LN are each equal to the (rectangle contained) by AG and GB. And since the (sum of the squares) on AG and GB is rational, and DM is equal to the (sum of the squares) on AG and GB, DM is thus rational. And it has been applied to the rational (straight-line) CD, producing CM as breadth. Thus, CM is rational, and commensurable in length with CD [Prop. 10.20]. Again, since twice the (rectangle contained) by AG and GB is medial, and F L (is) equal to twice the (rectangle contained) by AG and GB, F L (is) thus a medial (area). And it is applied to the rational (straight-line) CD, producing F M as breadth. F M is thus rational, and incommensurable in length with CD [Prop. 10.22]. And since the (sum of the squares) on AG and GB is rational, and twice the (rectangle contained) by AG and GB medial, the (sum of the squares) on AG and GB is thus incommensurable with twice the (rectangle contained) by AG and GB. And CL is equal to the (sum of the squares) on AG and GB, and F L to twice the

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ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

ÙπÕ τîν ΑΗ, ΗΒ τÕ ΝΛ, κሠτîν ΓΘ, ΚΛ ¥ρα µέσον ¢νάλογόν ™στι τÕ ΝΛ· œστιν ¥ρα æς τÕ ΓΘ πρÕς τÕ ΝΛ, οÛτως τÕ ΝΛ πρÕς τÕ ΚΛ. ¢λλ' æς µν τÕ ΓΘ πρÕς τÕ ΝΛ, οÛτως ™στˆν ¹ ΓΚ πρÕς τ¾ν ΝΜ· æς δ τÕ ΝΛ πρÕς τÕ ΚΛ, οÛτως ™στˆν ¹ ΝΜ πρÕς τ¾ν ΚΜ· τÕ ¥ρα ØπÕ τîν ΓΚ, ΚΜ ‡σον ™στˆ τù ¢πÕ τÁς ΝΜ, τουτέστι τù τετάρτJ µέρει τοà ¢πÕ τÁς ΖΜ. κሠεπεˆ σύµµετρόν ™στι τÕ ¢πÕ τÁς ΑΗ τù ¢πÕ τÁς ΗΒ, σύµµετρόν [™στι] κሠτÕ ΓΘ τù ΚΛ. æς δ τÕ ΓΘ πρÕς τÕ ΚΛ, οÛτως ¹ ΓΚ πρÕς τ¾ν ΚΜ· σύµµετρος ¥ρα ™στˆν ¹ ΓΚ τÍ ΚΜ. ™πεˆ οâν δύο εÙθε‹αι ¥νισοί ε„σιν αƒ ΓΜ, ΜΖ, κሠτù τετάρτJ µέρει τοà ¢πÕ τÁς ΖΜ ‡σον παρ¦ τ¾ν ΓΜ παραβέβληται ™λλε‹πον ε‡δει τετραγώνJ τÕ ØπÕ τîν ΓΚ, ΚΜ, καί ™στι σύµµετρος ¹ ΓΚ τÍ ΚΜ, ¹ ¥ρα ΓΜ τÁς ΜΖ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει. καί ™στιν ¹ ΓΜ σύµµετρος τÍ ™κκειµένV ·ητÍ τÍ Γ∆ µήκει· ¹ ¥ρα ΓΖ ¢ποτοµή ™στι πρώτη. ΤÕ ¥ρα ¢πÕ ¢ποτοµÁς παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν πρώτην· Óπερ œδει δε‹ξαι.

(rectangle contained) by AG and GB. DM is thus incommensurable with F L. And as DM (is) to F L, so CM is to F M [Prop. 6.1]. CM is thus incommensurable in length with F M [Prop. 10.11]. And both are rational (straightlines). Thus, CM and M F are rational (straight-lines which are) commensurable in square only. CF is thus an apotome [Prop. 10.73]. So, I say that (it is) also a first (apotome). For since the (rectangle contained) by AG and GB is the mean proportional to the (squares) on AG and GB [Prop. 10.21 lem.], and CH is equal to the (square) on AG, and KL equal to the (square) on BG, and N L to the (rectangle contained) by AG and GB, N L is thus also the mean proportional to CH and KL. Thus, as CH is to N L, so N L (is) to KL. But, as CH (is) to N L, so CK is to N M , and as N L (is) to KL, so N M is to KM [Prop. 6.1]. Thus, the (rectangle contained) by CK and KM is equal to the (square) on N M — that is to say, to the fourth part of the (square) on F M [Prop. 6.17]. And since the (square) on AG is commensurable with the (square) on GB, CH [is] also commensurable with KL. And as CH (is) to KL, so CK (is) to KM [Prop. 6.1]. CK is thus commensurable (in length) with KM [Prop. 10.11]. Therefore, since CM and M F are two unequal straight-lines, and the (rectangle contained) by CK and KM , equal to the fourth part of the (square) on F M , has been applied to CM , falling short by a square figure, and CK is commensurable (in length) with KM , the square on CM is thus greater than (the square on) M F by the (square) on (some straight-line) commensurable in length with (CM ) [Prop. 10.17]. And CM is commensurable in length with the (previously) laid down rational (straight-line) CD. Thus, CF is a first apotome [Def. 10.15]. Thus, the (square) on an apotome, applied to a rational (straight-line), produces a first apotome as breadth. (Which is) the very thing it was required to show.

&η΄.

Proposition 98

ΤÕ ¢πÕ µέσης ¢ποτοµÁς πρώτης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν δευτέραν. ”Εστω µέσης ¢ποτοµ¾ πρώτη ¹ ΑΒ, ·ητ¾ δ ¹ Γ∆, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΕ πλάτος ποιοàν τ¾ν ΓΖ· λέγω, Óτι ¹ ΓΖ ¢ποτοµή ™στι δευτέρα. ”Εστω γ¦ρ τÍ ΑΒ προσαρµόζουσα ¹ ΒΗ· αƒ ¥ρα ΑΗ, ΗΒ µέσαι ε„σˆ δυνάµει µόνον σύµµετροι ·ητÕν περιέχουσαι. κሠτù µν ¢πÕ τÁς ΑΗ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΘ πλάτος ποιοàν τ¾ν ΓΚ, τù δ ¢πÕ τÁς ΗΒ ‡σον τÕ ΚΛ πλάτος ποιοàν τ¾ν ΚΜ· Óλον

The (square) on a first apotome of a medial (straightline), applied to a rational (straight-line), produces a second apotome as breadth. Let AB be a first apotome of a medial (straight-line), and CD a rational (straight-line). And let CE, equal to the (square) on AB, have been applied to CD, producing CF as breadth. I say that CF is a second apotome. For let BG be an attachment to AB. Thus, AG and GB are medial (straight-lines which are) commensurable in square only, containing a rational (area) [Prop. 10.74]. And let CH, equal to the (square) on AG, have been ap-

398

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

¥ρα τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ· µέσον ¥ρα κሠτÕ ΓΛ. κሠπαρ¦ ·ητ¾ν τ¾ν Γ∆ παράκειται πλάτος ποιοàν τ¾ν ΓΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΓΜ κሠ¢σύµµετρος τÍ Γ∆ µήκει. κሠ™πεˆ τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ, ïν τÕ ¢πÕ τÁς ΑΒ ‡σον ™στˆ τù ΓΕ, λοιπÕν ¥ρα τÕ δˆς ØπÕ τîν ΑΗ, ΗΒ ‡σον ™στˆ τù ΖΛ. ·ητÕν δέ [™στι] τÕ δˆς ØπÕ τîν ΑΗ, ΗΒ· ·ητÕν ¥ρα τÕ ΖΛ. κሠπαρ¦ ·ητ¾ν τ¾ν ΖΕ παράκειται πλάτος ποιοàν τ¾ν ΖΜ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΖΜ κሠσύµµετρος τÍ Γ∆ µήκει. ™πεˆ οâν τ¦ µν ¢πÕ τîν ΑΗ, ΗΒ, τουτέστι τÕ ΓΛ, µέσον ™στίν, τÕ δ δˆς ØπÕ τîν ΑΗ, ΗΒ, τουτέστι τÕ ΖΛ, ·ητόν ¢σύµµετρον ¥ρα ™στˆ τÕ ΓΛ τù ΖΛ. æς δ τÕ ΓΛ πρÕς τÕ ΖΛ, οÛτως ™στˆν ¹ ΓΜ πρÕς τ¾ν ΖΜ· ¢σύµµετρος ¥ρα ¹ ΓΜ τÍ ΖΜ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ¥ρα ΓΜ, ΜΖ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¹ ΓΖ ¥ρα ¢ποτοµή ™στιν. λέγω δή, Óτι κሠδευτέρα.

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plied to CD, producing CK as breadth, and KL, equal to the (square) on GB, producing KM as breadth. Thus, the whole of CL is equal to the (sum of the squares) on AG and GB. Thus, CL (is) also a medial (area) [Props. 10.15, 10.23 corr.]. And it is applied to the rational (straight-line) CD, producing CM as breadth. CM is thus rational, and incommensurable in length with CD [Prop. 10.22]. And since CL is equal to the (sum of the squares) on AG and GB, of which the (square) on AB is equal to CE, twice the (rectangle contained) by AG and GB is thus equal to the remainder F L [Prop. 2.7]. And twice the (rectangle contained) by AG and GB [is] rational. Thus, F L (is) rational. And it is applied to the rational (straight-line) F E, producing F M as breadth. F M is thus also rational, and commensurable in length with CD [Prop. 10.20]. Therefore, since the (sum of the squares) on AG and GB—that is to say, CL—is medial, and twice the (rectangle contained) by AG and GB— that is to say, F L—(is) rational, CL is thus incommensurable with F L. And as CL (is) to F L, so CM is to F M [Prop. 6.1]. Thus, CM (is) incommensurable in length with F M [Prop. 10.11]. And they are both rational (straight-lines). Thus, CM and M F are rational (straight-lines which are) commensurable in square only. CF is thus an apotome [Prop. 10.73]. So, I say that (it is) also a second (apotome).

A

G

Z

N K

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X J

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Τετµήσθω γ¦ρ ¹ ΖΜ δίχα κατ¦ τÕ Ν, κሠ½χθω δι¦ τοà Ν τÍ Γ∆ παράλληλος ¹ ΝΞ· ˜κάτερον ¥ρα τîν ΖΞ, ΝΛ ‡σον ™στˆ τù ØπÕ τîν ΑΗ, ΗΒ. κሠ™πεˆ τîν ¢πÕ τîν ΑΗ, ΗΒ τετραγώνων µέσον ¢νάλογόν ™στι τÕ ØπÕ τîν ΑΗ, ΗΒ, καί ™στιν ‡σον τÕ µν ¢πÕ τÁς ΑΗ τù ΓΘ, τÕ δ ØπÕ τîν ΑΗ, ΗΒ τù ΝΛ, τÕ δ ¢πÕ τÁς ΒΗ τù ΚΛ, κሠτîν ΓΘ, ΚΛ ¥ρα µέσον ¢νάλογόν ™στι τÕ ΝΛ· œστιν ¥ρα æς τÕ ΓΘ πρÕς τÕ ΝΛ, οÛτως τÕ ΝΛ πρÕς τÕ ΚΛ. ¢λλ' æς µν τÕ ΓΘ πρÕς τÕ ΝΛ, οÛτως ™στˆν ¹ ΓΚ πρÕς τ¾ν ΝΜ, æς δ τÕ ΝΛ πρÕς τÕ ΚΛ, οÛτως ™στˆν ¹ ΝΜ πρÕς τ¾ν ΜΚ· æς ¥ρα ¹ ΓΚ πρÕς τ¾ν ΝΜ, οÛτως ™στˆν ¹ ΝΜ πρÕς τ¾ν ΚΜ· τÕ ¥ρα ØπÕ τîν ΓΚ, ΚΜ ‡σον ™στˆ τù ¢πÕ τÁς ΝΜ, τουτέστι τù τετάρτJ µέρει τοà ¢πÕ τÁς ΖΜ [κሠ™πεˆ σύµµετρόν ™στι τÕ ¢πÕ τÁς ΑΗ τù ¢πÕ τÁς ΒΗ, σύµµετρόν ™στι

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For let F M have been cut in half at N . And let N O have been drawn through (point) N , parallel to CD. Thus, F O and N L are each equal to the (rectangle contained) by AG and GB. And since the (rectangle contained) by AG and GB is the mean proportional to the squares on AG and GB [Prop. 10.21 lem.], and the (square) on AG is equal to CH, and the (rectangle contained) by AG and GB to N L, and the (square) on BG to KL, N L is thus also the mean proportional to CH and KL. Thus, as CH is to N L, so N L (is) to KL [Prop. 5.11]. But, as CH (is) to N L, so CK is to N M , and as N L (is) to KL, so N M is to M K [Prop. 6.1]. Thus, as CK (is) to N M , so N M is to KM [Prop. 5.11]. The (rectangle contained) by CK and KM is thus equal to the (square) on N M [Prop. 6.17]—that is to say, to

399

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

κሠτÕ ΓΘ τù ΚΛ, τουτέστιν ¹ ΓΚ τÍ ΚΜ]. ™πεˆ οâν δύο εÙθε‹αι ¥νισοί ε„σιν αƒ ΓΜ, ΜΖ, κሠτù τετάτρJ µέρει τοà ¢πÕ τÁς ΜΖ ‡σον παρ¦ τ¾ν µείζονα τ¾ν ΓΜ παραβέβληται ™λλε‹πον ε‡δει τετραγώνJ τÕ ØπÕ τîν ΓΚ, ΚΜ κሠε„ς σύµµετρα αÙτ¾ν διαιρε‹, ¹ ¥ρα ΓΜ τÁς ΜΖ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ µήκει. καί ™στιν ¹ προσαρµόζουσα ¹ ΖΜ σύµµετρος µήκει τÍ ™κκειµένV ·ητÍ τÍ Γ∆· ¹ ¥ρα ΓΖ ¢ποτοµή ™στι δευτέρα. ΤÕ ¥ρα ¢πÕ µέσης ¢ποτοµÁς πρώτης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν δευτέραν· Óπερ œδει δε‹ξαι.

the fourth part of the (square) on F M [and since the (square) on AG is commensurable with the (square) on BG, CH is also commensurable with KL—that is to say, CK with KM ]. Therefore, since CM and M F are two unequal straight-lines, and the (rectangle contained) by CK and KM , equal to the fourth part of the (square) on M F , has been applied to the greater CM , falling short by a square figure, and divides it into commensurable (parts), the square on CM is thus greater than (the square on) M F by the (square) on (some straight-line) commensurable in length with (CM ) [Prop. 10.17]. The attachment F M is also commensurable in length with the (previously) laid down rational (straight-line) CD. CF is thus a second apotome [Def. 10.16]. Thus, the (square) on a first apotome of a medial (straight-line), applied to a rational (straight-line), produces a second apotome as breadth. (Which is) the very thing it was required to show.

&θ΄.

Proposition 99

ΤÕ ¢πÕ µέσης ¢ποτοµÁς δευτέρας παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν τρίτην.

The (square) on a second apotome of a medial (straight-line), applied to a rational (straight-line), produces a third apotome as breadth.

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”Εστω µέσης ¢ποτοµ¾ δευτέρα ¹ ΑΒ, ·ητ¾ δ ¹ Γ∆, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΕ πλάτος ποιοàν τ¾ν ΓΖ· λέγω, Óτι ¹ ΓΖ ¢ποτοµή ™στι τρίτη. ”Εστω γ¦ρ τÍ ΑΒ προσαρµόζουσα ¹ ΒΗ· αƒ ¥ρα ΑΗ, ΗΒ µέσαι ε„σˆ δυνάµει µόνον σύµµετροι µέσον περιέχουσαι. κሠτù µν ¢πÕ τÁς ΑΗ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΘ πλάτος ποιοàν τ¾ν ΓΚ, τù δ ¢πÕ τÁς ΒΗ ‡σον παρ¦ τ¾ν ΚΘ παραβεβλήσθω τÕ ΚΛ πλάτος ποιοàν τ¾ν ΚΜ· Óλον ¥ρα τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ [καί ™στι µέσα τ¦ ¢πÕ τîν ΑΗ, ΗΒ]· µέσον ¥ρα κሠτÕ ΓΛ. κሠπαρ¦ ·ητ¾ν τ¾ν Γ∆ παραβέβληται πλάτος ποιοàν τ¾ν ΓΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΓΜ κሠ¢σύµµετρος τÍ Γ∆ µήκει. κሠ™πεˆ Óλον τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ, ïν τÕ ΓΕ ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ, λοιπÕν ¥ρα τÕ ΛΖ ‡σον ™στˆ τù δˆς ØπÕ τîν ΑΗ, ΗΒ. τετµήσθω οâν ¹ ΖΜ δίχα κατ¦ τÕ Ν σηµε‹ον,

Let AB be the second apotome of a medial (straightline), and CD a rational (straight-line). And let CE, equal to the (square) on AB, have been applied to CD, producing CF as breadth. I say that CF is a third apotome. For let BG be an attachment to AB. Thus, AG and GB are medial (straight-lines which are) commensurable in square only, containing a medial (area) [Prop. 10.75]. And let CH, equal to the (square) on AG, have been applied to CD, producing CK as breadth. And let KL, equal to the (square) on BG, have been applied to KH, producing KM as breadth. Thus, the whole of CL is equal to the (sum of the squares) on AG and GB [and the (sum of the squares) on AG and GB is medial]. CL (is) thus also medial [Props. 10.15, 10.23 corr.]. And it has been applied to the rational (straight-line) CD, producing CM as breadth. Thus, CM is rational, and incom-

400

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

κሠτÍ Γ∆ παράλληλος ½χθω ¹ ΝΞ· ˜κάτερον ¥ρα τîν ΖΞ, ΝΛ ‡σον ™στˆ τù ØπÕ τîν ΑΗ, ΗΒ. µέσον δ τÕ ØπÕ τîν ΑΗ, ΗΒ· µέσον ¥ρα ™στˆ κሠτÕ ΖΛ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται πλάτος ποιοàν τ¾ν ΖΜ· ·ητ¾ ¥ρα κሠ¹ ΖΜ κሠ¢σύµµετρος τÍ Γ∆ µήκει. κሠ™πεˆ αƒ ΑΗ, ΗΒ δυνάµει µόνον ε„σˆ σύµµετροι, ¢σύµµετρος ¥ρα [™στˆ] µήκει ¹ ΑΗ τÍ ΗΒ· ¢σύµµετρον ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς ΑΗ τù ØπÕ τîν ΑΗ, ΗΒ. ¢λλ¦ τù µν ¢πÕ τÁς ΑΗ σύµµετρά ™στι τ¦ ¢πÕ τîν ΑΗ, ΗΒ, τù δ ØπÕ τîν ΑΗ, ΗΒ τÕ δˆς ØπÕ τîν ΑΗ, ΗΒ· ¢σύµµετρα ¥ρα ™στˆ τ¦ ¢πÕ τîν ΑΗ, ΗΒ τù δˆς ØπÕ τîν ΑΗ, ΗΒ. ¢λλ¦ το‹ς µν ¢πÕ τîν ΑΗ, ΗΒ ‡σον ™στˆ τÕ ΓΛ, τù δ δˆς ØπÕ τîν ΑΗ, ΗΒ ‡σον ™στˆ τÕ ΖΛ· ¢σύµµετρον ¥ρα ™στˆ τÕ ΓΛ τù ΖΛ. æς δ τÕ ΓΛ πρÕς τÕ ΖΛ, οÛτως ™στˆν ¹ ΓΜ πρÕς τ¾ν ΖΜ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΓΜ τÍ ΖΜ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ¥ρα ΓΜ, ΜΖ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΓΖ. λέγω δή, Óτι κሠτρίτη. 'Επεˆ γ¦ρ σύµµετρόν ™στι τÕ ¢πÕ τÁς ΑΗ τù ¢πÕ τÁς ΗΒ, σύµµετρον ¥ρα κሠτÕ ΓΘ τù ΚΛ· éστε κሠ¹ ΓΚ τÍ ΚΜ. κሠ™πεˆ τîν ¢πÕ τîν ΑΗ, ΗΒ µέσον ¢νάλογόν ™στι τÕ ØπÕ τîν ΑΗ, ΗΒ, καί ™στι τù µν ¢πÕ τÁς ΑΗ ‡σον τÕ ΓΘ, τù δ ¢πÕ τÁς ΗΒ ‡σον τÕ ΚΛ, τù δ ØπÕ τîν ΑΗ, ΗΒ ‡σον τÕ ΝΛ, κሠτîν ΓΘ, ΚΛ ¥ρα µέσον ¢νάλογόν ™στι τÕ ΝΛ· œστιν ¥ρα æς τÕ ΓΘ πρÕς τÕ ΝΛ, οÛτως τÕ ΝΛ πρÕς τÕ ΚΛ. ¢λλ' æς µν τÕ ΓΘ πρÕς τÕ ΝΛ, οÛτως ™στˆν ¹ ΓΚ πρÕς τ¾ν ΝΜ, æς δ τÕ ΝΛ πρÕς τÕ ΚΛ, οÛτως ™στˆν ¹ ΝΜ πρÕς τ¾ν ΚΜ· æς ¥ρα ¹ ΓΚ πρÕς τ¾ν ΜΝ, οÛτως ™στˆν ¹ ΜΝ πρÕς τ¾ν ΚΜ· τÕ ¥ρα ØπÕ τîν ΓΚ, ΚΜ ‡σον ™στˆ τù [¢πÕ τÁς ΜΝ, τουτέστι τù] τετάρτJ µέρει τοà ¢πÕ τÁς ΖΜ. ™πεˆ οâν δύο εÙθε‹αι ¥νισοί ε„σιν αƒ ΓΜ, ΜΖ, κሠτù τετάρτJ µέρει τοà ¢πÕ τÁς ΖΜ ‡σον παρ¦ τ¾ν ΓΜ παραβέβληται ™λλε‹πον ε‡δει τετραγώνJ κሠε„ς σύµµετρα αÙτ¾ν διαιρε‹, ¹ ΓΜ ¥ρα τÁς ΜΖ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ. κሠοÙδετέρα τîν ΓΜ, ΜΖ σύµµετρός ™στι µήκει τÍ ™κκειµένV ·ητÍ τÍ Γ∆· ¹ ¥ρα ΓΖ ¢ποτοµή ™στι τρίτη. ΤÕ ¥ρα ¢πÕ µέσης ¢ποτοµÁς δευτέρας παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν τρίτην· Óπερ œδει δε‹ξαι.

mensurable in length with CD [Prop. 10.22]. And since the whole of CL is equal to the (sum of the squares) on AG and GB, of which CE is equal to the (square) on AB, the remainder LF is thus equal to twice the (rectangle contained) by AG and GB [Prop. 2.7]. Therefore, let F M have been cut in half at point N . And let N O have been drawn parallel to CD. Thus, F O and N L are each equal to the (rectangle contained) by AG and GB. And the (rectangle contained) by AG and GB (is) medial. Thus, F L is also medial. And it is applied to the rational (straight-line) EF , producing F M as breadth. F M is thus rational, and incommensurable in length with CD [Prop. 10.22]. And since AG and GB are commensurable in square only, AG [is] thus incommensurable in length with GB. Thus, the (square) on AG is also incommensurable with the (rectangle contained) by AG and GB [Props. 6.1, 10.11]. But, the (sum of the squares) on AG and GB is commensurable with the (square) on AG, and twice the (rectangle contained) by AG and GB with the (rectangle contained) by AG and GB. The (sum of the squares) on AG and GB is thus incommensurable with twice the (rectangle contained) by AG and GB [Prop. 10.13]. But, CL is equal to the (sum of the squares) on AG and GB, and F L is equal to the (rectangle contained) by AG and GB. Thus, CL is incommensurable with F L. And as CL (is) to F L, so CM is to F M [Prop. 6.1]. CM is thus incommensurable in length with F M [Prop. 10.11]. And they are both rational (straight-lines). Thus, CM and M F are rational (straight-lines which are) commensurable in square only. CF is thus an apotome [Prop. 10.73]. So, I say that (it is) also a third (apotome). For since the (square) on AG is commensurable with the (square) on GB, CH (is) thus also commensurable with KL. Hence, CK (is) also (commensurable in length) with KM [Props. 6.1, 10.11]. And since the (rectangle contained) by AG and GB is the mean proportional to the (squares) on AG and GB [Prop. 10.21 lem.], and CH is equal to the (square) on AG, and KL equal to the (square) on GB, and N L to the (rectangle contained) by AG and GB, N L is thus also the mean proportional to CH and KL. Thus, as CH is to N L, so N L (is) to KL. But, as CH (is) to N L, so CK is to N M , and as N L (is) to KL, so N M (is) to KM [Prop. 6.1]. Thus, as CK (is) to M N , so M N is to KM [Prop. 5.11]. Thus, the (rectangle contained) by CK and KM is equal to the [(square) on M N —that is to say, to the] fourth part of the (square) on F M [Prop. 6.17]. Therefore, since CM and M F are two unequal straight-lines, and (some area), equal to the fourth part of the (square) on F M , has been applied to CM , falling short by a square figure, and di-

401

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ELEMENTS BOOK 10 vides it into commensurable (parts), the square on CM is thus greater than (the square on) M F by the (square) on (some straight-line) commensurable (in length) with (CM ) [Prop. 10.17]. And neither of CM and M F is commensurable in length with the (previously) laid down rational (straight-line) CD. CF is thus a third apotome [Def. 10.13]. Thus, the (square) on a second apotome of a medial (straight-line), applied to a rational (straight-line), produces a third apotome as breadth. (Which is) the very thing it was required to show.

ρ΄.

Proposition 100

ΤÕ ¢πÕ ™λάσσονος παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν τετάρτην.

The (square) on a minor (straight-line), applied to a rational (straight-line), produces a fourth apotome as breadth.

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”Εστω ™λάσσων ¹ ΑΒ, ·ητ¾ δ ¹ Γ∆, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ ·ητ¾ν τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΕ πλάτος ποιοàν τ¾ν ΓΖ· λέγω, Óτι ¹ ΓΖ ¢ποτοµή ™στι τετάρτη. ”Εστω γ¦ρ τÍ ΑΒ προσαρµόζουσα ¹ ΒΗ· αƒ ¥ρα ΑΗ, ΗΒ δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΑΗ, ΗΒ τετραγώνων ·ητόν, τÕ δ δˆς ØπÕ τîν ΑΗ, ΗΒ µέσον. κሠτù µν ¢πÕ τÁς ΑΗ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΘ πλάτος ποιοàν τ¾ν ΓΚ, τù δ ¢πÕ τÁς ΒΗ ‡σον τÕ ΚΛ πλάτος ποιοàν τ¾ν ΚΜ· Óλον ¥ρα τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ. καί ™στι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΗ, ΗΒ ·ητόν· ·ητÕν ¥ρα ™στˆ κሠτÕ ΓΛ. κሠπαρ¦ ·ητ¾ν τ¾ν Γ∆ παράκειται πλάτος ποιοàν τ¾ν ΓΜ· ·ητ¾ ¥ρα κሠ¹ ΓΜ κሠσύµµετρος τÍ Γ∆ µήκει. κሠ™πεˆ Óλον τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ, ïν τÕ ΓΕ ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ, λοιπÕν ¥ρα τÕ ΖΛ ‡σον ™στˆ τù δˆς ØπÕ τîν ΑΗ, ΗΒ. τετµήσθω οâν ¹ ΖΜ δίχα κατ¦ τÕ Ν σηµε‹ον, κሠ½χθω δˆα τοà Ν Ðποτέρv τîν Γ∆, ΜΛ παράλληλος ¹ ΝΞ· ˜κάτερον ¥ρα τîν ΖΞ, ΝΛ ‡σον ™στˆ τù ØπÕ τîν ΑΗ, ΗΒ. κሠ™πεˆ τÕ δˆς ØπÕ τîν ΑΗ, ΗΒ µέσον ™στˆ καί ™στιν ‡σον τù ΖΛ, κሠτÕ ΖΛ ¥ρα µέσον ™στίν. κሠπαρ¦ ·ητ¾ν τ¾ν ΖΕ παράκειται πλάτος ποιοàν τ¾ν ΖΜ· ·ητ¾ ¥ρα ™στˆν

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Let AB be a minor (straight-line), and CD a rational (straight-line). And let CE, equal to the (square) on AB, have been applied to the rational (straight-line) CD, producing CF as breadth. I say that CF is a fourth apotome. For let BG be an attachment to AB. Thus, AG and GB are incommensurable in square, making the sum of the squares on AG and GB rational, and twice the (rectangle contained) by AG and GB medial [Prop. 10.76]. And let CH, equal to the (square) on AG, have been applied to CD, producing CK as breadth, and KL, equal to the (square) on BG, producing KM as breadth. Thus, the whole of CL is equal to the (sum of the squares) on AG and GB. And the sum of the (squares) on AG and GB is rational. CL is thus also rational. And it is applied to the rational (straight-line) CD, producing CM as breadth. Thus, CM (is) also rational, and commensurable in length with CD [Prop. 10.20]. And since the whole of CL is equal to the (sum of the squares) on AG and GB, of which CE is equal to the (square) on AB, the remainder F L is thus equal to twice the (rectangle contained) by AG and GB [Prop. 2.7]. Therefore, let F M have been cut in half at point N . And let N O have been drawn through N , parallel to either of CD or M L. Thus, F O and N L are each equal to the (rectangle con-

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ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

¹ ΖΜ κሠ¢σύµµετρος τÍ Γ∆ µήκει. κሠ™πεˆ τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΑΗ, ΗΒ ·ητόν ™στιν, τÕ δ δˆς ØπÕ τîν ΑΗ, ΗΒ µέσον, ¢σύµµετρα [¥ρα] ™στˆ τ¦ ¢πÕ τîν ΑΗ, ΗΒ τù δˆς ØπÕ τîν ΑΗ, ΗΒ. ‡σον δέ [™στι] τÕ ΓΛ το‹ς ¢πÕ τîν ΑΗ, ΗΒ, τù δ δˆς ØπÕ τîν ΑΗ, ΗΒ ‡σον τÕ ΖΛ· ¢σύµµετρον ¥ρα [™στˆ] τÕ ΓΛ τù ΖΛ. æς δ τÕ ΓΛ πρÕς τÕ ΖΛ, οÛτως ™στˆν ¹ ΓΜ πρÕς τ¾ν ΜΖ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΓΜ τÍ ΜΖ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ¥ρα ΓΜ, ΜΖ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΓΖ. λέγω [δή], Óτι κሠτετάρτη. 'Επεˆ γ¦ρ αƒ ΑΗ, ΗΒ δυνάµει ε„σˆν ¢σύµµετροι, ¢σύµµετρον ¥ρα κሠτÕ ¢πÕ τÁς ΑΗ τù ¢πÕ τÁς ΗΒ. καί ™στι τù µν ¢πÕ τÁς ΑΗ ‡σον τÕ ΓΘ, τù δ ¢πÕ τÁς ΗΒ ‡σον τÕ ΚΛ· ¢σύµµετρον ¥ρα ™στˆ τÕ ΓΘ τù ΚΛ. æς δ τÕ ΓΘ πρÕς τÕ ΚΛ, οÛτως ™στˆν ¹ ΓΚ πρÕς τ¾ν ΚΜ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΓΚ τÍ ΚΜ µήκει. κሠ™πεˆ τîν ¢πÕ τîν ΑΗ, ΗΒ µέσον ¢νάλογόν ™στι τÕ ØπÕ τîν ΑΗ, ΗΒ, καί ™στιν ‡σον τÕ µν ¢πÕ τÁς ΑΗ τù ΓΘ, τÕ δ ¢πÕ τÁς ΗΒ τù ΚΛ, τÕ δ ØπÕ τîν ΑΗ, ΗΒ τù ΝΛ, τîν ¥ρα ΓΘ, ΚΛ µέσον ¢νάλογόν ™στι τÕ ΝΛ· œστιν ¥ρα æς τÕ ΓΘ πρÕς τÕ ΝΛ, οÛτως τÕ ΝΛ πρÕς τÕ ΚΛ. ¢λλ' æς µν τÕ ΓΘ πρÕς τÕ ΝΛ, οÛτως ™στίν ¹ ΓΚ πρÕς τ¾ν ΝΜ, æς δ τÕ ΝΛ πρÕς τÕ ΚΛ, οÛτως ™στˆν ¹ ΝΜ πρÕς τ¾ν ΚΜ· æς ¥ρα ¹ ΓΚ πρÕς τ¾ν ΜΝ, οÛτως ™στˆν ¹ ΜΝ πρÕς τ¾ν ΚΜ· τÕ ¥ρα ØπÕ τîν ΓΚ, ΚΜ ‡σον ™στˆ τù ¢πÕ τÁς ΜΝ, τουτέστι τù τετάρτJ µέρει τοà ¢πÕ τÁς ΖΜ. ™πεˆ οâν δύο εÙθε‹αι ¥νισοί ε„σιν αƒ ΓΜ, ΜΖ, κሠτù τετράρτJ µέρει τοà ¢πÕ τÁς ΜΖ ‡σον παρ¦ τ¾ν ΓΜ παραβέβληται ™λλε‹πον ε‡δει τετραγώνJ τÕ ØπÕ τîν ΓΚ, ΚΜ κሠε„ς ¢σύµµετρα αÙτ¾ν διαιρε‹, ¹ ¥ρα ΓΜ τÁς ΜΖ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καί ™στιν Óλη ¹ ΓΜ σύµµετρος µήκει τÍ ™κκειµένV ·ητÍ τÍ Γ∆· ¹ ¥ρα ΓΖ ¢ποτοµή ™στι τετάρτη. ΤÕ ¥ρα ¢πÕ ™λάσσονος κሠτ¦ ˜ξÁς.

tained) by AG and GB. And since twice the (rectangle contained) by AG and GB is medial, and is equal to F L, F L is thus also medial. And it is applied to the rational (straight-line) F E, producing F M as breadth. Thus, F M is rational, and incommensurable in length with CD [Prop. 10.22]. And since the sum of the (squares) on AG and GB is rational, and twice the (rectangle contained) by AG and GB medial, the (sum of the squares) on AG and GB is [thus] incommensurable with twice the (rectangle contained) by AG and GB. And CL (is) equal to the (sum of the squares) on AG and GB, and F L equal to twice the (rectangle contained) by AG and GB. CL [is] thus incommensurable with F L. And as CL (is) to F L, so CM is to M F [Prop. 6.1]. CM is thus incommensurable in length with M F [Prop. 10.11]. And both are rational (straight-lines). Thus, CM and M F are rational (straight-lines which are) commensurable in square only. CF is thus an apotome [Prop. 10.73]. [So], I say that (it is) also a fourth (apotome). For since AG and GB are incommensurable in square, the (square) on AG (is) thus also incommensurable with the (square) on GB. And CH is equal to the (square) on AG, and KL to the (square) on GB. Thus, CH is incommensurable with KL. And as CH (is) to KL, so CK is to KM [Prop. 6.1]. CK is thus incommensurable in length with KM [Prop. 10.11]. And since the (rectangle contained) by AG and GB is the mean proportional to the (squares) on AG and GB [Prop. 10.21 lem.], and the (square) on AG is equal to CH, and the (square) on GB to KL, and the (rectangle contained) by AG and GB to N L, N L is thus the mean proportional to CH and KL. Thus, as CH is to N L, so N L (is) to KL. But, as CH (is) to N L, so CK is to N M , and as N L (is) to KL, so N M is to KM [Prop. 6.1]. Thus, as CK (is) to M N , so M N is to KM [Prop. 5.11]. The (rectangle contained) by CK and KM is thus equal to the (square) on M N —that is to say, to the fourth part of the (square) on F M [Prop. 6.17]. Therefore, since CM and M F are two unequal straight-lines, and the (rectangle contained) by CK and KM , equal to the fourth part of the (square) on M F , has been applied to CM , falling short by a square figure, and divides it into incommensurable (parts), the square on CM is thus greater than (the square on) M F by the (square) on (some straight-line) incommensurable (in length) with (CM ) [Prop. 10.18]. And the whole of CM is commensurable in length with the (previously) laid down rational (straight-line) CD. Thus, CF is a fourth apotome [Def. 10.14]. Thus, the (square) on a minor, and so on . . .

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ELEMENTS BOOK 10

ρα΄.

Proposition 101

ΤÕ ¢πÕ τÁς µετ¦ ·ητοà µέσον τÕ Óλον ποιούσης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν πέµπτην.

The (square) on that (straight-line) which with a rational (area) makes a medial whole, applied to a rational (straight-line), produces a fifth apotome as breadth.

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”Εστω ¹ µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσα ¹ ΑΒ, ·ητ¾ δ ¹ Γ∆, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΕ πλάτος ποιοàν τ¾ν ΓΖ· λέγω, Óτι ¹ ΓΖ ¢ποτοµή ™στι πέµπτη. ”Εστω γ¦ρ τÍ ΑΒ προσαρµόζουσα ¹ ΒΗ· αƒ ¥ρα ΑΗ, ΗΒ εÙθε‹αι δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον, τÕ δ δˆς Øπ' αÙτîν ·ητόν, κሠτù µν ¢πÕ τÁς ΑΗ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΘ, τù δ ¢πÕ τÁς ΗΒ †σον τÕ ΚΛ· Óλον ¥ρα τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ. τÕ δ συγκείµενον ™κ τîν ¢πÕ τîν ΑΗ, ΗΒ ¤µα µέσον ™στίν· µέσον ¥ρα ™στˆ τÕ ΓΛ. κሠπαρ¦ ·ητ¾ν τ¾ν Γ∆ παράκειται πλάτος ποιοàν τ¾ν ΓΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΓΜ κሠ¢σύµµετρος τÍ Γ∆. κሠ™πεˆ Óλον τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ, ïν τÕ ΓΕ ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ, λοιπÕν ¥ρα τÕ ΖΛ ‡σον ™στˆ τù δˆς ØπÕ τîν ΑΗ, ΗΒ. τετµήσθω οâν ¹ ΖΜ δίχα κατ¦ τÕ Ν, κሠ½χθω δι¦ τοà Ν Ðποτέρv τîν Γ∆, ΜΛ παράλληλος ¹ ΝΞ· ˜κάτερον ¥ρα τîν ΖΞ, ΝΛ ‡σον ™στˆ τù ØπÕ τîν ΑΗ, ΗΒ, κሠ™πεˆ τÕ δˆς ØπÕ τîν ΑΗ, ΗΒ ·ητόν ™στι καί [™στιν] ‡σον τù ΖΛ, ·ητÕν ¥ρα ™στˆ τÕ ΖΛ. κሠπαρ¦ ·ητ¾ν τ¾ν ΕΖ παράκειται πλάτος ποιοàν τ¾ν ΖΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΖΜ κሠσύµµετρος τÍ Γ∆ µήκει. κሠ™πεˆ τÕ µν ΓΛ µέσον ™στίν, τÕ δ ΖΛ ·ητόν, ¢σύµµετρον ¥ρα ™στˆ τÕ ΓΛ τù ΖΛ. æς δ τÕ ΓΛ πρÕς τÕ ΖΛ, οÛτως ¹ ΓΜ πρÕς τ¾ν ΜΖ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΓΜ τÍ ΜΖ µήκει. καί ε„σιν ¢µφότεραι ·ηταί· αƒ ¥ρα ΓΜ, ΜΖ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΓΖ. λγω δή, Óτι κሠπέµπτη. `Οµοίως γ¦ρ δείξοµεν, Óτι τÕ ØπÕ τîν ΓΚΜ ‡σον ™στˆ τù ¢πÕ τÁς ΝΜ, τουτέστι τù τετάρτJ µέρει τοà ¢πÕ τÁς ΖΜ. κሠ™πεˆ ¢σύµµετρόν ™στι τÕ ¢πÕ τÁς ΑΗ τù ¢πÕ τÁς ΗΒ, ‡σον δ τÕ µν ¢πÕ τÁς ΑΗ τù ΓΘ, τÕ δ ¢πÕ τÁς ΗΒ τù ΚΛ, ¢σύµµετρον ¥ρα τÕ

B

G

C

F

N K

M

D

E

O H

L

Let AB be that (straight-line) which with a rational (area) makes a medial whole, and CD a rational (straight-line). And let CE, equal to the (square) on AB, have been applied to CD, producing CF as breadth. I say that CF is a fifth apotome. Let BG be an attachment to AB. Thus, the straightlines AG and GB are incommensurable in square, making the sum of the squares on them medial, and twice the (rectangle contained) by them rational [Prop. 10.77]. And let CH, equal to the (square) on AG, have been applied to CD, and KL, equal to the (square) on GB. The whole of CL is thus equal to the (sum of the squares) on AG and GB. And the sum of the (squares) on AG and GB together is medial. Thus, CL is medial. And it has been applied to the rational (straight-line) CD, producing CM as breadth. CM is thus rational, and incommensurable (in length) with CD [Prop. 10.22]. And since the whole of CL is equal to the (sum of the squares) on AG and GB, of which CE is equal to the (square) on AB, the remainder F L is thus equal to twice the (rectangle contained) by AG and GB [Prop. 2.7]. Therefore, let F M have been cut in half at N . And let N O have been drawn through N , parallel to either of CD or M L. Thus, F O and N L are each equal to the (rectangle contained) by AG and GB. And since twice the (rectangle contained) by AG and GB is rational, and [is] equal to F L, F L is thus rational. And it is applied to the rational (straight-line) EF , producing F M as breadth. Thus, F M is rational, and commensurable in length with CD [Prop. 10.20]. And since CL is medial, and F L rational, CL is thus incommensurable with F L. And as CL (is) to F L, so CM (is) to M F [Prop. 6.1]. CM is thus incommensurable in length with M F [Prop. 10.11]. And both are rational. Thus, CM and M F are rational (straight-

404

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

ΓΘ τù ΚΛ. æς δ τÕ ΓΘ πρÕς τÕ ΚΛ, οÛτως ¹ ΓΚ πρÕς τ¾ν ΚΜ· ¢σύµµετρος ¥ρα ¹ ΓΚ τÍ ΚΜ µήκει. ™πεˆ οâν δύο εÙθε‹αι ¥νισοί ε„σιν αƒ ΓΜ, ΜΖ, κሠτù τετάρτJ µέρει τοà ¢πÕ τÁς ΖΜ ‡σον παρ¦ τ¾ν ΓΜ παραβέβληται ™λλε‹πον ε‡δει τετραγώνJ κሠε„ς ¢σύµµετρα αÙτ¾ν διαιρε‹, ¹ ¥ρα ΓΜ τÁς ΜΖ µε‹ζον δύναται τù ¢πÕ ¢σύµµέτρου ˜αυτÍ. καί ™στιν ¹ προσαρµόζουσα ¹ ΖΜ σύµµετρος τÍ ™κκειµένV ·ητÍ τÍ Γ∆· ¹ ¥ρα ΓΖ ¢ποτοµή ™στι πέµπτη· Óπερ œδει δε‹ξαι.

lines which are) commensurable in square only. CF is thus an apotome [Prop. 10.73]. So, I say that (it is) also a fifth (apotome). For, similarly (to the previous propositions), we can show that the (rectangle contained) by CKM is equal to the (square) on N M —that is to say, to the fourth part of the (square) on F M . And since the (square) on AG is incommensurable with the (square) on GB, and the (square) on AG (is) equal to CH, and the (square) on GB to KL, CH (is) thus incommensurable with KL. And as CH (is) to KL, so CK (is) to KM [Prop. 6.1]. Thus, CK (is) incommensurable in length with KM [Prop. 10.11]. Therefore, since CM and M F are two unequal straight-lines, and (some area), equal to the fourth part of the (square) on F M , has been applied to CM , falling short by a square figure, and divides it into incommensurable (parts), the square on CM is thus greater than (the square on) M F by the (square) on (some straight-line) incommensurable (in length) with (CM ) [Prop. 10.18]. And the attachment F M is commensurable with the (previously) laid down rational (straightline) CD. Thus, CF is a fifth apotome [Def. 10.15]. (Which is) the very thing it was required to show.

ρβ΄.

Proposition 102

ΤÕ ¢πÕ τÁς µετ¦ µέσου µέσον τÕ Óλον ποιούσης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν ›κτην.

The (square) on that (straight-line) which with a medial (area) makes a medial whole, applied to a rational (straight-line), produces a sixth apotome as breadth.

A

B

H

A

B

G

G

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N K

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C

F

N K

M

D

E

X J

L

D

E

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L

”Εστω ¹ µετ¦ µέσου µέσον τÕ Óλον ποιοàσα ¹ ΑΒ, ·ητ¾ δ ¹ Γ∆, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω τÕ ΓΕ πλάτος ποιοàν τ¾ν ΓΖ· λέγω, Óτι ¹ ΓΖ ¢ποτοµή ™στιν ›κτη. ”Εστω γ¦ρ τÍ ΑΒ προσαρµόζουσα ¹ ΒΗ· αƒ ¥ρα ΑΗ, ΗΒ δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τό τε συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον κሠτÕ δˆς ØπÕ τîν ΑΗ, ΗΒ µέσον κሠ¢σύµµετρον τ¦ ¢πÕ τîν ΑΗ, ΗΒ τù δˆς ØπÕ τîν ΑΗ, ΗΒ. παραβεβλήσθω οâν παρ¦ τ¾ν Γ∆ τù µν ¢πÕ τÁς ΑΗ ‡σον τÕ ΓΘ πλάτος ποιοàν τ¾ν ΓΚ, τù δ ¢πÕ τÁς ΒΗ τÕ ΚΛ· Óλον ¥ρα τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ· µέσον ¥ρα [™στˆ]

Let AB be that (straight-line) which with a medial (area) makes a medial whole, and CD a rational (straight-line). And let CE, equal to the (square) on AB, have been applied to CD, producing CF as breadth. I say that CF is a sixth apotome. For let BG be an attachment to AB. Thus, AG and GB are incommensurable in square, making the sum of the squares on them medial, and twice the (rectangle contained) by AG and GB medial, and the (sum of the squares) on AG and GB incommensurable with twice the (rectangle contained) by AG and GB [Prop. 10.78]. Therefore, let CH, equal to the (square) on AG, have

405

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

κሠτÕ ΓΛ. κሠπαρ¦ ·ητ¾ν τ¾ν Γ∆ παράκειται πλάτος ποιοàν τ¾ν ΓΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΓΜ κሠ¢σύµµετρος τÍ Γ∆ µήκει. ™πεˆ οâν τÕ ΓΛ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΗ, ΗΒ, ïν τÕ ΓΕ ‡σον τù ¢πÕ τÁς ΑΒ, λοιπÕν ¥ρα τÕ ΖΛ ‡σον ™στˆ τù δˆς ØπÕ τîν ΑΗ, ΗΒ. καί ™στι τÕ δˆς ØπÕ τîν ΑΗ, ΗΒ µέσον· κሠτÕ ΖΛ ¥ρα µέσον ™στίν. κሠπαρ¦ ·ητ¾ν τ¾ν ΖΕ παράκειται πλάτος ποιοàν τ¾ν ΖΜ· ·ητ¾ ¥ρα ™στˆν ¹ ΖΜ κሠ¢σύµµετρος τÍ Γ∆ µήκει. κሠ™πεˆ τ¦ ¢πÕ τîν ΑΗ, ΗΒ ¢σύµµετρά ™στι τù δˆς ØπÕ τîν ΑΗ, ΗΒ, καί ™στι το‹ς µν ¢πÕ τîν ΑΗ, ΗΒ ‡σον τÕ ΓΛ, τù δ δˆς ØπÕ τîν ΑΗ, ΗΒ ‡σον τÕ ΖΛ, ¢σύµµετρος ¥ρα [™στˆ] τÕ ΓΛ τù ΖΛ. æς δ τÕ ΓΛ πρÕς τÕ ΖΛ, οÛτως ™στˆν ¹ ΓΜ πρÕς τ¾ν ΜΖ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΓΜ τÍ ΜΖ µήκει. καί ε„σιν ¢µφότεραι ·ηταί. αƒ ΓΜ, ΜΖ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΓΖ. λέγω δή, Óτι κሠ›κτη. 'Επεˆ γ¦ρ τÕ ΖΛ ‡σον ™στˆ τù δˆς ØπÕ τîν ΑΗ, ΗΒ, τετµήσθω δίχα ¹ ΖΜ κατ¦ τÕ Ν, κሠ½χθω δι¦ τοà Ν τÍ Γ∆ παράλληλος ¹ ΝΞ· ˜κάτερον ¥ρα τîν ΖΞ, ΝΛ ‡σον ™στˆ τù ØπÕ τîν ΑΗ, ΗΒ. κሠ™πεˆ αƒ ΑΗ, ΗΒ δυνάµει ε„σˆν ¢σύµµετροι, ¢σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΗ τù ¢πÕ τÁς ΗΒ. ¢λλ¦ τù µν ¢πÕ τÁς ΑΗ ‡σον ™στˆ τÕ ΓΘ, τù δ ¢πÕ τÁς ΗΒ ‡σον ™στˆ τÕ ΚΛ· ¢σύµµετρον ¥ρα ™στˆ τÕ ΓΘ τù ΚΛ. æς δ τÕ ΓΘ πρÕς τÕ ΚΛ, οÛτως ™στˆν ¹ ΓΚ πρÕς τ¾ν ΚΜ· ¢σύµµετρος ¥ρα ™στˆν ¹ ΓΚ τÍ ΚΜ. κሠ™πεˆ τîν ¢πÕ τîν ΑΗ, ΗΒ µέσον ¢νάλογόν ™στι τÕ ØπÕ τîν ΑΗ, ΗΒ, κሠ™στι τù µν ¢πÕ τÁς ΑΗ ‡σον τÕ ΓΘ, τù δ ¢πÕ τÁς ΗΒ ‡σον τÕ ΚΛ, τù δ ØπÕ τîν ΑΗ, ΗΒ ‡σον τÕ ΝΛ, κሠτîν ¥ρα ΓΘ, ΚΛ µέσον ¢νάλογόν ™στι τÕ ΝΛ· œστιν ¥ρα æς τÕ ΓΘ πρÕς τÕ ΝΛ, οÛτως τÕ ΝΛ πρÕς τÕ ΚΛ. κሠδι¦ τ¦ αÙτ¦ ¹ ΓΜ τÁς ΜΖ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. κሠοÙδετέρα αÙτîν σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ Γ∆· ¹ ΓΖ ¥ρα ¢ποτοµή ™στιν ›κτη· Óπερ œδει δε‹ξαι.

been applied to CD, producing CK as breadth, and KL, equal to the (square) on BG. Thus, the whole of CL is equal to the (sum of the squares) on AG and GB. CL [is] thus also medial. And it is applied to the rational (straight-line) CD, producing CM as breadth. Thus, CM is rational, and incommensurable in length with CD [Prop. 10.22]. Therefore, since CL is equal to the (sum of the squares) on AG and GB, of which CE (is) equal to the (square) on AB, the remainder F L is thus equal to twice the (rectangle contained) by AG and GB [Prop. 2.7]. And twice the (rectangle contained) by AG and GB (is) medial. Thus, F L is also medial. And it is applied to the rational (straight-line) F E, producing F M as breadth. F M is thus rational, and incommensurable in length with CD [Prop. 10.22]. And since the (sum of the squares) on AG and GB is incommensurable with twice the (rectangle contained) by AG and GB, and CL equal to the (sum of the squares) on AG and GB, and F L to twice the (rectangle contained) by AG and GB, CL [is] thus incommensurable with F L. And as CL (is) to F L, so CM is to M F [Prop. 6.1]. Thus, CM is incommensurable in length with M F [Prop. 10.11]. And they are both rational. Thus, CM and M F are rational (straightlines which are) commensurable in square only. CF is thus an apotome [Prop. 10.73]. So, I say that (it is) also a sixth (apotome). For since F L is equal to twice the (rectangle contained) by AG and GB, let F M have been cut in half at N , and let N O have been drawn through N , parallel to CD. Thus, F O and N L are each equal to the (rectangle contained) by AG and GB. And since AG and GB are incommensurable in square, the (square) on AG is thus incommensurable with the (square) on GB. But, CH is equal to the (square) on AG, and KL is equal to the (square) on GB. Thus, CH is incommensurable with KL. And as CH (is) to KL, so CK is to KM [Prop. 6.1]. Thus, CK is incommensurable (in length) with KM [Prop. 10.11]. And since the (rectangle contained) by AG and GB is the mean proportional to the (squares) on AG and GB [Prop. 10.21 lem.], and CH is equal to the (square) on AG, and KL equal to the (square) on GB, and N L equal to the (rectangle contained) by AG and GB, N L is thus also the mean proportional to CH and KL. Thus, as CH is to N L, so N L (is) to KL. And for the same (reasons as the preceding propositions), the square on CM is greater than (the square on) M F by the (square) on (some straight-line) incommensurable (in length) with (CM ) [Prop. 10.18]. And neither of them is commensurable with the (previously) laid down rational (straight-line) CD. Thus, CF is a sixth apotome [Def. 10.16]. (Which is) the very thing

406

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 it was required to show.

ργ΄.

Proposition 103

`Η τÍ ¢ποτοµÍ µήκει σύµµετρος ¢ποτοµή ™στι κሠτÍ τάξει ¹ αÙτή.

A (straight-line) commensurable in length with an apotome is an apotome, and (is) the same in order.

Α Γ

Β ∆

Ε

A

Ζ

B C

”Εστω ¢ποτοµ¾ ¹ ΑΒ, κሠτÍ ΑΒ µήκει σύµµετρος œστω ¹ Γ∆· λέγω, Óτι κሠ¹ Γ∆ ¢ποτοµή ™στι κሠτÍ τάξει ¹ αÙτ¾ τÍ ΑΒ. 'Επεˆ γ¦ρ ¢ποτοµή ™στιν ¹ ΑΒ, œστω αÙτÍ προσαρµόζουσα ¹ ΒΕ· αƒ ΑΕ, ΕΒ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. κሠτù τÁς ΑΒ πρÕς τ¾ν Γ∆ λόγJ Ð αÙτÕς γεγονέτω Ð τÁς ΒΕ πρÕς τ¾ν ∆Ζ· κሠæς žν ¥ρα πρÕς ›ν, πάντα [™στˆ] πρÕς πάντα· œστιν ¥ρα κሠæς Óλη ¹ ΑΕ πρÕς Óλην τ¾ν ΓΖ, οÛτως ¹ ΑΒ πρÕς τ¾ν Γ∆. σύµµετρος δ ¹ ΑΒ τÍ Γ∆ µήκει· σύµµετρος ¥ρα κሠ¹ ΑΕ µν τÍ ΓΖ, ¹ δ ΒΕ τÍ ∆Ζ. καˆ αƒ ΑΕ, ΕΒ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· καˆ αƒ ΓΖ, Ζ∆ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι [¢ποτοµ¾ ¥ρα ™στˆν ¹ Γ∆. λέγω δή, Óτι κሠτÍ τάξει ¹ αÙτ¾ τÍ ΑΒ]. 'Επεˆ οâν ™στιν æς ¹ ΑΕ πρÕς τ¾ν ΓΖ, οÛτως ¹ ΒΕ πρÕς τ¾ν ∆Ζ, ™ναλλ¦ξ ¥ρα ™στˆν æς ¹ ΑΕ πρÕς τ¾ν ΕΒ, οÛτως ¹ ΓΖ πρÕς τ¾ν Ζ∆. ½τοι δ¾ ¹ ΑΕ τÁς ΕΒ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ À τù ¢πÕ ¢συµµέτρου. ε„ µν οâν ¹ ΑΕ τÁς ΕΒ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠ¹ ΓΖ τÁς Ζ∆ µε‹ζον δυνήσεται τù ¢πÕ συµµέτρου ˜αυτÍ. καˆ ε„ µν σύµµετρός ™στιν ¹ ΑΕ τÍ ™κκειµένV ·ητÍ µήκει, κሠ¹ ΓΖ, ε„ δ ¹ ΒΕ, κሠ¹ ∆Ζ, ε„ δ οÙδετέρα τîν ΑΕ, ΕΒ, κሠοÙδετέρα τîν ΓΖ, Ζ∆. ε„ δ ¹ ΑΕ [τÁς ΕΒ] µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ, κሠ¹ ΓΖ τÁς Ζ∆ µε‹ζον δυνήσεται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καˆ ε„ µν σύµµετρός ™στιν ¹ ΑΕ τÍ ™κκειµένV ·ητÍ µήκει, κሠ¹ ΓΖ, ε„ δ ¹ ΒΕ, κሠ¹ ∆Ζ, ε„ δ οÙδετέρα τîν ΑΕ, ΕΒ, οÙδετέρα τîν ΓΖ, Ζ∆. 'Αποτοµ¾ ¥ρα ™στˆν ¹ Γ∆ κሠτÍ τάξει ¹ αÙτ¾ τÍ ΑΒ· Óπερ œδει δε‹ξαι.

D

E F

Let AB be an apotome, and let CD be commensurable in length with AB. I say that CD is also an apotome, and (is) the same in order as AB. For since AB is an apotome, let BE be an attachment to it. Thus, AE and EB are rational (straight-lines which are) commensurable in square only [Prop. 10.73]. And let it have been contrived that the (ratio) of BE to DF is the same as the ratio of AB to CD [Prop. 6.12]. Thus also as one is to one, (so) all [are] to all [Prop. 5.12]. And thus as the whole AE is to the whole CF , so AB (is) to CD. And AB (is) commensurable in length with CD. AE (is) thus also commensurable (in length) with CF , and BE with DF [Prop. 10.11]. And AE and BE are rational (straight-lines which are) commensurable in square only. Thus, CF and F D are also rational (straight-lines which are) commensurable in square only [Prop. 10.13]. [CD is thus an apotome. So, I say that (it is) also the same in order as AB.] Therefore, since as AE is to CF , so BE (is) to DF , thus, alternately, as AE is to EB, so CF (is) to F D [Prop. 5.16]. So, the square on AE is greater than (the square on) EB either by the (square) on (some straight-line) commensurable, or by the (square) on (some straight-line) incommensurable, (in length) with (AE). Therefore, if the (square) on AE is greater than (the square on) EB by the (square) on (some straightline) commensurable (in length) with (AE), then the square on CF will also be greater than (the square on) F D by the (square) on (some straight-line) commensurable (in length) with (CF ) [Prop. 10.14]. And if AE is commensurable in length with a (previously) laid down rational (straight-line), then so (is) CF [Prop. 10.12], and if BE (is commensurable), so (is) DF , and if neither of AE or EB (are commensurable), neither (are) either of CF or F D [Prop. 10.13]. And if the (square) on AE is greater [than (the square on) EB] by the (square) on (some straight-line) incommensurable (in length) with (AE), then the (square) on CF will also be greater than (the square on) F D by the (square) on (some straight-line) incommensurable (in length) with (CF ) [Prop. 10.14]. And if AE is commensurable in length

407

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 with a (previously) laid down rational (straight-line), so (is) CF [Prop. 10.12], and if BE (is commensurable), so (is) DF , and if neither of AE or EB (are commensurable), neither (are) either of CF or F D [Prop. 10.13]. Thus, CD is an apotome, and (is) the same in order as AB [Defs. 10.11—10.16]. (Which is) the very thing it was required to show.

ρδ΄.

Proposition 104

`Η τÍ µέσης ¢ποτοµÍ σύµµετρος µέσης ¢ποτοµή ™στι κሠτÍ τάξει ¹ αÙτή.

A (straight-line) commensurable (in length) with an apotome of a medial (straight-line) is an apotome of a medial (straight-line), and (is) the same in order.

Α Γ

Β ∆

Ε

A

Ζ

B C

”Εστω µέσης ¢ποτοµ¾ ¹ ΑΒ, κሠτÍ ΑΒ µήκει σύµµετρος œστω ¹ Γ∆· λέγω, Óτι κሠ¹ Γ∆ µέσης ¢ποτοµή ™στι κሠτÍ τάξει ¹ αÙτ¾ τÍ ΑΒ. 'Επεˆ γ¦ρ µέσης ¢ποτοµή ™στιν ¹ ΑΒ, œστω αÙτÍ προσαρµόζουσα ¹ ΕΒ. αƒ ΑΕ, ΕΒ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι. κሠγεγονέτω æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΒΕ πρÕς τ¾ν ∆Ζ· σύµµετρος ¥ρα [™στˆ] κሠ¹ ΑΕ τÍ ΓΖ, ¹ δ ΒΕ τÍ ∆Ζ. αƒ δ ΑΕ, ΕΒ µέσαι ε„σˆ δυνάµει µόνον σύµµετροι· καˆ αƒ ΓΖ, Ζ∆ ¥ρα µέσαι ε„σˆ δυνάµει µόνον σύµµετροι· µέσης ¥ρα ¢ποτοµή ™στιν ¹ Γ∆. λέγω δή, Óτι κሠτÍ τάξει ™στˆν ¹ αÙτ¾ τÍ ΑΒ. 'Επεˆ [γάρ] ™στιν æς ¹ ΑΕ πρÕς τ¾ν ΕΒ, οÛτως ¹ ΓΖ πρÕς τ¾ν Ζ∆ [¢λλ' æς µν ¹ ΑΕ πρÕς τ¾ν ΕΒ, οÛτως τÕ ¢πÕ τÁς ΑΕ πρÕς τÕ ØπÕ τîν ΑΕ, ΕΒ, æς δ ¹ ΓΖ πρÕς τ¾ν Ζ∆, οÛτως τÕ ¢πÕ τÁς ΓΖ πρÕς τÕ ØπÕ τîν ΓΖ, Ζ∆], œστιν ¥ρα κሠæς τÕ ¢πÕ τÁς ΑΕ πρÕς τÕ ØπÕ τîν ΑΕ, ΕΒ, οÛτως τÕ ¢πÕ τÁς ΓΖ πρÕς τÕ ØπÕ τîν ΓΖ, Ζ∆ [κሠ™ναλλ¦ξ æς τÕ ¢πÕ τÁς ΑΕ πρÕς τÕ ¢πÕ τÁς ΓΖ, οÛτως τÕ ØπÕ τîν ΑΕ, ΕΒ πρÕς τÕ ØπÕ τîν ΓΖ, Ζ∆]. σύµµετρον δ τÕ ¢πÕ τÁς ΑΕ τù ¢πÕ τÁς ΓΖ· σύµµετρον ¥ρα ™στˆ κሠτÕ ØπÕ τîν ΑΕ, ΕΒ τù ØπÕ τîν ΓΖ, Ζ∆. ε‡τε οâν ·ητόν ™στι τÕ ØπÕ τîν ΑΕ, ΕΒ, ·ητÕν œσται κሠτÕ ØπÕ τîν ΓΖ, Ζ∆, ε‡τε µέσον [™στˆ] τÕ ØπÕ τîν ΑΕ, ΕΒ, µέσον [™στˆ] κሠτÕ ØπÕ τîν ΓΖ, Ζ∆. Μέσης ¥ρα ¢ποτοµή ™στιν ¹ Γ∆ κሠτÍ τάξει ¹ αÙτ¾ τÍ ΑΒ· Óπερ œδει δε‹ξαι.

D

E F

Let AB be an apotome of a medial (straight-line), and let CD be commensurable in length with AB. I say that CD is also an apotome of a medial (straight-line), and (is) the same in order as AB. For since AB is an apotome of a medial (straightline), let EB be an attachment to it. Thus, AE and EB are medial (straight-lines which are) commensurable in square only [Props. 10.74, 10.75]. And let it have been contrived that as AB is to CD, so BE (is) to DF [Prop. 6.12]. Thus, AE [is] also commensurable (in length) with CF , and BE with DF [Props. 5.12, 10.11]. And AE and EB are medial (straight-lines which are) commensurable in square only. CF and F D are thus also medial (straight-lines which are) commensurable in square only [Props. 10.23, 10.13]. Thus, CD is an apotome of a medial (straight-line) [Props. 10.74, 10.75]. So, I say that it is also the same in order as AB. [For] since as AE is to EB, so CF (is) to F D [Props. 5.12, 5.16] [but as AE (is) to EB, so the (square) on AE (is) to the (rectangle contained) by AE and EB, and as CF (is) to F D, so the (square) on CF (is) to the (rectangle contained) by CF and F D], thus as the (square) on AE is to the (rectangle contained) by AE and EB, so the (square) on CF also (is) to the (rectangle contained) by CF and F D [Prop. 10.21 lem.] [and, alternately, as the (square) on AE (is) to the (square) on CF , so the (rectangle contained) by AE and EB (is) to the (rectangle contained) by CF and F D]. And the (square) on AE (is) commensurable with the (square) on CF . Thus, the (rectangle contained) by AE and EB is also commensurable with the (rectangle contained) by CF and F D [Props. 5.16, 10.11]. Therefore, either the (rectangle contained) by AE and EB is rational, and the (rectangle contained) by CF and F D will also be ratio-

408

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 nal [Def. 10.4], or the (rectangle contained) by AE and EB [is] medial, and the (rectangle contained) by CF and F D [is] also medial [Prop. 10.23 corr.]. Therefore, CD is the apotome of a medial (straightline), and is the same in order as AB [Props. 10.74, 10.75]. (Which is) the very thing it was required to show.

ρε΄.

Proposition 105

`Η τÍ ™λάσσονι σύµµετρος ™λάσσων ™στίν.

Α Γ

Β ∆

A (straight-line) commensurable (in length) with a minor (straight-line) is a minor (straight-line).

Ε

A

Ζ

B C

”Εστω γ¦ρ ™λάσσων ¹ ΑΒ κሠτÍ ΑΒ σύµµετρος ¹ Γ∆· λέγω, Óτι κሠ¹ Γ∆ ™λάσσων ™στίν. Γεγονέτω γ¦ρ τ¦ αÙτά· κሠ™πεˆ αƒ ΑΕ, ΕΒ δυνάµει ε„σˆν ¢σύµµετροι, καˆ αƒ ΓΖ, Ζ∆ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι. ™πεˆ οâν ™στιν æς ¹ ΑΕ πρÕς τ¾ν ΕΒ, οÛτως ¹ ΓΖ πρÕς τ¾ν Ζ∆, œστιν ¥ρα κሠæς τÕ ¢πÕ τÁς ΑΕ πρÕς τÕ ¢πÕ τÁς ΕΒ, οÛτως τÕ ¢πÕ τÁς ΓΖ πρÕς τÕ ¢πÕ τÁς Ζ∆. συνθέντι ¥ρα ™στˆν æς τ¦ ¢πÕ τîν ΑΕ, ΕΒ πρÕς τÕ ¢πÕ τÁς ΕΒ, οÛτως τ¦ ¢πÕ τîν ΓΖ, Ζ∆ πρÕς τÕ ¢πÕ τÁς Ζ∆ [κሠ™ναλλάξ]· σύµµετρον δέ ™στι τÕ ¢πÕ τÁς ΒΕ τù ¢πÕ τÁς ∆Ζ· σύµµετρον ¥ρα κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΕ, ΕΒ τετραγώνων τù συγκειµένJ ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆ τετραγώνων. ·ητÕν δέ ™στι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΕ, ΕΒ τετραγώνων· ·ητÕν ¥ρα ™στˆ κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆ τετραγώνων. πάλιν, ™πεί ™στιν æς τÕ ¢πÕ τÁς ΑΕ πρÕς τÕ ØπÕ τîν ΑΕ, ΕΒ, οÛτως τÕ ¢πÕ τÁς ΓΖ πρÕς τÕ ØπÕ τîν ΓΖ, Ζ∆, σύµµετρον δ τÕ ¢πÕ τÁς ΑΕ τετράγωνον τù ¢πÕ τÁς ΓΖ τετραγώνJ, σύµµετρον ¥ρα ™στˆ κሠτÕ ØπÕ τîν ΑΕ, ΕΒ τù ØπÕ τîν ΓΖ, Ζ∆. µέσον δ τÕ ØπÕ τîν ΑΕ, ΕΒ· µέσον ¥ρα κሠτÕ ØπÕ τîν ΓΖ, Ζ∆· αƒ ΓΖ, Ζ∆ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων ·ητόν, τÕ δ' Øπ' αÙτîν µέσον. 'Ελάσσων ¥ρα ™στˆν ¹ Γ∆· Óπερ œδει δε‹ξαι.

D

E F

For let AB be a minor (straight-line), and (let) CD (be) commensurable (in length) with AB. I say that CD is also a minor (straight-line). For let the same things have been contrived (as in the former proposition). And since AE and EB are (straight-lines which are) incommensurable in square [Prop. 10.76], CF and F D are thus also (straight-lines which are) incommensurable in square [Prop. 10.13]. Therefore, since as AE is to EB, so CF (is) to F D [Props. 5.12, 5.16], thus also as the (square) on AE is to the (square) on EB, so the (square) on CF (is) to the (square) on F D [Prop. 6.22]. Thus, via composition, as the (sum of the squares) on AE and EB is to the (square) on EB, so the (sum of the squares) on CF and F D (is) to the (square) on F D [Prop. 5.18], [also alternately]. And the (square) on BE is commensurable with the (square) on DF [Prop. 10.104]. The sum of the squares on AE and EB (is) thus also commensurable with the sum of the squares on CF and F D [Prop. 5.16, 10.11]. And the sum of the (squares) on AE and EB is rational [Prop. 10.76]. Thus, the sum of the (squares) on CF and F D is also rational [Def. 10.4]. Again, since as the (square) on AE is to the (rectangle contained) by AE and EB, so the (square) on CF (is) to the (rectangle contained) by CF and F D [Prop. 10.21 lem.], and the square on AE (is) commensurable with the square on CF , the (rectangle contained) by AE and EB is thus also commensurable with the (rectangle contained) by CF and F D. And the (rectangle contained) by AE and EB (is) medial [Prop. 10.76]. Thus, the (rectangle contained) by CF and F D (is) also medial [Prop. 10.23 corr.]. CF and F D are thus (straight-lines which are) incommensurable in square, making the sum of the squares on them rational, and the (rectangle contained) by them medial. Thus, CD is a minor (straight-line) [Prop. 10.76].

409

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 (Which is) the very thing it was required to show.

ρ$΄.

Proposition 106

`Η τÍ µετ¦ ·ητοà µέσον τÕ Óλον ποιούσV σύµµετρος µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσά ™στιν.

A (straight-line) commensurable (in length) with a (straight-line) which with a rational (area) makes a medial whole is a (straight-line) which with a rational (area) makes a medial whole.

Α Γ

Β ∆

Ε

A

Ζ

B C

D

E F

”Εστω µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσα ¹ ΑΒ κሠτÍ ΑΒ σύµµετρος ¹ Γ∆· λέγω, Óτι κሠ¹ Γ∆ µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσά ™στιν. ”Εστω γ¦ρ τÍ ΑΒ προσαρµόζουσα ¹ ΒΕ· αƒ ΑΕ, ΕΒ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΑΕ, ΕΒ τετραγώνων µέσον, τÕ δ' Øπ' αÙτîν ·ητόν. κሠτ¦ αÙτ¦ κατεσκευάσθω. еοίως δ¾ δείξοµεν το‹ς πρότερον, Óτι αƒ ΓΖ, Ζ∆ ™ν τù αÙτù λόγJ ε„σˆ τα‹ς ΑΕ, ΕΒ, κሠσύµµετρόν ™στι τÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΕ, ΕΒ τετραγώνων τù συγκειµένJ ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆ τετραγώνων, τÕ δ ØπÕ τîν ΑΕ, ΕΒ τù ØπÕ τîν ΓΖ, Ζ∆· éστε καˆ αƒ ΓΖ, Ζ∆ δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τÕ µν συγκείµενον ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆ τετραγώνων µέσον, τÕ δ' Øπ' αÙτîν ·ητόν. `Η Γ∆ ¥ρα µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσά ™στιν· Óπερ œδει δε‹ξαι.

Let AB be a (straight-line) which with a rational (area) makes a medial whole, and (let) CD (be) commensurable (in length) with AB. I say that CD is also a (straight-line) which with a rational (area) makes a medial (whole). For let BE be an attachment to AB. Thus, AE and EB are (straight-lines which are) incommensurable in square, making the sum of the squares on AE and EB medial, and the (rectangle contained) by them rational [Prop. 10.77]. And let the same construction have been made (as in the previous propositions). So, similarly to the previous (propositions), we can show that CF and F D are in the same ratio as AE and EB, and the sum of the squares on AE and EB is commensurable with the sum of the squares on CF and F D, and the (rectangle contained) by AE and EB with the (rectangle contained) by CF and F D. Hence, CF and F D are also (straightlines which are) incommensurable in square, making the sum of the squares on CF and F D medial, and the (rectangle contained) by them rational. CD is thus a (straight-line) which with a rational (area) makes a medial whole [Prop. 10.77]. (Which is) the very thing it was required to show.

ρζ΄.

Proposition 107

`Η τÍ µετ¦ µέσου µέσον τÕ Óλον ποιούσV σύµµετρος κሠαÙτ¾ µετ¦ µέσου µέσον τÕ Óλον ποιοàσά ™στιν.

A (straight-line) commensurable (in length) with a (straight-line) which with a medial (area) makes a medial whole is itself also a (straight-line) which with a medial (area) makes a medial whole.

Α Γ

Β ∆

Ε

A

Ζ

B C

D

E F

”Εστω µετ¦ µέσου µέσον τÕ Óλον ποιοàσα ¹ ΑΒ, κሠLet AB be a (straight-line) which with a medial (area) τÍ ΑΒ œστω σύµµετρος ¹ Γ∆· λέγω, Óτι κሠ¹ Γ∆ µετ¦ makes a medial whole, and let CD be commensurable (in µέσου µέσον τÕ Óλον ποιοàσά ™στιν. length) with AB. I say that CD is also a (straight-line) ”Εστω γ¦ρ τÍ ΑΒ προσαρµόζουσα ¹ ΒΕ, κሠτ¦ which with a medial (area) makes a medial whole. αÙτ¦ κατεσκευάσθω· αƒ ΑΕ, ΕΒ ¥ρα δυνάµει εƒσˆν For let BE be an attachment to AB. And let the same 410

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

¢σύµµετροι ποιοàσαι τό τε συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον κሠτÕ Øπ' αÙτîν µέσον κሠœτι ¢σύµµετρον τÕ συγκέιµενον ™κ τîν ¢π' αÙτîν τετραγώνων τù Øπ' αÙτîν. καί ε„σιν, æς ™δείχθη, αƒ ΑΕ, ΕΒ σύµµετροι τα‹ς ΓΖ, Ζ∆, κሠτÕ συγκείµενον ™κ τîν ¢πÕ τîν ΑΕ, ΕΒ τετραγώνων τù συγκειµένJ ™κ τîν ¢πÕ τîν ΓΖ, Ζ∆, τÕ δ ØπÕ τîν ΑΕ, ΕΒ τù ØπÕ τîν ΓΖ, Ζ∆· καˆ αƒ ΓΖ, Ζ∆ ¥ρα δυνάµει ε„σˆν ¢σύµµετροι ποιοàσαι τό τε συγκείµενον ™κ τîν ¢π' αÙτîν τετραγώνων µέσον κሠτÕ Øπ' ¢Ùτîν µέσον κሠœτι ¢σύµµετρον τÕ συγκείµενον ™κ τîν ¢π' αÙτîν [τετραγώνων] τù Øπ' αÙτîν. `Η Γ∆ ¥ρα µετ¦ µέσου µέσον τÕ Óλον ποιοàσά ™στιν· Óπερ œδει δε‹ξαι.

construction have been made (as in the previous propositions). Thus, AE and EB are (straight-lines which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them medial, and, further, the sum of the squares on them incommensurable with the (rectangle contained) by them [Prop. 10.78]. And, as was shown (previously), AE and EB are commensurable (in length) with CF and F D (respectively), and the sum of the squares on AE and EB with the sum of the squares on CF and F D, and the (rectangle contained) by AE and EB with the (rectangle contained) by CF and F D. Thus, CF and F D are also (straight-lines which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them medial, and, further, the sum of the [squares] on them incommensurable with the (rectangle contained) by them. Thus, CD is a (straight-line) which with a medial (area) makes a medial whole [Prop. 10.78]. (Which is) the very thing it was required to show.

ρη΄.

Proposition 108

'ΑπÕ ·ητοà µέσου ¢φαιρουµένου ¹ τÕ λοιπÕν χωρίον A medial (area) being subtracted from a rational δυναµένη µία δύο ¢λόγων γίνεται ½τοι ¢ποτοµ¾ À (area), one of two irrational (straight-lines) arise (as) the ™λάσσων. square-root of the remaining area—either an apotome, or a minor (straight-line). A E B Α Ε Β

Θ Γ

Λ

Η

Κ

Ζ

H

∆ 'ΑπÕ γ¦ρ ·ητοà τοà ΒΓ µέσον ¢φVρήσθω τÕ Β∆· λέγω, Óτι ¹ τÕ λοιπÕν δυναµένη τÕ ΕΓ µία δύο ¢λόγων γίνεται ½τοι ¢ποτοµ¾ À ™λάσσων. 'Εκκείσθω γ¦ρ ·ητ¾ ¹ ΖΗ, κሠτù µν ΒΓ ‡σον παρ¦ τ¾ν ΖΗ παραβεβλήσθω Ñρθογώνιον παραλληλόγραµµον τÕ ΗΘ, τù δ ∆Β ‡σον ¢φVρήσθω τÕ ΗΚ· λοιπÕν ¥ρα τÕ ΕΓ ‡σον ™στˆ τù ΛΘ. ™πεˆ οâν ·ητÕν µέν ™στι τÕ ΒΓ, µέσον δ τÕ Β∆, ‡σον δ τÕ µν ΒΓ τù ΗΘ, τÕ δ Β∆ τù ΗΚ, ·ητÕν µν ¥ρα ™στˆ τÕ ΗΘ, µέσον δ τÕ ΗΚ. κሠπαρ¦ ·ητ¾ν τ¾ν ΖΗ παράκειται· ·ητ¾ µν ¥ρα ¹ ΖΘ κሠσύµµετρος τÍ ΖΗ µήκει, ·ητ¾ δ ¹ ΖΚ κሠ¢σύµµετρος τÍ ΖΗ µήκει· ¢σύµµετρος ¥ρα ™στˆν ¹ ΖΘ τÍ ΖΚ µήκει. αƒ ΖΘ, ΖΚ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΚΘ, προσαρµόζουσα δ αÙτÍ ¹ ΚΖ. ½τοι δ¾ ¹

C

L

G

K

F

D For let the medial (area) BD have been subtracted from the rational (area) BC. I say that one of two irrational (straight-lines) arise (as) the square-root of the remaining (area), EC—either an apotome, or a minor (straight-line). For let the rational (straight-line) F G have been laid out, and let the right-angled parallelogram GH, equal to BC, have been applied to F G, and let GK, equal to DB, have been subtracted (from GH). Thus, the remainder EC is equal to LH. Therefore, since BC is a rational (area), and BD a medial (area), and BC (is) equal to GH, and BD to GK, GH is thus a rational (area), and GK a medial (area). And they are applied to the rational (straight-line) F G. Thus, F H (is) rational, and commensurable in length with F G [Prop. 10.20], and F K (is)

411

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

ΘΖ τÁς ΖΚ µε‹ζον δύναται τù ¢πÕ συµµέτρου À οÜ. ∆υνάσθω πρότερον τù ¢πÕ συµµέτρου. καί ™στιν Óλη ¹ ΘΖ σύµµετρος τÍ ™κκειµένV ·ητÍ µήκει τÍ ΖΗ· ¢ποτοµ¾ ¥ρα πρώτη ™στˆν ¹ ΚΘ. τÕ δ' ØπÕ ·ητÁς κሠ¢ποτοµÁς πρώτης περιεχόµενον ¹ δυναµένη ¢ποτοµή ™στιν. ¹ ¥ρα τÕ ΛΘ, τουτέστι τÕ ΕΓ, δυναµένη ¢ποτοµή ™στιν. Ε„ δ ¹ ΘΖ τÁς ΖΚ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ, καί ™στιν Óλη ¹ ΖΘ σύµµετρος τÍ ™κκειµένV ·ητÍ µήκει τÍ ΖΗ, ¢ποτοµ¾ τετάρτη ™στˆν ¹ ΚΘ. τÕ δ' ØπÕ ·ητÁς κሠ¢ποτοµÁς τετάρτης περιεχόµενον ¹ δυναµένη ™λάσσων ™στίν· Óπερ œδει δε‹ξαι.

also rational, and incommensurable in length with F G [Prop. 10.22]. Thus, F H is incommensurable in length with F K [Prop. 10.13]. F H and F K are thus rational (straight-lines which are) commensurable in square only. Thus, KH is an apotome [Prop. 10.73], and KF an attachment to it. So, the square on HF is greater than (the square on) F K by the (square) on (some straight-line which is) either commensurable (in length with HF ), or not (commensurable). First, let the square (on it) be (greater) by the (square) on (some straight-line which is) commensurable (in length with HF ). And the whole of HF is commensurable in length with the (previously) laid down rational (straight-line) F G. Thus, KH is a first apotome [Def. 10.1]. And the square-root of an (area) contained by a rational (straight-line) and a first apotome is an apotome [Prop. 10.91]. Thus, the square-root of LH—that is to say, (of) EC—is an apotome. And if the square on HF is greater than (the square on) F K by the (square) on (some straight-line which is) incommensurable (in length) with (HF ), and (since) the whole of F H is commensurable in length with the (previously) laid down rational (straight-line) F G, KH is a fourth apotome [Prop. 10.14]. And the square-root of an (area) contained by a rational (straight-line) and a fourth apotome is a minor (straight-line) [Prop. 10.94]. (Which is) the very thing it was required to show.

ρθ΄.

Proposition 109

'ΑπÕ µέσου ·ητοà ¢φαιρουµένου ¥λλαι δύο ¥λοA rational (area) being subtracted from a medial γοι γίνονται ½τοι µέσης ¢ποτοµ¾ πρώτη À µετ¦ ·ητοà (area), two other irrational (straight-lines) arise (as the µέσον τÕ Óλον ποιοàσα. square-root of the remaining area)—either a first apotome of a medial (straight-line), or that (straight-line) which with a rational (area) makes a medial whole. B E F K H Ζ Κ Θ Β Ε

Α

∆

Γ

A

Η Λ 'ΑπÕ γ¦ρ µέσου τοà ΒΓ ·ητÕν ¢φVρήσθω τÕ Β∆. λέγω, Óτι ¹ τÕ λοιπÕν τÕ ΕΓ δυναµένη µία δύο ¢λόγων γίνεται ½τοι µέσης ¢ποτοµ¾ πρώτη À µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσα.

D

C

G L For let the rational (area) BD have been subtracted from the medial (area) BC. I say that one of two irrational (straight-lines) arise (as) the square-root of the remaining (area), EC—either a first apotome of a medial

412

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

'Εκκείσθω γ¦ρ ·ητ¾ ¹ ΖΗ, κሠπαραβεβλήσθω еοίως τ¦ χωρία. œστι δ¾ ¢κολούθως ·ητ¾ µν ¹ ΖΘ κሠ¢σύµµετρος τÍ ΖΗ µήκει, ·ητ¾ δ ¹ ΚΖ κሠσύµµετρος τÍ ΖΗ µήκει· αƒ ΖΘ, ΖΚ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΚΘ, προσαρµόζουσα δ ταύτV ¹ ΖΚ. ½τοι δ¾ ¹ ΘΖ τÁς ΖΚ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ À τù ¢πÕ ¢συµµέτρου. Ε„ µν οâν ¹ ΘΖ τÁς ΖΚ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, καί ™στιν ¹ προσαρµόζουσα ¹ ΖΚ σύµµετρος τÍ ™κκειµένV ·ητÍ µήκει τÍ ΖΗ, ¢ποτοµ¾ δευτέρα ™στˆν ¹ ΚΘ. ·ητ¾ δ ¹ ΖΗ· éστε ¹ τÕ ΛΘ, τουτέστι τÕ ΕΓ, δυναµένη µέσης ¢ποτοµ¾ πρώτη ™στίν. Ε„ δ ¹ ΘΖ τÁς ΖΚ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου, καί ™στιν ¹ προσαρµόζουσα ¹ ΖΚ σύµµετρος τÍ ™κκειµένV ·ητÍ µήκει τÍ ΖΗ, ¢ποτοµ¾ πέµπτη ™στˆν ¹ ΚΘ· éστε ¹ τÕ ΕΓ δυναµένη µετ¦ ·ητοà µέσον τÕ Óλον ποιοàσά ™στιν· Óπερ œδει δε‹ξαι.

(straight-line), or that (straight-line) which with a rational (area) makes a medial whole. For let the rational (straight-line) F G be laid down, and let similar areas (to the preceding proposition) have been applied (to it). So, analogously, F H is rational, and incommensurable in length with F G, and KF (is) also rational, and commensurable in length with F G. Thus, F H and F K are rational (straight-lines which are) commensurable in square only [Prop. 10.13]. KH is thus an apotome [Prop. 10.73], and F K an attachment to it. So, the square on HF is greater than (the square on) F K either by the (square) on (some straight-line) commensurable (in length) with (HF ), or by the (square) on (some straight-line) incommensurable (in length with HF ). Therefore, if the square on HF is greater than (the square on) F K by the (square) on (some straight-line) commensurable (in length) with (HF ), and (since) the attachment F K is commensurable in length with the (previously) laid down rational (straight-line) F G, KH is a second apotome [Def. 10.12]. And F G (is) rational. Hence, the square-root of LH—that is to say, (of) EC—is a first apotome of a medial (straight-line) [Prop. 10.92]. And if the square on HF is greater than (the square on) F K by the (square) on (some straight-line) incommensurable (in length with HF ), and (since) the attachment F K is commensurable in length with the (previously) laid down rational (straight-line) F G, KH is a fifth apotome [Def. 10.15]. Hence, the square-root of EC is that (straight-line) which with a rational (area) makes a medial whole [Prop. 10.95]. (Which is) the very thing it was required to show.

ρι΄.

Proposition 110

'ΑπÕ µέσου µέσου ¢φαιρουµένου ¢συµµέτρου τù ÓλJ αƒ λοιπሠδύο ¥λογοι γίνονται ½τοι µέσης ¢ποτοµ¾ δευτέρα À µετ¦ µέσου µέσον τÕ Óλον ποιοàσα. 'ΑφVρήσθω γ¦ρ æς ™πˆ τîν προκειµένων καταγραφîν ¢πÕ µέσου τοà ΒΓ µέσον τÕ Β∆ ¢σύµµετρον τù ÓλJ· λέγω, Óτι ¹ τÕ ΕΓ δυναµένη µία ™στˆ δύο ¢λόγων ½τοι µέσης ¢ποτοµ¾ δευτέρα À µετ¦ µέσου µέσον τÕ Óλον ποιοàσα.

A medial (area), incommensurable with the whole, being subtracted from a medial (area), the two remaining irrational (straight-lines) arise (as) the (square-root of the area)—either a second apotome of a medial (straightline), or that (straight-line) which with a medial (area) makes a medial whole. For, as in the previous figures, let the medial (area) BD, incommensurable with the whole, have been subtracted from the medial (area) BC. I say that the squareroot of EC is one of two irrational (straight-lines)— either a second apotome of a medial (straight-line), or that (straight-line) which with a medial (area) makes a medial whole.

413

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

Β

Ε

Α

∆

Ζ

Κ Θ

Γ Η

Λ

'Επεˆ γ¦ρ µέσον ™στˆν ˜κάτερον τîν ΒΓ, Β∆, κሠ¢σύµµετρον τÕ ΒΓ τù Β∆, œσται ¢κολούθως ·ητ¾ ˜κατέρα τîν ΖΘ, ΖΚ κሠ¢σύµµετρος τÍ ΖΗ µήκει. κሠ™πεˆ ¢σύµµετρόν ™στι τÕ ΒΓ τù Β∆, τουτέστι τÕ ΗΘ τù ΗΚ, ¢σύµµετρος κሠ¹ ΘΖ τÍ ΖΚ· αƒ ΖΘ, ΖΚ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΚΘ [προσαρµόζουσα δ ¹ ΖΚ. ½τοι δ¾ ¹ ΖΘ τÁς ΖΚ µε‹ζον δύναται τù ¢πÕ συµµέτρου À τù ¢πÕ ¢συµµέτρου ˜αυτÍ]. Ε„ µν δ¾ ¹ ΖΘ τÁς ΖΚ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠοÙθετέρα τîν ΖΘ, ΖΚ σύµµετρός ™στι τÍ ™κκειµέµνV ·ητÍ µήκει τÍ ΖΗ, ¢ποτοµ¾ τρίτη ™στˆν ¹ ΚΘ. ·ητ¾ δ ¹ ΚΛ, τÕ δ' ØπÕ ·ητÁς κሠ¢ποτοµÁς τρίτης περιεχόµενον Ñρθογώνιον ¥λογόν ™στιν, κሠ¹ δυναµένη αÙτÕ ¥λογός ™στιν, καλε‹ται δ µέσης ¢ποτοµ¾ δευτέρα· éστε ¹ τÕ ΛΘ, τουτέστι τÕ ΕΓ, δυναµένη µέσης ¢ποτοµή ™στι δευτερά. Ε„ δ ¹ ΖΘ τÁς ΖΚ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ [µήκει], κሠοÙθετέρα τîν ΘΖ, ΖΚ σύµµετρός ™στι τÍ ΖΗ µήκει, ¢ποτοµ¾ ›κτη ™στˆν ¹ ΚΘ. τÕ δ' ØπÕ ·ητÁς κሠ¢ποτοµÁς ›κτης ¹ δυναµένη ™στˆ µετ¦ µέσου µέσον τÕ Óλον ποιοàσα. ¹ τÕ ΛΘ ¥ρα, τουτέστι τÕ ΕΓ, δυναµένη µετ¦ µέσου µέσον τÕ Óλον ποιοàσά ™στιν· Óπερ œδει δε‹ξαι.

B

E

A

D

F

K

G

L

H

C

For since BC and BD are each medial (areas), and BC (is) incommensurable with BD, analogously (to the previous propositions), F H and F K will each be rational (straight-lines), and incommensurable in length with F G [Prop. 10.22]. And since BC is incommensurable with BD—that is to say, GH with GK—HF (is) also incommensurable (in length) with F K [Props. 6.1, 10.11]. Thus, F H and F K are rational (straight-lines which are) commensurable in square only. KH is thus as apotome [Prop. 10.73], [and F K an attachment (to it). So, the square on F H is greater than (the square on) F K either by the (square) on (some straight-line) commensurable, or by the (square) on (some straight-line) incommensurable, (in length) with (F H).] So, if the square on F H is greater than (the square on) F K by the (square) on (some straight-line) commensurable (in length) with (F H), and (since) neither of F H and F K is commensurable in length with the (previously) laid down rational (straight-line) F G, KH is a third apotome [Def. 10.3]. And KL (is) rational. And the rectangle contained by a rational (straight-line) and a third apotome is irrational, and the square-root of it is that irrational (straight-line) called a second apotome of a medial (straight-line) [Prop. 10.93]. Hence, the squareroot of LH—that is to say, (of) EC—is a second apotome of a medial (straight-line). And if the square on F H is greater than (the square on) F K by the (square) on (some straight-line) incommensurable [in length] with (F H), and (since) neither of HF and F K is commensurable in length with F G, KH is a sixth apotome [Def. 10.16]. And the square-root of the (rectangle contained) by a rational (straight-line) and a sixth apotome is that (straight-line) which with a medial (area) makes a medial whole [Prop. 10.96]. Thus, the square-root of LH—that is to say, (of) EC—is that (straight-line) which with a medial (area) makes a medial whole. (Which is) the very thing it was required to

414

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 show.

ρια΄.

Proposition 111

`Η ¢ποτοµ¾ οÙκ œστιν ¹ αÙτ¾ τÍ ™κ δύο Ñνοµάτων.

Α ∆

An apotome is not the same as a binomial.

Β Η

Ε

A Ζ

D

Γ

B G

E

F

C

”Εστω ¢ποτοµ¾ ¹ ΑΒ· λέγω, Óτι ¹ ΑΒ οÙκ œστιν ¹ αÙτ¾ τÍ ™κ δύο Ñνοµάτων. Ε„ γ¦ρ δυνατόν, œστω· κሠ™κκείσθω ·ητ¾ ¹ ∆Γ, κሠτù ¢πÕ τÁς ΑΒ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω Ñρθογώνιον τÕ ΓΕ πλάτος ποιοàν τ¾ν ∆Ε. ™πεˆ οâν ¢ποτοµή ™στιν ¹ ΑΒ, ¢ποτοµ¾ πρώτη ™στˆν ¹ ∆Ε. œστω αÙτÍ προσαρµόζουσα ¹ ΕΖ· αƒ ∆Ζ, ΖΕ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ ∆Ζ τÁς ΖΕ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠ¹ ∆Ζ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ µήκει τÍ ∆Γ. πάλιν, ™πεˆ ™κ δύο Ñνοµάτων ™στˆν ¹ ΑΒ, ™κ δύο ¥ρα Ñνοµάτων πρώτη ™στˆν ¹ ∆Ε. διVρήσθω ε„ς τ¦ Ñνόµατα κατ¦ τÕ Η, κሠœστω µε‹ζον Ôνοµα τÕ ∆Η· αƒ ∆Η, ΗΕ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι, κሠ¹ ∆Η τÁς ΗΕ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠτÕ µε‹ζον ¹ ∆Η σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ µήκει τÍ ∆Γ. κሠ¹ ∆Ζ ¥ρα τÍ ∆Η σύµµετρός ™στι µήκει· κሠλοιπ¾ ¥ρα ¹ ΗΖ σύµµετρός ™στι τÍ ∆Ζ µήκει. [™πεˆ οàν σύµµετρός ™στιν ¹ ∆Ζ τÍ ΗΖ, ·ητ¾ δέ ™στιν ¹ ∆Ζ, ·ητ¾ ¥ρα ™στˆ κሠ¹ ΗΖ. ™πεˆ οâν σύµµετρός ™στιν ¹ ∆Ζ τÍ ΗΖ µήκει] ¢σύµµετρος δ ¹ ∆Ζ τÍ ΕΖ µήκει. ¢σύµµετρος ¥ρα ™στˆ κሠ¹ ΖΗ τÍ ΕΖ µήκει. αƒ ΗΖ, ΖΕ ¥ρα ·ηταί [ε„σι] δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΕΗ. ¢λλ¦ κሠ·ητή· Óπερ ™στˆν ¢δύνατον. `Η ¥ρα ¢ποτοµ¾ οÙκ œστιν ¹ αÙτ¾ τÍ ™κ δύο Ñνοµάτων· Óπερ œδει δε‹ξαι.

Let AB be an apotome. I say that AB is not the same as a binomial. For, if possible, let it be (the same). And let a rational (straight-line) DC be laid down. And let the rectangle CE, equal to the (square) on AB, have been applied to CD, producing DE as breadth. Therefore, since AB is an apotome, DE is a first apotome [Prop. 10.97]. Let EF be an attachment to it. Thus, DF and F E are rational (straight-lines which are) commensurable in square only, and the square on DF is greater than (the square on) F E by the (square) on (some straight-line) commensurable (in length) with (DE), and DF is commensurable in length with the (previously) laid down rational (straightline) DC [Def. 10.10]. Again, since AB is a binomial, DE is thus a first binomial [Prop. 10.60]. Let (DE) have been divided into its (component terms at G, and let DG be the greater term. Thus, DG and GE are rational (straight-lines which are) commensurable in square only, and the square on DG is greater than (the square on) GE by the (square) on (some straight-line) commensurable (in length) with (DG), and the greater (term) DG is commensurable in length with the (previously) laid down rational (straight-line) DC [Def. 10.5]. Thus, DF is also commensurable in length with DG [Prop. 10.12]. The remainder GF is thus commensurable in length with DF [Prop. 10.15]. [Therefore, since DF is commensurable with GF , and DF is rational, GF is thus also rational. Therefore, since DF is commensurable in length with GF ,] DF (is) incommensurable in length with EF . Thus, F G is also incommensurable in length with EF [Prop. 10.13]. GF and F E [are] thus rational (straight-

415

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 lines which are) commensurable in square only. Thus, EG is an apotome [Prop. 10.73]. But, (it is) also rational. The very thing is impossible. Thus, an apotome is not the same as a binomial. (Which is) the very thing it was required to show.

[Πόρισµα.]

[Corollary]

`Η ¢ποτοµ¾ καˆ αƒ µετ' αÙτ¾ν ¥λογοι οÜτε τÍ µέσV οÜτε ¢λλήλαις ε„σˆν αƒ αÙταί. ΤÕ µν γ¦ρ ¢πÕ µέσης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ·ητ¾ν κሠ¢σύµµετρον τÍ, παρ' ¿ν παράκειται, µήκει, τÕ δ ¢πÕ ¢ποτοµÁς παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν πρώτην, τÕ δ ¢πÕ µέσης ¢ποτοµÁς πρώτης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν δευτέραν, τÕ δ ¢πÕ µέσης ¢ποτοµÁς δευτέρας παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν τρίτην, τÕ δ ¢πÕ ™λάσσονος παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν τετάρτην, τÕ δ ¢πÕ τÁς µετ¦ ·ητοà µέσον τÕ Óλον ποιούσης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν πέµπτην, τÕ δ ¢πÕ τÁς µετ¦ µέσου µέσον τÕ Óλον ποιούσης παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν ›κτην. ™πεˆ οâν τ¦ ε„ρηµένα πλάτη διαφέρει τοà τε πρώτου κሠ¢λλήλων, τοà µν πρώτου, Óτι ·ητή ™στιν, ¢λλήλων δ, ™πεˆ τÍ τάξει οÙκ ε„σˆν αƒ αÙταί, δÁλον, æς κሠαÙταˆ αƒ ¥λογοι διαφέρουσιν ¢λλήλων. κሠ™πεˆ δέδεικται ¹ ¢ποτοµ¾ οÙκ οâσα ¹ αÙτ¾ τÍ ™κ δύο Ñνοµάτων, ποιοàσι δ πλάτη παρ¦ ·ητ¾ν παραβαλλόµεναι αƒ µετ¦ τ¾ν ¢ποτοµ¾ν ¢ποτﵦς ¢κολούθως ˜κάστη τÍ τάξει τÍ καθ' αØτήν, αƒ δ µετ¦ τ¾ν ™κ δύο Ñνοµάτων τ¦ς ™κ δύο Ñνοµάτων κሠαÙτሠτÍ τάξει ¢κολούθως, ›τεραι ¥ρα ε„σˆν αƒ µετ¦ τ¾ν ¢ποτοµ¾ν κሠ›τεραι αƒ µετ¦ τ¾ν ™κ δύο Ñνοµάτων, æς εναι τÍ τάξει πάσας ¢λόγους ιγ,

The apotome, and the irrational (straight-lines) after it, are neither the same as a medial (straight-line), nor (the same) as one another. For the (square) on a medial (straight-line), applied to a rational (straight-line), produces as breadth a rational (straight-line which is) incommensurable in length with the (straight-line) to which (the area) is applied [Prop. 10.22]. And the (square) on an apotome, applied to a rational (straight-line), produces as breadth a first apotome [Prop. 10.97]. And the (square) on a first apotome of a medial (straight-line), applied to a rational (straight-line), produces as breadth a second apotome [Prop. 10.98]. And the (square) on a second apotome of a medial (straight-line), applied to a rational (straightline), produces as breadth a third apotome [Prop. 10.99]. And (square) on a minor (straight-line), applied to a rational (straight-line), produces as breadth a fourth apotome [Prop. 10.100]. And (square) on that (straight-line) which with a rational (area) produces a medial whole, applied to a rational (straight-line), produces as breadth a fifth apotome [Prop. 10.101]. And (square) on that (straight-line) which with a medial (area) produces a medial whole, applied to a rational (straight-line), produces as breadth a sixth apotome [Prop. 10.102]. Therefore, since the aforementioned breadths differ from the first (breadth), and from one another—from the first, because it is rational, and from one another since they are not the same in order—clearly, the irrational (straightlines) themselves also differ from one another. And since it has been shown that an apotome is not the same as a binomial [Prop. 10.111], and (that) the (irrational straight-lines) after the apotome, being applied to a rational (straight-line), produce as breadth, each according to its own (order), apotomes, and (that) the (irrational straight-lines) after the binomial also themselves (produce), according (to their) order, binomials, the (irrational straight-lines) after the apotome are thus different, and the (irrational straight-lines) after the binomial (are also) different, so that there are, in order, 13 irrational (straight-lines) in all:

416

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

Μέσην, 'Εκ δύο Ñνοµάτων,

Medial, Binomial,

'Εκ δύο µέσων πρώτην, 'Εκ δύο µέσων δευτέραν,

First bimedial, Second bimedial,

Μείζονα, `ΡητÕν κሠµέσον δυναµένην,

Major, Square-root of a rational plus a medial (area),

∆ύο µέσα δυναµένην, 'Αποτοµήν,

Square-root of (the sum of) two medial (areas), Apotome,

̘σης ¢ποτοµ¾ν πρώτην,

First apotome of a medial,

̘σης ¢ποτοµ¾ν δευτέραν, 'Ελάσσονα,

Second apotome of a medial, Minor,

Μετ¦ ·ητοà µέσον τÕ Óλον ποιοàσαν,

That which with a rational (area) produces a medial whole,

Μετ¦ µέσου µέσον τÕ Óλον ποιοàσαν.

That which with a medial (area) produces a medial whole.

ριβ΄.

Proposition 112†

ΤÕ ¢πÕ ·ητÁς παρ¦ τ¾ν ™κ δύο Ñνοµάτων παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµήν, Âς τ¦ Ñνόµατα σύµµετρά ™στι το‹ς τÁς ™κ δύο Ñνοµάτων Ñνόµασι κሠœτι ™ν τù αÙτù λόγJ, κሠœτι ¹ γινοµένη ¢ποτοµ¾ τ¾ν αÙτ¾ν ›ξει τάξιν τÍ ™κ δύο Ñνοµάτων.

The (square) on a rational (straight-line), applied to a binomial (straight-line), produces as breadth an apotome whose terms are commensurable (in length) with the terms of the binomial, and, furthermore, in the same ratio. Moreover, the created apotome will have the same order as the binomial.

Α Β Κ

A ∆

Γ Ε

B

Η Ζ

Θ

K

”Εστω ·ητ¾ µν ¹ Α, ™κ δύο Ñνοµάτων δ ¹ ΒΓ, Âς µε‹ζον Ôνοµα œστω ¹ ∆Γ, κሠτù ¢πÕ τÁς Α ‡σον œστω τÕ ØπÕ τîν ΒΓ, ΕΖ· λέγω, Óτι ¹ ΕΖ ¢ποτοµή ™στιν, Âς τ¦ Ñνόµατα σύµµετρά ™στι το‹ς Γ∆, ∆Β, κሠ™ν τù αÙτù λόγJ, κሠœτι ¹ ΕΖ τ¾ν αÙτ¾ν ›ξει τάξιν τÍ ΒΓ. ”Εστω γ¦ρ πάλιν τù ¢πÕ τÁς Α ‡σον τÕ ØπÕ τîν Β∆, Η. ™πεˆ οâν τÕ ØπÕ τîν ΒΓ, ΕΖ ‡σον ™στˆ τù ØπÕ τîν Β∆, Η, œστιν ¥ρα æς ¹ ΓΒ πρÕς τ¾ν Β∆, οÛτως ¹ Η πρÕς τ¾ν ΕΖ. µείζων δ ¹ ΓΒ τÁς Β∆· µείζων ¥ρα ™στˆ κሠ¹ Η τÁς ΕΖ. œστω τÍ Η ‡ση ¹ ΕΘ· œστιν ¥ρα æς ¹ ΓΒ πρÕς τ¾ν Β∆, οÛτως ¹ ΘΕ πρÕς τ¾ν ΕΖ· διελόντι ¥ρα ™στˆν æς ¹ Γ∆ πρÕς τ¾ν Β∆, οÛτως ¹ ΘΖ πρÕς τ¾ν ΖΕ. γεγονέτω æς ¹ ΘΖ πρÕς τ¾ν ΖΕ, οÛτως ¹ ΖΕ πρÕς τ¾ν ΚΕ· κሠÓλη ¥ρα ¹ ΘΚ πρÕς Óλην τ¾ν ΚΖ ™στιν, æς ¹ ΖΚ πρÕς ΚΕ· æς γ¦ρ žν τîν ¹γουµένων πρÕς žν τîν ˜ποµένων, οÛτως ¤παντα τ¦ ¹γούµενα πρÕς ¤παντα τ¦ ˜πόµενα. æς δ ¹ ΖΚ πρÕς ΚΕ, οÛτως ™στˆν ¹ Γ∆ πρÕς τ¾ν ∆Β· κሠæς ¥ρα ¹ ΘΚ πρÕς ΚΖ, οÛτως ¹ Γ∆ πρÕς τ¾ν ∆Β. σύµµετρον δ τÕ ¢πÕ τÁς Γ∆ τù ¢πÕ τÁς

D

C E

G F

H

Let A be a rational (straight-line), and BC a binomial (straight-line), of which let DC be the greater term. And let the (rectangle contained) by BC and EF be equal to the (square) on A. I say that EF is an apotome whose terms are commensurable (in length) with CD and DB, and in the same ratio, and, moreover, that EF will have the same order as BC. For, again, let the (rectangle contained) by BD and G be equal to the (square) on A. Therefore, since the (rectangle contained) by BC and EF is equal to the (rectangle contained) by BD and G, thus as CB is to BD, so G (is) to EF [Prop. 6.16]. And CB (is) greater than BD. Thus, G is also greater than EF [Props. 5.16, 5.14]. Let EH be equal to G. Thus, as CB is to BD, so HE (is) to EF . Thus, via separation, as CD is to BD, so HF (is) to F E [Prop. 5.17]. Let it have been contrived that as HF (is) to F E, so F K (is) to KE. And, thus, the whole HK is to the whole KF , as F K (is) to KE. For as one of the leading (proportional magnitudes is) to one of the

417

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

∆Β· σύµµετρον ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς ΘΚ τù ¢πÕ τÁς ΚΖ. καί ™στιν æς τÕ ¢πÕ τÁς ΘΚ πρÕς τÕ ¢πÕ τÁς ΚΖ, οÛτως ¹ ΘΚ πρÕς τ¾ν ΚΕ, ™πεˆ αƒ τρε‹ς αƒ ΘΚ, ΚΖ, ΚΕ ¢νάλογόν ε„σιν. σύµµετρος ¥ρα ¹ ΘΚ τÍ ΚΕ µήκει. éστε κሠ¹ ΘΕ τÍ ΕΚ σύµµετρός ™στι µήκει. κሠ™πεˆ τÕ ¢πÕ τÁς Α ‡σον ™στˆ τù ØπÕ τîν ΕΘ, Β∆, ·ητÕν δέ ™στι τÕ ¢πÕ τÁς Α, ·ητÕν ¥ρα ™στˆ κሠτÕ ØπÕ τîν ΕΘ, Β∆. κሠπαρ¦ ·ητ¾ν τ¾ν Β∆ παράκειται· ·ητ¾ ¥ρα ™στˆν ¹ ΕΘ κሠσύµµετρος τÍ Β∆ µήκει· éστε κሠ¹ σύµµετρος αÙτÍ ¹ ΕΚ ·ητή ™στι κሠσύµµετρος τÍ Β∆ µήκει. ™πεˆ οâν ™στιν æς ¹ Γ∆ πρÕς ∆Β, οÛτως ¹ ΖΚ πρÕς ΚΕ, αƒ δ Γ∆, ∆Β δυνάµει µόνον ε„σˆ σύµµετροι, καˆ αƒ ΖΚ, ΚΕ δυνάµει µόνον ε„σˆ σύµµετροι. ·ητ¾ δέ ™στιν ¹ ΚΕ· ·ητ¾ ¥ρα ™στˆ κሠ¹ ΖΚ. αƒ ΖΚ, ΚΕ ¥ρα ·ητሠδυνάµει µόνον ε„σˆ σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΕΖ. ”Ητοι δ ¹ Γ∆ τÁς ∆Β µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ À τù ¢πÕ ¢συµµέτρου. Ε„ µν οâν ¹ Γ∆ τÁς ∆Β µε‹ζον δύναται τù ¢πÕ συµµέτρου [˜αυτÍ], κሠ¹ ΖΚ τÁς ΚΕ µε‹ζον δυνήσεται τù ¢πÕ συµµέτρου ˜αυτÍ. καˆ ε„ µν σύµµετρός ™στιν ¹ Γ∆ τÍ ™κκειµένV ·ητÍ µήκει, κሠ¹ ΖΚ· ε„ δ ¹ Β∆, κሠ¹ ΚΕ· ε„ δ οÙδετέρα τîν Γ∆, ∆Β, κሠοÙδετέρα τîν ΖΚ, ΚΕ. Ε„ δ ¹ Γ∆ τÁς ∆Β µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ, κሠ¹ ΖΚ τÁς ΚΕ µε‹ζον δυνήσεται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καˆ ε„ µν ¹ Γ∆ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ µήκει, κሠ¹ ΖΚ· ε„ δ ¹ Β∆, κሠ¹ ΚΕ· ε„ δ οÙδετέρα τîν Γ∆, ∆Β, κሠοÙδετέρα τîν ΖΚ, ΚΕ· éστε ¢ποτοµή ™στιν ¹ ΖΕ, Âς τ¦ Ñνόµατα τ¦ ΖΚ, ΚΕ σύµµετρά ™στι το‹ς τÁς ™κ δύο Ñνοµάτων Ñνόµασι το‹ς Γ∆, ∆Β κሠ™ν τù αÙτù λόγJ, κሠτ¾ν αÙτÁν τάξιν œχει τÍ ΒΓ· Óπερ œδει δε‹ξαι.

following, so all of the leading (magnitudes) are to all of the following [Prop. 5.12]. And as F K (is) to KE, so CD is to DB [Prop. 5.11]. And, thus, as HK (is) to KF , so CD is to DB [Prop. 5.11]. And the (square) on CD (is) commensurable with the (square) on DB [Prop. 10.36]. The (square) on HK is thus also commensurable with the (square) on KF [Props. 6.22, 10.11]. And as the (square) on HK is to the (square) on KF , so HK (is) to KE, since the three (straight-lines) HK, KF , and KE are proportional [Def. 5.9]. HK is thus commensurable in length with KE [Prop. 10.11]. Hence, HE is also commensurable in length with EK [Prop. 10.15]. And since the (square) on A is equal to the (rectangle contained) by EH and BD, and the (square) on A is rational, the (rectangle contained) by EH and BD is thus also rational. And it is applied to the rational (straight-line) BD. Thus, EH is also rational, and commensurable in length with BD [Prop. 10.20]. And, hence, the (straight-line) commensurable (in length) with it, EK, is also rational [Def. 10.3], and commensurable in length with BD [Prop. 10.12]. Therefore, since as CD is to DB, so F K (is) to KE, and CD and DB are (straight-lines which are) commensurable in square only, F K and KE are also commensurable in square only [Prop. 10.11]. And KE is rational. Thus, F K is also rational. F K and KE are thus rational (straight-lines which are) commensurable in square only. Thus, EF is an apotome [Prop. 10.73]. And the square on CD is greater than (the square on) DB either by the (square) on (some straight-line) commensurable, or by the (square) on (some straight-line) incommensurable, (in length) with (CD). Therefore, if the square on CD is greater than (the square on) DB by the (square) on (some straight-line) commensurable (in length) with [CD], then the square on F K will also be greater than (the square on) KE by the (square) on (some straight-line) commensurable (in length) with (F K) [Prop. 10.14]. And if CD is commensurable in length with a (previously) laid down rational (straight-line), (so) also (is) F K [Props. 10.11, 10.12]. And if BD (is commensurable), (so) also (is) KE [Prop. 10.12]. And if neither of CD or DB (is commensurable), neither also (are) either of F K or KE. And if the square on CD is greater than (the square on) DB by the (square) on (some straight-line) incommensurable (in length) with (CD), then the square on F K will also be greater than (the square on) KE by the (square) on (some straight-line) incommensurable (in length) with (F K) [Prop. 10.14]. And if CD is commensurable in length with a (previously) laid down rational (straight-line), (so) also (is) F K [Props. 10.11, 10.12]. And if BD (is commensurable), (so) also (is) KE

418

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10 [Prop. 10.12]. And if neither of CD or DB (is commensurable), neither also (are) either of F K or KE. Hence, F E is an apotome whose terms, F K and KE, are commensurable (in length) with the terms, CD and DB, of the binomial, and in the same ratio. And (F E) has the same order as BC [Defs. 10.5—10.10]. (Which is) the very thing it was required to show.

†

Heiberg considers this proposition, and the succeeding ones, to be relatively early interpolations into the original text.

ριγ΄.

Proposition 113

ΤÕ ¢πÕ ·ητÁς παρ¦ ¢ποτοµ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ™κ δύο Ñνοµάτων, Âς τ¦ Ñνόµατα σύµµετρά ™στι το‹ς τÁς ¢ποτοµÁς Ñνόµασι κሠ™ν τù αÙτù λόγJ, œτι δ ¹ γινοµένη ™κ δύο Ñνοµάτων τ¾ν αÙτ¾ν τάξιν œχει τÍ ¢ποτοµÍ.

The (square) on a rational (straight-line), applied to an apotome, produces as breadth a binomial whose terms are commensurable with the terms of the apotome, and in the same ratio. Moreover, the created binomial has the same order as the apotome.

Α Β Κ

A ∆

Γ

B

Η Ε Ζ

Θ

K

”Εστω ·ητ¾ µν ¹ Α, ¢ποτοµ¾ δ ¹ Β∆, κሠτù ¢πÕ τÁς Α ‡σον œστω τÕ ØπÕ τîν Β∆, ΚΘ, éστε τÕ ¢πÕ τÁς Α ·ητÁς παρ¦ τ¾ν Β∆ ¢ποτοµ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν ΚΘ· λέγω, Óτι ™κ δύο Ñνοµάτων ™στˆν ¹ ΚΘ, Âς τ¦ Ñνόµατα σύµµετρά ™στι το‹ς τÁς Β∆ Ñνόµασι κሠ™ν τù αÙτù λόγJ, κሠœτι ¹ ΚΘ τ¾ν αÙτ¾ν œχει τάξιν τÍ Β∆. ”Εστω γ¦ρ τÍ Β∆ προσαρµόζουσα ¹ ∆Γ· αƒ ΒΓ, Γ∆ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. κሠτù ¢πÕ τÁς Α ‡σον œστω κሠτÕ ØπÕ τîν ΒΓ, Η. ·ητÕν δ τÕ ¢πÕ τÁς Α· ·ητÕν ¥ρα κሠτÕ ØπÕ τîν ΒΓ, Η. κሠπαρ¦ ·ητ¾ν τ¾ν ΒΓ παραβέβληται· ·ητ¾ ¥ρα ™στˆν ¹ Η κሠσύµµετρος τÍ ΒΓ µήκει. ™πεˆ οâν τÕ ØπÕ τîν ΒΓ, Η ‡σον ™στˆ τù ØπÕ τîν Β∆, ΚΘ, ¢νάλογον ¥ρα ™στˆν æς ¹ ΓΒ πρÕς Β∆, οÛτως ¹ ΚΘ πρÕς Η. µείζων δ ¹ ΒΓ τÁς Β∆· µείζων ¥ρα κሠ¹ ΚΘ τÁς Η. κείσθω τÍ Η ‡ση ¹ ΚΕ· σύµµετρος ¥ρα ™στˆν ¹ ΚΕ τÍ ΒΓ µήκει. κሠ™πεί ™στιν æς ¹ ΓΒ πρÕς Β∆, οÛτως ¹ ΘΚ πρÕς ΚΕ, ¢ναστρέψαντι ¥ρα ™στˆν æς ¹ ΒΓ πρÕς τ¾ν Γ∆, οÛτως ¹ ΚΘ πρÕς ΘΕ. γεγονέτω æς ¹ ΚΘ πρÕς ΘΕ, οÛτως ¹ ΘΖ πρÕς ΖΕ· κሠλοιπ¾ ¥ρα ¹ ΚΖ πρÕς ΖΘ ™στιν, æς ¹ ΚΘ πρÕς ΘΕ, τουτέστιν [æς] ¹ ΒΓ πρÕς Γ∆. αƒ δ ΒΓ, Γ∆ δυνάµει µόνον [ε„σˆ] σύµµετροι· καˆ αƒ ΚΖ, ΖΘ ¥ρα δυνάµει µόνον ε„σˆ σύµµετροι· κሠ™πεί ™στιν æς ¹ ΚΘ πρÕς ΘΕ, ¹ ΚΖ πρÕς ΖΘ, ¢λλ' æς ¹ ΚΘ πρÕς ΘΕ, ¹ ΘΖ πρÕς ΖΕ, κሠæς ¥ρα ¹ ΚΖ πρÕς ΖΘ, ¹ ΘΖ πρÕς ΖΕ· éστε κሠæς ¹ πρώτη πρÕς τ¾ν τρίτην, τÕ ¢πÕ τÁς πρώτης πρÕς τÕ ¢πÕ τÁς δευτέρας· κሠæς ¥ρα ¹ ΚΖ πρÕς ΖΕ, οÛτως τÕ ¢πÕ τÁς ΚΖ πρÕς

D

C

G E F

H

Let A be a rational (straight-line), and BD an apotome. And let the (rectangle contained) by BD and KH be equal to the (square) on A, such that the square on the rational (straight-line) A, applied to the apotome BD, produces KH as breadth. I say that KH is a binomial whose terms are commensurable with the terms of BD, and in the same ratio, and, moreover, that KH has the same order as BD. For let DC be an attachment to BD. Thus, BC and CD are rational (straight-lines which are) commensurable in square only [Prop. 10.73]. And let the (rectangle contained) by BC and G also be equal to the (square) on A. And the (square) on A (is) rational. The (rectangle contained) by BC and G (is) thus also rational. And it has been applied to the rational (straight-line) BC. Thus, G is rational, and commensurable in length with BC [Prop. 10.20]. Therefore, since the (rectangle contained) by BC and G is equal to the (rectangle contained) by BD and KH, thus, proportionally, as CB is to BD, so KH (is) to G [Prop. 6.16]. And BC (is) greater than BD. Thus, KH (is) also greater than G [Prop. 5.16, 5.14]. Let KE be made equal to G. KE is thus commensurable in length with BC. And since as CB is to BD, so HK (is) to KE, thus, via conversion, as BC (is) to CD, so KH (is) to HE [Prop. 5.19 corr.]. Let it have been contrived that as KH (is) to HE, so HF (is) to F E. And thus the remainder KF is to F H, as KH (is) to HE—that is to say, [as] BC (is) to CD [Prop. 5.19]. And BC and CD [are] commensurable in square only.

419

ΣΤΟΙΧΕΙΩΝ ι΄.

ELEMENTS BOOK 10

τÕ ¢πÕ τÁς ΖΘ. σύµµετρον δέ ™στι τÕ ¢πÕ τÁς ΚΖ τù ¢πÕ τÁς ΖΘ· αƒ γ¦ρ ΚΖ, ΖΘ δυνάµει ε„σˆ σύµµετροι· σύµµετρος ¥ρα ™στˆ κሠ¹ ΚΖ τÍ ΖΕ µήκει· éστε ¹ ΚΖ κሠτÍ ΚΕ σύµµετρός [™στι] µήκει. ·ητ¾ δέ ™στιν ¹ ΚΕ κሠσύµµετρος τÍ ΒΓ µήκει. ·ητ¾ ¥ρα κሠ¹ ΚΖ κሠσύµµετρος τÍ ΒΓ µήκει. κሠ™πεί ™στιν æς ¹ ΒΓ πρÕς Γ∆, οÛτως ¹ ΚΖ πρÕς ΖΘ, ™ναλλ¦ξ æς ¹ ΒΓ πρÕς ΚΖ, οÛτως ¹ ∆Γ πρÕς ΖΘ. σύµµετρος δ ¹ ΒΓ τÍ ΚΖ· σύµµετρος ¥ρα κሠ¹ ΖΘ τÍ Γ∆ µήκει. αƒ ΒΓ, Γ∆ δ ·ηταί ε„σι δυνάµει µόνον σύµµετροι· καˆ αƒ ΚΖ, ΖΘ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ™κ δύο Ñνοµάτων ™στˆν ¥ρα ¹ ΚΘ. Ε„ µν οâν ¹ ΒΓ τÁς Γ∆ µε‹ζον δύναται τù ¢πÕ συµµέτρου ˜αυτÍ, κሠ¹ ΚΖ τÁς ΖΘ µε‹ζον δυνήσεται τù ¢πÕ συµµέτρου ˜αυτÍ. καˆ ε„ µν σύµµετρός ™στιν ¹ ΒΓ τÍ ™κκειµένV ·ητÍ µήκει, κሠ¹ ΚΖ, ε„ δ ¹ Γ∆ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ µήκει, κሠ¹ ΖΘ, ε„ δ οÙδετέρα τîν ΒΓ, Γ∆, οÙδετέρα τîν ΚΖ, ΖΘ. Ε„ δ ¹ ΒΓ τÁς Γ∆ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ, κሠ¹ ΚΖ τÁς ΖΘ µε‹ζον δυνήσεται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. καˆ ε„ µν σύµµετρός ™στιν ¹ ΒΓ τÍ ™κκειµένV ·ητÍ µήκει, κሠ¹ ΚΖ, ε„ δ ¹ Γ∆, κሠ¹ ΖΘ, ε„ δ οÙδετέρα τîν ΒΓ, Γ∆, οÙδετέρα τîν ΚΖ, ΖΘ. 'Εκ δύο ¥ρα Ñνοµάτων ™στˆν ¹ ΚΘ, Âς τ¦ Ñνόµατα τ¦ ΚΖ, ΖΘ σύµµετρά [™στι] το‹ς τÁς ¢ποτοµÁς Ñνόµασι το‹ς ΒΓ, Γ∆ κሠ™ν τù αÙτù λόγJ, κሠœτι ¹ ΚΘ τÍ ΒΓ τ¾ν αÙτ¾ν ›ξει τάξιν· Óπερ œδει δε‹ξαι.

KF and F H are thus also commensurable in square only [Prop. 10.11]. And since as KH is to HE, (so) KF (is) to F H, but as KH (is) to HE, (so) HF (is) to F E, thus, also as KF (is) to F H, (so) HF (is) to F E [Prop. 5.11]. And hence as the first (is) to the third, so the (square) on the first (is) to the (square) on the second [Def. 5.9]. And thus as KF (is) to F E, so the (square) on KF (is) to the (square) on F H. And the (square) on KF is commensurable with the (square) on F H. For KF and F H are commensurable in square. Thus, KF is also commensurable in length with F E [Prop. 10.11]. Hence, KF [is] also commensurable in length with KE [Prop. 10.15]. And KE is rational, and commensurable in length with BC. Thus, KF (is) also rational, and commensurable in length with BC [Prop. 10.12]. And since as BC is to CD, (so) KF (is) to F H, alternately, as BC (is) to KF , so DC (is) to F H [Prop. 5.16]. And BC (is) commensurable (in length) with KF . Thus, F H (is) also commensurable in length with CD [Prop. 10.11]. And BC and CD are rational (straight-lines which are) commensurable in square only. KF and F H are thus also rational (straight-lines which are) commensurable in square only [Def. 10.3, Prop. 10.13]. Thus, KH is a binomial [Prop. 10.36]. Therefore, if the square on BC is greater than (the square on) CD by the (square) on (some straight-line) commensurable (in length) with (BC), then the square on KF will also be greater than (the square on) F H by the (square) on (some straight-line) commensurable (in length) with (KF ) [Prop. 10.14]. And if BC is commensurable in length with a (previously) laid down rational (straight-line), (so) also (is) KF [Prop. 10.12]. And if CD is commensurable in length with a (previously) laid down rational (straight-line), (so) also (is) F H [Prop. 10.12]. And if neither of BC or CD (are commensurable), neither also (are) either of KF or F H [Prop. 10.13]. And if the square on BC is greater than (the square on) CD by the (square) on (some straight-line) incommensurable (in length) with (BC), then the square on KF will also be greater than (the square on) F H by the (square) on (some straight-line) incommensurable (in length) with (KF ) [Prop. 10.14]. And if BC is commensurable in length with a (previously) laid down rational (straight-line), (so) also (is) KF [Prop. 10.12]. And if CD is commensurable, (so) also (is) F H [Prop. 10.12]. And if neither of BC or CD (are commensurable), neither also (are) either of KF or F H [Prop. 10.13]. KH is thus a binomial whose terms, KF and F H, [are] commensurable (in length) with the terms, BC and CD, of the apotome, and in the same ratio. Moreover,

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ELEMENTS BOOK 10 KH will have the same order as BC [Defs. 10.5—10.10]. (Which is) the very thing it was required to show.

ριδ΄.

Proposition 114

'Ε¦ν χωρίον περιέχηται ØπÕ ¢ποτοµÁς κሠτÁς ™κ If an area is contained by an apotome, and a binomial δύο Ñνοµάτων, Âς τ¦ Ñνόµατα σύµµετρά τέ ™στι το‹ς τÁς whose terms are commensurable with, and in the same ¢ποτοµÁς Ñνόµασι κሠ™ν τù αÙτù λόγJ, ¹ τÕ χωρίον ratio as, the terms of the apotome, then the square-root δυναµένη ·ητή ™στιν. of the area is a rational (straight-line).

Α Γ

Β Ε

Ζ

A

∆

C

Η

G

Θ

H

Κ

Λ

Μ

K

Περιεχέσθω γ¦ρ χωρίον τÕ ØπÕ τîν ΑΒ, Γ∆ ØπÕ ¢ποτοµÁς τÁς ΑΒ κሠτÁς ™κ δύο Ñνοµάτων τÁς Γ∆, Âς µε‹ζον Ôνοµα œστω τÕ ΓΕ, κሠœστω τ¦ Ñνόµατα τÁς ™κ δύο Ñνοµάτων τ¦ ΓΕ, Ε∆ σύµµετρά τε το‹ς τÁς ¢ποτοµÁς Ñνόµασι το‹ς ΑΖ, ΖΒ κሠ™ν τù αÙτù λόγJ, κሠœστω ¹ τÕ ØπÕ τîν ΑΒ, Γ∆ δυναµένη ¹ Η· λέγω, Óτι ·ητή ™στιν ¹ Η. 'Εκκείσθω γ¦ρ ·ητ¾ ¹ Θ, κሠτù ¢πÕ τÁς Θ ‡σον παρ¦ τ¾ν Γ∆ παραβεβλήσθω πλάτος ποιοàν τ¾ν ΚΛ· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΚΛ, Âς τ¦ Ñνόµατα œστω τ¦ ΚΜ, ΜΛ σύµµετρα το‹ς τÁς ™κ δύο Ñνοµάτων Ñνόµασι το‹ς ΓΕ, Ε∆ κሠ™ν τù αÙτù λόγJ. ¢λλ¦ καˆ αƒ ΓΕ, Ε∆ σύµµετροί τέ ε„σι τα‹ς ΑΖ, ΖΒ κሠ™ν τù αÙτù λόγJ· œστιν ¥ρα æς ¹ ΑΖ πρÕς τ¾ν ΖΒ, οÛτως ¹ ΚΜ πρÕς ΜΛ. ™ναλλ¦ξ ¥ρα ™στˆν æς ¹ ΑΖ πρÕς τ¾ν ΚΜ, οÛτως ¹ ΒΖ πρÕς τ¾ν ΛΜ· κሠλοιπ¾ ¥ρα ¹ ΑΒ πρÕς λοιπ¾ν τ¾ν ΚΛ ™στιν æς ¹ ΑΖ πρÕς ΚΜ. σύµµετρος δ ¹ ΑΖ τÍ ΚΜ· σύµµετρος ¥ρα ™στˆ κሠ¹ ΑΒ τÍ ΚΛ. καί ™στιν æς ¹ ΑΒ πρÕς ΚΛ, οÛτως τÕ ØπÕ τîν Γ∆, ΑΒ πρÕς τÕ ØπÕ τîν Γ∆, ΚΛ· σύµµετρον ¥ρα ™στˆ κሠτÕ ØπÕ τîν Γ∆, ΑΒ τù ØπÕ τîν Γ∆, ΚΛ. ‡σον δ τÕ ØπÕ τîν Γ∆, ΚΛ τù ¢πÕ τÁς Θ· σύµµετρον ¥ρα ™στˆ τÕ ØπÕ τîν Γ∆, ΑΒ τù ¢πÕ τÁς Θ. τù δ ØπÕ τîν Γ∆, ΑΒ ‡σον ™στˆ τÕ ¢πÕ τÁς Η· σύµµετρον ¥ρα ™στˆ τÕ ¢πÕ τÁς Η τù ¢πÕ τÁς Θ. ·ητÕν δ τÕ ¢πÕ τÁς Θ· ·ητÕν ¥ρα ™στˆ κሠτÕ ¢πÕ τÁς Η· ·ητ¾ ¥ρα ™στˆν ¹ Η. κሠδύναται τÕ ØπÕ τîν Γ∆, ΑΒ. 'Ε¦ν ¥ρα χωρίον περιέχηται ØπÕ ¢ποτοµÁς κሠτÁς ™κ δύο Ñνοµάτων, Âς τ¦ Ñνόµατα σύµµετρά ™στι το‹ς τÁς ¢ποτοµÁς Ñνόµασι κሠ™ν τù αÙτù λόγJ, ¹ τÕ χωρίον δυναµένη ·ητή ™στιν.

B

F

E

D

L

M

For let an area, the (rectangle contained) by AB and CD, have been contained by the apotome AB, and the binomial CD, of which let the greater term be CE. And let the terms of the binomial, CE and ED, be commensurable with the terms of the apotome, AF and F B (respectively), and in the same ratio. And let the square-root of the (rectangle contained) by AB and CD be G. I say that G is a rational (straight-line). For let the rational (straight-line) H be laid down. And let (some rectangle), equal to the (square) on H, have been applied to CD, producing KL as breadth. Thus, KL is an apotome, of which let the terms, KM and M L, be commensurable with the terms of the binomial, CE and ED (respectively), and in the same ratio [Prop. 10.112]. But, CE and ED are also commensurable with AF and F B (respectively), and in the same ratio. Thus, as AF is to F B, so KM (is) to M L. Thus, alternately, as AF is to KM , so BF (is) to LM [Prop. 5.16]. Thus, the remainder AB is also to the remainder KL as AF (is) to KM [Prop. 5.19]. And AF (is) commensurable with KM [Prop. 10.12]. AB is thus also commensurable with KL [Prop. 10.11]. And as AB is to KL, so the (rectangle contained) by CD and AB (is) to the (rectangle contained) by CD and KL [Prop. 6.1]. Thus, the (rectangle contained) by CD and AB is also commensurable with the (rectangle contained) by CD and KL [Prop. 10.11]. And the (rectangle contained) by CD and KL (is) equal to the (square) on H. Thus, the (rectangle contained) by CD and AB is commensurable with the (square) on H. And the (square) on G is equal to the (rectangle contained) by CD and AB. The (square) on G

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ELEMENTS BOOK 10 is thus commensurable with the (square) on H. And the (square) on H (is) rational. Thus, the (square) on G is also rational. G is thus rational. And it is the square-root of the (rectangle contained) by CD and AB. Thus, if an area is contained by an apotome, and a binomial whose terms are commensurable with, and in the same ratio as, the terms of the apotome, then the square-root of the area is a rational (straight-line).

Πόρισµα.

Corollary

Κሠγέγονεν ¹µ‹ν κሠδι¦ τούτου φανερόν, Óτι δυAnd it has also been made clear to us, through this, νατόν ™στι ·ητÕν χωρίον ØπÕ ¢λόγων εÙθειîν περιέχε- that it is possible for a rational area to be contained by σθαι. Óπερ œδει δε‹ξαι. irrational straight-lines. (Which is) the very thing it was required to show.

ριε΄.

Proposition 115

'ΑπÕ µέσης ¥πειροι ¥λογοι γίνονται, κሠοÙδεµία οÙδεµι´ τîν πρότερον ¹ αÙτή.

An infinite (series) of irrational (straight-lines) can be created from a medial (straight-line), and none of them is the same as any of the preceding (straight-lines).

Α Β Γ ∆

A B C D

”Εστω µέση ¹ Α· λέγω, Óτι ¢πÕ τÁς Α ¥πειροι ¥λογοι γίνονται, κሠοÙδεµία οÙδεµι´ τîν πρότερον ¹ αÙτή. 'Εκκείσθω ·ητ¾ ¹ Β, κሠτù ØπÕ τîν Β, Α ‡σον œστω τÕ ¢πÕ τÁς Γ· ¥λογος ¥ρα ™στˆν ¹ Γ· τÕ γ¦ρ ØπÕ ¢λόγου κሠ·ητÁς ¥λογόν ™στιν. κሠοÙδεµι´ τîν πρότερον ¹ αÙτή· τÕ γ¦ρ ¢π' οÙδεµι©ς τîν πρότερον παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ µέσην. πάλιν δ¾ τù ØπÕ τîν Β, Γ ‡σον œστω τÕ ¢πÕ τÁς ∆· ¥λογον ¥ρα ™στˆ τÕ ¢πÕ τÁς ∆. ¥λογος ¥ρα ™στˆν ¹ ∆· κሠοÙδεµι´ τîν πρότερον ¹ αÙτή· τÕ γ¦ρ ¢π' οÙδεµι©ς τîν πρότερον παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ τ¾ν Γ. еοίως δ¾ τÁς τοιαύτης τάξεως ™π' ¥πειρον προβαινούσης φανερόν, Óτι ¢πÕ τÁς µέσης ¥πειροι ¥λογοι γίνονται, κሠοÙδεµία οÙδεµι´ τîν πρότερον ¹ αÙτή· Óπερ œδει δε‹ξαι.

Let A be a medial (straight-line). I say that an infinite (series) of irrational (straight-lines) can be created from A, and that none of them is the same as any of the preceding (straight-lines). Let the rational (straight-line) B be laid down. And let the (square) on C be equal to the (rectangle contained) by B and A. Thus, C is irrational [Def. 10.4]. For an (area contained) by an irrational and a rational (straight-line) is irrational [Prop. 10.20]. And (C is) not the same as any of the preceding (straight-lines). For the (square) on none of the preceding (straight-lines), applied to a rational (straight-line), produces a medial (straight-line) as breadth. So, again, let the (square) on D be equal to the (rectangle contained) by B and C. Thus, the (square) on D is irrational [Prop. 10.20]. D is thus irrational [Def. 10.4]. And (D is) not the same as any of the preceding (straight-lines). For the (square) on none of the preceding (straight-lines), applied to a rational (straight-line), produces C as breadth. So, similarly, this arrangement being advanced to infinity, it is clear that an infinite (series) of irrational (straight-lines) can be created from a medial (straight-line), and that none of them is the same as any of the preceding (straight-lines). (Which is) the very thing it was required to show.

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ELEMENTS BOOK 11 Elementary stereometry

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ELEMENTS BOOK 11

“Οροι.

Definitions

α΄. Στερεόν ™στι τÕ µÁκος κሠπλάτος κሠβάθος œχον. β΄. Στερεοà δ πέρας ™πιφάνεια. γ΄. ΕÙθε‹α πρÕς ™πίπεδον Ñρθή ™στιν, Óταν πρÕς πάσας τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù [ØποκειµένJ] ™πιπέδJ Ñρθ¦ς ποιÍ γωνίας. δ΄. 'Επίπεδον πρÕς ™πίπεδον Ñρθόν ™στιν, Óταν αƒ τÍ κοινÍ τοµÍ τîν ™πιπέδων πρÕς Ñρθ¦ς ¢γόµεναι εÙθε‹αι ™ν ˜νˆ τîν ™πιπέδων τù λοιπù ™πιπέδJ πρÕς Ñρθ¦ς ðσιν. ε΄. ΕÙθείας πρÕς ™πίπεδον κλίσις ™στίν, Óταν ¢πÕ τοà µετεώρου πέρατος τÁς εÙθείας ™πˆ τÕ ™πίπεδον κάθετος ¢χθÍ, κሠ¢πÕ τοà γενοµένου σηµείου ™πˆ τÕ ™ν τù ™πιπέδJ πέρας τÁς εÙθείας εÙθε‹α ™πιζευχθÍ, ¹ περιεχοµένη γωνία ØπÕ τÁς ¢χθείσης κሠτÁς ™φεστώσης. $΄. 'Επιπέδου πρÕς ™πίπεδον κλίσις ™στˆν ¹ περιεχοµένη Ñξε‹α γωνία ØπÕ τîν πρÕς Ñρθ¦ς τÍ κοινÍ τοµÍ ¢γοµένων πρÕς τù αÙτù σηµείJ ™ν ˜κατέρJ τîν ™πιπέδων. ζ΄. 'Επίπεδον πρÕς ™πίπεδον еοίως κεκλίσθαι λέγεται κሠ›τερον πρÕς ›τερον, Óταν αƒ ε„ρηµέναι τîν κλίσεων γωνίαι ‡σαι ¢λλήλαις ðσιν. η΄. Παράλληλα ™πίπεδά ™στι τ¦ ¢σύµπτωτα. θ΄. “Οµοια στερε¦ σχήµατά ™στι τ¦ ØπÕ Ðµοίων ™πιπέδων περιεχόµενα ‡σων τÕ πλÁθος. ι΄. ”Ισα δ καˆ Óµοια στερε¦ σχήµατά ™στι τ¦ ØπÕ Ðµοίων ™πιπέδων περιεχόµενα ‡σων τù πλήθει κሠτù µεγέθει. ια΄. Στερε¦ γωνία ™στˆν ¹ ØπÕ πλειόνων À δύο γραµµîν ¡πτοµένων ¢λλήλων κሠµ¾ ™ν τÍ αÙτÍ ™πιφανείv οÙσîν πρÕς πάσαις τα‹ς γραµµα‹ς κλίσις. ¥λλως· στερε¦ γωνία ™στˆν ¹ ØπÕ πλειόνων À δύο γωνιîν ™πιπέδων περιεχοµένη µ¾ οÙσîν ™ν τù αÙτù ™πιπέδJ πρÕς ˜νˆ σηµείJ συνισταµένων. ιβ΄. Πυραµίς ™στι σχÁµα στερεÕν ™πιπέδοις περιχόµενον ¢πÕ ˜νÕς ™πιπέδου πρÕς ˜νˆ σηµείJ συνεστώς. ιγ΄. Πρίσµα ™στˆ σχÁµα στερεÕν ™πιπέδοις περιεχόµενον, ïν δύο τ¦ ¢πεναντίον ‡σα τε καˆ Óµοιά ™στι κሠπαράλληλα, τ¦ δ λοιπ¦ παραλληλόγραµµα. ιδ΄. Σφα‹ρά ™στιν, Óταν ¹µικυκλίου µενούσης τÁς διαµέτρου περιενεχθν τÕ ¹µικύκλιον ε„ς τÕ αÙτÕ πάλιν ¢ποκατασταθÍ, Óθεν ½ρξατο φέρεσθαι, τÕ περιληφθν σχÁµα. ιε΄. ”Αξων δ τÁς σφαίρας ™στˆν ¹ µένουσα εÙθε‹α, περˆ ¿ν τÕ ¹µικύκλιον στρέφεται. ι$΄. Κέντρον δ τÁς σφαίρας ™στˆ τÕ αÙτό, Ö κሠτοà ¹µικυκλίου. ιζ΄. ∆ιάµετρος δ τÁς σφαίρας ™στˆν εÙθε‹ά τις δι¦ τοà κέντρου ºγµένη κሠπερατουµένη ™φ' ˜κάτερα τ¦ µέρη ØπÕ τÁς ™πιφανείας τÁς σφαίρας.

1. A solid is a (figure) having length and breadth and depth. 2. The extremity of a solid (is) a surface. 3. A straight-line is at right-angles to a plane when it makes right-angles with all of the straight-lines joined to it which are also in the plane. 4. A plane is at right-angles to a(nother) plane when (all of) the straight-lines drawn in one of the planes, at right-angles to the common section of the planes, are at right-angles to the remaining plane. 5. The inclination of a straight-line to a plane is the angle contained by the drawn and standing (straightlines), when a perpendicular is lead to the plane from the end of the (standing) straight-line raised (out of the plane), and a straight-line is (then) joined from the point (so) generated to the end of the (standing) straight-line (lying) in the plane. 6. The inclination of a plane to a(nother) plane is the acute angle contained by the (straight-lines), (one) in each of the planes, drawn at right-angles to the common segment (of the planes), at the same point. 7. A plane is said to have been similarly inclined to a plane, as another to another, when the aforementioned angles of inclination are equal to one another. 8. Parallel planes are those which do not meet (one another). 9. Similar solid figures are those contained by equal numbers of similar planes (which are similarly arranged). 10. But equal and similar solid figures are those contained by similar planes equal in number and in magnitude (which are similarly arranged). 11. A solid angle is the inclination (constituted) by more than two lines joining one another (at the same point), and not being in the same surface, to all of the lines. Otherwise, a solid angle is that contained by more than two plane angles, not being in the same plane, and constructed at one point. 12. A pyramid is a solid figure, contained by planes, (which is) constructed from one plane to one point. 13. A prism is a solid figure, contained by planes, of which the two opposite (planes) are equal, similar, and parallel, and the remaining (planes are) parallelograms. 14. A sphere is the figure enclosed when, the diameter of a semicircle remaining (fixed), the semicircle is carried around, and again established at the same (position) from which it began to be moved. 15. And the axis of the sphere is the fixed straight-line about which the semicircle is turned.

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ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

ιη΄. Κîνός ™στιν, Óταν Ñρθογωνίου τριγώνου µενούσης µι©ς πλευρ©ς τîν περˆ τ¾ν Ñρθ¾ν γωνίαν περιενεχθν τÕ τρίγωνον ε„ς τÕ αÙτÕ πάλιν ¢ποκατασταθÍ, Óθεν ½ρξατο φέρεσθαι, τÕ περιληφθν σχÁµα. κ¨ν µν ¹ µένουσα εÙθε‹α ‡ση Ï τÍ λοιπÍ [τÍ] περˆ τ¾ν Ñρθ¾ν περιφεροµένV, Ñρθογώνιος œσται Ð κîνος, ™¦ν δ ™λάττων, ¢µβλυγώνιος, ™¦ν δ µείζων, Ñξυγώνιος. ιθ΄. ”Αξων δ τοà κώνου ™στˆν ¹ µένουσα εÙθε‹α, περˆ ¿ν τÕ τρίγωνον στρέφεται. κ΄. Βάσις δ Ð κύκλος Ð ØπÕ τÁς περιφερουµένης εÙθείας γραφόµενος. κα΄. Κύλινδρός ™στιν, Óταν Ñρθογωνίου παραλληλογράµµου µενούσης µι©ς πλευρ©ς τîν περˆ τ¾ν Ñρθ¾ν γωνίαν περιενεχθν τÕ παραλληλόγραµµον ε„ς τÕ αÙτÕ πάλιν ¢ποκατασταθÍ, Óθεν ½ρξατο φέρεσθαι, τÕ περιληφθν σχÁµα. κβ΄. ”Αξων δ τοà κυλίνδρου ™στˆν ¹ µένουσα εÙθε‹α, περˆ ¿ν τÕ παραλληλόγραµµον στρέφεται. κγ΄. Βάσεις δ οƒ κύκλοι οƒ ØπÕ τîν ¢πεναντίον περιαγοµένων δύο πλευρîν γραφόµενοι. κδ΄. “Οµοιοι κîνοι κሠκύλινδροί ε„σιν, ïν ο† τε ¥ξονες καˆ αƒ διάµετροι τîν βάσεων ¢νάλογόν ε„σιν. κε΄. Κύβος ™στˆ σχÁµα στερεÕν ØπÕ žξ τετραγώνων ‡σων περιεχόµενον. κ$΄. 'Οκτάεδρόν ™στˆ σχÁµα στερεÕν ØπÕ Ñκτë τριγώνων ‡σων κሠ„σοπλεύρων περιεχόµενον. κζ΄. Ε„κοσάεδρόν ™στι σχÁµα στερεÕν ØπÕ ε‡κοσι τριγώνων ‡σων κሠ„σοπλεύρων περιεχόµενον. κη΄. ∆ωδεκάεδρόν ™στι σχÁµα στερεÕν ØπÕ δώδεκα πενταγώνων ‡σων κሠ„σοπλεύρων κሠ„σογωνίων περιεχόµενον.

16. And the center of the sphere is the same as that of the semicircle. 17. And the diameter of the sphere is any straightline which is drawn through the center and terminated in both directions by the surface of the sphere. 18. A cone is the figure enclosed when, one of the sides of a right-angled triangle about the right-angle remaining (fixed), the triangle is carried around, and again established at the same (position) from which it began to be moved. And if the stationary straight-line is equal to the remaining (straight-line) about the right-angle, (which is) carried around, then the cone will be rightangled, and if less, obtuse-angled, and if greater, acuteangled. 19. And the axis of the cone is the fixed straight-line about which the triangle is turned. 20. And the base (of the cone is) the circle described by the (remaining) straight-line (about the right-angle which is) carried around (the axis). 21. A cylinder is the figure enclosed when, one of the sides of a right-angled parallelogram about the rightangle remaining (fixed), the parallelogram is carried around, and again established at the same (position) from which it began to be moved. 22. And the axis of the cylinder is the stationary straight-line about which the parallelogram is turned. 23. And the bases (of the cylinder are) the circles described by the two opposite sides (which are) carried around. 24. Similar cones and cylinders are those for which the axes and the diameters of the bases are proportional. 25. A cube is a solid figure contained by six equal squares. 26. An octahedron is a solid figure contained by eight equal and equilateral triangles. 27. An icosahedron is a solid figure contained by twenty equal and equilateral triangles. 28. A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.

α΄.

Proposition 1†

ΕÙθείας γραµµÁς µέρος µέν τι οÙκ œστιν ™ν τù ØποκειµένJ ™πιπέδJ, µέρος δέ τι ™ν µετεωροτέρJ. Ε„ γ¦ρ δυνατόν, εÙθείας γραµµÁς τÁς ΑΒΓ µέρος µέν τι τÕ ΑΒ œστω ™ν τù ØποκειµένJ ™πιπέδJ, µέρος δέ τι τÕ ΒΓ ™ν µετεωροτέρJ. ”Εσται δή τις τÍ ΑΒ συνεχ¾ς εÙθε‹α ™π' εÙθείας ™ν τù ØποκειµένJ ™πιπέδJ. œστω ¹ Β∆· δύο ¥ρα εÙθειîν τîν ΑΒΓ, ΑΒ∆ κοινÕν τµÁµά ™στιν ¹ ΑΒ· Óπερ ™στˆν ¢δύνατον, ™πειδήπερ ™¦ν κέντρJ τù Β καˆ

Some part of a straight-line cannot be in a reference plane, and some part in a more elevated (plane). For, if possible, let some part, AB, of the straight-line ABC be in a reference plane, and some part, BC, in a more elevated (plane). In the reference plane, there will be some straight-line continuous with, and straight-on to, AB.‡ Let it be BD. Thus, AB is a common segment of the two (different) straight-lines ABC and ABD. The very thing is impos-

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ELEMENTS BOOK 11

διαστήµατι τù ΑΒ κύκλον γράψωµεν, αƒ διάµετροι ¢νίσους ¢πολήψονται τοà κύκλου περιφερείας.

sible, inasmuch as if we draw a circle with center B and radius AB then the diameters (ABD and ABC) will cut off unequal circumferences of the circle.

Γ

C

∆

D

Β

B

Α

A

ΕÙθείας ¥ρα γραµµÁς µέρος µέν τι οÙκ œστιν ™ν τù Thus, some part of a straight-line cannot be in a referØποκειµένJ ™πιπέδJ, τÕ δ ™ν µετεωροτέρJ· Óπερ œδει ence plane, and (some part) in a more elevated (plane). δε‹ξαι. (Which is) the very thing it was required to show. †

The proofs of the first three propositions in this book are not at all rigorous. Hence, these three propositions should really be regarded as

additional axioms. ‡

This assumption essentially presupposes the validity of the proposition under discussion.

β΄.

Proposition 2

'Ε¦ν δύο εÙθε‹αι τέµνωσιν ¢λλήλας, ™ν ˜νί ε„σιν If two straight-lines cut one another then they are in ™πιπέδJ, κሠπ©ν τρίγωνον ™ν ˜νί ™στιν ™πιπέδJ. one plane, and all triangles (formed using segments of both lines) are in one plane.

Α

∆

A

D

Ε Ζ

Γ

E Η

Θ

Κ

F

Β

C

∆ύο γ¦ρ εÙθε‹αι αƒ ΑΒ, Γ∆ τεµνέτωσαν ¢λλήλας κατ¦ τÕ Ε σηµε‹ον. λέγω, Óτι αƒ ΑΒ, Γ∆ ™ν ˜νί ε„σιν ™πιπέδJ, κሠπ©ν τρίγωνον ™ν ˜νί ™στιν ™πιπέδJ. Ε„λήφθω γ¦ρ ™πˆ τîν ΕΓ, ΕΒ τυχόντα σηµε‹α τ¦ Ζ, Η, κሠ™πεζεύχθωσαν αƒ ΓΒ, ΖΗ, κሠδιήχθωσαν αƒ ΖΘ, ΗΚ· λέγω πρîτον, Óτι τÕ ΕΓΒ τρίγωνον ™ν ˜νί ™στιν ™πιπέδJ. ε„ γάρ ™στι τοà ΕΓΒ τριγώνου µέρος ½τοι τÕ ΖΘΓ À τÕ ΗΒΚ ™ν τù ØποκειµένJ [™πιπέδJ],

G

H

K

B

For let the two straight-lines AB and CD have cut one another at point E. I say that AB and CD are in one plane, and that all triangles (formed using segments of both lines) are in one plane. For let the random points F and G have been taken on EC and EB (respectively). And let CB and F G have been joined, and let F H and GK have been drawn across. I say, first of all, that triangle ECB is in one (ref-

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ELEMENTS BOOK 11

τÕ δ λοιπÕν ™ν ¥λλJ, œσται κሠµι©ς τîν ΕΓ, ΕΒ εÙθειîν µέρος µέν τι ™ν τù ØποκειµένJ ™πιπέδJ, τÕ δ ™ν αλλJ. ε„ δ τοà ΕΓΒ τριγώνου τÕ ΖΓΒΗ µέρος Ï ™ν τù ØποκειµένJ ™πιπέδJ, τÕ δ λοιπÕν ™ν ¥λλJ, œσται κሠ¢µφοτέρων τîν ΕΓ, ΕΒ εÙθειîν µέρος µέν τι ™ν τù ØποκειµένJ ™πιπέδJ, τÕ δ ™ν ¥λλω· Óπερ ¥τοπον ™δείχθη. τÕ ¥ρα ΕΓΒ τρίγωνον ™ν ˜νί ™στιν ™πιπέδJ. ™ν ú δέ ™στι τÕ ΕΓΒ τρίγωνον, ™ν τούτJ κሠ˜κατέρα τîν ΕΓ, ΕΒ, ™ν ú δ ˜κατέρα τîν ΕΓ, ΕΒ, ™ν τούτJ καˆ αƒ ΑΒ, Γ∆. αƒ ΑΒ, Γ∆ ¥ρα εÙθε‹αι ™ν ˜νί ε„σιν ™πιπέδJ, κሠπ©ν τρίγωνον ™ν ˜νί ™στιν ™πιπέδJ· Óπερ œδει δε‹ξαι.

erence) plane. For if part of triangle ECB, either F HC or GBK, is in the reference [plane], and the remainder in a different (plane), then a part of one the straightlines EC and EB will also be in the reference plane, and (a part) in a different (plane). And if the part F CBG of triangle ECB is in the reference plane, and the remainder in a different (plane), then parts of both of the straight-lines EC and EB will also be in the reference plane, and (parts) in a different (plane). The very thing was shown to be absurb [Prop. 11.1]. Thus, triangle ECB is in one plane. And in whichever (plane) triangle ECB is (found), in that (plane) EC and EB (will) each also (be found). And in whichever (plane) EC and EB (are) each (found), in that (plane) AB and CD (will) also (be found) [Prop. 11.1]. Thus, the straight-lines AB and CD are in one plane, and all triangles (formed using segments of both lines) are in one plane. (Which is) the very thing it was required to show.

γ΄.

Proposition 3

'Ε¦ν δύο ™πίπεδα τεµνÍ ¥λληλα, ¹ κοιν¾ αÙτîν τοµ¾ εÙθε‹ά ™στιν.

If two planes cut one another then their common section is a straight-line.

Β

B Ε

Ζ ∆

E F Α

D

Γ

A

C

∆ύο γ¦ρ ™πίπεδα τ¦ ΑΒ, ΒΓ τεµνέτω ¥λληλα, κοιν¾ δ αÙτîν τοµ¾ œστω ¹ ∆Β γραµµή· λέγω, Óτι ¹ ∆Β γραµµ¾ εÙθε‹ά ™στιν. Ε„ γ¦ρ µή, ™πεζεύχθω ¢πÕ τοà ∆ ™πˆ τÕ Β ™ν µν τù ΑΒ ™πιπέδJ εÙθε‹α ¹ ∆ΕΒ, ™ν δ τù ΒΓ ™πιπέδJ εÙθε‹α ¹ ∆ΖΒ. œσται δ¾ δύο εÙθειîν τîν ∆ΕΒ, ∆ΖΒ τ¦ αÙτ¦ πέρατα, κሠπεριέξουσι δηλαδ¾ χωρίον· Óπερ ¥τοπον. οÜκ ¥ρα α„ ∆ΕΒ, ∆ΖΒ εÙθε‹αί ε„σιν. еοίως δ¾ δείξοµεν, Óτι οÙδ ¥λλη τις ¢πÕ τοà ∆ ™πˆ τÕ Β ™πιζευγνυµένη εÙθε‹α œσται πλ¾ν τÁς ∆Β κοινÁς τοµÁς τîν ΑΒ, ΒΓ ™πιπέδων. 'Ε¦ν ¥ρα δύο ™πίπεδα τέµνV ¥λληλα, ¹ κοιν¾ αÙτîν τοµ¾ εÙθε‹ά ™στιν· Óπερ œδει δε‹ξαι.

For let the two planes AB and BC cut one another, and let their common section be the line DB. I say that the line DB is straight. For, if not, let the straight-line DEB have been joined from D to B in the plane AB, and the straight-line DF B in the plane BC. So two straight-lines, DEB and DF B, will have the same ends, and they will clearly enclose an area. The very thing (is) absurd. Thus, DEB and DF B are not straight-lines. So, similarly, we can show than no other straight-line can be joined from D to B except DB, the common section of the planes AB and BC. Thus, if two planes cut one another then their common section is a straight-line. (Which is) the very thing it was required to show.

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ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

δ΄.

Proposition 4

'Ε¦ν εÙθε‹α δύο εÙθείαις τεµνούσαις ¢λλήλας πρÕς If a straight-line is set up at right-angles to two Ñρθ¦ς ™πˆ τÁς κοινÁς τοµÁς ™πισταθÍ, κሠτù δι' αÙτîν straight-lines cutting one another, at the common point ™πιπέδJ πρÕς Ñρθ¦ς œσται. of section, then it will also be at right-angles to the plane (passing) through them (both).

Ζ

Α Η ∆

Ε

F

Γ

A

Θ

G

Β

D

C H E

ΕÙθε‹α γάρ τις ¹ ΕΖ δύο εÙθείαις τα‹ς ΑΒ, Γ∆ τεµνούσαις ¢λλήλας κατ¦ τÕ Ε σηµε‹ον ¢πÕ τοà Ε πρÕς Ñρθ¦ς ™φεστάτω· λέγω, Óτι ¹ ΕΖ κሠτù δι¦ τîν ΑΒ, Γ∆ ™πιπέδJ πρÕς Ñρθάς ™στιν. 'Απειλήφθωσαν γ¦ρ αƒ ΑΕ, ΕΒ, ΓΕ, Ε∆ ‡σαι ¢λλήλαις, κሠδιήχθω τις δι¦ τοà Ε, æς œτυχεν, ¹ ΗΕΘ, κሠ™πεζεύχθωσαν αƒ Α∆, ΓΒ, κሠœτι ¢πÕ τυχόντος τοà Ζ ™πεζεύχθωσαν αƒ ΖΑ, ΖΗ, Ζ∆, ΖΓ, ΖΘ, ΖΒ. Κሠ™πεˆ δύο αƒ ΑΕ, Ε∆ δυσˆ τα‹ς ΓΕ, ΕΒ ‡σαι ε„σˆ κሠγωνίας ‡σας περιέχουσιν, βάσις ¥ρα ¹ Α∆ βάσει τÍ ΓΒ ‡ση ™στίν, κሠτÕ ΑΕ∆ τρίγωνον τù ΓΕΒ τριγώνJ ‡σον œσται· éστε κሠγωνία ¹ ØπÕ ∆ΑΕ γωνίv τÍ ØπÕ ΕΒΓ ‡ση [™στίν]. œστι δ κሠ¹ ØπÕ ΑΕΗ γωνία τÍ ØπÕ ΒΕΘ ‡ση. δύο δ¾ τρίγωνά ™στι τ¦ ΑΗΕ, ΒΕΘ τ¦ς δύο γωνίας δυσˆ γωνίαις ‡σας œχοντα ˜κατέραν ˜κατέρv κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην τ¾ν πρÕς τα‹ς ‡σαις γωνίαις τ¾ν ΑΕ τÍ ΕΒ· κሠτ¦ς λοιπ¦ς ¥ρα πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξουσιν. ‡ση ¥ρα ¹ µν ΗΕ τÍ ΕΘ, ¹ δ ΑΗ τÍ ΒΘ. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΕ τÍ ΕΒ, κοιν¾ δ κሠπρÕς Ñρθ¦ς ¹ ΖΕ, βάσις ¥ρα ¹ ΖΑ βάσει τÍ ΖΒ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΖΓ τÍ Ζ∆ ™στιν ‡ση. κሠ™πεˆ ‡ση ™στˆν ¹ Α∆ τÍ ΓΒ, œστι δ κሠ¹ ΖΑ τÍ ΖΒ ‡ση, δύο δ¾ αƒ ΖΑ, Α∆ δυσˆ τα‹ς ΖΒ, ΒΓ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠβάσις ¹ Ζ∆ βάσει τÍ ΖΓ ™δείχθη ‡ση· κሠγωνία ¥ρα ¹ ØπÕ ΖΑ∆ γωνίv τÍ ØπÕ ΖΒΓ ‡ση ™στίν. κሠ™πεˆ πάλιν ™δείχθη ¹ ΑΗ τÍ ΒΘ ‡ση, ¢λλ¦ µ¾ν κሠ¹ ΖΑ τÍ ΖΒ ‡ση, δύο δ¾ αƒ ΖΑ, ΑΗ δυσˆ τα‹ς ΖΒ, ΒΘ ‡σαι ε„σίν. κሠγωνία ¹ ØπÕ ΖΑΗ ™δείχθη ‡ση τÍ ØπÕ ΖΒΘ· βάσις ¥ρα ¹ ΖΗ βάσει τÍ ΖΘ ™στιν ‡ση. κሠ™πεˆ πάλιν ‡ση ™δείχθη ¹ ΗΕ τÍ ΕΘ, κοιν¾ δ ¹ ΕΖ, δύο δ¾ αƒ ΗΕ, ΕΖ δυσˆ τα‹ς ΘΕ, ΕΖ ‡σαι ε„σίν· κሠβάσις ¹ ΖΗ

B

For let some straight-line EF have (been) set up at right-angles to two straight-lines, AB and CD, cutting one another at point E, at E. I say that EF is also at right-angles to the plane (passing) through AB and CD. For let AE, EB, CE and ED have been cut off from (the two straight-lines so as to be) equal to one another. And let GEH have been drawn, at random, through E (in the plane passing through AB and CD). And let AD and CB have been joined. And, furthermore, let F A, F G, F D, F C, F H, and F B have been joined from the random (point) F (on EF ). For since the two (straight-lines) AE and ED are equal to the two (straight-lines) CE and EB, and they enclose equal angles [Prop. 1.15], the base AD is thus equal to the base CB, and triangle AED will be equal to triangle CEB [Prop. 1.4]. Hence, the angle DAE [is] equal to the angle EBC. And the angle AEG (is) equal to the angle BEH [Prop. 1.15]. So AGE and BEH are two triangles having two angles equal to two angles, respectively, and one side equal to one side—(namely), those by the equal angles, AE and EB. Thus, they will also have the remaining sides equal to the remaining sides [Prop. 1.26]. Thus, GE (is) equal to EH, and AG to BH. And since AE is equal to EB, and F E is common and at right-angles, the base F A is thus equal to the base F B [Prop. 1.4]. So, for the same (reasons), F C is also equal to F D. And since AD is equal to CB, and F A is also equal to F B, the two (straight-lines) F A and AD are equal to the two (straight-lines) F B and BC, respectively. And the base F D was shown (to be) equal to the base F C. Thus, the angle F AD is also equal to

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ELEMENTS BOOK 11

βάσει τÍ ΖΘ ‡ση· γωνία ¥ρα ¹ ØπÕ ΗΕΖ γωνίv τÍ ØπÕ ΘΕΖ ‡ση ™στίν. Ñρθ¾ ¥ρα ˜κατέρα τîν ØπÕ ΗΕΖ, ΘΕΖ γωνιîν. ¹ ΖΕ ¥ρα πρÕς τ¾ν ΗΘ τυχόντως δι¦ τοà Ε ¢χθε‹σαν Ñρθή ™στιν. еοίως δ¾ δείξοµεν, Óτι ¹ ΖΕ κሠπρÕς πάσας τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù ØποκειµένJ ™πιπέδJ Ñρθ¦ς ποιήσει γωνίας. εÙθε‹α δ πρÕς ™πίπεδον Ñρθή ™στιν, Óταν πρÕς πάσας τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù αÙτù ™πιπέδJ Ñρθ¦ς ποιÍ γωνίας· ¹ ΖΕ ¥ρα τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς ™στιν. τÕ δ Øποκείµενον ™πίπεδόν ™στι τÕ δι¦ τîν ΑΒ, Γ∆ εÙθειîν. ¹ ΖΕ ¥ρα πρÕς Ñρθάς ™στι τù δι¦ τîν ΑΒ, Γ∆ ™πιπέδJ. 'Ε¦ν ¥ρα εÙθε‹α δύο εÙθείαις τεµνούσαις ¢λλήλας πρÕς Ñρθ¦ς ™πˆ τÁς κοινÁς τοµÁς ™πισταθÍ, κሠτù δι' αÙτîν ™πιπέδJ πρÕς Ñρθ¦ς œσται· Óπερ œδει δε‹ξαι.

the angle F BC [Prop. 1.8]. And, again, since AG was shown (to be) equal to BH, but F A (is) also equal to F B, the two (straight-lines) F A and AG are equal to the two (straight-lines) F B and BH (respectively). And the angle F AG was shown (to be) equal to the angle F BH. Thus, the base F G is equal to the base F H [Prop. 1.4]. And, again, since GE was shown (to be) equal to EH, and EF (is) common, the two (straight-lines) GE and EF are equal to the two (straight-lines) HE and EF (respectively). And the base F G (is) equal to the base F H. Thus, the angle GEF is equal to the angle HEF [Prop. 1.8]. Each of the angles GEF and HEF (are) thus right-angles [Def. 1.10]. Thus, F E is at right-angles to GH, which was drawn at random through E (in the reference plane passing though AB and AC). So, similarly, we can show that F E will make right-angles with all straight-lines joined to it which are in the reference plane. And a straight-line is at right-angles to a plane when it makes right-angles with all straight-lines joined to it which are in the plane [Def. 11.3]. Thus, F E is at right-angles to the reference plane. And the reference plane is that (passing) through the straight-lines AB and CD. Thus, F E is at right-angles to the plane (passing) through AB and CD. Thus, if a straight-line is set up at right-angles to two straight-lines cutting one another, at the common point of section, then it will also be at right-angles to the plane (passing) through them (both). (Which is) the very thing it was required to show.

ε΄.

Proposition 5

'Ε¦ν εÙθε‹α τρισˆν εÙθείαις ¡πτοµέναις ¢λλήλων πρÕς If a straight-line is set up at right-angles to three Ñρθ¦ς ™πˆ τÁς κοινÁς τοµÁς ™πισταθÍ, αƒ τρε‹ς εÙθε‹αι straight-lines cutting one another, at the common point ™ν ˜νί ε„σιν ™πιπέδJ. of section, then the three straight-lines are in one plane.

Α

A Γ

C Ζ

Β

F ∆

B

Ε

D E

ΕÙθε‹α γάρ τις ¹ ΑΒ τρισˆν εÙθείαις τα‹ς ΒΓ, Β∆, For let some straight-line AB have been set up at ΒΕ πρÕς Ñρθ¦ς ™πˆ τÁς κατ¦ τÕ Β ¡φÁς ™φεστάτω· λέγω, right-angles to three straight-lines BC, BD, and BE, at Óτι αƒ ΒΓ, Β∆, ΒΕ ™ν ˜νί ε„σιν ™πιπέδJ. the (common) point of section B. I say that BC, BD, Μ¾ γάρ, ¢λλ' ε„ δυνατόν, œστωσαν αƒ µν Β∆, ΒΕ and BE are in one plane. 429

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

™ν τù ØποκειµένJ ™πιπέδJ, ¹ δ ΒΓ ™ν µετεωροτέρJ, κሠ™κβεβλήσθω τÕ δˆα τîν ΑΒ, ΒΓ ™πίπεδον· κοιν¾ν δ¾ τοµ¾ν ποιήσει ™ν τù ØποκειµένJ ™πιπέδJ εÙθε‹αν. ποιείτω τ¾ν ΒΖ. ™ν ˜νˆ ¥ρα ε„σˆν ™πιπέδJ τù διηγµένJ δι¦ τîν ΑΒ, ΒΓ αƒ τρε‹ς εÙθε‹αι αƒ ΑΒ, ΒΓ, ΒΖ. κሠ™πεˆ ¹ ΑΒ Ñρθή ™στι πρÕς ˜κατέραν τîν Β∆, ΒΕ, κሠτù δι¦ τîν Β∆, ΒΕ ¥ρα ™πιπέδJ Ñρθή ™στιν ¹ ΑΒ. τÕ δ δι¦ τîν Β∆, ΒΕ ™πίπεδον τÕ Øποκείµενόν ™στιν· ¹ ΑΒ ¥ρα Ñρθή ™στι πρÕς τÕ Øποκείµενον ™πίπεδον. éστε κሠπρÕς πάσας τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù ØποκειµένJ ™πιπέδJ Ñρθ¦ς ποιήσει γωνίας ¹ ΑΒ. ¤πτεται δ αÙτÁς ¹ ΒΖ οâσα ™ν τù ØποκειµένJ ™πιπέδJ· ¹ ¥ρα ØπÕ ΑΒΖ γωνία Ñρθή ™στιν. Øπόκειται δ κሠ¹ ØπÕ ΑΒΓ Ñρθή· ‡ση ¥ρα ¹ ØπÕ ΑΒΖ γωνία τÍ ØπÕ ΑΒΓ. καί ε„σιν ™ν ˜νˆ ™πιπέδJ· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ¹ ΒΓ εÙθε‹α ™ν µετεωροτέρJ ™στˆν ™πιπέδJ· αƒ τρε‹ς ¥ρα εÙθε‹αι αƒ ΒΓ, Β∆, ΒΕ ™ν ˜νί ε„σιν ™πιπέδJ. 'Ε¦ν ¥ρα εÙθε‹α τρισˆν εÙθείαις ¡πτοµέναις ¢λλήλων ™πˆ τÁς ¡φÁς πρÕς Ñρθ¦ς ™πισταθÍ, αƒ τρε‹ς εÙθε‹αι ™ν ˜νί ε„σιν ™πιπέδJ· Óπερ œδει δε‹ξαι.

For (if) not, and if possible, let BD and BE be in the reference plane, and BC in a more elevated (plane). And let the plane through AB and BC have been produced. So it will make a straight-line as a common section with the reference plane [Def. 11.3]. Let it make BF . Thus, the three straight-lines AB, BC, and BF are in one plane—(namely), that drawn through AB and BC. And since AB is at right-angles to each of BD and BE, AB is thus also at right-angles to the plane (passing) through BD and BE [Prop. 11.4]. And the plane (passing) through BD and BE is the reference plane. Thus, AB is at right-angles to the reference plane. Hence, AB will also make right-angles with all straight-lines joined to it which are also in the reference plane [Def. 11.3]. And BF , which is in the reference plane, is joined to it. Thus, the angle ABF is a right-angle. And ABC was also assumed to be a right-angle. Thus, angle ABF (is) equal to ABC. And they are in one plane. The very thing is impossible. Thus, BC is not in a more elevated plane. Thus, the three straight-lines BC, BD, and BE are in one plane. Thus, if a straight-line is set up at right-angles to three straight-lines cutting one another, at the (common) point of section, then the three straight-lines are in one plane. (Which is) the very thing it was required to show.

$΄.

Proposition 6

'Ε¦ν δύο εÙθε‹αι τù αÙτù ™πιπέδJ πρÕς Ñρθ¦ς ðσιν, παράλληλοι œσονται αƒ εÙθε‹αι.

If two straight-lines are at right-angles to the same plane then the straight-lines will be parallel.†

G

A

C

A

B

D

B D

E

E

∆ύο γ¦ρ εÙθε‹αι αƒ ΑΒ, Γ∆ τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθ¦ς œστωσαν· λέγω, Óτι παράλληλός ™στιν ¹ ΑΒ τÍ Γ∆. Συµβαλλέτωσαν γ¦ρ τù ØποκειµένJ ™πιπέδJ κατ¦ τ¦ Β, ∆ σηµε‹α, κሠ™πεζεύχθω ¹ Β∆ εÙθε‹α, κሠ½χθω τÍ Β∆ πρÕς Ñρθ¦ς ™ν τù ØποκειµένJ ™πιπέδJ ¹ ∆Ε, κሠκείσθω τÍ ΑΒ ‡ση ¹ ∆Ε, κሠ™πεζεύχθωσαν αƒ ΒΕ, ΑΕ, Α∆. Κሠ™πεˆ ¹ ΑΒ Ñρθή ™στι πρÕς τÕ Øποκείµενον

For let the two straight-lines AB and CD be at rightangles to a reference plane. I say that AB is parallel to CD. For let them meet the reference plane at points B and D (respectively). And let the straight-line BD have been joined. And let DE have been drawn at right-angles to BD in the reference plane. And let DE be made equal to AB. And let BE, AE, and AD have been joined. And since AB is at right-angles to the reference plane,

430

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

™πίπεδον, κሠπρÕς πάσας [¥ρα] τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù ØποκειµένJ ™πιπέδJ Ñρθ¦ς ποιήσει γωνίας. ¤πτεται δ τÁς ΑΒ ˜κατέρα τîν Β∆, ΒΕ οâσα ™ν τù ØποκειµένJ ™πιπέδJ· Ñρθ¾ ¥ρα ™στˆν ˜κατέρα τîν ØπÕ ΑΒ∆, ΑΒΕ γωνιîν. δι¦ τ¦ αÙτ¦ δ¾ κሠ˜κατέρα τîν ØπÕ Γ∆Β, Γ∆Ε Ñρθή ™στιν. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΒ τÍ ∆Ε, κοιν¾ δ ¹ Β∆, δύο δ¾ αƒ ΑΒ, Β∆ δυσˆ τα‹ς Ε∆, ∆Β ‡σαι ε„σίν· κሠγωνίας Ñρθ¦ς περιέχουσιν· βάσις ¥ρα ¹ Α∆ βάσει τÍ ΒΕ ™στιν ‡ση. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΒ τÍ ∆Ε, ¢λλ¦ κሠ¹ Α∆ τÍ ΒΕ, δύο δ¾ αƒ ΑΒ, ΒΕ δυσˆ τα‹ς Ε∆, ∆Α ‡σαι ε„σίν· κሠβάσις αÙτîν κοιν¾ ¹ ΑΕ· γωνία ¥ρα ¹ ØπÕ ΑΒΕ γωνι® τÍ ØπÕ Ε∆Α ™στιν ‡ση. Ñρθ¾ δ ¹ ØπÕ ΑΒΕ· Ñρθ¾ ¥ρα κሠ¹ ØπÕ Ε∆Α· ¹ Ε∆ ¥ρα πρÕς τ¾ν ∆Α Ñρθή ™στιν. œστι δ κሠπρÕς ˜κατέραν τîν Β∆, ∆Γ Ñρθή. ¹ Ε∆ ¥ρα τρισˆν εÙθείαις τα‹ς Β∆, ∆Α, ∆Γ πρÕς Ñρθ¦ς ™πˆ τÁς ¡φÁς ™φέστηκεν· αƒ τρε‹ς ¥ρα εÙθε‹αι αƒ Β∆, ∆Α, ∆Γ ™ν ˜νί ε„σιν ™πιπέδJ. ™ν ú δ αƒ ∆Β, ∆Α, ™ν τούτJ κሠ¹ ΑΒ· π©ν γ¦ρ τρίγωνον ™ν ˜νί ™στιν ™πιπέδJ· αƒ ¥ρα ΑΒ, Β∆, ∆Γ εÙθε‹αι ™ν ˜νί ε„σιν ™πιπέδJ. καί ™στιν Ñρθ¾ ˜κατέρα τîν ØπÕ ΑΒ∆, Β∆Γ γωνιîν· παράλληλος ¥ρα ™στˆν ¹ ΑΒ τÍ Γ∆. 'Ε¦ν ¥ρα δύο εÙθε‹αι τù αÙτù ™πιπέδJ πρÕς Ñρθ¦ς ðσιν, παράλληλοι œσονται αƒ εÙθε‹αι· Óπερ œδει δε‹ξαι.

†

it will [thus] also make right-angles with all straight-lines joined to it which are in the reference plane [Def. 11.3]. And BD and BE, which are in the reference plane, are each joined to AB. Thus, each of the angles ABD and ABE are right-angles. So, for the same (reasons), each of the angles CDB and CDE are also right-angles. And since AB is equal to DE, and BD (is) common, the two (straight-lines) AB and BD are equal to the two (straight-lines) ED and DB (respectively). And they contain right-angles. Thus, the base AD is equal to the base BE [Prop. 1.4]. And since AB is equal to DE, and AD (is) also (equal) to BE, the two (straight-lines) AB and BE are thus equal to the two (straight-lines) ED and DA (respectively). And their base AE (is) common. Thus, angle ABE is equal to angle EDA [Prop. 1.8]. And ABE (is) a right-angle. Thus, EDA (is) also a rightangle. ED is thus at right-angles to DA. And it is also at right-angles to each of BD and DC. Thus, ED is standing at right-angles to the three straight-lines BD, DA, and DC at the (common) point of section. Thus, the three straight-lines BD, DA, and DC are in one plane [Prop. 11.5]. And in which(ever) plane DB and DA (are found), in that (plane) AB (will) also (be found). For all triangles are in one plane [Prop. 11.2]. And each of the angles ABD and BDC is a right-angle. Thus, AB is parallel to CD [Prop. 1.28]. Thus, if two straight-lines are at right-angles to the same plane then the straight-lines will be parallel. (Which is) the very thing it was required to show.

In other words, the two straight-lines lie in the same plane, and never meet when produced in either direction.

ζ΄.

Proposition 7

'Ε¦ν ðσι δύο εÙθε‹αι παράλληλοι, ληφθÍ δ ™φ' If there are two parallel straight-lines, and random ˜κατέρας αÙτîν τυχόντα σηµε‹α, ¹ ™πˆ τ¦ σηµε‹α ™πι- points are taken on each of them, then the straight-line ζευγνυµένη εÙθε‹α ™ν τù αÙτù ™πιπέδJ ™στˆ τα‹ς πα- joining the two points is in the same plane as the parallel ραλλήλοις. (straight-lines).

A

E

B

A

H G

E

B G

Z

D

C

F

D

”Εστωσαν δύο εÙθε‹αι παράλληλοι αƒ ΑΒ, Γ∆, κሠLet AB and CD be two parallel straight-lines, and let ε„λήφθω ™φ' ˜κατέρας αÙτîν τυνχόντα σηµε‹α τ¦ Ε, Ζ· the random points E and F have been taken on each of λέγω, Óτι ¹ ™πˆ τ¦ Ε, Ζ σηµε‹α ™πιζευγνυµένη εÙθε‹α ™ν them (respectively). I say that the straight-line joining τù αÙτù ™πιπέδJ ™στˆ τα‹ς παραλλήλοις. points E and F is in the same (reference) plane as the 431

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

Μ¾ γάρ, ¢λλ' ε„ δυνατόν, œστω ™ν µετεωροτέρJ æς ¹ ΕΗΖ, κሠδιήχθω δι¦ τÁς ΕΗΖ ™πίπεδον· τοµ¾ν δ¾ ποιήσει ™ν τù ØποκειµένJ ™πιπέδJ εÙθε‹αν. ποιείτω æς τ¾ν ΕΖ· δύο ¥ρα εÙθε‹αι αƒ ΕΗΖ, ΕΖ χωρίον περιέξουσιν· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ¹ ¢πÕ τοà Ε ™πˆ τÕ Ζ ™πιζευγνυµένη εÙθε‹αι ™ν µετεωροτέρJ ™στˆν ™πιπέδJ· ™ν τù δι¦ τîν ΑΒ, ΓΒ ¥ρα παραλλήλων ™στˆν ™πιπέδJ ¹ ¢πÕ τοà Ε ™πˆ τÕ Ζ ™πιζευγνυµένη εÙθε‹α. 'Ε¦ν ¥ρα ðσι δύο εÙθε‹αι παράλληλοι, ληφθÍ δ ™φ' ˜κατέρας αÙτîν τυχόντα σηµε‹α, ¹ ™πˆ τ¦ σηµε‹α ™πιζευγνυµένη εÙθε‹α ™ν τù αÙτù ™πιπέδJ ™στˆ τα‹ς παραλλήλοις· Óπερ œδει δε‹ξαι.

parallel (straight-lines). For (if) not, and if possible, let it be in a more elevated (plane), such as EGF . And let a plane have been drawn through EGF . So it will make a straight cutting in the reference plane [Prop. 11.3]. Let it make EF . Thus, two straight-lines (with the same end-points), EGF and EF , will enclose an area. The very thing is impossible. Thus, the straight-line joining E to F is not in a more elevated plane. The straight-line joining E to F is thus in the plane through the parallel (straight-lines) AB and CD. Thus, if there are two parallel straight-lines, and random points are taken on each of them, then the straightline joining the two points is in the same plane as the parallel (straight-lines). (Which is) the very thing it was required to show.

η΄.

Proposition 8

'Ε¦ν ðσι δύο εÙθε‹αι παράλληλοι, ¹ δ ˜τέρα αÙτîν If two straight-lines are parallel, and one of them is at ™πιπέδJ τινˆ πρÕς Ñρθ¦ς Ï, κሠ¹ λοιπ¾ τù αÙτù ™πιπέδJ right-angles to some plane, then the remaining (one) will πρÕς Ñρθ¦ς œσται. also be at right-angles to the same plane.

G

A

B

D E

C

A

B

D

E

”Εστωσαν δύο εÙθε‹αι παράλληλοι αƒ ΑΒ, Γ∆, ¹ δ ˜τέρα αÙτîν ¹ ΑΒ τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθ¦ς œστω· λέγω, Óτι κሠ¹ λοιπ¾ ¹ Γ∆ τù αÙτù ™πιπέδJ πρÕς Ñρθ¦ς œσται. Συµβαλλέτωσαν γ¦ρ αƒ ΑΒ, Γ∆ τù ØποκειµένJ ™πιπέδJ κατ¦ τ¦ Β, ∆ σηµε‹α, κሠ™πεζέυχθω ¹ Β∆· αƒ ΑΒ, Γ∆, Β∆ ¥ρα ™ν ˜νί ε„σιν ™πιπέδJ. ½χθω τÍ ΒΑ πρÕς Ñρθ¦ς ™ν τù ØποκειµένJ ™πιπέδJ ¹ ∆Ε, κሠκείσθω τÍ ΑΒ ‡ση ¹ ∆Ε, κሠ™πεζεύχθωσαν αƒ ΒΕ, ΑΕ, Α∆. Κሠ™πεˆ ¹ ΑΒ Ðρθή ™στι πρÕς τÕ Øποκείµενον ™πίπεδον, κሠπρÕς πάσας ¥ρα τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς ™στιν ¹ ΑΒ· Ñρθ¾ ¥ρα [™στˆν] ˜κατέρα τîν ØπÕ ΑΒ∆, ΑΒΕ γωνιîν. κሠ™πεˆ ε„ς παραλλήλους τ¦ς ΑΒ, Γ∆ εÙθε‹α ™µπέπτωκεν ¹ Β∆, αƒ ¥ρα ØπÕ ΑΒ∆, Γ∆Β γωνίαι

Let AB and CD be two parallel straight-lines, and let one of them, AB, be at right-angles to a reference plane. I say that the remaining (one), CD, will also be at rightangles to the same plane. For let AB and CD meet the reference plane at points B and D (respectively). And let BD have been joined. AB, CD, and BD are thus in one plane [Prop. 11.7]. Let DE have been drawn at right-angles to BD in the reference plane, and let DE be made equal to AB, and let BE, AE, and AD have been joined. And since AB is at right-angles to the reference plane, AB is thus also at right-angles to all of the straight-lines joined to it which are in the reference plane [Def. 11.3]. Thus, the angles ABD and ABE [are] each right-angles. And since the straight-line BD has met the

432

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

δυσˆν Ñρθα‹ς ‡σαι ε„σίν. Ñρθ¾ δ ¹ ØπÕ ΑΒ∆· Ñρθ¾ ¥ρα κሠ¹ ØπÕ Γ∆Β· ¹ Γ∆ ¥ρα πρÕς τ¾ν Β∆ Ñρθή ™στιν. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΒ τÍ ∆Ε, κοιν¾ δ ¹ Β∆, δύο δ¾ αƒ ΑΒ, Β∆ δυσˆ τα‹ς Ε∆, ∆Β ‡σαι ε„σίν· κሠγωνία ¹ ØπÕ ΑΒ∆ γωνίv τÍ ØπÕ Ε∆Β ‡ση· Ñρθ¾ γ¦ρ ˜κατέρα· βάσις ¥ρα ¹ Α∆ βάσει τÍ ΒΕ ‡ση. κሠ™πεˆ ‡ση ™στˆν ¹ µν ΑΒ τÍ ∆Ε, ¹ δ ΒΕ τÍ Α∆, δύο δ¾ αƒ ΑΒ, ΒΕ δυσˆ τα‹ς Ε∆, ∆Α ‡σαι ε„σˆν ˜κατέρα ˜κατέρv. κሠβάσις αÙτîν κοιν¾ ¹ ΑΕ· γωνία ¥ρα ¹ ØπÕ ΑΒΕ γωνίv τÍ ØπÕ Ε∆Α ™στιν ‡ση. Ñρθ¾ δ ¹ ØπÕ ΑΒΕ· Ñρθ¾ ¥ρα κሠ¹ ØπÕ Ε∆Α· ¹ Ε∆ ¥ρα πρÕς τ¾ν Α∆ Ñρθή ™στιν. œστι δ κሠπρÕς τ¾ν ∆Β Ñρθή· ¹ Ε∆ ¥ρα κሠτö δι¦ τîν Β∆, ∆Α ™πιπέδJ Ñρθή ™στιν. κሠπρÕς πάσας ¥ρα τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù δι¦ τîν Β∆Α ™πιπέδJ Ñρθ¦ς ποιήσει γωνίας ¹ Ε∆. ™ν δ τù δι¦ τîν Β∆Α ™πιπέδJ ™στˆν ¹ ∆Γ, ™πειδήπερ ™ν τù δι¦ τîν Β∆Α ™πιπέδJ ™στˆν αƒ ΑΒ, Β∆, ™ν ú δ αƒ ΑΒ, Β∆, ™ν τούτJ ™στˆ κሠ¹ ∆Γ. ¹ Ε∆ ¥ρα τÍ ∆Γ πρÕς Ñρθάς ™στιν· éστε κሠ¹ Γ∆ τÍ ∆Ε πρÕς Ñρθάς ™στιν. œστι δ κሠ¹ Γ∆ τÍ Β∆ πρÕς Ñρθάς. ¹ Γ∆ ¥ρα δύο εÙθείαις τεµνούσαις ¢λλήλας τα‹ς ∆Ε, ∆Β ¢πÕ τÁς κατ¦ τÕ ∆ τοµÁς πρÕς Ñρθ¦ς ™φέστηκεν· éστε ¹ Γ∆ κሠτù δι¦ τîν ∆Ε, ∆Β ™πιπέδJ πρÕς Ñρθάς ™στιν. τÕ δ δι¦ τîν ∆Ε, ∆Β ™πίπεδον τÕ Øποκείµενόν ™στιν· ¹ Γ∆ ¥ρα τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς ™στιν. 'Ε¦ν ¥ρα ðσι δύο εÙθε‹α παράλληλοι, ¹ δ µία αÙτîν ™πιπέδJ τινˆ πρÕς Ñρθ¦ς Ï, κሠ¹ λοιπ¾ τù αÙτù ™πιπέδJ πρÕς Ñρθ¦ς œσται· Óπερ œδει δε‹ξαι.

parallel (straight-lines) AB and CD, the (sum of the) angles ABD and CDB is thus equal to two right-angles [Prop. 1.29]. And ABD (is) a right-angle. Thus, CDB (is) also a right-angle. CD is thus at right-angles to BD. And since AB is equal to DE, and BD (is) common, the two (straight-lines) AB and BD are equal to the two (straight-lines) ED and DB (respectively). And angle ABD (is) equal to angle EDB. For each (is) a rightangle. Thus, the base AD (is) equal to the base BE [Prop. 1.4]. And since AB is equal to DE, and BE to AD, the two (sides) AB, BE are equal to the two (sides) ED, DA, respectively. And their base AE is common. Thus, angle ABE is equal to angle EDA [Prop. 1.8]. And ABE (is) a right-angle. EDA (is) thus also a rightangle. Thus, ED is at right-angles to AD. And it is also at right-angles to DB. Thus, ED is also at right-angles to the plane through BD and AD [Prop. 11.4]. And ED will thus make right-angles with all of the straightlines joined to it which are also in the plane through BDA. And DC is in the plane through BDA, inasmuch as AB and BD are in the plane through BDA [Prop. 11.2], and in which(ever plane) AB and BD (are found), DC is also (found). Thus, ED is at right-angles to DC. Hence, CD is also at right-angles to DE. And CD is also at right-angles to BD. Thus, CD is standing at right-angles to two straight-lines, DE and DB, which meet one another, at the (point) of section, D. Hence, CD is also at right-angles to the plane through DE and DB [Prop. 11.4]. And the plane through DE and DB is the reference (plane). CD is thus at right-angles to the reference plane. Thus, if two straight-lines are parallel, and one of them is at right-angles to some plane, then the remaining (one) will also be at right-angles to the same plane. (Which is) the very thing it was required to show.

θ΄.

Proposition 9

Αƒ τÍ αÙτÍ εÙθείv παράλληλοι κሠµ¾ οâσαι αÙτÍ ™ν (Straight-lines) parallel to the same straight-line, and τù αÙτù ™πιπέδJ κሠ¢λλήλαις ε„σˆ παράλληλοι. which are not in the same plane as it, are also parallel to one another.

B Z

D

J

H K

A

B

E

F

G

D

H

G

E

C K

433

A

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

”Εστω γ¦ρ ˜κατέρα τîν ΑΒ, Γ∆ τÍ ΕΖ παράλληλος µ¾ οâσαι αÙτÍ ™ν τù αÙτù ™πιπέδJ· λέγω, Óτι παράλληλός ™στιν ¹ ΑΒ τÍ Γ∆. Ε„λήφθω γ¦ρ ™πˆ τÁς ΕΖ τυχÕν σηµε‹ον τÕ Η, κሠ¢π' αÙτοà τÍ ΕΖ ™ν µν τù δι¦ τîν ΕΖ, ΑΒ ™πιπέδJ πρÕς Ñρθ¦ς ½χθω ¹ ΗΘ, ™ν δ τù δι¦ τîν ΖΕ, Γ∆ τÍ ΕΖ πάλιν πρÕς Ñρθ¦ς ½χθω ¹ ΗΚ. Κሠ™πεˆ ¹ ΕΖ πρÕς ˜κατέραν τîν ΗΘ, ΗΚ Ñρθή ™στιν, ¹ ΕΖ ¥ρα κሠτù δι¦ τîν ΗΘ, ΗΚ ™πιπέδJ πρÕς Ñρθάς ™στιν. καί ™στιν ¹ ΕΖ τÍ ΑΒ παράλληλος· κሠ¹ ΑΒ ¥ρα τù δι¦ τîν ΘΗΚ ™πιπέδJ πρÕς Ñρθάς ™στιν. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ Γ∆ τù δι¦ τîν ΘΗΚ ™πιπέδJ πρÕς Ñρθάς ™στιν· ˜κατέρα ¥ρα τîν ΑΒ, Γ∆ τù δι¦ τîν ΘΗΚ ™πιπέδJ πρÕς Ñρθάς ™στιν. ™¦ν δ δύο εÙθε‹αι τù αÙτù ™πιπέδJ πρÕς Ñρθ¦ς ðσιν, παράλληλοί ε„σιν αƒ εÙθε‹αι· παράλληλος ¥ρα ™στˆν ¹ ΑΒ τÍ Γ∆· Óπερ œδει δε‹ξαι.

For let AB and CD each be parallel to EF , not being in the same plane as it. I say that AB is parallel to CD. For let some point G have been taken at random on EF . And from it let GH have been drawn at right-angles to EF in the plane through EF and AB. And let GK have been drawn, again at right-angles to EF , in the plane through F E and CD. And since EF is at right-angles to each of GH and GK, EF is thus also at right-angles to the plane through GH and GK [Prop. 11.4]. And EF is parallel to AB. Thus, AB is also at right-angles to the plane through HGK [Prop. 11.8]. So, for the same (reasons), CD is also at right-angles to the plane through HGK. Thus, AB and CD are each at right-angles to the plane through HGK. And if two straight-lines are at right–angles to the same plane then the straight-lines are parallel [Prop. 11.6]. Thus, AB is parallel to CD. (Which is) the very thing it was required to show.

ι΄.

Proposition 10

'Ε¦ν δύο εÙθε‹αι ¡πτόµεναι ¢λλήλων παρ¦ δύο If two straight-lines joined to one another are (respecεÙθείας ¡πτοµένας ¢λλήλων ðσι µ¾ ™ν τù αÙτù ™πιπέδJ, tively) parallel to two straight-lines joined to one another, †σας γωνίας περιέξουσιν. (but are) not in the same plane, then they will contain equal angles.

B

G

B

A

C

A

E

Z

E

D

F

D

∆ύο γ¦ρ εÙθε‹αι αƒ ΑΒ, ΒΓ ¡πτόµεναι ¢λλήλων παρ¦ δύο εÙθείας τ¦ς ∆Ε, ΕΖ ¡πτοµένας ¢λλήλων œστωσαν µ¾ ™ν τù αÙτù ™πιπέδJ· λέγω, Óτι ‡ση ™στˆν ¹ ØπÕ ΑΒΓ γωνία τÍ ØπÕ ∆ΕΖ. 'Απειλήφθωσαν γ¦ρ αƒ ΒΑ, ΒΓ, Ε∆, ΕΖ ‡σαι ¢λλήλαις, κሠ™πεζεύχθωσαν αƒ Α∆, ΓΖ, ΒΕ, ΑΓ, ∆Ζ. Κሠ™πεˆ ¹ ΒΑ τÍ Ε∆ ‡ση ™στˆ κሠπαράλληλος, κሠ¹ Α∆ ¥ρα τÍ ΒΕ ‡ση ™στˆ κሠπαράλληλος. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΓΖ τÍ ΒΕ ‡ση ™στˆ κሠπαράλληλος· ˜κατέρα ¥ρα τîν Α∆, ΓΖ τÍ ΒΕ ‡ση ™στˆ κሠπαράλληλος. αƒ δ τÍ αÙτÍ εÙθείv παράλληλοι κሠµ¾ οâσαι αÙτÍ ™ν τù αÙτù ™πιπέδJ κሠ¢λλήλαις ε„σˆ παράλληλοι· παράλληλος ¥ρα ™στˆν ¹ Α∆ τÍ ΓΖ κሠ‡ση. κሠ™πιζευγνύουσιν αÙτ¦ς αƒ ΑΓ, ∆Ζ· κሠ¹ ΑΓ ¥ρα τÍ ∆Ζ ‡ση ™στˆ κሠπαράλληλος.

For let the two straight-lines joined to one another, AB and BC, be (respectively) parallel to the two straight-lines joined to one another, DE and EF , (but) not in the same plane. I say that angle ABC is equal to (angle) DEF . For let BA, BC, ED, and EF have been cut off (so as to be, respectively) equal to one another. And let AD, CF , BE, AC, and DF have been joined. And since BA is equal and parallel to ED, AD is thus also equal and parallel to BE [Prop. 1.33]. So, for the same reasons, CF is also equal and parallel to BE. Thus, AD and CF are each equal and parallel to BE. And straight-lines parallel to the same straight-line, and which are not in the same plane as it, are also parallel to one an-

434

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

κሠ™πεˆ δύο αƒ ΑΒ, ΒΓ δυσˆ τα‹ς ∆Ε, ΕΖ ‡σαι ε„σίν, κሠβάσις ¹ ΑΓ βάσει τÍ ∆Ζ ‡ση, γωνία ¥ρα ¹ ØπÕ ΑΒΓ γωνίv τÍ ØπÕ ∆ΕΖ ™στιν ‡ση. 'Ε¦ν ¥ρα δύο εÙθε‹αι ¡πτόµεναι ¢λλήλων παρ¦ δύο εÙθείας ¡πτοµένας ¢λλήλων ðσι µ¾ ™ν τù αÙτù ™πιπέδJ, †σας γωνίας περιέξουσιν· Óπερ œδει δε‹ξαι.

other [Prop. 11.9]. Thus, AD is parallel and equal to CF . And AC and DF join them. Thus, AC is also equal and parallel to DF [Prop. 1.33]. And since the two (straightlines) AB and BC are equal to the two (straight-lines) DE and EF (respectvely), and the base AC (is) equal to the base DF , the angle ABC is thus equal to the (angle) DEF [Prop. 1.8]. Thus, if two straight-lines joined to one another are (respectively) parallel to two straight-lines joined to one another, (but are) not in the same plane, then they will contain equal angles. (Which is) the very thing it was required to show.

ια΄.

Proposition 11

'ΑπÕ τοà δοθέντος σηµείου µετεώρου ™πˆ τÕ δοθν ™πίπεδον κάθετον εÙθε‹αν γραµµ¾ν ¢γαγε‹ν.

To draw a perpendicular straight-line from a given raised point to a given plane.

A

A

E H

J Z B

D

E

H

G

C

F G

”Εστω τÕ µν δοθν σηµε‹ον µετέωρον τÕ Α, τÕ δ δοθν ™πίπεδον τÕ Øποκείµενον· δε‹ δ¾ ¢πÕ τοà Α σηµείου ™πˆ τÕ Øποκείµενον ™πίπεδον κάθετον εÙθε‹αν γραµµ¾ν ¢γαγε‹ν. ∆ιήχθω γάρ τις ™ν τù ØποκειµένJ ™πιπέδJ εÙθε‹α, æς œτυχεν, ¹ ΒΓ, κሠ½χθω ¢πÕ τοà Α σηµείου ™πˆ τ¾ν ΒΓ κάθετος ¹ Α∆. ε„ µν οâν ¹ Α∆ κάθετός ™στι κሠ™πˆ τÕ Øποκείµενον ™πίπεδον, γεγονÕς ¨ν ε‡η τÕ ™πιταχθέν. ε„ δ οÜ, ½χθω ¢πÕ τοà ∆ σηµείου τÍ ΒΓ ™ν τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθ¦ς ¹ ∆Ε, κሠ½χθω ¢πÕ τοà Α ™πˆ τ¾ν ∆Ε κάθετος ¹ ΑΖ, κሠδι¦ τοà Ζ σηµείου τÍ ΒΓ παράλληλος ½χθω ¹ ΗΘ. Κሠ™πεˆ ¹ ΒΓ ˜κατέρv τîν ∆Α, ∆Ε πρÕς Ñρθάς ™στιν, ¹ ΒΓ ¥ρα κሠτù δι¦ τîν Ε∆Α ™πιπέδJ πρÕς Ñρθάς ™στιν. καί ™στιν αÙτÍ παράλληλος ¹ ΗΘ· ™¦ν δ ðσι δύο εÙθε‹αι παράλληλοι, ¹ δ µία αÙτîν ™πιπέδJ τινˆ πρÕς Ñρθ¦ς Ï, κሠ¹ λοιπ¾ τù αÙτù ™πιπέδJ πρÕς Ñρθ¦ς œσται· κሠ¹ ΗΘ ¥ρα τù δι¦ τîν Ε∆, ∆Α ™πιπέδJ πρÕς Ñρθάς ™στιν. κሠπρÕς πάσας ¥ρα τ¦ς ¡πτοµένας αÙτÁς

D B

Let A be the given raised point, and the given plane the reference (plane). So, it is required to draw a perpendicular straight-line from point A to the reference plane. Let some random straight-line BC have been drawn across in the reference plane, and let the (straight-line) AD have been drawn from point A perpendicular to BC [Prop. 1.12]. If, therefore, AD is also perpendicular to the reference plane then that which was prescribed will have occurred. And, if not, let DE have been drawn in the reference plane from point D at right-angles to BC [Prop. 1.11], and let the (straight-line) AF have been drawn from A perpendicular to DE [Prop. 1.12], and let GH have been drawn through point F , parallel to BC [Prop. 1.31]. And since BC is at right-angles to each of DA and DE, BC is thus also at right-angles to the plane through EDA [Prop. 11.4]. And GH is parallel to it. And if two straight-lines are parallel, and one of them is at rightangles to some plane, then the remaining (straight-line)

435

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

εÙθείας κሠοÜσας ™ν τù δι¦ τîν Ε∆, ∆Α ™πιπέδJ Ñρθή ™στιν ¹ ΗΘ. ¤πτεται δ αÙτÁς ¹ ΑΖ οâσα ™ν τù δι¦ τîν Ε∆, ∆Α ™πιπέδJ· ¹ ΗΘ ¥ρα Ñρθή ™στι πρÕς τ¾ν ΖΑ· éστε κሠ¹ ΖΑ Ñρθή ™στι πρÕς τ¾ν ΘΗ. œστι δ ¹ ΑΖ κሠπρÕς τ¾ν ∆Ε Ñρθή· ¹ ΑΖ ¥ρα πρÕς ˜κατέραν τîν ΗΘ, ∆Ε Ñρθή ™στιν. ™¦ν δ εÙθε‹α δυσˆν εÙθείαις τεµνούσαις ¢λλήλας ™πˆ τÁς τοµÁς πρÕς Ñρθ¦ς ™πισταθÍ, κሠτù δι' αÙτîν ™πιπέδJ πρÕς Ñρθ¦ς œσται· ¹ ΖΑ ¥ρα τù δι¦ τîν Ε∆, ΗΘ ™πιπέδJ πρÕς Ñρθάς ™στιν. τÕ δ δι¦ τîν Ε∆, ΗΘ ™πίπεδόν ™στι τÕ Øποκείµενον· ¹ ΑΖ ¥ρα τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς ™στιν. 'ΑπÕ τοà ¥ρα δοθέντος σηµείου µετεώρου τοà Α ™πˆ τÕ Øποκείµενον ™πίπεδον κάθετος εÙθε‹α γραµµ¾ Ãκται ¹ ΑΖ· Óπερ œδει ποιÁσαι.

will also be at right-angles to the same plane [Prop. 11.8]. Thus, GH is also at right-angles to the plane through ED and DA. And GH is thus at right-angles to all of the straight-lines joined to it which are also in the plane through ED and AD [Def. 11.3]. And AF , which is in the plane through ED and DA, is joined to it. Thus, GH is at right-angles to F A. Hence, F A is also at right-angles to HG. And AF is also at right-angles to DE. Thus, AF is at right-angles to each of GH and DE. And if a straightline is set up at right-angles to two straight-lines cutting one another, at the point of section, then it will also be at right-angles to the plane through them [Prop. 11.4]. Thus, F A is at right-angles to the plane through ED and GH. And the plane through ED and GH is the reference (plane). Thus, AF is at right-angles to the reference plane. Thus, the perpendicular straight-line AF has been drawn from the given raised point A to the reference plane. (Which is) the very thing it was required to do.

ιβ΄.

Proposition 12

Τù δοθέντι ™πιπέδJ ¢πÕ τοà πρÕς αÙτù δοθέντος σηµείου πρÕς Ñρθ¦ς εÙθε‹αν γραµµ¾ν ¢ναστÁσαι.

To set up a straight-line at right-angles to a given plane from a given point in it.

B

B

D

D

G

C

A

A

”Εστω τÕ µν δοθν ™πίπεδον τÕ Øποκείµενον, τÕ δ πρÕς αÙτù σηµε‹ον τÕ Α· δε‹ δ¾ ¢πÕ τοà Α σηµείου τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθ¦ς εÙθε‹αν γραµµ¾ν ¢ναστÁσαι. Νενοήσθω τι σηµε‹ον µετέωρον τÕ Β, κሠ¢πÕ τοà Β ™πˆ τÕ Øποκείµενον ™πίπεδον κάθετος ½χθω ¹ ΒΓ, κሠδι¦ τοà Α σηµείου τÍ ΒΓ παράλληλος ½χθω ¹ Α∆. 'Επεˆ οâν δύο εÙθε‹αι παράλληλοί ε„σιν αƒ Α∆, ΓΒ, ¹ δ µία αÙτîν ¹ ΒΓ τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς ™στιν, κሠ¹ λοιπ¾ ¥ρα ¹ Α∆ τù ØποκειµένJ ™πιπέδJ

Let the given plane be the reference (plane), and A a point in it. So, it is required to set up a straight-line at right-angles to the reference plane at point A. Let some raised point B have been assumed, and let the perpendicular (straight-line) BC have been drawn from B to the reference plane [Prop. 11.11]. And let AD have been drawn from point A parallel to BC [Prop. 1.31]. Therefore, since AD and CB are two parallel straightlines, and one of them, BC, is at right-angles to the refer-

436

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

πρÕς Ñρθ¦ς ™στιν. Τù ¥ρα δοθέντι ™πιπέδJ ¢πÕ τοà πρÕς αÙτù σηµείου τοà Α πρÕς Ñρθ¦ς ¢νέσταται ¹ Α∆· Óπερ œδει ποιÁσαι.

ence plane, the remaining (one) AD is thus also at rightangles to the reference plane [Prop. 11.8]. Thus, AD has been set up at right-angles to the given plane, from the point in it, A. (Which is) the very thing it was required to do.

ιγ΄.

Proposition 13

'ΑπÕ τοà αÙτοà σηµείου τù αÙτù ™πιπέδJ δύο εÙθε‹αι πρÕς Ñρθ¦ς οÙκ ¢ναστήσονται ™πˆ τ¦ αÙτ¦ µέρη.

Two (different) straight-lines cannot be set up at the same point at right-angles to the same plane, on the same side.

B

B

G

C

D

D

A

A

E

E

Ε„ γ¦ρ δυνατόν, ¢πÕ τοà αÙτοà σηµείου τοà Α τù ØποκειµένJ ™πιπέδJ δύο εÙθε‹αι αƒ ΑΒ, ΒΓ πρÕς Ñρθ¦ς ¢νεστάτωσαν ™πˆ τ¦ αÙτ¦ µέρη, κሠδιήχθω τÕ δι¦ τîν ΒΑ, ΑΓ ™πˆπεδον· τοµ¾ν δ¾ ποιήσει δι¦ τοà Α ™ν τù ØποκειµένJ ™πιπέδJ εÙθε‹αν. ποιείτω τ¾ν ∆ΑΕ· αƒ ¥ρα ΑΒ, ΑΓ, ∆ΑΕ εÙθε‹αι ™ν ˜νι ε„σιν ™πιπέδJ. κሠ™πεˆ ¹ ΓΑ τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς ™στιν, κሠπρÕς πάσας ¥ρα τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù ØποκειµένJ ™πιπέδJ Ñρθ¦ς ποιήσει γωνίας. ¤πτεται δ αÙτÁς ¹ ∆ΑΕ οâσα ™ν τù ØποκειµένJ ™πιπέδJ· ¹ ¥ρα ØπÕ ΓΑΕ γωνία Ñρθή ™στιν. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ ΒΑΕ Ñρθή ™στιν· ‡ση ¥ρα ¹ ØπÕ ΓΑΕ τÍ ØπÕ ΒΑΕ καί ε„σιν ™ν ˜νˆ ™πιπέδJ· Óπερ ™στˆν ¢δύνατον. ΟÙκ ¥ρα ¢πÕ τοà αÙτοà σηµείου τù αÙτù ™πιπέδJ δύο εÙθε‹αι πρÕς Ñρθ¦ς ¢νασταθήσονται ™πˆ τ¦ αÙτ¦ µέρη· Óπερ œδει δε‹ξαι.

For, if possible, let the two straight-lines AB and AC have been set up at the same point A at right-angles to the reference plane, on the same side. And let the plane through BA and AC have been drawn. So it will make a straight cutting (passing) through (point) A in the reference plane [Prop. 11.3]. Let it have made DAE. Thus, AB, AC, and DAE are straight-lines in one plane. And since CA is at right-angles to the reference plane, it will thus also make right-angles with all of the straightlines joined to it which are also in the reference plane [Def. 11.3]. And DAE, which is in the reference plane, is joined to it. Thus, angle CAE is a right-angle. So, for the same (reasons), BAE is also a right-angle. Thus, CAE (is) equal to BAE. And they are in one plane. The very thing is impossible. Thus, two (different) straight-lines cannot be set up at the same point at right-angles to the same plane, on the same side. (Which is) the very thing it was required to show.

ιδ΄.

Proposition 14

ΠρÕς § ™πίπεδα ¹ αÙτ¾ εÙθε‹α Ñρθή ™στιν, παράλληλα Planes to which the same straight-line is at rightœσται τ¦ ™πίπεδα. angles will be parallel planes. ΕÙθε‹α γάρ τις ¹ ΑΒ πρÕς ˜κάτερον τîν Γ∆, ΕΖ For let some straight-line AB be at right-angles to

437

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

™πιπέδων πρÕς Ñρθ¦ς œστω· λέγω, Óτι παράλληλά ™στι each of the planes CD and EF . I say that the planes τ¦ ™πίπεδα. are parallel.

H

G

G C

A D

E

A

K

Z B

K D F

J

B E

H

Ε„ γ¦ρ µή, ™κβαλλόµενα συµπεσοàνται. συµπιπτέτωσαν· ποιήσουσι δ¾ κοιν¾ν τοµ¾ν εÙθε‹αν. ποιείτωσαν τ¾ν ΗΘ, κሠε„λήφθω ™πˆ τÁς ΗΘ τυχÕν σηµε‹ον τÕ Κ, κሠ™πεζεύχθωσαν αƒ ΑΚ, ΒΚ. Κሠ™πεˆ ¹ ΑΒ Ñρθή ™στι πρÕς τÕ ΕΖ ™πίπεδον, κሠπρÕς τ¾ν ΒΚ ¥ρα εÙθε‹αν οâσαν ™ν τù ΕΖ ™κβληθέντι ™πιπέδJ Ñρθή ™στιν ¹ ΑΒ· ¹ ¥ρα ØπÕ ΑΒΚ γωνία Ñρθή ™στιν. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ ΒΑΚ Ñρθή ™στιν. τριγώνου δ¾ τοà ΑΒΚ αƒ δύο γωνίαι αƒ ØπÕ ΑΒΚ, ΒΑΚ δυσˆν Ñρθα‹ς ε„σιν ‡σαι· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τ¦ Γ∆, ΕΖ ™πίπεδα ™κβαλλόµενα συµπεσοàνται· παράλληλα ¥ρα ™στˆ τ¦ Γ∆, ΕΖ ™πίπεδα. ΠρÕς § ™πίπεδα ¥ρα ¹ αÙτ¾ εÙθε‹α Ñρθή ™στιν, παράλληλά ™στι τ¦ ™πίπεδα· Óπερ œδει δε‹ξαι.

For, if not, being produced, they will meet. Let them have met. So they will make a straight-line as a common section [Prop. 11.3]. Let them have made GH. And let some random point K have been taken on GH. And let AK and BK have been joined. And since AB is at right-angles to the plane EF , AB is thus also at right-angles to the straight-line BK, which is in the produced plane EF [Def. 11.3]. Thus, angle ABK is a right-angle. So, for the same (reasons), BAK is also a right-angle. So the (sum of the) two angles ABK and BAK in the triangle ABK is equal to two right-angles. The very thing is impossible [Prop. 1.17]. Thus, planes CD and EF , being produced, will not meet. Planes CD and EF are thus parallel [Def. 11.8]. Thus, planes to which the same straight-line is at right-angles are parallel planes. (Which is) the very thing it was required to show.

ιε΄.

Proposition 15

'Ε¦ν δύο εÙθε‹αι ¡πτόµεναι ¢λλήλων παρ¦ δύο εÙθείας ¡πτοµένας ¢λλήλων ðσι µ¾ ™ν τù αÙτù ™πιπέδJ οâσαι, παράλληλά ™στι τ¦ δι' αÙτîν ™πίπεδα. ∆ύο γ¦ρ εÙθε‹αι ¡πτόµεναι ¢λλήλων αƒ ΑΒ, ΒΓ παρ¦ δύο εÙθείας ¡πτοµένας ¢λλήλων τ¦ς ∆Ε, ΕΖ œστωσαν µ¾ ™ν τù αÙτù ™πιπέδJ οâσαι· λέγω, Óτι ™κβαλλόµενα τ¦ δι¦ τîν ΑΒ, ΒΓ, ∆Ε, ΕΖ ™πίπεδα οÙ συµπεσε‹ται ¢λλήλοις. ”Ηχθω γ¦ρ ¢πÕ τοà Β σηµείου ™πˆ τÕ δι¦ τîν ∆Ε, ΕΖ ™πίπεδον κάθετος ¹ ΒΗ κሠσυµβαλλέτω τù ™πιπέδJ κατ¦ τÕ Η σηµε‹ον, κሠδι¦ τοà Η τÍ µν Ε∆ παράλληλος ½χθω ¹ ΗΘ, τÍ δ ΕΖ ¹ ΗΚ.

If two straight-lines joined to one another are parallel (respectively) to two straight-lines joined to one another, which are not in the same plane, then the planes through them are parallel (to one another). For let the two straight-lines joined to one another, AB and BC, be parallel to the two straight-lines joined to one another, DE and EF (respectively), not being in the same plane. I say that the planes through AB, BC and DE, EF will not meet one another (when) produced. For let BG have been drawn from point B perpendicular to the plane through DE and EF [Prop. 11.11], and let it meet the plane at point G. And let GH have been drawn through G parallel to ED, and GK (parallel) to EF [Prop. 1.31].

438

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

B

B

G

A E

D J

H

C

A

E G

Z K

D H

F K

Κሠ™πεˆ ¹ ΒΗ Ñρθή ™στι πρÕς τÕ δι¦ τîν ∆Ε, ΕΖ ™πίπεδον, κሠπρÕς πάσας ¥ρα τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù δι¦ τîν ∆Ε, ΕΖ ™πιπέδJ Ñρθ¦ς ποιήσει γωνίας. ¤πτεται δ αÙτÁς ˜κατέρα τîν ΗΘ, ΗΚ οâσα ™ν τù δι¦ τîν ∆Ε, ΕΖ ™πιπέδJ· Ñρθ¾ ¥ρα ™στˆν ˜κατέρα τîν ØπÕ ΒΗΘ, ΒΗΚ γωνιîν. κሠ™πεˆ παράλληλός ™στιν ¹ ΒΑ τÍ ΗΘ, αƒ ¥ρα ØπÕ ΗΒΑ, ΒΗΘ γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν. Ñρθ¾ δ ¹ ØπÕ ΒΗΘ· Ñρθ¾ ¥ρα κሠ¹ ØπÕ ΗΒΑ· ¹ ΗΒ ¥ρα τÍ ΒΑ πρÕς Ñρθάς ™στιν. δι¦ τ¦ αÙτ¦ δ¾ ¹ ΗΒ κሠτÍ ΒΓ ™στι πρÕς Ñρθάς. ™πεˆ οâν εÙθε‹α ¹ ΗΒ δυσˆν εÙθείαις τα‹ς ΒΑ, ΒΓ τεµνούσαις ¢λλήλας πρÕς Ñρθ¦ς ™φέστηκεν, ¹ ΗΒ ¥ρα κሠτù δι¦ τîν ΒΑ, ΒΓ ™πιπέδJ πρÕς Ñρθάς ™στιν. [δι¦ τ¦ αÙτ¦ δ¾ ¹ ΒΗ κሠτù δι¦ τîν ΗΘ, ΗΚ ™πιπέδJ πρÕς Ñρθάς ™στιν. τÕ δ δι¦ τîν ΗΘ, ΗΚ ™πίπεδόν ™στι τÕ δι¦ τîν ∆Ε, ΕΖ· ¹ ΒΗ ¥ρα τù δι¦ τîν ∆Ε, ΕΖ ™πιπέδJ ™στˆ πρÕς Ñρθάς. ™δείχθη δ ¹ ΗΒ κሠτù δι¦ τîν ΑΒ, ΒΓ ™πιπέδJ πρÕς Ñρθάς]. πρÕς § δ ™πίπεδα ¹ αÙτ¾ εÙθε‹α Ñρθή ™στιν, παράλληλά ™στι τ¦ ™πίπεδα· παράλληλον ¥ρα ™στˆ τÕ δι¦ τîν ΑΒ, ΒΓ ™πίπεδον τù δι¦ τîν ∆Ε, ΕΖ. 'Ε¦ν ¥ρα δύο εÙθε‹αι ¡πτόµεναι ¢λλήλων παρ¦ δύο εÙθείας ¡πτοµένας ¢λλήλων ðσι µ¾ ™ν τù αÙτù ™πιπέδJ, παράλληλά ™στι τ¦ δι' αÙτîν ™πίπεδα· Óπερ œδει δε‹ξαι.

And since BG is at right-angles to the plane through DE and EF , it will thus also make right-angles with all of the straight-lines joined to it, which are also in the plane through DE and EF [Def. 11.3]. And each of GH and GK, which are in the plane through DE and EF , are joined to it. Thus, each of the angles BGH and BGK are right-angles. And since BA is parallel to GH [Prop. 11.9], the (sum of the) angles GBA and BGH is equal to two right-angles [Prop. 1.29]. And BGH (is) a right-angle. GBA (is) thus also a right-angle. Thus, GB is at right-angles to BA. So, for the same (reasons), GB is also at right-angles to BC. Therefore, since the straight-line GB has been set up at right-angles to two straight-lines, BA and BC, cutting one another, GB is thus at right-angles to the plane through BA and BC [Prop. 11.4]. [So, for the same (reasons), BG is also at right-angles to the plane through GH and GK. And the plane through GH and GK is the (plane) through DE and EF . And it was also shown that GB is at rightangles to the plane through AB and BC.] And planes to which the same straight-line is at right-angles are parallel planes [Prop. 11.14]. Thus, the plane through AB and BC is parallel to the (plane) through DE and EF . Thus, if two straight-lines joined to one another are parallel (respectively) to two straight-lines joined to one another, which are not in the same plane, then the planes through them are parallel (to one another). (Which is) the very thing it was required to show.

ι$΄.

Proposition 16

'Ε¦ν δύο ™πίπεδα παράλληλα ØπÕ ™πιπέδου τινÕς If two parallel planes are cut by some plane then their τέµνηται, αƒ κοινሠαÙτîν τοµαˆ παράλληλοί ε„σιν. common sections are parallel. ∆ύο γ¦ρ ™πίπεδα παράλληλα τ¦ ΑΒ, Γ∆ ØπÕ For let the two parallel planes AB and CD have been ™πιπέδου τοà ΕΖΗΘ τεµνέσθω, κοιναˆ δ αÙτîν τοµαˆ cut by the plane EF GH. And let EF and GH be their œστωσαν αƒ ΕΖ, ΗΘ· λέγω, Óτι παράλληλός ™στιν ¹ ΕΖ common sections. I say that EF is parallel to GH. τÍ ΗΘ.

439

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

B Z

E A

B

D H

F

E

K

K

A D

J

G

G

H

C

Ε„ γ¦ρ µή, ™κβαλλόµεναι αƒ ΕΖ, ΗΘ ½τοι ™πˆ τ¦ Ζ, Θ µέρη À ™πˆ τ¦ Ε, Η συµπεσοàνται. ™κβεβλήσθωσαν æς ™πˆ τ¦ Ζ, Θ µέρη κሠσυµπιπτέτωσαν πρότερον κατ¦ τÕ Κ. κሠ™πεˆ ¹ ΕΖΚ ™ν τù ΑΒ ™στιν ™πιπέδJ, κሠπάντα ¥ρα τ¦ ™πˆ τÁς ΕΖΚ σηµε‹α ™ν τù ΑΒ ™στιν ™πιπέδJ. žν δ τîν ™πˆ τÁς ΕΖΚ εÙθείας σηµείων ™στˆ τÕ Κ· τÕ Κ ¥ρα ™ν τù ΑΒ ™στιν ™πιπέδJ. δι¦ τ¦ αÙτ¦ δ¾ τÕ Κ κሠ™ν τù Γ∆ ™στιν ™πιπέδJ· τ¦ ΑΒ, Γ∆ ¥ρα ™πίπεδα ™κβαλλόµενα συνπεσοàνται. οÙ συµπίπτουσι δ δι¦ τÕ παράλληλα Øποκε‹σθαι· οÙκ ¥ρα αƒ ΕΖ, ΗΘ εÙθε‹αι ™κβαλλόµεναι ™πˆ τ¦ Ζ, Θ µέρη συµπεσοàνται. еοίως δ¾ δείξοµεν, Óτι αƒ ΕΖ, ΗΘ εÙθε‹αι οÙδέ ™πˆ τ¦ Ε, Η µέρη ™κβαλλόµεναι συµπεσοàνται. αƒ δ ™πˆ µηδέτερα τ¦ µέρη συµπίπτουσαι παράλληλοί ε„σιν. παράλληλος ¥ρα ™στˆν ¹ ΕΖ τÍ ΗΘ. 'Ε¦ν ¥ρα δύο ™πίπεδα παράλληλα ØπÕ ™πιπέδου τινÕς τέµνηται, αƒ κοινሠαÙτîν τοµαˆ παράλληλοί ε„σιν· Óπερ œδει δε‹ξαι.

For, if not, being produced, EF and GH will meet either in the direction of F , H, or of E, G. Let them be produced, as in the direction of F , H, and let them, first of all, have met at K. And since EF K is in the plane AB, all of the points on EF K are thus also in the plane AB [Prop. 11.1]. And K is one of the points on EF K. Thus, K is in the plane AB. So, for the same (reasons), K is also in the plane CD. Thus, the planes AB and CD, being produced, will meet. But they do not meet, on account of being (initially) assumed (to be mutually) parallel. Thus, the straight-lines EF and GH, being produced in the direction of F , H, will not meet. So, similarly, we can show that the straight-lines EF and GH, being produced in the direction of E, G, will not meet either. And (straight-lines in one plane which), being produced, do not meet in either direction are parallel [Def. 1.23]. EF is thus parallel to GH. Thus, if two parallel planes are cut by some plane then their common sections are parallel. (Which is) the very thing it was required to show.

ιζ΄.

Proposition 17

'Ε¦ν δύο εÙθε‹αι ØπÕ παραλλήλων ™πιπεδων τέµνωνται, ε„ς τοÝς αÙτοÝς λόγους τµηθήσονται. ∆ύο γ¦ρ εÙθε‹αι αƒ ΑΒ, Γ∆ ØπÕ παραλλήλων ™πιπέδων τîν ΗΘ, ΚΛ, ΜΝ τεµνέσθωσαν κατ¦ τ¦ Α, Ε, Β, Γ, Ζ, ∆ σηµε‹α· λέγω, Óτι ™στˆν æς ¹ ΑΕ εÙθε‹α πρÕς τ¾ν ΕΒ, οÛτως ¹ ΓΖ πρÕς τ¾ν Ζ∆. 'Επεζεύθχωσαν γ¦ρ αƒ ΑΓ, Β∆, Α∆, κሠσυµβαλλέτω ¹ Α∆ τù ΚΛ ™πιπέδJ κατ¦ τÕ Ξ σηµε‹ον, κሠ™πεζεύχθωσαν αƒ ΕΞ, ΞΖ. Κሠ™πεˆ δύο ™πίπεδα παράλληλα τ¦ ΚΛ, ΜΝ ØπÕ ™πιπέδου τοà ΕΒ∆Ξ τέµνεται, αƒ κοινሠαÙτîν τοµαˆ αƒ ΕΞ, Β∆ παράλληλοί ε„σιν. δι¦ τ¦ αÙτ¦ δ¾ ™πεˆ δύο ™πίπεδα παράλληλα τ¦ ΗΘ, ΚΛ ØπÕ ™πιπέδου τοà ΑΞΖΓ τέµνεται, αƒ κοινሠαÙτîν τοµαˆ αƒ ΑΓ, ΞΖ παράλληλοί ε„σιν. κሠ™πεˆ τριγώνου τοà ΑΒ∆ παρ¦ µίαν τîν πλευρîν τ¾ν Β∆ εÙθε‹α Ãκται ¹ ΕΞ, ¢νάλογον

If two straight-lines are cut by parallel planes then they will be cut in the same ratios. For let the two straight-lines AB and CD be cut by the parallel planes GH, KL, and M N at the points A, E, B, and C, F , D (respectively). I say that as the straight-line AE is to EB, so CF (is) to F D. For let AC, BD, and AD have been joined, and let AD meet the plane KL at point O, and let EO and OF have been joined. And since two parallel planes KL and M N are cut by the plane EBDO, their common sections EO and BD are parallel [Prop. 11.16]. So, for the same (reasons), since two parallel planes GH and KL are cut by the plane AOF C, their common sections AC and OF are parallel [Prop. 11.16]. And since EO has been drawn parallel to one of the sides BD of triangle ABD, thus, proportion-

440

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

¥ρα ™στˆν æς ¹ ΑΕ πρÕς ΕΒ, οÛτως ¹ ΑΞ πρÕς Ξ∆. πάλιν ™πεˆ τριγώνου τοà Α∆Γ παρ¦ µίαν τîν πλευρîν τ¾ν ΑΓ εÙθε‹α Ãκται ¹ ΞΖ, ¢νάλογόν ™στιν æς ¹ ΑΞ πρÕς Ξ∆, οÛτως ¹ ΓΖ πρÕς Ζ∆. ™δείχθη δ κሠæς ¹ ΑΞ πρÕς Ξ∆, οÛτως ¹ ΑΕ πρÕς ΕΒ· κሠæς ¥ρα ¹ ΑΕ πρÕς ΕΒ, οÛτως ¹ ΓΖ πρÕς Ζ∆.

H

A

G

J

ally, as AE is to EB, so AO (is) to OD [Prop. 6.2]. Again, since OF has been drawn parallel to one of the sides AC of triangle ADC, proportionally, as AO is to OD, so CF (is) to F D [Prop. 6.2]. And it was also shown that as AO (is) to OD, so AE (is) to EB. And thus as AE (is) to EB, so CF (is) to F D [Prop. 5.11].

H A C

Z

E

X

G

L

L E

O F

K

K

N

N

B M

B

D

D M

'Ε¦ν ¥ρα δύο εÙθε‹αι ØπÕ παραλλήλων ™πιπέδων τέµνωνται, ε„ς τοÝς αÙτοÝς λόγους τµηθήσονται· Óπερ œδει δειξαι.

Thus, if two straight-lines are cut by parallel planes then they will be cut in the same ratios. (Which is) the very thing it was required to show.

ιη΄.

Proposition 18

'Ε¦ν εÙθε‹α ™πιπέδJ τινˆ πρÕς Ñρθ¦ς Ï, κሠπάντα τ¦ δι' αÙτÁς ™πίπεδα τù αÙτù ™πιπέδJ πρÕς Ñρθ¦ς œσται. ΕÙθε‹α γάρ τις ¹ ΑΒ τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθ¦ς œστω· λέγω, Óτι κሠπάντα τ¦ δι¦ τÁς ΑΒ ™πίπεδα τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς ™στιν. 'Εκβεβλήσθω γ¦ρ δι¦ τÁς ΑΒ ™πίπεδον τÕ ∆Ε, κሠœστω κοιν¾ τοµ¾ τοà ∆Ε ™πιπέδου κሠτοà Øποκειµένου ¹ ΓΕ, κሠε„λήφθω ™πˆ τÁς ΓΕ τυχÕν σηµε‹ον τÕ Ζ, κሠ¢πÕ τοà Ζ τÍ ΓΕ πρÕς Ñρθ¦ς ½χθω ™ν τù ∆Ε ™πιπέδJ ¹ ΖΗ. Κሠ™πεˆ ¹ ΑΒ πρÕς τÕ Øποκείµενον ™πίπεδον Ñρθή ™στιν, κሠπρÕς πάσας ¥ρα τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù ØποκαιµένJ ™πιπέδJ Ñρθή ™στιν ¹ ΑΒ· éστε κሠπρÕς τ¾ν ΓΕ Ñρθή ™στιν· ¹ ¥ρα ØπÕ ΑΒΖ γωνία Ñρθή ™στιν. œστι δ κሠ¹ ØπÕ ΗΖΒ Ñρθ¾· παράλληλος ¥ρα ™στˆν ¹ ΑΒ τÍ ΖΗ. ¹ δ ΑΒ τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς ™στιν. κሠ¹ ΖΗ ¥ρα τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς ™στιν. κሠ™πίπεδον πρÕς ™πίπεδον Ñρθόν ™στιν, Óταν αƒ τÍ κοινÍ τοµÍ τîν ™πιπέδων πρÕς Ñρθ¦ς ¢γόµεναι εÙθε‹αι ™ν ˜νˆ τîν ™πιπέδων τù λοιπù ™πιπέδJ πρÕς Ñρθ¦ς ðσιν. κሠτÍ

If a straight-line is at right-angles to some plane then all of the planes (passing) through it will also be at rightangles to the same plane. For let some straight-line AB be at right-angles to a reference plane. I say that all of the planes (passing) through AB are also at right-angles to the reference plane. For let the plane DE have been produced through AB. And let CE be the common section of the plane DE and the reference (plane). And let some random point F have been taken on CE. And let F G have been drawn from F , at right-angles to CE, in the plane DE [Prop. 1.11]. And since AB is at right-angles to the reference plane, AB is thus also at right-angles to all of the straightlines joined to it which are also in the reference plane [Def. 11.3]. Hence, it is also at right-angles to CE. Thus, angle ABF is a right-angle. And GF B is also a rightangle. Thus, AB is parallel to F G [Prop. 1.28]. And AB is at right-angles to the reference plane. Thus, F G is also at right-angles to the reference plane [Prop. 11.8]. And

441

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

κοινÍ τοµÍ τîν ™πιπέδων τÍ ΓΕ ™ν ˜νˆ τîν ™πιπέδων τù ∆Ε πρÕς Ñρθ¦ς ¢χθε‹σα ¹ ΖΗ ™δείχθη τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς· τÕ ¥ρα ∆Ε ™πίπεδον Ñρθόν ™στι πρÕς τÕ Øποκείµενον. еοίως δ¾ δειχθήσεται κሠπάντα τ¦ δι¦ τÁς ΑΒ ™πίπεδα Ñρθ¦ τυγχανοντα πρÕς τÕ Øποκείµενον ™πίπεδον.

D

G

H

A

Z

B

a plane is at right-angles to a(nother) plane when the straight-lines drawn at right-angles to the common section of the planes, (and lying) in one of the planes, are at right-angles to the remaining plane [Def. 11.4]. And F G, (which was) drawn at right-angles to the common section of the planes, CE, in one of the planes, DE, was shown to be at right-angles to the reference plane. Thus, plane DE is at right-angles to the reference (plane). So, similarly, it can be shown that all of the planes (passing) at random through AB (are) at right-angles to the reference plane.

D

E

G

A

F

B

C

E

'Ε¦ν ¥ρα εÙθε‹α ™πιπέδJ τινˆ πρÕς Ñρθ¦ς Ï, κሠπάντα τ¦ δι' αÙτÁς ™πίπεδα τù αÙτù ™πιπέδJ πρÕς Ñρθ¦ς œσται· Óπερ œδει δε‹ξαι.

Thus, if a straight-line is at right-angles to some plane then all of the planes (passing) through it will also be at right-angles to the same plane. (Which is) the very thing it was required to show.

ιθ΄.

Proposition 19

'Ε¦ν δύο ™πίπεδα τέµνοντα ¥λληλα ™πιπέδJ τινˆ πρÕς If two planes cutting one another are at right-angles Ñρθ¦ς Ï, κሠ¹ κοιν¾ αÙτîν τοµ¾ τù αÙτù ™πιπέδJ πρÕς to some plane then their common section will also be at Ñρθ¦ς œσται. right-angles to the same plane.

B

B

Z

E D A

F

E

D

G

A

C

∆ύο γ¦ρ ™πίπεδα τ¦ ΑΒ, ΒΓ τù ØποκειµένJ ™πιπέδJ For let the two planes AB and BC be at right-angles πρÕς Ñρθ¦ς œστω, κοιν¾ δ αÙτîν τοµ¾ œστω ¹ Β∆· to a reference plane, and let their common section be λέγω, Óτι ¹ Β∆ τù ØποκειµένJ ™πιπέδJ πρÕς Ñρθάς BD. I say that BD is at right-angles to the reference ™στιν. plane. 442

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

Μ¾ γάρ, κሠ½χθωσαν ¢πÕ τοà ∆ σηµείου ™ν µν τù ΑΒ ™πιπέδJ τÍ Α∆ εÙθείv πρÕς Ñρθ¦ς ¹ ∆Ε, ™ν δ τù ΒΓ ™πιπέδJ τÍ Γ∆ πρÕς Ñρθ¦ς ¹ ∆Ζ. Κሠ™πεˆ τÕ ΑΒ ™πίπεδον Ñρθόν ™στι πρÕς τÕ Øποκείµενον, κሠτÍ κοινÍ αÙτîν τοµÍ τÍ Α∆ πρÕς Ñρθ¦ς ™ν τù ΑΒ ™πιπέδJ Ãκται ¹ ∆Ε, ¹ ∆Ε ¥ρα Ñρθή ™στι πρÕς τÕ Øποκείµενον ™πίπεδον. еοίως δ¾ δείξοµεν, Óτι κሠ¹ ∆Ζ Ñρθή ™στι πρÕς τÕ Øποκείµενον ™πίπεδον. ¢πÕ τοà αÙτοà ¥ρα σηµείου τοà ∆ τù ØποκειµένJ ™πιπέδJ δύο εÙθε‹α πρÕς Ñρθ¦ς ¢νεσταµέναι ε„σˆν ™πˆ τ¦ αÙτ¦ µέρη· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα τù ØποκειµένJ ™πιπέδJ ¢πÕ τοà ∆ σηµείου ¢νασταθήσεται πρÕς Ñρθ¦ς πλ¾ν τÁς ∆Β κοινÁς τοµÁς τîν ΑΒ, ΒΓ ™πιπέδων. 'Ε¦ν ¥ρα δύο ™πίπεδα τέµνοντα ¥λληλα ™πιπέδJ τινˆ πρÕς Ñρθ¦ς Ï, κሠ¹ κοιν¾ αÙτîν τοµ¾ τù αÙτù ™πιπέδJ πρÕς Ñρθ¦ς œσται· Óπερ œδει δε‹ξαι.

For (if) not, let DE also have been drawn from point D, in the plane AB, at right-angles to the straight-line AD, and DF , in the plane BC, at right-angles to CD. And since the plane AB is at right-angles to the reference (plane), and DE has been drawn at right-angles to their common section AD, in the plane AB, DE is thus at right-angles to the reference plane [Def. 11.4]. So, similarly, we can show that DF is also at right-angles to the reference plane. Thus, two (different) straight-lines are set up, at the same point D, at right-angles to the reference plane, on the same side. The very thing is impossible [Prop. 11.13]. Thus, no (other straight-line) except the common section DB of the planes AB and BC can be set up at point D, at right-angles to the reference plane. Thus, if two planes cutting one another are at rightangles to some plane then their common section will also be at right-angles to the same plane. (Which is) the very thing it was required to show.

κ΄.

Proposition 20

'Ε¦ν στερε¦ γωνία ØπÕ τριîν γωνιîν ™πιπέδων περιέχηται, δύο Ðποιαιοàν τÁς λοιπÁς µείζονές ε„σι πάντV µεταλαµβανόµεναι.

If a solid angle is contained by three plane angles then (the sum of) any two (angles) is greater than the remaining (one), (the angles) being taken up in any (possible way).

D

D

A B

A

E

G

B

Στερε¦ γ¦ρ γωνία ¹ πρÕς τù Α ØπÕ τριîν γωνιîν ™πιπέδων τîν ØπÕ ΒΑΓ, ΓΑ∆, ∆ΑΒ περιεχέσθω· λέγω, Óτι τîν ØπÕ ΒΑΓ, ΓΑ∆, ∆ΑΒ γωνιîν δύο Ðποιαιοàν τÁς λοιπÁς µείζονές ε„σι πάντV µεταλαµβανόµεναι. Ε„ µν οâν αƒ ØπÕ ΒΑΓ, ΓΑ∆, ∆ΑΒ γωνίαι ‡σαι ¢λλήλαις ε„σίν, φανερόν, Óτι δύο Ðποιαιοàν τÁς λοιπÁς µείζονές ε„σιν. ε„ δ οÜ, œστω µείζων ¹ ØπÕ ΒΑΓ, κሠσυνεστάτω πρÕς τÍ ΑΒ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ ØπÕ ∆ΑΒ γωνίv ™ν τù δι¦ τîν ΒΑΓ ™πιπέδJ ‡ση ¹ ØπÕ ΒΑΕ, κሠκείσθω τÍ Α∆ ‡ση ¹ ΑΕ, κሠδι¦ τοà Ε σηµείου διαχθε‹σα ¹ ΒΕΓ τεµνέτω τ¦ς ΑΒ, ΑΓ εÙθείας κατ¦ τ¦ Β, Γ σηµε‹α, κሠ™πεζεύχθωσαν αƒ ∆Β, ∆Γ. Κሠ™πεˆ ‡ση ™στˆν ¹ ∆Α τÍ ΑΕ, κοιν¾ δ ¹ ΑΒ, δύο δυσˆν ‡σαι· κሠγωνία ¹ ØπÕ ∆ΑΒ γωνίv τÍ ØπÕ ΒΑΕ ‡ση· βάσις ¥ρα ¹ ∆Β βάσει τÍ ΒΕ ™στιν ‡ση. κሠ™πεˆ δύο αƒ Β∆, ∆Γ τÁς ΒΓ µείζονές ε„σιν, ïν ¹ ∆Β τÍ ΒΕ ™δείχθη

E

C

For let the solid angle A have been contained by the three plane angles BAC, CAD, and DAB. I say that (the sum of) any two of BAC, CAD, and DAB is greater than the remaining (one), (the angles) being taken up in any (possible way). For if the angles BAC, CAD, and DAB are equal to one another then (it is) clear that (the sum of) any two is greater than the remaining (one). But, if not, let BAC be greater (than CAD or DAB). And let (angle) BAE, equal to the angle DAB, have been constructed in the plane through BAC, on the straight-line AB, at the point A on it. And let AE be made equal to AD. And BEC being drawn across through point E, let it cut the straightlines AB and AC at points B and C (respectively). And let DB and DC have been joined. And since DA is equal to AE, and AB (is) common, the two (straight-lines AD and AB are) equal to the

443

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

‡ση, λοιπ¾ ¥ρα ¹ ∆Γ λοιπÁς τÁς ΕΓ µείζων ™στίν. κሠ™πεˆ ‡ση ™στˆν ¹ ∆Α τÍ ΑΕ, κοιν¾ δ ¹ ΑΓ, κሠβάσις ¹ ∆Γ βάσεως τÁς ΕΓ µείζων ™στίν, γωνία ¥ρα ¹ ØπÕ ∆ΑΓ γωνάις τÁς ØπÕ ΕΑΓ µείζων ™στίν. ™δείχθη δ κሠ¹ ØπÕ ∆ΑΒ τÍ ØπÕ ΒΑΕ ‡ση· αƒ ¥ρα ØπÕ ∆ΑΒ, ∆ΑΓ τÁς ØπÕ ΒΑΓ µείζονές ε„σιν. еοίως δ¾ δείξοµεν, Óτι καˆ αƒ λοιπሠσύνδυο λαµβανόµεναι τÁς λοιπÁς µείζονές ε„σιν. 'Ε¦ν ¥ρα στερε¦ γωνία ØπÕ τριîν γωνιîν ™πιπέδων περιέχηται, δύο Ðποιαιοàν τÁς λοιπÁς µείζονές ε„σι πάντV µεταλαµβανόµεναι· Óπερ œδει δε‹ξαι.

two (straight-lines EA and AB, respectively). And angle DAB (is) equal to angle BAE. Thus, the base DB is equal to the base BE [Prop. 1.4]. And since the (sum of the) two (straight-lines) BD and DC is greater than BC [Prop. 1.20], of which DB was shown (to be) equal to BE, the remainder DC is thus greater than the remainder EC. And since DA is equal to AE, but AC (is) common, and the base DC is greater than the base EC, the angle DAC is thus greater than the angle EAC [Prop. 1.25]. And DAB was also shown (to be) equal to BAE. Thus, (the sum of) DAB and DAC is greater than BAC. So, similarly, we can also show that the remaining (angles), being taken in pairs, are greater than the remaining (one). Thus, if a solid angle is contained by three plane angles then (the sum of) any two (angles) is greater than the remaining (one), (the angles) being taken up in any (possible way). (Which is) the very thing it was required to show.

κα΄.

Proposition 21

“Απασα στερε¦ γωνία ØπÕ ™λασσόνων [À] τεσσάρων All solid angles are contained by plane angles (whose Ñρθîν γωνιîν ™πιπέδων περιέχεται. sum is) less [than] four right-angles.†

G

A B

C

D

A

D

B

”Εστω στερε¦ γωνία ¹ πρÕς τù Α περιεχοµένη ØπÕ ™πιπέδων γωνιîν τîν ØπÕ ΒΑΓ, ΓΑ∆, ∆ΑΒ· λέγω, Óτι αƒ ØπÕ ΒΑΓ, ΓΑ∆, ∆ΑΒ τεσσάρων Ñρθîν ™λάσσονές ε„σιν. Ε„λήφθω γ¦ρ ™φ' ˜κάστης τîν ΑΒ, ΑΓ, Α∆ τυχόντα σηµε‹α τ¦ Β, Γ, ∆, κሠ™πεζεύχθωσαν αƒ ΒΓ, Γ∆, ∆Β. κሠ™πεˆ στερε¦ γωνία ¹ πρÕς τù Β ØπÕ τριîν γωνιîν ™πιπέδων περιέχεται τîν ØπÕ ΓΒΑ, ΑΒ∆, ΓΒ∆, δύο Ðποιαιοàν τÁς λοιπÁς µείζονές ε„σιν· αƒ ¥ρα ØπÕ ΓΒΑ, ΑΒ∆ τÁς ØπÕ ΓΒ∆ µείζονές ε„σιν. δι¦ τ¦ αÙτ¦ δ¾ καˆ αƒ µν ØπÕ ΒΓΑ, ΑΓ∆ τÁς ØπÕ ΒΓ∆ µείζονές ε„σιν, αƒ δ ØπÕ Γ∆Α, Α∆Β τÁς ØπÕ Γ∆Β µείζονές ε„σιν· αƒ žξ ¥ρα γωνίαι αƒ ØπÕ ΓΒΑ, ΑΒ∆, ΒΓΑ, ΑΓ∆, Γ∆Α, Α∆Β τριîν τîν ØπÕ ΓΒ∆, ΒΓΑ, Γ∆Β µείζονές ε„σιν. ¢λλ¦ αƒ τρε‹ς αƒ ØπÕ ΓΒ∆, Β∆Γ, ΒΓ∆ δυσˆν Ñρθα‹ς ‡σαι ε„σίν· αƒ žξ ¥ρα αƒ ØπÕ ΓΒΑ, ΑΒ∆, ΒΓΑ, ΑΓ∆, Γ∆Α, Α∆Β δύο Ñρθîν µείζονές ε„σιν. κሠ™πεˆ ˜κάστου τîν ΑΒΓ, ΑΓ∆, Α∆Β τριγώνων αƒ τρε‹ς γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν, αƒ ¥ρα τîν τριîν τριγώνων ™ννέα γωνίαι αƒ ØπÕ

Let the solid angle A be contained by the plane angles BAC, CAD, and DAB. I say that (the sum of) BAC, CAD, and DAB is less than four right-angles. For let the random points B, C, and D have been taken on each of (the straight-lines) AB, AC, and AD (respectively). And let BC, CD, and DB have been joined. And since the solid angle at B is contained by the three plane angles CBA, ABD, and CBD, (the sum of) any two is greater than the remaining (one) [Prop. 11.20]. Thus, (the sum of) CBA and ABD is greater than CBD. So, for the same (reasons), (the sum of) BCA and ACD is also greater than BCD, and (the sum of) CDA and ADB is greater than CDB. Thus, the (sum of the) six angles CBA, ABD, BCA, ACD, CDA, and ADB is greater than the (sum of the) three (angles) CBD, BCD, and CDB. But, the (sum of the) three (angles) CBD, BDC, and BCD is equal to two right-angles [Prop. 1.32]. Thus, the (sum of the) six an-

444

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

ΓΒΑ, ΑΓΒ, ΒΑΓ, ΑΓ∆, Γ∆Α, ΓΑ∆, Α∆Β, ∆ΒΑ, ΒΑ∆ žξ Ñρθα‹ς ‡σαι ε„σίν, ïν αƒ ØπÕ ΑΒΓ, ΒΓΑ, ΑΓ∆, Γ∆Α, Α∆Β, ∆ΒΑ žξ γωνίαι δύο Ñρθîν ε„σι µείζονες· λοιπሠ¥ρα αƒ ØπÕ ΒΑΓ, ΓΑ∆, ∆ΑΒ τρε‹ς [γωνίαι] περιέχουσαι τ¾ν στερε¦ν γωνίαν τεσσάρων Ñρθîν ™λάσσονές ε„σιν. “Απασα ¥ρα στερε¦ γωνία ØπÕ ™λασσόνων [½] τεσσάρων Ñρθîν γωνιîν ™πιπέδων περιέχεται· Óπερ œδει δε‹ξαι.

†

gles CBA, ABD, BCA, ACD, CDA, and ADB is greater than two right-angles. And since the (sum of the) three angles of each of the triangles ABC, ACD, and ADB is equal to two right-angles, the (sum of the) nine angles CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, and BAD of the three triangles is equal to six right-angles, of which the (sum of the) six angles ABC, BCA, ACD, CDA, ADB, and DBA is greater than two right-angles. Thus, the (sum of the) remaining three [angles] BAC, CAD, and DAB, containing the solid angle, is less than four right-angles. Thus, all solid angles are contained by plane angles (whose sum is) less [than] four right-angles. (Which is) the very thing it was required to show.

This proposition is only proved for the case of a solid angle contained by three plane angles. However, the generalization to a solid angle

contained by more than three plane angles is straightforward.

κβ΄.

Proposition 22

'Ε¦ν ïσι τρε‹ς γωνίαι ™πίπεδοι, ïν αƒ δύο τÁς λοιπÁς If there are three plane angles, of which (the sum of µείζονές ε„σι πάντV µεταλαµβανόµεναι, περιέχωσι δ any) two is greater than the remaining (one), (the anαÙτ¦ς ‡σαι εÙθε‹αι, δυνατόν ™στιν ™κ τîν ™πιζευγνυουσîν gles) being taken up in any (possible way), and if equal τ¦ς ‡σας εÙθείας τρίγωνον συστήσασθαι. straight-lines contain them, then it is possible to construct a triangle from (the straight-lines created by) joining the (ends of the) equal straight-lines.

B

A

J

E

G D

Z H

H

L

B

L

E

K

A

”Εστωσαν τρε‹ς γωνίαι ™πίπεδοι αƒ ØπÕ ΑΒΓ, ∆ΕΖ, ΗΘΚ, ïν αƒ δύο τÁς λοιπÁς µείζονές ε„σι πάντV µεταλαµβανόµεναι, αƒ µν ØπÕ ΑΒΓ, ∆ΕΖ τÁς ØπÕ ΗΘΚ, αƒ δ ØπÕ ∆ΕΖ, ΗΘΚ τÁς ØπÕ ΑΒΓ, κሠœτι αƒ ØπÕ ΗΘΚ, ΑΒΓ τÁς ØπÕ ∆ΕΖ, κሠœστωσαν ‡σαι αƒ ΑΒ, ΒΓ, ∆Ε, ΕΖ, ΗΘ, ΘΚ εÙθε‹αι, κሠ™πεζεύχθωσαν αƒ ΑΓ, ∆Ζ, ΗΚ· λέγω, Óτι δυνατόν ™στιν ™κ τîν ‡σων τα‹ς ΑΓ, ∆Ζ, ΗΚ τρίγωνον συστήσασθαι, τουτέστιν Óτι τîν ΑΓ, ∆Ζ, ΗΚ δύο Ðποιαιοàν τÁς λοιπÁς µείζονές ε„σιν. Ε„ µν οâν αƒ ØπÕ ΑΒΓ, ∆ΕΖ, ΗΘΚ γωνίαι ‡σαι ¢λλήλαις ε„σίν, φανερόν, Óτι κሠτîν ΑΓ, ∆Ζ, ΗΚ ‡σων γινοµένων δυνατόν ™στιν ™κ τîν ‡σων τα‹ς ΑΓ, ∆Ζ, ΗΚ τρίγωνον συστήσασθαι. ε„ δ οÜ, œστωσαν ¥νισοι, κሠσυνεστάτω πρÕς τÍ ΘΚ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Θ τÍ ØπÕ ΑΒΓ γωνίv ‡ση ¹ ØπÕ ΚΘΛ· κሠκείσθω µι´ τîν ΑΒ, ΒΓ, ∆Ε, ΕΖ, ΗΘ, ΘΚ ‡ση ¹ ΘΛ, κሠ™πεζεύχθωσαν αƒ ΚΛ, ΗΛ. κሠ™πεˆ δύο αƒ ΑΒ, ΒΓ δυσˆ τα‹ς ΚΘ, ΘΛ ‡σαι ε„σίν, κሠγωνία ¹ πρÕς τù Β γωνίv τÍ ØπÕ ΚΘΛ ‡ση, βάσις ¥ρα ¹ ΑΓ βάσει τÍ ΚΛ ‡ση. κሠ™πεˆ αƒ ØπÕ ΑΒΓ, ΗΘΚ τÁς ØπÕ ∆ΕΖ µείζονές ε„σιν,

C D

F G

K

Let ABC, DEF , and GHK be three plane angles, of which the sum of any) two is greater than the remaining (one), (the angles) being taken up in any (possible way)—(that is), ABC and DEF (greater) than GHK, DEF and GHK (greater) than ABC, and, further, GHK and ABC (greater) than DEF . And let AB, BC, DE, EF , GH, and HK be equal straight-lines. And let AC, DF , and GK have been joined. I say that that it is possible to construct a triangle out of (straight-lines) equal to AC, DF , and GK—that is to say, that (the sum of) any two of AC, DF , and GK is greater than the remaining (one). Now, if the angles ABC, DEF , and GHK are equal to one another, (it is) clear that, (with) AC, DF , and GK also becoming equal, it is possible to construct a triangle from (straight-lines) equal to AC, DF , and GK. And if not, let them be unequal, and let KHL, equal to angle ABC, have been constructed on the straight-line HK, at the point H on it. And let HL be made equal to one of AB, BC, DE, EF , GH, and HK. And let KL

445

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

‡ση δ ¹ ØπÕ ΑΒΓ τÍ ØπÕ ΚΘΛ, ¹ ¥ρα ØπÕ ΗΘΛ τÁς ØπÕ ∆ΕΖ µείζων ™στίν. κሠ™πεˆ δύο αƒ ΗΘ, ΘΛ δύο τα‹ς ∆Ε, ΕΖ ‡σαι ε„σίν, κሠγωνία ¹ ØπÕ ΗΘΛ γωνίας τÁς ØπÕ ∆ΕΖ µείζων, βάσις ¥ρα ¹ ΗΛ βάσεως τÁς ∆Ζ µείζων ™στίν. ¢λλ¦ αƒ ΗΚ, ΚΛ τÁς ΗΛ µείζονές ε„σιν. πολλù ¥ρα αƒ ΗΚ, ΚΛ τÁς ∆Ζ µείζονές ε„σιν. ‡ση δ ¹ ΚΛ τÍ ΑΓ· αƒ ΑΓ, ΗΚ ¥ρα τÁς λοιπÁς τÁς ∆Ζ µείζονές ε„σιν. еοίως δ¾ δείξοµεν, Óτι καˆ αƒ µν ΑΓ, ∆Ζ τÁς ΗΚ µείζονές ε„σιν, κሠœτι αƒ ∆Ζ, ΗΚ τÁς ΑΓ µείζονές ε„σιν. δυνατÕν ¥ρα ™στˆν ™κ τîν ‡σων τα‹ς ΑΓ, ∆Ζ, ΗΚ τρίγωνον συστήσασθαι· Óπερ œδει δε‹ξαι.

and GL have been joined. And since the two (straightlines) AB and BC are equal to the two (straight-lines) KH and HL (respectively), and the angle B (is) equal to KHL, the base AC is thus equal to the base KL [Prop. 1.4]. And since (the sum of) ABC and GHK is greater than DEF , and ABC equal to KHL, GHL is thus greater than DEF . And since the two (straightlines) GH and HL are equal to the two (straight-lines) DE and EF (respectively), and angle GHL (is) greater than DEF , the base GL is thus greater than the base DF [Prop. 1.24]. But, (the sum of) GK and KL is greater than GL [Prop. 1.20]. Thus, (the sum of) GK and KL is much greater than DF . And KL (is) equal to AC. Thus, (the sum of) AC and GK is greater than the remaining (straight-line) DF . So, similarly, we can show that (the sum of) AC and DF is greater than GK, and, further, that (the sum of) DF and GK is greater than AC. Thus, it is possible to construct a triangle from (straight-lines) equal to AC, DF , and GK. (Which is) the very thing it was required to show.

κγ΄.

Proposition 23

'Εκ τριîν γωνιîν ™πιπέδων, ïν αƒ δύο τÁς λοιπÁς To construct a solid angle from three (given) plane µείζονές ε„σι πάντV µεταλαµβανόµεναι, στερε¦ν γωνίαν angles, (the sum of) two of which is greater than the reσυστήσασθαι· δε‹ δ¾ τ¦ς τρε‹ς τεσσάρων Ñρθîν ™λάσσ- maining (one, the angles) being taken up in any (possible ονας εναι. way). So, it is necessary for the (sum of the) three (angles) to be less than four right-angles [Prop. 11.21]. H J B B E E

A

G D

Z H

K

”Εστωσαν αƒ δοθε‹σαι τρε‹ς γωνίαι ™πίπεδοι αƒ ØπÕ ΑΒΓ, ∆ΕΖ, ΗΘΚ, ïν αƒ δύο τÁς λοιπÁς µείζονες œστωσαν πάντV µεταλαµβανόµεναι, œτι δ αƒ τρε‹ς τεσσάρων Ñρθîν ™λάσσονες· δε‹ δ¾ ™κ τîν ‡σων τα‹ς ØπÕ ΑΒΓ, ∆ΕΖ, ΗΘΚ στερε¦ν γωνίαν συστήσασθαι. 'Απειλήφθωσαν ‡σαι αƒ ΑΒ, ΒΓ, ∆Ε, ΕΖ, ΗΘ, ΘΚ, κሠ™πεζεύχθωσαν αƒ ΑΓ, ∆Ζ, ΗΚ· δυνατÕν ¥ρα ™στˆν ™κ τîν ‡σων τα‹ς ΑΓ, ∆Ζ, ΗΚ τρίγωνον συστήσασθαι. συνεστάτω τÕ ΛΜΝ, éστε ‡σην εναι τ¾ν µν ΑΓ τÍ ΛΜ, τ¾ν δ ∆Ζ τÍ ΜΝ, κሠœτι τ¾ν ΗΚ τÍ ΝΛ, κሠπεριγεγράφθω περˆ τÕ ΛΜΝ τρίγωνον κύκλος Ð ΛΜΝ, κሠε„λήφθω αÙτοà τÕ κέντρον κሠœστω τÕ Ξ, κሠ™πεζεύχθωσαν αƒ ΛΞ, ΜΞ, ΝΞ·

A

C D F G K Let ABC, DEF , and GHK be the three given plane angles, of which let (the sum of) two be greater than the remaining (one, the angles) being taken up in any (possible way), and, further, (let) the (sum of the) three (be) less than four right-angles. So, it is necessary to construct a solid angle from (plane angles) equal to ABC, DEF , and GHK. Let AB, BC, DE, EF , GH, and HK be cut off (so as to be) equal (to one another). And let AC, DF , and GK have been joined. It is, thus, possible to construct a triangle from (straight-lines) equal to AC, DF , and GK [Prop. 11.22]. Let (such a triangle), LM N , have be constructed, such that AC is equal to LM , DF to M N , and, further, GK to N L. And let the circle LM N have been circumscribed about triangle LM N [Prop. 4.5]. And let its center have been found, and let it be (at) O. And let

446

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11 LO, M O, and N O have been joined.

M

R

M

R

P

L

O

X

Q

O

N

L

Λέγω, Óτι ¹ ΑΒ µείζων ™στˆ τÁς ΛΞ. ε„ γ¦ρ µή, ½τοι ‡ση ™στˆν ¹ ΑΒ τÍ ΛΞ À ™λάττων. œστω πρότερον ‡ση. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΒ τÍ ΛΞ, ¢λλ¦ ¹ µν ΑΒ τÍ ΒΓ ™στιν ‡ση, ¹ δ ΞΛ τÍ ΞΜ, δύο δ¾ αƒ ΑΒ, ΒΓ δύο τα‹ς ΛΞ, ΞΜ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠβάσις ¹ ΑΓ βάσει τÍ ΛΜ Øπόκειται ‡ση· γωνία ¥ρα ¹ ØπÕ ΑΒΓ γωνίv τÍ ØπÕ ΛΞΜ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ µν ØπÕ ∆ΕΖ τÍ ØπÕ ΜΞΝ ™στιν ‡ση, κሠœτι ¹ ØπÕ ΗΘΚ τÍ ØπÕ ΝΞΛ· αƒ ¥ρα τρε‹ς αƒ ØπÕ ΑΒΓ, ∆ΕΖ, ΗΘΚ γωνίαι τρισˆ τα‹ς ØπÕ ΛΞΜ, ΜΞΝ, ΝΞΛ ε„σιν ‡σαι. ¢λλ¦ αƒ τρε‹ς αƒ ØπÕ ΛΞΜ, ΜΞΝ, ΝΞΛ τέτταρσιν Ñρθα‹ς ε„σιν ‡σαι· καˆ αƒ τρε‹ς ¥ρα αƒ ØπÕ ΑΒΓ, ∆ΕΖ, ΗΘΚ τέτταρσιν Ñρθα‹ς ‡σαι ε„σίν. Øπόκεινται δ κሠτεσσάρων Ñρθîν ™λάσσονες· Óπερ ¥τοπον. οÙκ ¥ρα ¹ ΑΒ τÍ ΛΞ ‡ση ™στίν. λέγω δή, Óτι οØδ ™λάττων ™στˆν ¹ ΑΒ τÁς ΛΞ. ε„ γ¦ρ δυνατόν, œστω· κሠκείσθω τÍ µν ΑΒ ‡ση ¹ ΞΟ, τÍ δ ΒΓ ‡ση ¹ ΞΠ, κሠ™πεζεύχθω ¹ ΟΠ. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΒ τÍ ΒΓ, ‡ση ™στˆ κሠ¹ ΞΟ τÍ ΞΠ· éστε κሠλοιπ¾ ¹ ΛΟ τÍ ΠΜ ™στιν ‡ση. παράλληλος ¥ρα ™στˆν ¹ ΛΜ τÍ ΟΠ, κሠ„σογώνιον τÕ ΛΜΞ τù ΟΠΞ· œστιν ¥ρα æς ¹ ΞΛ πρÕς ΛΜ, οÛτως ¹ ΞΟ πρÕς ΟΠ· ™ναλλ¦ξ æς ¹ ΛΞ πρÕς ΞΟ, οÛτως ¹ ΛΜ πρÕς ΟΠ. µείζων δ ¹ ΛΞ τÁς ΞΟ· µείζων ¥ρα κሠ¹ ΛΜ τÁς ΟΠ. ¢λλ¦ ¹ ΛΜ κε‹ται τÍ ΑΓ ‡ση· κሠ¹ ΑΓ ¥ρα τÁς ΟΠ µείζων ™στίν. ™πεˆ οâν δύο αƒ ΑΒ, ΒΓ δυσˆ τα‹ς ΟΞ, ΞΠ ‡σαι ε„σίν, κሠβάσις ¹ ΑΓ βάσεως τÁς ΟΠ µείζων ™στίν, γωνία ¥ρα ¹ ØπÕ ΑΒΓ γωνίας τÁς ØπÕ ΟΞΠ µε‹ζων ™στίν. еοίως δ¾ δείξοµεν, Óτι κሠ¹ µν ØπÕ ∆ΕΖ τÁς ØπÕ ΜΞΝ µείζων ™στίν, ¹ δ ØπÕ ΗΘΚ τÁς ØπÕ ΝΞΛ. αƒ ¥ρα τρε‹ς γωνίαι αƒ ØπÕ ΑΒΓ, ∆ΕΖ, ΗΘΚ τριîν τîν ØπÕ ΛΞΜ, ΜΞΝ, ΝΞΛ µείζονές ε„σιν. ¢λλ¦ αƒ ØπÕ ΑΒΓ, ∆ΕΖ, ΗΘΚ τεσσάρων Ñρθîν ™λάσσονες Øπόκεινται· πολλù ¥ρα αƒ ØπÕ ΛΞΜ, ΜΞΝ, ΝΞΛ τεσσάρων Ñρθîν ™λάσσονές ε„σιν. ¢λλ¦ κሠ‡σαι· Óπερ ™στˆν ¥τοπον. οÙκ ¥ρα ¹ ΑΒ ™λάσσων ™στˆ τÁς ΛΞ. ™δείχθη δέ, Óτι οÙδ ‡ση· µείζων ¥ρα ¹ ΑΒ τÁς ΛΞ.

P

N

I say that AB is greater than LO. For, if not, AB is either equal to, or less than, LO. Let it, first of all, be equal. And since AB is equal to LO, but AB is equal to BC, and OL to OM , so the two (straight-lines) AB and BC are equal to the two (straight-lines) LO and OM , respectively. And the base AC was assumed (to be) equal to the base LM . Thus, angle ABC is equal to angle LOM [Prop. 1.8]. So, for the same (reasons), DEF is also equal to M ON , and, further, GHK to N OL. Thus, the three angles ABC, DEF , and GHK are equal to the three angles LOM , M ON , and N OL, respectively. But, the (sum of the) three angles LOM , M ON , and N OL is equal to four right-angles. Thus, the (sum of the) three angles ABC, DEF , and GHK is also equal to four rightangles. And it was also assumed (to be) less than four right-angles. The very thing (is) absurd. Thus, AB is not equal to LO. So, I say that AB is not less than LO either. For, if possible, let it be (less). And let OP be made equal to AB, and OQ equal to BC, and let P Q have been joined. And since AB is equal to BC, OP is also equal to OQ. Hence, the remainder LP is also equal to (the remainder) QM . LM is thus parallel to P Q [Prop. 6.2], and (triangle) LM O (is) equiangular with (triangle) P QO [Prop. 1.29]. Thus, as OL is to LM , so OP (is) to P Q [Prop. 6.4]. Alternately, as LO (is) to OP , so LM (is) to P Q [Prop. 5.16]. And LO (is) greater than OP . Thus, LM (is) also greater than P Q [Prop. 5.14]. But LM was made equal to AC. Thus, AC is also greater than P Q. Therefore, since the two (straight-lines) AB and BC are equal to the two (straight-lines) P O and OQ (respectively), and the base AC is greater than the base P Q, the angle ABC is thus greater than the angle P OQ [Prop. 1.25]. So, similarly, we can show that DEF is also greater than M ON , and GHK than N OL. Thus, the (sum of the) three angles ABC, DEF , and GHK is greater than the (sum of the) three angles LOM , M ON , and N OL. But, (the sum of) ABC, DEF , and GHK was

447

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

'Ανεστάτω δ¾ ¢πÕ τοà Ξ σηµείου τù τοà ΛΜΝ κύκλου ™πιπέδJ πρÕς Ñρθ¦ς ¹ ΞΡ, κሠú µε‹ζόν ™στι τÕ ¢πÕ τÁς ΑΒ τετράγωνον τοà ¢πÕ τÁς ΛΞ, ™κείνJ ‡σον œστω τÕ ¢πÕ τÁς ΞΡ, κሠ™πεζεύχθωσαν αƒ ΡΛ, ΡΜ, ΡΝ. Κሠ™πεˆ ¹ ΡΞ Ñρθ¾ ™στι πρÕς τÕ τοà ΛΜΝ κέκλου ™πίπεδον, κሠπρÕς ˜κάστην ¥ρα τîν ΛΞ, ΜΞ, ΝΞ Ñρθή ™στιν ¹ ΡΞ. κሠ™πεˆ ‡ση ™στˆν ¹ ΛΞ τÍ ΞΜ, κοιν¾ δ κሠπρÕς Ñρθ¦ς ¹ ΞΡ, βάσις ¥ρα ¹ ΡΛ βάσει τÊ ΡΜ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΡΝ ˜κατέρv τîν ΡΛ, ΡΜ ™στιν ‡ση· αƒ τρε‹ς ¥ρα αƒ ΡΛ, ΡΜ, ΡΝ ‡σαι ¢λλήλαις ε„σίν. κሠ™πεˆ ú µε‹ζόν ™στι τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς ΛΞ, ™κείνJ ‡σον Øπόκειται τÕ ¢πÕ τÁς ΞΡ, τÕ ¥ρα ¢πÕ τÁς ΑΒ ‡σον ™στˆ το‹ς ¢πÕ τîν ΛΞ, ΞΡ. το‹ς δ ¢πÕ τîν ΛΞ, ΞΡ ‡σον ™στˆ τÕ ¢πÕ τÁς ΛΡ· Ñρθ¾ γ¦ρ ¹ ØπÕ ΛΞΡ· τÕ ¥ρα ¢πÕ τÁς ΑΒ ‡σον ™στˆ τù ¢πÕ τÁς ΡΛ· ‡ση ¥ρα ¹ ΑΒ τÍ ΡΛ. ¢λλ¦ τÍ µν ΑΒ ‡ση ™στˆν ˜κάστη τîν ΒΓ, ∆Ε, ΕΖ, ΗΘ, ΘΚ, τÍ δ ΡΛ ‡ση ˜κατέρα τîν ΡΜ, ΡΝ· ˜κάστη ¥ρα τîν ΑΒ, ΒΓ, ∆Ε, ΕΖ, ΗΘ, ΘΚ ˜κάστV τîν ΡΛ, ΡΜ, ΡΝ ‡ση ™στίν. κሠ™πεˆ δύο αƒ ΛΡ, ΡΜ δυσˆ τα‹ς ΑΒ, ΒΓ ‡σαι ε„σίν, κሠβάσις ¹ ΛΜ βάσει τÍ ΑΓ Øπόκειται ‡ση, γωνία ¥ρα ¹ ØπÕ ΛΡΜ γωνίv τÍ ØπÕ ΑΒΓ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ µν ØπÕ ΜΡΝ τÍ ØπÕ ∆ΕΖ ™στιν ‡ση, ¹ δ ØπÕ ΛΡΝ τÍ ØπÕ ΗΘΚ. 'Εκ τριîν ¥ρα γωνιîν ™πιπέδων τîν ØπÕ ΛΡΜ, ΜΡΝ, ΛΡΝ, α† ε„σιν ‡σαι τρισˆ τα‹ς δοθείσαις τα‹ς ØπÕ ΑΒΓ, ∆ΕΖ, ΗΘΚ, στερε¦ γωνία συνέσταται ¹ πρÕς τù Ρ περιεχοµένη ØπÕ τîν ΛΡΜ, ΜΡΝ, ΛΡΝ γωνιîν· Óπερ œδει ποιÁσαι.

assumed (to be) less than four right-angles. Thus, (the sum of) LOM , M ON , and N OL is much less than four right-angles. But, (it is) also equal (to four right-angles). The very thing is absurd. Thus, AB is not less than LO. And it was shown (to be) not equal either. Thus, AB (is) greater than LO. So let OR have been set up at point O at rightangles to the plane of circle LM N [Prop. 11.12]. And let the (square) on OR be equal to that (area) by which the square on AB is greater than the (square) on LO [Prop. 11.23 lem.]. And let RL, RM , and RN have been joined. And since RO is at right-angles to the plane of circle LM N , RO is thus also at right-angles to each of LO, M O, and N O. And since LO is equal to OM , and OR is common and at right-angles, the base RL is thus equal to the base RM [Prop. 1.4]. So, for the same (reasons), RN is also equal to each of RL and RM . Thus, the three (straight-lines) RL, RM , and RN are equal to one another. And since the (square) on OR was assumed to be equal to that (area) by which the (square) on AB is greater than the (square) on LO, the (square) on AB is thus equal to the (sum of the squares) on LO and OR. And the (square) on LR is equal to the (sum of the squares) on LO and OR. For LOR (is) a right-angle [Prop. 1.47]. Thus, the (square) on AB is equal to the (square) on RL. Thus, AB (is) equal to RL. But, each of BC, DE, EF , GH, and HK is equal to AB, and each of RM and RN equal to RL. Thus, each of AB, BC, DE, EF , GH, and HK is equal to each of RL, RM , and RN . And since the two (straight-lines) LR and RM are equal to the two (straight-lines) AB and BC (respectively), and the base LM was assumed (to be) equal to the base AC, the angle LRM is thus equal to the angle ABC [Prop. 1.8]. So, for the same (reasons), M RN is also equal to DEF , and LRN to GHK. Thus, the solid angle R, contained by the angles LRM , M RN , and LRN , has been constructed out of the three plane angles LRM , M RN , and LRN , which are equal to the three given (plane angles) ABC, DEF , and GHK (respectively). (Which is) the very thing it was required to do.

448

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

G A

C

B

ΛÁµµα.

A

B Lemma

•Ον δ τρόπον, ú µε‹ζόν ™στι τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς ΛΞ, ™κείνJ ‡σον λαβε‹ν œστι τÕ ¢πÕ τÁς ΞΡ, δείξοµεν οÛτως. ™κκείσθωσαν αƒ ΑΒ, ΛΞ εÙθε‹αι, κሠœστω µείζων ¹ ΑΒ, κሠγεγράφθω ™π' αÙτÁς ¹µικύκλιον τÕ ΑΒΓ, κሠε„ς τÕ ΑΒΓ ¹µικύκλιον ™νηρµόσθω τÍ ΛΞ εÙθείv µ¾ µείζονι οÜσV τÁς ΑΒ διαµέτρου ‡ση ¹ ΑΓ, κሠ™πεζεύχθω ¹ ΓΒ. ™πεˆ οâν ™ν ¹µικυκλίJ τù ΑΓΒ γωνία ™στˆν ¹ ØπÕ ΑΓΒ, Ñρθ¾ ¥ρα ™στˆν ¹ ØπÕ ΑΓΒ. τÕ ¥ρα ¢πÕ τÁς ΑΒ ‡σον ™στˆ το‹ς ¢πÕ τîν ΑΓ, ΓΒ. éστε τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς ΑΓ µε‹ζόν ™στι τù ¢πÕ τÁς ΓΒ. ‡ση δ ¹ ΑΓ τÍ ΛΞ. τÕ ¥ρα ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς ΛΞ µε‹ζόν ™στι τù ¢πÕ τÁς ΓΒ. ™¦ν οâν τÍ ΒΓ ‡σην τ¾ν ΞΡ ¢πολάβωµεν, œσται τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς ΛΞ µε‹ζον τù ¢πÕ τÁς ΞΡ· Óπερ προέκειτο ποιÁσαι.

And we can demonstrate, thusly, in which manner to take the (square) on OR equal to that (area) by which the (square) on AB is greater than the (square) on LO. Let the straight-lines AB and LO be set out, and let AB be greater, and let the semicircle ABC have been drawn around it. And let AC, equal to the straight-line LO, which is not greater than the diameter AB, have been inserted into the semicircle ABC [Prop. 4.1]. And let CB have been joined. Therefore, since the angle ACB is in the semicircle ACB, ACB is thus a right-angle [Prop. 3.31]. Thus, the (square) on AB is equal to the (sum of the) squares on AC and CB [Prop. 1.47]. Hence, the (square) on AB is greater than the (square) on AC by the (square) on CB. And AC (is) equal to LO. Thus, the (square) on AB is greater than the (square) on LO by the (square) on CB. Therefore, if we take OR equal to BC, then the (square) on AB will be greater than the (square) on LO by the (square) on OR. (Which is) the very thing it was prescribed to do.

κδ΄.

Proposition 24

'Ε¦ν στερεÕν ØπÕ παραλλήλων ™πιπέδων περιέχηται, If a solid (figure) is contained by parallel planes then τ¦ ¢πεναντίον αÙτοà ™πίπεδα ‡σα τε κሠπαραλ- its opposite planes are both equal and parallelogrammic. ληλόγραµµά ™στιν.

B

J

A

G D

H

B

H

Z E

A

C

D

ΣτερεÕν γ¦ρ τÕ Γ∆ΘΗ ØπÕ παραλλήλων ™πιπέδων περιεχέσθω τîν ΑΓ, ΗΖ, ΑΘ, ∆Ζ, ΒΖ, ΑΕ· λέγω, Óτι τ¦ ¢πεναντίον αÙτοà ™πίπεδα ‡σα τε κሠπαραλληλόγραµµά ™στιν. 'Επεˆ γ¦ρ δύο ™πίπεδα παράλληλα τ¦ ΒΗ, ΓΕ ØπÕ ™πιπέδου τοà ΑΓ τέµνεται, αƒ κοινሠαÙτîν τοµαˆ παράλληλοί ε„σιν. παράλληλος ¥ρα ™στˆν ¹ ΑΒ τÍ

G

F

E

For let the solid (figure) CDHG have been contained by the parallel planes AC, GF , and AH, DF , and BF , AE. I say that its opposite planes are both equal and parallelogrammic. For since the two parallel planes BG and CE are cut by the plane AC, their common sections are parallel [Prop. 11.16]. Thus, AB is parallel to DC. Again, since

449

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

∆Γ. πάλιν, ˜πεˆ δύο ™πίπεδα παράλληλα τ¦ ΒΖ, ΑΕ ØπÕ ™πιπέδου τοà ΑΓ τέµνεται, αƒ κοινሠαÙτîν τοµαˆ παράλληλοί ε„σιν. παράλληλος ¥ρα ™στˆν ¹ ΒΓ τÍ Α∆. ™δείχθη δ κሠ¹ ΑΒ τÍ ∆Γ παράλληλος· παραλληλόγραµµον ¥ρα ™στˆ τÕ ΑΓ. еοίως δ¾ δείξοµεν, Óτι κሠ›καστον τîν ∆Ζ, ΖΗ, ΗΒ, ΒΖ, ΑΕ παραλληλόγραµµόν ™στιν. 'Επεζεύχθωσαν αƒ ΑΘ, ∆Ζ. κሠ™πεˆ παράλληλός ™στιν ¹ µν ΑΒ τÍ ∆Γ, ¹ δ ΒΘ τÍ ΓΖ, δύο δ¾ αƒ ΑΒ, ΒΘ ¡πτόµεναι ¢λλήλων παρ¦ δύο εÙθείας τ¦ς ∆Γ, ΓΖ ¡πτοµένας ¢λλήλων ε„σˆν οÙκ ™ν τù αÙτù ™πιπέδJ· ‡σας ¥ρα γωνίας περιέξουσιν· ‡ση ¥ρα ¹ ØπÕ ΑΒΘ γωνία τÍ ØπÕ ∆ΓΖ. κሠ™πεˆ δύο αƒ ΑΒ, ΒΘ δυσˆ τα‹ς ∆Γ, ΓΖ ‡σαι ε„σίν, κሠγωνία ¹ ØπÕ ΑΒΘ γωνίv τÍ ØπÕ ∆ΓΖ ™στιν ‡ση, βάσις ¥ρα ¹ ΑΘ βάσει τÍ ∆Ζ ™στιν ‡ση, κሠτÕ ΑΒΘ τρίγωνον τù ∆ΓΖ τριγώνJ ‡σον ™στίν. καί ™στι τοà µν ΑΒΘ διπλάσιον τÕ ΒΗ παραλληλόγραµµον, τοà δ ∆ΓΖ διπλάσιον τÕ ΓΕ παραλληλόγραµµον· ‡σον ¥ρα τÕ ΒΗ παραλληλόγραµµον τù ΓΕ παραλληλογράµµJ· еοίως δ¾ δείξοµεν, Óτι κሠτÕ µν ΑΓ τù ΗΖ ™στιν ‡σον, τÕ δ ΑΕ τù ΒΖ. 'Ε¦ν ¥ρα στερεÕν ØπÕ παραλλήλων ™πιπέδων περιέχηται, τ¦ ¢πεναντίον αÙτοà ™πίπεδα ‡σα τε κሠπαραλληλόγραµµά ™στιν· Óπερ œδει δε‹ξαι.

the two parallel planes BF and AE are cut by the plane AC, their common sections are parallel [Prop. 11.16]. Thus, BC is parallel to AD. And AB was also shown (to be) parallel to DC. Thus, AC is a parallelogram. So, similarly, we can also show that DF , F G, GB, BF , and AE are each parallelograms. Let AH and DF have been joined. And since AB is parallel to DC, and BH to CF , so the two (straight-lines) joining one another, AB and BH, are parallel to the two straight-lines joining one another, DC and CF (respectively), not (being) in the same plane. Thus, they will contain equal angles [Prop. 11.10]. Thus, angle ABH (is) equal to (angle) DCF . And since the two (straightlines) AB and BH are equal to the two (straight-lines) DC and CF (respectively) [Prop. 1.34], and angle ABH is equal to angle DCF , the base AH is thus equal to the base DF , and triangle ABH is equal to triangle DCF [Prop. 1.4]. And parallelogram BG is double (triangle) ABH, and parallelogram CE double (triangle) DCF [Prop. 1.34]. Thus, parallelogram BG (is) equal to parallelogram CE. So, similarly, we can show that AC is also equal to GF , and AE to BF . Thus, if a solid (figure) is contained by parallel planes then its opposite planes are both equal and parallelogrammic. (Which is) the very thing it was required to show.

κε΄.

Proposition 25

'Ε¦ν στερεÕν παραλληλεπίπεδον ™πιπέδJ τµηθÍ παIf a parallelipiped solid is cut by a plane which is parραλλήλJ Ôντι το‹ς ¢πεναντίον ™πιπέδοις, œσται æς ¹ allel to the opposite planes (of the parallelipiped) then as βάσις πρÕς τ¾ν βάσιν, οÛτως τÕ στερεÕν πρÕς τÕ the base (is) to the base, so the solid will be to the solid. στερεόν.

Y

R

P

X

B O

L

K

U D W H I

F A

Z E

G

Q

J M N

T

Q

X

R

S

P

L

ΣτερεÕν γ¦ρ παραλληλεπίπεδον τÕ ΑΒΓ∆ ™πιπέδJ τù ΖΗ τετµήσθω παραλλήλJ Ôντι το‹ς ¢πεναντίον ™πιπέδοις το‹ς ΡΑ, ∆Θ· λέγω, Óτι ™στˆν æς ¹ ΑΕΖΦ βάσις πρÕς τ¾ν ΕΘΓΖ βάσιν, οÛτως τÕ ΑΒΖΥ στερεÕν πρÕς τÕ ΕΗΓ∆ στερεόν. 'Εκβεβλήσθω γ¦ρ ¹ ΑΘ ™φ' ˜κάτερα τ¦ µέρη, κሠκείσθωσαν τÍ µν ΑΕ ‡σαι Ðσαιδηποτοàν αƒ ΑΚ, ΚΛ, τÍ δ ΕΘ ‡σαι Ðσαιδηποτοàν αƒ ΘΜ, ΜΝ, κሠσυµπε-

K

V

A

D

F

E

Y

T

I

G

B

O

U

C

H

W

M

S

N

For let the parallelipiped solid ABCD have been cut by the plane F G which is parallel to the opposite planes RA and DH. I say that as the base AEF V (is) to the base EHCF , so the solid ABF U (is) to the solid EGCD. For let AH have been produced in each direction. And let any number whatsoever (of lengths), AK and KL, be made equal to AE, and any number whatsoever (of lengths), HM and M N , equal to EH. And let the paral-

450

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

πληρώσθω τ¦ ΛΟ, ΚΦ, ΘΧ, ΜΣ παραλληλόγραµµα κሠτ¦ ΛΠ, ΚΡ, ∆Μ, ΜΤ στερεά. Κሠ™πεˆ ‡σαι ε„σˆν αƒ ΛΚ, ΚΑ, ΑΕ εÙθε‹αι ¢λλήλαις, ‡σα ™στˆ κሠτ¦ µν ΛΟ, ΚΦ, ΑΖ παραλληλόγραµµα ¢λλήλοις, τ¦ δ ΚΞ, ΚΒ, ΑΗ ¢λλήλοις κሠœτι τ¦ ΛΨ, ΚΠ, ΑΡ ¢λλήλοις· ¢πεναντίον γάρ. δι¦ τ¦ αÙτ¦ δ¾ κሠτ¦ µν ΕΓ, ΘΧ, ΜΣ παραλληλόγραµµα ‡σα ε„σˆν ¢λλήλοις, τ¦ δ ΘΗ, ΘΙ, ΙΝ ‡σα ε„σˆν ¢λλήλοις, κሠœτι τ¦ ∆Θ, ΜΩ, ΝΤ· τρία ¥ρα ™πίπεδα τîν ΛΠ, ΚΡ, ΑΥ στερεîν τρισˆν ™πιπέδοις ™στˆν ‡σα. ¢λλ¦ τ¦ τρία τρισˆ το‹ς ¢πεναντίον ™στˆν ‡σα· τ¦ ¥ρα τρία στερε¦ τ¦ ΛΠ, ΚΡ, ΑΥ ‡σα ¢λλήλοις ™στίν. δι¦ τ¦ αÙτ¦ δ¾ κሠτ¦ τρία στερε¦ τ¦ Ε∆, ∆Μ, ΜΤ ‡σα ¢λλήλοις ™στίν· Ðσαπλασίων ¥ρα ™στˆν ¹ ΛΖ βάσις τÁς ΑΖ βάσεως, τοσαυταπλάσιόν ™στι κሠτÕ ΛΥ στερεÕν τοà ΑΥ στερεοà. δι¦ τ¦ αÙτ¦ δ¾ Ðσαπλασίων ™στˆν ¹ ΝΖ βάσις τÁς ΖΘ βάσεως, τοσαυταπλασίον ™στι κሠτÕ ΝΥ στερεÕν τοà ΘΥ στερεοà. καˆ ε„ ‡ση ™στˆν ¹ ΛΖ βάσις τÍ ΝΖ βάσει, ‡σον ™στˆ κሠτÕ ΛΥ στερεÕν τù ΝΥ στερεù, καˆ ε„ Øπερέχει ¹ ΛΖ βάσις τÁς ΝΖ βάσεως, Øπερέχει κሠτÕ ΛΥ στερεÕν τοà ΝΥ στερεοà, καˆ ε„ ™λλείπει, ™λλείπει. τεσσάρων δ¾ Ôντων µεγεθîν, δύο µν βάσεων τîν ΑΖ, ΖΘ, δύο δ στερεîν τîν ΑΥ, ΥΘ, ε‡ληπται „σάκις πολλαλάσια τÁς µν ΑΖ βάσεως κሠτοà ΑΥ στερεοà ¼ τε ΛΖ βάσις κሠτÕ ΛΥ στερεόν, τÁς δ ΘΖ βάσεως κሠτοà ΘΥ στερεοà ¼ τε ΝΖ βάσις κሠτÕ ΝΥ στερεόν, κሠδέδεικται, Óτι εƒ Øπερέχει ¹ ΛΖ βάσις τÁς ΖΝ βάσεως, Øπερέχει κሠτÕ ΛΥ στερεÕν τοà ΝΥ [στερεοà], καˆ ε„ ‡ση, ‡σον, καˆ ε„ ™λλείπει, ™λλείπει. œστιν ¥ρα æς ¹ ΑΖ βάσις πρÕς τ¾ν ΖΘ βάσιν, οÛτως τÕ ΑΥ στερεÕν πρÕς τÕ ΥΘ στερεόν· Óπερ œδει δε‹ξαι.

†

lelograms LP , KV , HW , and M S have been completed, and the solids LQ, KR, DM , and M T . And since the straight-lines LK, KA, and AE are equal to one another, the parallelograms LP , KV , and AF are also equal to one another, and KO, KB, and AG (are equal) to one another, and, further, LX, KQ, and AR (are equal) to one another. For (they are) opposite [Prop. 11.24]. So, for the same (reasons), the parallelograms EC, HW , and M S are also equal to one another, and HG, HI, and IN are equal to one another, and, further, DH, M Y , and N T (are equal to one another). Thus, three planes of (one of) the solids LQ, KR, and AU are equal to the (corresponding) three planes (of the others). But, the three planes (in one of the soilds) are equal to the three opposite planes [Prop. 11.24]. Thus, the three solids LQ, KR, and AU are equal to one another [Def. 11.10]. So, for the same (reasons), the three solids ED, DM , and M T are also equal to one another. Thus, as many multiples as the base LF is of the base AF , so many multiples is the solid LU also of the the solid AU . So, for the same (reasons), as many multiples as the base N F is of the base F H, so many multiples is the solid N U also of the solid HU . And if the base LF is equal to the base N F then the solid LU is also equal to the solid N U .† And if the base LF exceeds the base N F then the solid LU also exceeds the solid N U . And if (LF ) is less than (N F ) then (LU ) is (also) less than (N U ). So, there are four magnitudes, the two bases AF and F H, and the two solids AU and U H, and equal multiples have been taken of the base AF and the solid AU — (namely), the base LF and the solid LU —and of the base HF and the solid HU —(namely), the base N F and the solid N U . And it has been shown that if the base LF exceeds the base F N then the solid LU also exceeds the [solid] N U , and if (LF is) equal (to F N ) then (LU is) equal (to N U ), and if (LF is) less than (F N ) then (LU is) less than (N U ). Thus, as the base AF is to the base F H, so the solid AU (is) to the solid U H [Def. 5.5]. (Which is) the very thing it was required to show.

Here, Euclid assumes that LF T N F implies LU T N U . This is easily demonstrated.

κ$΄.

Proposition 26

ΠρÕς τÍ δοθείσV εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τÍ δοθείσV στερε´ γωνίv ‡σην στερε¦ν γωνίαν συστήσασθαι. ”Εστω ¹ µν δοθε‹σα εÙθε‹α ¹ ΑΒ, τÕ δ πρÕς αÙτÍ δοθν σηµε‹ον τÕ Α, ¹ δ δοθε‹σα στερε¦ γωνία ¹ πρÕς τù ∆ περιεχοµένη ØπÕ τîν ØπÕ Ε∆Γ, Ε∆Ζ, Ζ∆Γ γωνιîν ™πιπέδων· δε‹ δ¾ πρÕς τÍ ΑΒ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ πρÕς τù ∆ στερε´ γωνίv ‡σην στερε¦ν

To construct a solid angle equal to a given solid angle on a given straight-line, and at a given point on it. Let AB be the given straight-line, and A the given point on it, and D the given solid angle, contained by the plane angles EDC, EDF , and F DC. So, it is necessary to construct a solid angle equal to the solid angle D on the straight-line AB, and at the point A on it.

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ELEMENTS BOOK 11

γωνίαν συστήσασθαι.

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Ε„λήφθω γ¦ρ ™πˆ τÁς ∆Ζ τυχÕν σηµε‹ον τÕ Ζ, κሠ½χθω ¢πÕ τοà Ζ ™πˆ τÕ δι¦ τîν Ε∆, ∆Γ ™πίπεδον κάθετος ¹ ΖΗ, κሠσυµβαλλέτω τù ™πιπέδJ κατ¦ τÕ Η, κሠ™πεζεύχθω ¹ ∆Η, κሠσυνεστάτω πρÕς τÍ ΑΒ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ µν ØπÕ Ε∆Γ γωνίv ‡ση ¹ ØπÕ ΒΑΛ, τÍ δ ØπÕ Ε∆Η ‡ση ¹ ØπÕ ΒΑΚ, κሠκείσθω τÍ ∆Η ‡ση ¹ ΑΚ, κሠ¢νεστάτω ¢πÕ τοà Κ σηµείου τù δι¦ τîν ΒΑΛ ™πιπέδJ πρÕς Ñρθ¦ς ¹ ΚΘ, κሠκείσθω ‡ση τÍ ΗΖ ¹ ΚΘ, κሠ™πεζεύχθω ¹ ΘΑ· λέγω, Óτι ¹ πρÕς τù Α στερε¦ γωνία περιεχοµένη ØπÕ τîν ΒΑΛ, ΒΑΘ, ΘΑΛ γωνιîν ‡ση ™στˆ τÍ πρÕς τù ∆ στερε´ γωνίv τÍ περιεχοµένV ØπÕ τîν Ε∆Γ, Ε∆Ζ, Ζ∆Γ γωνιîν. 'Απειλήφθωσαν γ¦ρ ‡σαι αƒ ΑΒ, ∆Ε, κሠ™πεζεύχθωσαν αƒ ΘΒ, ΚΒ, ΖΕ, ΗΕ. κሠ™πεˆ ¹ ΖΗ Ñρθή ™στι πρÕς τÕ Øποκείµενον ™πίπεδον, κሠπρÕς πάσας ¥ρα τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù ØποκειµένJ ™πιπέδJ Ñρθ¦ς ποιήσει γωνίας· Ñρθ¾ ¥ρα ™στˆν ˜κατέρα τîν ØπÕ ΖΗ∆, ΖΗΕ γωνιîν. δι¦ τ¦ αÙτ¦ δ¾ κሠ˜κατέρα τîν ØπÕ ΘΚΑ, ΘΚΒ γωνιîν Ñρθή ™στιν. κሠ™πεˆ δύο αƒ ΚΑ, ΑΒ δύο τα‹ς Η∆, ∆Ε ‡σαι ε„σˆν ˜κατέρα ˜κατέρv, κሠγωνίας ‡σας περιέχουσιν, βάσις ¥ρα ¹ ΚΒ βάσει τÍ ΗΕ ‡ση ™στίν. œστι δ κሠ¹ ΚΘ τÍ ΗΖ ‡ση· κሠγωνίας Ñρθ¦ς περιέχουσιν· ‡ση ¥ρα κሠ¹ ΘΒ τÍ ΖΕ. πάλιν ™πεˆ δύο αƒ ΑΚ, ΚΘ δυσˆ τα‹ς ∆Η, ΗΖ ‡σαι ε„σίν, κሠγωνίας Ñρθ¦ς περιέχουσιν, βάσις ¥ρα ¹ ΑΘ βάσει τÍ Ζ∆ ‡ση ™στίν. œστι δ κሠ¹ ΑΒ τÍ ∆Ε ‡ση· δύο δ¾ αƒ ΘΑ, ΑΒ δύο τα‹ς ∆Ζ, ∆Ε ‡σαι ε„σίν. κሠβάσις ¹ ΘΒ βάσει τÍ ΖΕ ‡ση· γωνία ¥ρα ¹ ØπÕ ΒΑΘ γωνίv τÍ ØπÕ Ε∆Ζ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ ΘΑΛ τÍ ØπÕ Ζ∆Γ ™στιν ‡ση. œστι δ κሠ¹ ØπÕ ΒΑΛ τÍ ØπÕ Ε∆Γ ‡ση. ΠρÕς ¥ρα τÍ δοθείσV εÙθείv τÍ ΑΒ κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ δοθείσV στερε´ γωνίv τÍ πρÕς τù ∆ ‡ση συνέσταται· Óπερ œδει ποιÁσαι.

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For let some random point F have been taken on DF , and let F G have been drawn from F perpendicular to the plane through ED and DC [Prop. 11.11], and let it meet the plane at G, and let DG have been joined. And let BAL, equal to the angle EDC, and BAK, equal to EDG, have been constructed on the straight-line AB at the point A on it [Prop. 1.23]. And let AK be made equal to DG. And let KH have been set up at the point K at right-angles to the plane through BAL [Prop. 11.12]. And let KH be made equal to GF . And let HA have been joined. I say that the solid angle at A, contained by the (plane) angles BAL, BAH, and HAL, is equal to the solid angle at D, contained by the (plane) angles EDC, EDF , and F DC. For let AB and DE have been cut off (so as to be) equal, and let HB, KB, F E, and GE have been joined. And since F G is at right-angles to the reference plane (EDC), it will also make right-angles with all of the straight-lines joined to it which are also in the reference plane [Def. 11.3]. Thus, the angles F GD and F GE are right-angles. So, for the same (reasons), the angles HKA and HKB are also right-angles. And since the two (straight-lines) KA and AB are equal to the two (straight-lines) GD and DE, respectively, and they contain equal angles, the base KB is thus equal to the base GE [Prop. 1.4]. And KH is also equal to GF . And they contain right-angles (with the respective bases). Thus, HB (is) also equal to F E [Prop. 1.4]. Again, since the two (straight-lines) AK and KH are equal to the two (straight-lines) DG and GF (respectively), and they contain right-angles, the base AH is thus equal to the base F D [Prop. 1.4]. And AB (is) also equal to DE. So, the two (straight-lines) HA and AB are equal to the two (straight-lines) DF and DE (respectively). And the base HB (is) equal to the base F E. Thus, the angle BAH is equal to the angle EDF [Prop. 1.8]. So, for the same (reasons), HAL is also equal to F DC. And BAL is also equal to EDC. Thus, (a solid angle) has been constructed, equal to

452

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11 the given solid angle at D, on the given straight-line AB, at the given point A on it. (Which is) the very thing it was required to do.

κζ΄.

Proposition 27

'ΑπÕ τÁς δοθείσης εÙθείας τù δοθέντι στερεù παραλληλεπιπέδJ Óµοιόν τε καˆ Ðµοίως κείµενον στερεÕν παραλληλεπίπεδον ¢ναγράψαι. ”Εστω ¹ µν δοθε‹σα εÙθε‹α ¹ ΑΒ, τÕ δ δοθν στερεÕν παραλληλεπίπεδον τÕ Γ∆· δε‹ δ¾ ¢πÕ τÁς δοθείσης εÙθείας τÁς ΑΒ τù δοθέντι στερεù παραλληλεπιπέδJ τù Γ∆ Óµοιόν τε καˆ Ðµοίως κείµενον στερεÕν παραλληλεπίπεδον ¢ναγράψαι. Συνεστάτω γ¦ρ πρÕς τÍ ΑΒ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Α τÍ πρÕς τù Γ στερε´ γωνίv ‡ση ¹ περιεχοµένη ØπÕ τîν ΒΑΘ, ΘΑΚ, ΚΑΒ, éστε ‡σην εναι τ¾ν µν ØπÕ ΒΑΘ γωνίαν τÍ ØπÕ ΕΓΖ, τ¾ν δ ØπÕ ΒΑΚ τÍ ØπÕ ΕΓΗ, τ¾ν δ ØπÕ ΚΑΘ τÍ ØπÕ ΗΓΖ· κሠγεγονέτω æς µν ¹ ΕΓ πρÕς τ¾ν ΓΗ, οÛτως ¹ ΒΑ πρÕς τ¾ν ΑΚ, æς δ ¹ ΗΓ πρÕς τ¾ν ΓΖ, οÛτως ¹ ΚΑ πρÕς τ¾ν ΑΘ. κሠδι' ‡σου ¥ρα ™στˆν æς ¹ ΕΓ πρÕς τ¾ν ΓΖ, οÛτως ¹ ΒΑ πρÕς τ¾ν ΑΘ. κሠσυµπεπληρώσθω τÕ ΘΒ παραλληλόγραµµον κሠτÕ ΑΛ στερεόν.

To describe a parallelepiped solid similar, and similarly laid out, to a given parallelepiped solid on a given straight-line. Let the given straight-line be AB, and the given parallelepiped solid CD. So, it is necessary to describe a parallelepiped solid similar, and similarly laid out, to the given parallelepiped solid CD on the given straight-line AB. For, let a (solid angle) contained by the (plane angles) BAH, HAK, and KAB have been constructed, equal to solid angle at C, on the straight-line AB at the point A on it [Prop. 11.26], such that angle BAH is equal to ECF , and BAK to ECG, and KAH to GCF . And let it have been contrived that as EC (is) to CG, so BA (is) to AK, and as GC (is) to CF , so KA (is) to AH [Prop. 6.12]. And thus, via equality, as EC is to CF , so BA (is) to AH [Prop. 5.22]. And let the parallelogram HB have been completed, and the solid AL.

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Κሠ™πεί ™στιν æς ¹ ΕΓ πρÕς τ¾ν ΓΗ, οÛτως ¹ ΒΑ πρÕς τ¾ν ΑΚ, κሠπερˆ ‡σας γωνίας τ¦ς ØπÕ ΕΓΗ, ΒΑΚ αƒ πλευρሠ¢νάλογόν ε„σιν, Óµοιον ¥ρα ™στˆ τÕ ΗΕ παραλληλόγραµµον τù ΚΒ παραλληλογράµµJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ µν ΚΘ παραλληλόγραµµον τù ΗΖ παραλληλογράµµJ Óµοιόν ™στι κሠœτι τÕ ΖΕ τù ΘΒ· τρία ¥ρα παραλληλόγραµµα τοà Γ∆ στερεοà τρισˆ παραλληλογράµµοις τοà ΑΛ στερεοà Óµοιά ™στιν. ¢λλ¦ τ¦ µν τρία τρισˆ το‹ς ¢πεναντίον ‡σα τέ ™στι καˆ Óµοια, τ¦ δ τρία τρισˆ το‹ς ¢πεναντίον ‡σα τέ ™στι καˆ Óµοια· Óλον ¥ρα τÕ Γ∆ στερεÕν ÓλJ τù ΑΛ στερεJ Óµοιόν ™στιν. 'ΑπÕ τÁς δοθείσης ¥ρα εÙθείας τÁς ΑΒ τù δοθέντι στερεù παραλληλεπιπέδJ τù Γ∆ Óµοιόν τε καˆ Ðµοίως κείµενον ¢ναγέγραπται τÕ ΑΛ· Óπερ œδει ποιÁσαι.

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And since as EC is to CG, so BA (is) to AK, and the sides about the equal angles ECG and BAK are (thus) proportional, the parallelogram GE is thus similar to the parallelogram KB. So, for the same (reasons), the parallelogram KH is also similar to the parallelogram GF , and, further, F E (is similar) to HB. Thus, three of the parallelograms of solid CD are similar to three of the parallelograms of solid AL. But, the (former) three are equal and similar to the three opposite, and the (latter) three are equal and similar to the three opposite. Thus, the whole solid CD is similar to the whole solid AL [Def. 11.9]. Thus, AL, similar, and similarly laid out, to the given parallelepiped solid CD, has been described on the given straight-lines AB, at the given point A on it. (Which is) the very thing it was required to do.

453

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ELEMENTS BOOK 11

κη΄.

Proposition 28

'Ε¦ν στερεÕν παραλληλεπίπεδον ™πιπέδJ τµηθÍ κατ¦ τ¦ς διαγωνίους τîν ¢πεναντίον ™πιπέδων, δίχα τµηθήσεται τÕ στερεÕν ØπÕ τοà ™πιπέδου.

If a parallelepiped solid is cut by a plane (passing) through the diagonals of (a pair of) opposite planes then the solid will be cut in half by the plane.

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ΣτερεÕν γ¦ρ παραλληλεπίπεδον τÕ ΑΒ ™πιπέδJ τù Γ∆ΕΖ τετµήσθω κατ¦ τ¦ς διαγωνίους τîν ¢πεναντίον ™πιπέδων τ¦ς ΓΖ, ∆Ε· λέγω, Óτι δίχα τµηθήσεται τÕ ΑΒ στερεÕν ØπÕ τοà Γ∆ΕΖ ™πιπέδου. 'Επεˆ γ¦ρ ‡σον ™στˆ τÕ µν ΓΗΖ τρίγωνον τù ΓΖΒ τριγώνJ, τÕ δ Α∆Ε τù ∆ΕΘ, œστι δ κሠτÕ µν ΓΑ παραλληλόγραµµον τù ΕΒ ‡σον· ¢πεναντίον γάρ· τÕ δ ΗΕ τù ΓΘ, κሠτÕ πρίσµα ¥ρα τÕ περιεχόµενον ØπÕ δύο µν τριγώνων τîν ΓΗΖ, Α∆Ε, τριîν δ παραλληλογράµµων τîν ΗΕ, ΑΓ, ΓΕ ‡σον ™στˆ τù πρίσµατι τù περιεχοµένJ ØπÕ δύο µν τριγώνων τîν ΓΖΒ, ∆ΕΘ, τριîν δ παραλληλογράµµων τîν ΓΘ, ΒΕ, ΓΕ· ØπÕ γ¦ρ ‡σων ™πιπέδων περιέχονται τù τε πλήθει κሠτù µεγέθει. éστε Óλον τÕ ΑΒ στερεÕν δίχα τέτµηται ØπÕ τοà Γ∆ΕΖ ™πιπέδου· Óπερ œδει δε‹ξαι.

For let the parallelepiped solid AB have been cut by the plane CDEF through the diagonals of the opposite planes CF and DE. I say that the solid AB will be cut in half by the plane CDEF . For since triangle CGF is equal to triangle CF B, and ADE (is equal) to DEH [Prop. 1.34], and parallelogram CA is also equal to EB—for (they are) opposite [Prop. 11.24]—and GE (equal) to CH, thus the prism contained by the two triangles CGF and ADE, and the three parallelograms GE, AC, and CE, is also equal to the prism contained by the two triangles CF B and DEH, and the three parallelograms CH, BE, and CE. For they are contained by planes equal in number and in magnitude [Def. 11.10]. Thus, the whole of solid AB is cut in half by the plane CDEF . (Which is) the very thing it was required to show.

κθ΄.

Proposition 29

Τ¦ ™πˆ τÁς αÙτÁς βάσεως Ôντα στερε¦ παραλληParallelepiped solids which are on the same base, and λεπίπεδα κሠØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι ™πˆ τîν (have) the same height, and in which the (ends of the αÙτîν ε„σιν εÙθειîν, ‡σα ¢λλήλοις ™στίν. straight-lines) standing up are on the same straight-lines, are equal to one another.

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ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

”Εστω ™πˆ τÁς αÙτÁς βάσεως τÁς ΑΒ στερε¦ παραλληλεπίπεδα τ¦ ΓΜ, ΓΝ ØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι αƒ ΑΗ, ΑΖ, ΛΜ, ΛΝ, Γ∆, ΓΕ, ΒΘ, ΒΚ ™πˆ τîν αÙτîν εÙθειîν œστωσαν τîν ΖΝ, ∆Κ· λέγω, Óτι ‡σον ™στˆ τÕ ΓΜ στερεÕν τù ΓΝ στερεù. 'Επεˆ γ¦ρ παραλληλόγραµµόν ™στιν ˜κάτερον τîν ΓΘ, ΓΚ, ‡ση ™στˆν ¹ ΓΒ ˜κατέρv τîν ∆Θ, ΕΚ· éστε κሠ¹ ∆Θ τÍ ΕΚ ™στιν ‡ση. κοιν¾ ¢φVρήσθω ¹ ΕΘ· λοιπ¾ ¥ρα ¹ ∆Ε λοιπÍ τÍ ΘΚ ™στιν ‡ση. éστε κሠτÕ µν ∆ΓΕ τρίγωνον τù ΘΒΚ τριγώνJ ‡σον ™στίν, τÕ δ ∆Η παραλληλόγραµµον τù ΘΝ παραλληλογράµµJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ΑΖΗ τρίγωνον τù ΜΛΝ τριγώνJ ‡σον ™στίν. œστι δ κሠτÕ µν ΓΖ παραλληλόγραµµον τù ΒΜ παραλληλογράµµJ ‡σον, τÕ δ ΓΗ τù ΒΝ· ¢πεναντίον γάρ· κሠτÕ πρίσµα ¥ρα τÕ περιεχόµενον ØπÕ δύο µν τριγώνων τîν ΑΖΗ, ∆ΓΕ, τριîν δ παραλληλογράµµων τîν Α∆, ∆Η, ΓΗ ‡σον ™στˆ τù πρίσµατι τù περιεχοµένJ ØπÕ δύο µν τριγώνων τîν ΜΛΝ, ΘΒΚ, τριîν δ παραλληλογράµµων τîν ΒΜ, ΘΝ, ΒΝ. κοινÕν προσκείσθω τÕ στερεÕν, οá βάσις µν τÕ ΑΒ παραλληλόγραµµον, ¢πεναντίον δ τÕ ΗΕΘΜ· Óλον ¥ρα τÕ ΓΜ στερεÕν παραλληλεπίπεδον ÓλJ τù ΓΝ στερεù παραλληλεπιπέδJ ‡σον ™στίν. Τ¦ ¥ρα ™πˆ τÁς αÙτÁς βάσεως Ôντα στερε¦ παραλληλεπίπεδα κሠØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι ™πˆ τîν αÙτîν ε„σιν εÙθειîν, ‡σα ¢λλήλοις ™στίν· Óπερ œδει δε‹ξαι.

For let the parallelepiped solids CM and CN be on the same base AB, and (have) the same height, and let the (ends of the straight-lines) standing up in them, AG, AF , LM , LN , CD, CE, BH, and BK, be on the same straight-lines, F N and DK. I say that solid CM is equal to solid CN . For since CH and CK are each parallelograms, CB is equal to each DH and EK [Prop. 1.34]. Hence, DH is also equal to EK. Let EH have been subtracted from both. Thus, the remainder DE is equal to the remainder HK. Hence, triangle DCE is also equal to triangle HBK [Props. 1.4, 1.8], and parallelogram DG to parallelogram HN [Prop. 1.36]. So, for the same (reasons), traingle AF G is also equal to triangle M LN . And parallelogram CF is also equal to parallelogram BM , and CG to BN [Prop. 11.24]. For they are opposite. Thus, the prism contained by the two triangles AF G and DCE, and the three parallelograms AD, DG, and CG, is equal to the prism contained by the two triangles M LN and HBK, and the three parallelograms BM , HN , and BN . Let the solid whose base (is) parallelogram AB, and (whose) opposite (face is) GEHM , have been added to both (prisms). Thus, the whole parallelepiped solid CM is equal to the whole parallelepiped solid CN . Thus, parallelepiped solids which are on the same base, and (have) the same height, and in which the (ends of the straight-lines) standing up (are) on the same straight-lines, are equal to one another. (Which is) the very thing it was required to show.

λ΄.

Proposition 30

Τ¦ ™πˆ τÁς αÙτÁς βάσεως Ôντα στερε¦ παραλληParallelepiped solids which are on the same base, and λεπίπεδα κሠØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι οÙκ (have) the same height, and in which the (ends of the ε„σˆν ™πˆ τîν αÙτîν εÙθειîν, ‡σα ¢λλήλοις ™στίν. straight-lines) standing up are not on the same straightlines, are equal to one another.

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Let the parallelepiped solids CM and CN be on the

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

ραλληλεπίπεδα τ¦ ΓΜ, ΓΝ ØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι αƒ ΑΖ, ΑΗ, ΛΜ, ΛΝ, Γ∆, ΓΕ, ΒΘ, ΒΚ µ¾ œστωσαν ™πˆ τîν αÙτîν εÙθειîν· λέγω, Óτι ‡σον ™στˆ τÕ ΓΜ στερεÕν τù ΓΝ στερεù. 'Εκβεβλήσθωσαν γ¦ρ αƒ ΝΚ, ∆Θ κሠσυµπιπτέτωσαν ¢λλήλαις κατ¦ τÕ Ρ, κሠœτι ™κβεβλήσθωσαν αƒ ΖΜ, ΗΕ ™πˆ τ¦ Ο, Π, κሠ™πεζεύχθωσαν αƒ ΑΞ, ΛΟ, ΓΠ, ΒΡ. ‡σον δή ™στι τÕ ΓΜ στερεόν, οá βάσις µν τÕ ΑΓΒΛ παραλληλόγραµµον, ¢πεναντίον δ τÕ Ζ∆ΘΜ, τù ΓΟ στερεù, οá βάσις µν τÕ ΑΓΒΛ παραλληλόγραµµον, ¢πεναντίον δ τÕ ΞΠΡΟ· ™πί τε γ¦ρ τÁς αÙτÁς βάσεώς ε„σι τÁς ΑΓΒΛ κሠØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι αƒ ΑΖ, ΑΞ, ΛΜ, ΛΟ, Γ∆, ΓΠ, ΒΘ, ΒΡ ™πˆ τîν αÙτîν ε„σιν εÙθειîν τîν ΖΟ, ∆Ρ. ¢λλ¦ τÕ ΓΟ στερεόν, οá βάσις µέν ™στι τÕ ΑΓΒΛ παραλληλόγραµµον, ¢πεναντίον δ τÕ ΞΠΡΟ, ‡σον ™στˆ τù ΓΝ στερεù, οá βάσις µν τÕ ΑΓΒΛ παραλληλόγραµµον, ¢πεναντίον δ τÕ ΗΕΚΝ· ™πί τε γ¦ρ πάλιν τÁς αÙτÁς βάσεώς ε„σι τÁς ΑΓΒΛ κሠØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι αƒ ΑΗ, ΑΞ, ΓΕ, ΓΠ, ΛΝ, ΛΟ, ΒΚ, ΒΡ ™πˆ τîν αÙτîν ε„σιν εÙθειîν τîν ΗΠ, ΝΡ. éστε κሠτÕ ΓΜ στερεÕν ‡σον ™στˆ τù ΓΝ στερεù. Τ¦ ¥ρα ™πˆ τÁς αÙτÁς βάσεως στερε¦ παραλληλεπίπεδα κሠØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι οÙκ ε„σˆν ™πˆ τîν αÙτîν εÙθειîν, ‡σα ¢λλήλοις ™στίν· Óπερ œδει δε‹ξαι.

same base, AB, and (have) the same height, and let the (ends of the straight-lines) standing up in them, AF , AG, LM , LN , CD, CE, BH, and BK, not be on the same straight-lines. I say that the solid CM is equal to the solid CN . For let N K and DH have been produced, and let them have joined one another at R. And, further, let F M and GE have been produced to P and Q (respectively). And let AO, LP , CQ, and BR have been joined. So, solid CM , whose base (is) parallelogram ACBL, and opposite (face) F DHM , is equal to solid CP , whose base (is) parallelogram ACBL, and opposite (face) OP RQ. For they are on the same base, ACBL, and (have) the same height, and the (ends of the straight-lines) standing up in them, AF , AO, LM , LP , CD, CQ, BH, and BR, are on the same straight-lines, F P and DR [Prop. 11.29]. But, solid CP , whose base is parallelogram ACBL, and opposite (face) OQRP , is equal to solid CN , whose base (is) parallelogram ACBL, and opposite (face) GEKN . For, again, they are on the same base, ACBL, and (have) the same height, and the (ends of the straight-lines) standing up in them, AG, AO, CE, CQ, LN , LP , BK, and BR, are on the same straight-lines, GQ and N R [Prop. 11.29]. Hence, solid CM is also equal to solid CN . Thus, parallelepiped solids (which are) on the same base, and (have) the same height, and in which the (ends of the straight-lines) standing up are not on the same straight-lines, are equal to one another. (Which is) the very thing it was required to show.

λα΄.

Proposition 31

Τ¦ ™πˆ ‡σων βάσεων Ôντα στερε¦ παραλληλεπίπεδα κሠØπÕ τÕ αÙτÕ Ûψος ‡σα ¢λλήλοις ™στίν. ”Εστω ™πˆ ‡σων βάσεων τîν ΑΒ, Γ∆ στερε¦ παραλληλεπίπεδα τ¦ ΑΕ, ΓΖ ØπÕ τÕ αÙτÕ Ûψος. λέγω, Óτι ‡σον ™στˆ τÕ ΑΕ στερεÕν τù ΓΖ στερεù. ”Εστωσαν δ¾ πρότερον αƒ ™φεστηκυ‹αι αƒ ΘΚ, ΒΕ, ΑΗ, ΛΜ, ΟΠ, ∆Ζ, ΓΞ, ΡΣ πρÕς Ñρθ¦ς τα‹ς ΑΒ, Γ∆ βάσεσιν, κሠ™κβεβλήσθω ™π' εÙθείας τÍ ΓΡ εÙθε‹α ¹ ΡΤ, κሠσυνεστάτω πρÕς τÍ ΡΤ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Ρ τÍ ØπÕ ΑΛΒ γωνίv ‡ση ¹ ØπÕ ΤΡΥ, κሠκείσθω τÍ µν ΑΛ ‡ση ¹ ΡΤ, τÍ δ ΛΒ ‡ση ¹ ΡΥ, κሠσυµπεπληρώσθω ¼ τε ΡΧ βάσις κሠτÕ ΨΥ στερεόν.

Parallelepiped solids which are on equal bases, and (have) the same height, are equal to one another. Let the parallelepiped solids AE and CF be on the equal bases AB and CD (respectively), and (have) the same height. I say that solid AE is equal to solid CF . So, let the (straight-lines) standing up, HK, BE, AG, LM , P Q, DF , CO, and RS, first of all, be at right-angles to the bases AB and CD. And let RT have been produced in a straight-line with CR. And let (angle) T RU , equal to angle ALB, have been constructed on the straight-line RT , at the point R on it [Prop. 1.23]. And let RT be made equal to AL, and RU to LB. And let the base RW , and the solid XU , have been completed.

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ΣΤΟΙΧΕΙΩΝ ια΄.

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ELEMENTS BOOK 11

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Κሠ™πεˆ δύο αƒ ΤΡ, ΡΥ δυσˆ τα‹ς ΑΛ, ΛΒ ‡σαι ε„σίν, κሠγωνίας ‡σας περιέχουσιν, ‡σον ¥ρα καˆ Óµοιον τÕ ΡΧ παραλληλόγραµµον τù ΘΛ παραλληλογράµµJ. κሠ™πεˆ πάλιν ‡ση µν ¹ ΑΛ τÍ ΡΤ, ¹ δ ΛΜ τÍ ΡΣ, κሠγωνίας Ñρθ¦ς περιέχουσιν, ‡σον ¥ρα καˆ Óµοιόν ™στι τÕ ΡΨ παραλληλόγραµµον τù ΑΜ παραλληλογράµµJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ΛΕ τù ΣΥ ‡σον τέ ™στι καˆ Óµοιον· τρία ¥ρα παραλληλόγραµµα τοà ΑΕ στερεοà τρισˆ παραλληλογράµµοις τοà ΨΥ στερεοà ‡σα τέ ™στι καˆ Óµοια. ¢λλ¦ τ¦ µν τρία τρισˆ το‹ς ¢πεναντίον ‡σα τέ ™στι καˆ Óµοια, τ¦ δ τρία τρισˆ το‹ς ¢πεναντίον· Óλον ¥ρα τÕ ΑΕ στερεÕν παραλληλεπίπεδον ÓλJ τù ΨΥ στερεù παραλληλεπιπέδJ ‡σον ™στίν. διήχθωσαν αƒ ∆Ρ, ΧΥ κሠσυµπιπέτωσαν ¢λλήλαις κατ¦ τÕ Ω, κሠδι¦ τοà Τ τÍ ∆Ω παράλληλος ½χθω ¹ αΤ", κሠ™κβεβλήσθω ¹ Ο∆ κατ¦ τÕ α, κሠσυµπεπληρώσθω τ¦ ΩΨ, ΡΙ στερεά. ‡σον δή ™στι τÕ ΨΩ στερεόν, οá βάσις µέν ™στι τÕ ΡΨ παραλληλόγραµµον, ¢πεναντίον δ τÕ Ω&, τù ΨΥ στερεù, οá βάσις µν τÕ ΡΨ παραλληλόγραµµον, ¢πεναντίον δ τÕ ΥΦ· ™πί τε γ¦ρ τÁς αÙτÁς βάσεώς ε„σι τÁς ΡΨ κሠØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι αƒ ΡΩ, ΡΥ, Τ", ΤΧ, Σ$, Σ˜ ο, Ψ&, ΨΦ ™πˆ τîν αÙτîν ε„σιν εÙθειîν τîν ΩΧ, $Φ. ¢λλ¦ τÕ ΨΥ στερεÕν τù ΑΕ ™στιν ‡σον· κሠτÕ ΨΩ ¥ρα στερεÕν τù ΑΕ στερεù ™στιν ‡σον. κሠ™πεˆ ‡σον ™στˆ τÕ ΡΥΧΤ παραλληλόγραµµον τù ΩΤ παραλληλογράµµJ· ™πί τε γ¦ρ τÁς αÙτÁς βάσεώς ε„σι τÁς ΡΤ κሠ™ν τα‹ς αÙτα‹ς παραλλήλοις τα‹ς ΡΤ, ΩΧ· ¢λλ¦

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And since the two (straight-lines) T R and RU are equal to the two (straight-lines) AL and LB (respectively), and they contain equal angles, parallelogram RW is thus equal and similar to parallelogram HL [Prop. 6.14]. And, again, since AL is equal to RT , and LM to RS, and they contain right-angles, parallelogram RX is thus equal and similar to parallelogram AM [Prop. 6.14]. So, for the same (reasons), LE is also equal and similar to SU . Thus, three parallelograms of solid AE are equal and similar to three parallelograms of solid XU . But, the three (faces of the former solid) are equal and similar to the three opposite (faces), and the three (faces of the latter solid) to the three opposite (faces) [Prop. 11.24]. Thus, the whole parallelepiped solid AE is equal to the whole parallelepiped solid XU [Def. 11.10]. Let DR and W U have been drawn across, and let them have met one another at Y . And let aT b have been drawn through T parallel to DY . And let P D have been produced to a. And let the solids Y X and RI have been completed. So, solid XY , whose base is parallelogram RX, and opposite (face) Y c, is equal to solid XU , whose base (is) parallelogram RX, and opposite (face) U V . For they are on the same base RX, and (have) the same height, and the (ends of the straightlines) standing up in them, RY , RU , T b, T W , Se, Sd, Xc and XV , are on the same straight-lines, Y W and eV [Prop. 11.29]. But, solid XU is equal to AE. Thus,

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ELEMENTS BOOK 11

τÕ ΡΥΧΤ τù Γ∆ ™στιν ‡σον, ™πεˆ κሠτù ΑΒ, κሠτÕ ΩΤ ¥ρα παραλληλόγραµµον τù Γ∆ ™στιν ‡σον. ¥λλο δ τÕ ∆Τ· œστιν ¥ρα æς ¹ Γ∆ βάσις πρÕς τ¾ν ∆Τ, οÛτως ¹ ΩΤ πρÕς τ¾ν ∆Τ. κሠ™πεˆ στερεÕν παραλληλεπίπεδον τÕ ΓΙ ™πιπέδJ τù ΡΖ τέτµηται παραλλήλJ Ôντι το‹ς ¢πεναντίον ™πιπέδοις, œστιν æς ¹ Γ∆ βάσις πρÕς τ¾ν ∆Τ βάσιν, οÛτως τÕ ΓΖ στερεÕν πρÕς τÕ ΡΙ στερεόν. δι¦ τ¦ αÙτ¦ δή, ™πεˆ στερεÕν παραλληλεπίπεδον τÕ ΩΙ ™πιπέδJ τù ΡΨ τέτµηται παραλλήλJ Ôντι το‹ς ¢πεναντίον ™πιπέδοις, œστιν æς ¹ ΩΤ βάσις πρÕς τ¾ν ΤΛ βάσιν, οÛτως τÕ ΩΨ στερεÕν πρÕς τÕ ΡΙ. ¢λλ' æς ¹ Γ∆ βάσις πρÕς τ¾ν ∆Τ, οÛτως ¹ ΩΤ πρÕς τ¾ν ∆Τ· κሠæς ¥ρα τÕ ΓΖ στερεÕν πρÕς τÕ ΡΙ στερεόν, οÛτως τÕ ΩΨ στερεÕν πρÕς τÕ ΡΙ. ˜κάτερον ¥ρα τîν ΓΖ, ΩΨ στερεîν πρÕς τÕ ΡΙ τÕν αÙτÕν œχει λόγον· ‡σον ¥ρα ™στˆ τÕ ΓΖ στερεÕν τù ΩΨ στερεù. ¢λλ¦ τÕ ΩΨ τù ΑΕ ™δείχθη ‡σον· κሠτÕ ΑΕ ¥ρα τù ΓΖ ™στιν ‡σον.

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solid XY is also equal to solid AE. And since parallelogram RU W T is equal to parallelogram Y T . For they are on the same base RT , and between the same parallels RT and Y W [Prop. 1.35]. But, RU W T is equal to CD, since (it is) also (equal) to AB. Parallelogram Y T is thus also equal to CD. And DT is another (parallelogram). Thus, as base CD is to DT , so Y T (is) to DT [Prop. 5.7]. And since the parallelepiped solid CI has been cut by the plane RF , which is parallel to the opposite planes (of CI), as base CD is to base DT , so solid CF (is) to solid RI [Prop. 11.25]. So, for the same (reasons), since the parallelepiped solid Y I has been cut by the plane RX, which is parallel to the opposite planes (of Y I), as base Y T is to base T D, so solid Y X (is) to solid RI [Prop. 11.25]. But, as base CD (is) to DT , so Y T (is) to DT . And, thus, as solid CF (is) to solid RI, so solid Y X (is) to solid RI. Thus, solids CF and Y X each have the same ratio to RI [Prop. 5.11]. Thus, solid CF is equal to solid Y X [Prop. 5.9]. But, Y X was show (to be) equal to AE. Thus, AE is also equal to CF .

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Μ¾ œστωσαν δ¾ αƒ ™φεστηκυ‹αι αƒ ΑΗ, ΘΚ, ΒΕ, ΛΜ, ΓΞ, ΟΠ, ∆Ζ, ΡΣ πρÕς Ñρθ¦ς τα‹ς ΑΒ, Γ∆ βάσεσιν· λέγω πάλιν, Óτι †σον τÕ ΑΕ στερεÕν τù ΓΖ στερεù. ½χθωσαν γ¦ρ ¢πÕ τîν Κ, Ε, Η, Μ, Π, Ζ, Ξ, Σ σηµείων ™πˆ τÕ Øποκείµενον ™πίπεδον κάθετοι αƒ ΚΝ, ΕΤ, ΗΥ, ΜΦ, ΠΧ, ΖΨ, ΞΩ, ΣΙ, κሠσυµβαλλέτωσαν τù ™πιπέδJ κατ¦ τ¦ Ν, Τ, Υ, Φ, Χ, Ψ, Ω, Ι σηµε‹α, κሠ™πεζεύχθωσαν αƒ ΝΤ, ΝΥ, ΥΦ, ΤΦ, ΧΨ, ΧΩ, ΩΙ, ΙΨ. ‡σον δή ™στι τÕ ΚΦ στερεÕν τù ΠΙ στερεù· ™πί τε γ¦ρ ‡σων βάσεών ε„σι τîν ΚΜ, ΠΣ κሠØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι πρÕς Ñρθάς ε„σι τα‹ς βάσεσιν. ¢λλ¦ τÕ µν ΚΦ στερεÕν τù ΑΕ στερεù ™στιν ‡σον, τÕ δ ΠΙ τù ΓΖ· ™πί τε γ¦ρ τÁς αÙτÁς βάσεώς ε„σι κሠØπÕ τÕ αÙτÕ Ûψος, ïν αƒ ™φεστîσαι οÜκ ε„σιν ™πˆ τîν αÙτîν εÙθειîν. κሠτÕ ΑΕ ¥ρα στερεÕν τù ΓΖ στερεù ™στιν ‡σον. Τ¦ ¥ρα ™πˆ ‡σων βάσεων Ôντα στερε¦ παραλληλεπίπεδα κሠØπÕ τÕ αÙτÕ Ûψος ‡σα ¢λλήλοις ™στίν· Óπερ œδει δε‹ξαι.

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And so let the (straight-lines) standing up, AG, HK, BE, LM , CO, P Q, DF , and RS, not be at right-angles to the bases AB and CD. Again, I say that solid AE (is) equal to solid CF . For let KN , ET , GU , M V , QW , F X, OY , and SI have been drawn from points K, E, G, M , Q, F , O, and S (respectively) perpendicular to the reference plane (i.e., the plane of the bases AB and CD), and let them have met the plane at points N , T , U , V , W , X, Y , and I (respectively). And let N T , N U , U V , T V , W X, W Y , Y I, and IX have been joined. So solid KV is equal to solid QI. For they are on the equal bases KM and QS, and (have) the same height, and the (straight-lines) standing up in them are at right-angles to their bases (see first part of proposition). But, solid KV is equal to solid AE, and QI to CF . For they are on the same base, and (have) the same height, and the (straight-lines) standing up in them are not on the same straight-lines [Prop. 11.30]. Thus, solid AE is also equal to solid CF . Thus, parallelepiped solids which are on equal bases,

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ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11 and (have) the same height, are equal to one another. (Which is) the very thing it was required to show.

λβ΄.

Proposition 32

Τ¦ ØπÕ τÕ αÙτÕ Ûψος Ôντα στερε¦ παραλληλεπίπεδα Parallelepiped solids which (have) the same height πρÕς ¥λληλά ™στιν æς αƒ βάσεις. are to one another as their bases.

B

D

B

Z

EL A

K

G

D

F

E L

H

J

A

K

C

G

H

”Εστω ØπÕ τÕ αÙτÕ Ûψος στερε¦ παραλληλεπίπεδα τ¦ ΑΒ, Γ∆· λέγω, Óτι τ¦ ΑΒ, Γ∆ στερε¦ παραλληλεπίπεδα πρÕς ¥λληλά ™στιν æς αƒ βάσεις, τουτέστιν Óτι ™στˆν æς ¹ ΑΕ βάσις πρÕς τ¾ν ΓΖ βάσιν, οÛτως τÕ ΑΒ στερεÕν πρÕς τÕ Γ∆ στερεόν. Παραβεβλήσθω γ¦ρ παρ¦ τ¾ν ΖΗ τù ΑΕ ‡σον τÕ ΖΘ, κሠ¢πÕ βάσεως µν τÁς ΖΘ, Ûψους δ τοà αÙτοà τù Γ∆ στερεÕν παραλληλεπίπεδον συµπεπληρώσθω τÕ ΗΚ. ‡σον δή ™στι τÕ ΑΒ στερεÕν τù ΗΚ στερεù· ™πί τε γ¦ρ ‡σων βάσεών ε„σι τîν ΑΕ, ΖΘ κሠØπÕ τÕ αÙτο Ûψος. κሠ™πεˆ στερεÕν παραλληλεπίπεδον τÕ ΓΚ ™πιπέδJ τù ∆Η τέτµηται παραλλήλJ Ôντι το‹ς ¢πεναντίον ™πιπέδοις, œστιν ¥ρα æς ¹ ΓΖ βάσις πρÕς τ¾ν ΖΘ βάσιν, οÛτως τÕ Γ∆ στερεÕν πρÕς τÕ ∆Θ στερεόν. ‡ση δ ¹ µν ΖΘ βάσις τÍ ΑΕ βάσει, τÕ δ ΗΚ στερεÕν τù ΑΒ στερεù· œστιν ¥ρα κሠæς ¹ ΑΕ βάσις πρÕς τ¾ν ΓΖ βάσιν, οÛτως τÕ ΑΒ στερεÕν πρÕς τÕ Γ∆ στερεόν. Τ¦ ¥ρα ØπÕ τÕ αÙτÕ Ûψος Ôντα στερε¦ παραλληλεπίπεδα πρÕς ¥λληλά ™στιν æς αƒ βάσεις· Óπερ œδει δε‹ξαι.

Let AB and CD be parallelepiped solids (having) the same height. I say that the parallelepiped solids AB and CD are to one another as their bases. That is to say, as base AE is to base CF , so solid AB (is) to solid CD. For let F H, equal to AE, have been applied to F G (in the angle F GH equal to angle LCG) [Prop. 1.45]. And let the parallelepiped solid GK, (having) the same height as CD, have been completed on the base F H. So solid AB is equal to solid GK. For they are on the equal bases AE and F H, and (have) the same height [Prop. 11.31]. And since the parallelepiped solid CK has been cut by the plane DG, which is parallel to the opposite planes (of CK), thus as the base CF is to the base F H, so the solid CD (is) to the solid DH [Prop. 11.25]. And base F H (is) equal to base AE, and solid GK to solid AB. And thus as base AE is to base CF , so solid AB (is) to solid CD. Thus, parallelepiped solids which (have) the same height are to one another as their bases. (Which is) the very thing it was required to show.

λγ΄.

Proposition 33

Τ¦ Óµοια στερε¦ παραλληλεπίπεδα πρÕς ¥λληλα ™ν τριπλασίονι λόγJ ε„σˆ τîν еολόγων πλευρîν. ”Εστω Óµοια στερε¦ παραλληλεπίπεδα τ¦ ΑΒ, Γ∆, еόλογος δ œστω ¹ ΑΕ τÍ ΓΖ· λέγω, Óτι τÕ ΑΒ στερεÕν πρÕς τÕ Γ∆ στερεÕν τριπλασίονα λόγον œχει, ½περ ¹ ΑΕ πρÕς τ¾ν ΓΖ. 'Εκβεβλήσθωσαν γ¦ρ ™π' εÙθείας τα‹ς ΑΕ, ΗΕ, ΘΕ αƒ ΕΚ, ΕΛ, ΕΜ, κሠκείσθω τÍ µν ΓΖ ‡ση ¹ ΕΚ, τÍ δ ΖΝ ‡ση ¹ ΕΛ, κሠœτι τÍ ΖΡ ‡ση ¹ ΕΜ, κሠσυµπεπληρώσθω τÕ ΚΛ παραλληλόγραµµον κሠτÕ ΚΟ στερεόν.

Similar parallelepiped solids are to one another as the cubed ratio of their corresponding sides. Let AB and CD be similar parallelepiped solids, and let AE correspond to CF . I say that solid AB has to solid CD the cubed ratio that AE (has) to CF . For let EK, EL, and EM have been produced in a straight-line with AE, GE, and HE (respectively). And let EK be made equal to CF , and EL equal to F N , and, further, EM equal to F R. And let the parallelogram KL have been completed, and the solid KP .

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ELEMENTS BOOK 11

G

Z

C

N D

F N

R

R D

Κሠ™πεˆ δύο αƒ ΚΕ, ΕΛ δυσˆ τα‹ς ΓΖ, ΖΝ ‡σαι ε„σίν, ¢λλ¦ κሠγωνία ¹ ØπÕ ΚΕΛ γωνίv τÍ ØπÕ ΓΖΝ ™στιν ‡ση, ™πειδήπερ κሠ¹ ØπÕ ΑΕΗ τÍ ØπÕ ΓΖΝ ™στιν ‡ση δι¦ τ¾ν еοιότητα τîν ΑΒ, Γ∆ στερεîν, ‡σον ¥ρα ™στˆ [καˆ Óµοιον] τÕ ΚΛ παραλληλόγραµµον τù ΓΝ παραλληλογράµµJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ µν ΚΜ παραλληλόγραµµον ‡σον ™στˆ καˆ Óµοιον τù ΓΡ [παραλληλογράµµJ] κሠœτι τÕ ΕΟ τù ∆Ζ· τρία ¥ρα παραλληλόγραµµα τοà ΚΟ στερεοà τρισˆ παραλληλογράµµοις τοà Γ∆ στερεοà ‡σα ™στˆ καˆ Óµοια. ¢λλ¦ τ¦ µν τρία τρισˆ το‹ς ¢πεναντίον ‡σα ™στˆ καˆ Óµοια, τ¦ δ τρία τρισˆ το‹ς ¢πεναντίον ‡σα ™στˆ καˆ Óµοια· Óλον ¥ρα τÕ ΚΟ στερεÕν ÓλJ τù Γ∆ στερεù ‡σον ™στˆ καˆ Óµοιον. συµπεπληρώσθω τÕ ΗΚ παραλληλόγραµµον, κሠ¢πÕ βάσεων µν τîν ΗΚ, ΚΛ παραλληλόγραµµον, Ûψους δ τοà αÙτοà τù ΑΒ στερε¦ συµπεπληρώσθω τ¦ ΕΞ, ΛΠ. κሠ™πεˆ δι¦ τ¾ν еοιότητα τîν ΑΒ, Γ∆ στερεîν ™στιν æς ¹ ΑΕ πρÕς τ¾ν ΓΖ, οÛτως ¹ ΕΗ πρÕς τ¾ν ΖΝ, κሠ¹ ΕΘ πρÕς τ¾ν ΖΡ, ‡ση δ ¹ µν ΓΖ τÍ ΕΚ, ¹ δ ΖΝ τÍ ΕΛ, ¹ δ ΖΡ τÍ ΕΜ, œστιν ¥ρα æς ¹ ΑΕ πρÕς τ¾ν ΕΚ, οÛτως ¹ ΗΕ πρÕς τ¾ν ΕΛ κሠ¹ ΘΕ πρÕς τ¾ν ΕΜ. ¢λλ' æς µν ¹ ΑΕ πρÕς τ¾ν ΕΚ, οÛτως τÕ ΑΗ [παραλληλόγραµµον] πρÕς τÕ ΗΚ παραλληλόγραµµον, æς δ ¹ ΗΕ πρÕς τ¾ν ΕΛ, οÛτως τÕ ΗΚ πρÕς τÕ ΚΛ, æς δ ¹ ΘΕ πρÕς ΕΜ, οÛτως τÕ ΠΕ πρÕς τÕ ΚΜ· κሠæς ¥ρα τÕ ΑΗ παραλληλόγραµµον πρÕς τÕ ΗΚ, οÛτως τÕ ΗΚ πρÕς τÕ ΚΛ κሠτÕ ΠΕ πρÕς τÕ ΚΜ. ¢λλ' æς µν τÕ ΑΗ πρÕς τÕ ΗΚ, οÛτως τÕ ΑΒ στερεÕν πρÕς τÕ ΕΞ στερεόν, æς δ τÕ ΗΚ πρÕς τÕ ΚΛ, οÛτως τÕ ΞΕ στερεÕν πρÕς τÕ ΠΛ στερεÕν, æς δ τÕ ΠΕ πρÕς τÕ ΚΜ, οÛτως τÕ ΠΛ στερεÕν πρÕς τÕ ΚΟ στερεόν· κሠæς ¥ρα τÕ ΑΒ στερεÕν πρÕς τÕ ΕΞ, οÛτως τÕ ΕΞ πρÕς τÕ ΠΛ κሠτÕ ΠΛ πρÕς τÕ ΚΟ. ™¦ν δ τέσσαρα µεγέθη κατ¦ τÕ συνεχς ¢νάλογον Ï, τÕ πρîτον πρÕς τÕ τέταρτον τριπλασίονα λόγον œχει ½περ πρÕς τÕ δεύτερον· τÕ ΑΒ ¥ρα στερεÕν πρÕς τÕ ΚΟ τριπλασίονα λόγον œχει ½περ τÕ ΑΒ πρÕς τÕ ΕΞ. ¢λλ' æς τÕ ΑΒ πρÕς τÕ ΕΞ, οÛτως τÕ ΑΗ παραλληλόγραµµον πρÕς τÕ ΗΚ κሠ¹ ΑΕ εÙθε‹α πρÕς τ¾ν

And since the two (straight-lines) KE and EL are equal to the two (straight-lines) CF and F N , but angle KEL is also equal to angle CF N , inasmuch as AEG is also equal to CF N , on account of the similarity of the solids AB and CD, parallelogram KL is thus equal [and similar] to parallelogram CN . So, for the same (reasons), parallelogram KM is equal and similar to [parallelogram] CR, and, further, EP to DF . Thus, three parallelograms of solid KP are equal and similar to three parallelograms of solid CD. But the three (former parallelograms) are equal and similar to the three opposite (parallelograms), and the three (latter parallelograms) are equal and similar to the three opposite (parallelograms) [Prop. 11.24]. Thus, the whole of solid KP is equal and similar to the whole of solid CD [Def. 11.10]. Let parallelogram GK have been completed. And let the the solids EO and LQ, with bases the parallelograms GK and KL (respectively), and with the same height as AB, have been completed. And since, on account of the similarity of solids AB and CD, as AE is to CF , so EG (is) to F N , and EH to F R [Defs. 6.1, 11.9], and CF (is) equal to EK, and F N to EL, and F R to EM , thus as AE is to EK, so GE (is) to EL, and HE to EM . But, as AE (is) to EK, so [parallelogram] AG (is) to parallelogram GK, and as GE (is) to EL, so GK (is) to KL, and as HE (is) to EM , so QE (is) to KM [Prop. 6.1]. And thus as parallelogram AG (is) to GK, so GK (is) to KL, and QE (is) to KM . But, as AG (is) to GK, so solid AB (is) to solid EO, and as GK (is) to KL, so solid OE (is) to solid QL, and as QE (is) to KM , so solid QL (is) to solid KP [Prop. 11.32]. And, thus, as solid AB is to EO, so EO (is) to QL, and QL to KP . And if four magnitudes are continuously proportional then the first has to the fourth the cubed ratio that (it has) to the second [Def. 5.10]. Thus, solid AB has to KP the cubed ratio which AB (has) to EO. But, as AB (is) to EO, so parallelogram AG (is) to GK, and the straight-line AE to EK [Prop. 6.1]. Hence, solid AB also has to KP the cubed ratio that AE (has) to EK. And solid KP (is)

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ELEMENTS BOOK 11

ΕΚ· éστε κሠτÕ ΑΒ στερεÕν πρÕς τÕ ΚΟ τριπλασίονα λόγον œχει ½περ ¹ ΑΕ πρÕς τ¾ν ΕΚ. ‡σον δ τÕ [µν] ΚΟ στερεÕν τù Γ∆ στερεù, ¹ δ ΕΚ εÙθε‹α τÍ ΓΖ· κሠτÕ ΑΒ ¥ρα στερεÕν πρÕς τÕ Γ∆ στερεÕν τριπλασίονα λόγον œχει ½περ ¹ еόλογος αÙτοà πλευρ¦ ¹ ΑΕ πρÕς τ¾ν еόλογον πλευρ¦ν τ¾ν ΓΖ.

B

X

J

H

equal to solid CD, and straight-line EK to CF . Thus, solid AB also has to solid CD the cubed ratio which its corresponding side AE (has) to the corresponding side CF .

B

P

O H

Q

G

E

A

L

K

A

M

K

E

L

M

O

P

Τ¦ ¥ρα Óµοια στερε¦ παραλληλεπίπεδα ™ν τριThus, similar parallelepiped solids are to one another πλασίονι λόγJ ™στˆ τîν еολόγων πλευρîν· Óπερ œδει as the cubed ratio of their corresponding sides. (Which δε‹ξαι. is) the very thing it was required to show.

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι ™¦ν τέσσαρες εÙθε‹αι ¢νάλογον ðσιν, œσται æς ¹ πρώτη πρÕς τ¾ν τετάρτην, οÛτω τÕ ¢πÕ τÁς πρώτης στερεÕν παραλληλεπίπεδον πρÕς τÕ ¢πÕ τÁς δευτέρας τÕ Óµοιον καˆ Ðµοίως ¢ναγραφόµενον, ™πείπερ κሠ¹ πρώτη πρÕς τ¾ν τετάρτην τριπλασίονα λόγον œχει ½περ πρÕς τ¾ν δευτέραν.

So, (it is) clear, from this, that if four straight-lines are (continuously) proportional then as the first is to the fourth, so the parallelepiped solid on the first will be to the similar, and similarly described, parallelepiped solid on the second, since the first also has to the fourth the cubed ratio that (it has) to the second.

λδ΄.

Proposition 34†

Τîν ‡σων στερεîν παραλληλεπιπέδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν· κሠïν στερεîν παραλληλεπιπέδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν, †σα ™στˆν ™κε‹να. ”Εστω ‡σα στερε¦ παραλληλεπίπεδα τ¦ ΑΒ, Γ∆· λέγω, Óτι τîν ΑΒ, Γ∆ στερεîν παραλληλεπιπέδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν, καί ™στιν æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως τÕ τοà Γ∆ στερεοà Ûψος πρÕς τÕ τοà ΑΒ στερεοà Ûψος. ”Εστωσαν γ¦ρ πρότερον αƒ ™φεστηκυ‹αι αƒ ΑΗ, ΕΖ, ΛΒ, ΘΚ, ΓΜ, ΝΞ, Ο∆, ΠΡ πρÕς Ñρθ¦ς τα‹ς βάσεσιν αÙτîν· λέγω, Óτι ™στˆν æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως ¹ ΓΜ πρÕς τ¾ν ΑΗ.

The bases of equal parallelepiped solids are reciprocally proportional to their heights. And those parallelepiped solids whose bases are reciprocally proportional to their heights are equal. Let AB and CD be equal parallelepiped solids. I say that the bases of the parallelepiped solids AB and CD are reciprocally proportional to their heights, and (so) as base EH is to base N Q, so the height of solid CD (is) to the height of solid AB. For, first of all, let the (straight-lines) standing up, AG, EF , LB, HK, CM , N O, P D, and QR, be at rightangles to their bases. I say that as base EH is to base N Q, so CM (is) to AG.

461

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

R H

K

A

J

Z E

B

M T

L P

G

D F

D

R

X

K

B

M V

F

G

O

T

O

N

H A

Ε„ µν οâν ‡ση ™στˆν ¹ ΕΘ βάσιν τÍ ΝΠ βάσει, œστι δ κሠτÕ ΑΒ στερεÕν τù Γ∆ στερεù ‡σον, œσται κሠ¹ ΓΜ τÍ ΑΗ ‡ση. τ¦ γ¦ρ ØπÕ τÕ αÙτÕ Ûψος στερε¦ παραλληλεπίπεδα πρÕς ¥λληλά ™στιν æς αƒ βάσεις. κሠœσται æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ, οÛτως ¹ ΓΜ πρÕς τ¾ν ΑΗ, κሠφανερόν, Óτι τîν ΑΒ, Γ∆ στερεîν παραλληλεπιπέδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν. Μ¾ œστω δ¾ ‡ση ¹ ΕΘ βάσις τÍ ΝΠ βάσει, ¢λλ' œστω µείζων ¹ ΕΘ. œστι δ κሠτÕ ΑΒ στερεÕν τù Γ∆ στερεù ‡σον· µείζων ¥ρα ™στˆ κሠ¹ ΓΜ τÁς ΑΗ. κείσθω οâν τÍ ΑΗ ‡ση ¹ ΓΤ, κሠσυµπεπληρώσθω ¢πÕ βάσεως µν τÁς ΝΠ, Ûψους δ τοà ΓΤ, στερεÕν παραλληλεπίπεδον τÕ ΦΓ. κሠ™πεˆ ‡σον ™στˆ τÕ ΑΒ στερεÕν τù Γ∆ στερεù, œξωθεν δ τÕ ΓΦ, τ¦ δ ‡σα πρÕς τÕ αÙτÕ τÕν αÙτÕν œχει λόγον, œστιν ¥ρα æς τÕ ΑΒ στερεÕν πρÕς τÕ ΓΦ στερεόν, οÛτως τÕ Γ∆ στερεÕν πρÕς τÕ ΓΦ στερεόν. ¢λλ' æς µν τÕ ΑΒ στερεÕν πρÕς τÕ ΓΦ στερεόν, οÛτως ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν· „σοϋψÁ γ¦ρ τ¦ ΑΒ, ΓΦ στερεά· æς δ τÕ Γ∆ στερεÕν πρÕς τÕ ΓΦ στερεόν, οÛτως ¹ ΜΠ βάσις πρÕς τ¾ν ΤΠ βάσιν κሠ¹ ΓΜ πρÕς τ¾ν ΓΤ· κሠæς ¥ρα ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως ¹ ΜΓ πρÕς τ¾ν ΓΤ. ‡ση δ ¹ ΓΤ τÍ ΑΗ· κሠæς ¥ρα ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως ¹ ΜΓ πρÕς τ¾ν ΑΗ. τîν ΑΒ, Γ∆ ¥ρα στερεîν παραλληλεπιπέδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν. Πάλιν δ¾ τîν ΑΒ, Γ∆ στερεîν παραλληλεπιπέδων ¢ντιπεπονθέτωσαν αƒ βάσεις το‹ς Ûψεσιν, κሠœστω æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως τÕ τοà Γ∆ στερεοà Ûψος πρÕς τÕ τοà ΑΒ στερεοà Ûψος· λέγω, Óτι ‡σον ™στˆ τÕ ΑΒ στερεÕν τù Γ∆ στερεù. ”Εστωσαν [γ¦ρ] πάλιν αƒ ™φεστηκυ‹αι πρÕς Ñρθ¦ς τα‹ς βάσεσιν. καˆ ε„ µν ‡ση ™στˆν ¹ ΕΘ βάσις τÍ ΝΠ βάσει, καί ™στιν æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως τÕ τοà Γ∆ στερεοà Ûψος πρÕς τÕ τοà ΑΒ στερεοà Ûψος, ‡σον ¥ρα ™στˆ κሠτÕ τοà Γ∆ στερεοà Ûψος τù τοà ΑΒ στερεοà Ûψει. τ¦ δ ™πˆ ‡σων βάσεων στερεά παραλληλεπίπεδα κሠØπÕ τÕ αÙτÕ Ûψος ‡σα ¢λλήλοις ™στίν· ‡σον ¥ρα ™στˆ τÕ ΑΒ στερεÕν τù Γ∆ στερεù. Μ¾ œστω δ¾ ¹ ΕΘ βάσις τÍ ΝΠ [βάσει] ‡ση, ¢λλ'

L Q E

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Therefore, if base EH is equal to base N Q, and solid AB is also equal to solid CD, CM will also be equal to AG. For parallelepiped solids of the same height are to one another as their bases [Prop. 11.32]. And as base EH (is) to N Q, so CM will be to AG. And (so it is) clear that the bases of the parallelepiped solids AB and CD are reciprocally proportional to their heights. So let base EH not be equal to base N Q, but let EH be greater. And solid AB is also equal to solid CD. Thus, CM is also greater than AG. Therefore, let CT be made equal to AG. And let the parallelepiped solid V C have been completed on the base N Q, with height CT . And since solid AB is equal to solid CD, and CV (is) extrinsic (to them), and equal (magnitudes) have the same ratio to the same (magnitude) [Prop. 5.7], thus as solid AB is to solid CV , so solid CD (is) to solid CV . But, as solid AB (is) to solid CV , so base EH (is) to base N Q. For the solids AB and CV (are) of equal height [Prop. 11.32]. And as solid CD (is) to solid CV , so base M Q (is) to base T Q [Prop. 11.25], and CM to CT [Prop. 6.1]. And, thus, as base EH is to base N Q, so M C (is) to AG. And CT (is) equal to AG. And thus as base EH (is) to base N Q, so M C (is) to AG. Thus, the bases of the parallelepiped solids AB and CD are reciprocally proportional to their heights. So, again, let the bases of the parallelepipid solids AB and CD be reciprocally proportional to their heights, and let base EH be to base N Q, as the height of solid CD (is) to the height of solid AB. I say that solid AB is equal to solid CD. [For] let the (straight-lines) standing up again be at right-angles to the bases. And if base EH is equal to base N Q, and as base EH is to base N Q, so the height of solid CD (is) to the height of solid AB, the height of solid CD is thus also equal to the height of solid AB. And parallelepiped solids on equal bases, and also with the same height, are equal to one another [Prop. 11.31]. Thus, solid AB is equal to solid CD. So, let base EH not be equal to [base] N Q, but let EH be greater. Thus, the height of solid CD is also

462

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ELEMENTS BOOK 11

œστω µείζων ¹ ΕΘ· µε‹ζον ¥ρα ™στˆ κሠτÕ τοà Γ∆ στερεοà Ûψος τοà τοà ΑΒ στερεοà Ûψους, τουτέστιν ¹ ΓΜ τÁς ΑΗ. κείσθω τÍ ΑΗ ‡ση πάλιν ¹ ΓΤ, κሠσυµπεπληρώσθω еοίως τÕ ΓΦ στερεόν. ™πεί ™στιν æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως ¹ ΜΓ πρÕς τ¾ν ΑΗ, ‡ση δ ¹ ΑΗ τÍ ΓΤ, œστιν ¥ρα æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως ¹ ΓΜ πρÕς τ¾ν ΓΤ. ¢λλ' æς µν ¹ ΕΘ [βάσις] πρÕς τ¾ν ΝΠ βάσιν, οÛτως τÕ ΑΒ στερεÕν πρÕς τÕ ΓΦ στερεόν· „σοϋψÁ γάρ ™στι τ¦ ΑΒ, ΓΦ στερεά· æς δ ¹ ΓΜ πρÕς τ¾ν ΓΤ, οÛτως ¼ τε ΜΠ βάσις πρÕς τ¾ν ΠΤ βάσιν κሠτÕ Γ∆ στερεÕν πρÕς τÕ ΓΦ στερεόν. κሠæς ¥ρα τÕ ΑΒ στερεÕν πρÕς τÕ ΓΦ στερεόν, οÛτως τÕ Γ∆ στερεÕν πρÕς τÕ ΓΦ στερεόν· ˜κάτερον ¥ρα τîν ΑΒ, Γ∆ πρÕς τÕ ΓΦ τÕν αÙτÕν œχει λόγον. ‡σον ¥ρα ™στˆ τÕ ΑΒ στερεÕν τù Γ∆ στερεù.

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greater than the height of solid AB, that is to say CM (greater) than AG. Let CT again be made equal to AG, and let the solid CV have been similarly completed. Since as base EH is to base N Q, so M C (is) to AG, and AG (is) equal to CT , thus as base EH (is) to base N Q, so CM (is) to CT . But, as [base] EH (is) to base N Q, so solid AB (is) to solid CV . For solids AB and CV are of equal heights [Prop. 11.32]. And as CM (is) to CT , so (is) base M Q to base QT [Prop. 6.1], and solid CD to solid CV [Prop. 11.25]. And thus as solid AB (is) to solid CV , so solid CD (is) to solid CV . Thus, AB and CD each have the same ratio to CV . Thus, solid AB is equal to solid CD [Prop. 5.9].

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Μ¾ œστωσαν δ¾ αƒ ™φεστηκυ‹αι αƒ ΖΕ, ΒΛ, ΗΑ, ΚΘ, ΞΝ, ∆Ο, ΜΓ, ΡΠ πρÕς Ñρθ¦ς τα‹ς βάσεσιν αÙτîν, κሠ½χθωσαν ¢πÕ τîν Ζ, Η, Β, Κ, Ξ, Μ, Ρ, ∆ σηµείων ™πˆ τ¦ δι¦ τîν ΕΘ, ΝΠ ™πίπεδα κάθετοι κሠσυµβαλλέτωσαν το‹ς ™πιπέδοις κατ¦ τ¦ Σ, Τ, Υ, Φ, Χ, Ψ, Ω, ς, κሠσυµπεπληρώσθω τ¦ ΖΦ, ΞΩ στερεά· λέγω, Óτι κሠοÛτως ‡σων Ôντων τîν ΑΒ, Γ∆ στερεîν ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν, καί ™στιν æς ¹ ΕΘ β¡σιν πρÕς τ¾ν ΝΠ βάσιν, οÛτως τÕ τοà Γ∆ στερεοà Ûψος πρÕς τÕ τοà ΑΒ στερεοà Ûψος. 'Επεˆ ‡σον ™στˆ τÕ ΑΒ στερεÕν τù Γ∆ στερεù, ¢λλ¦ τÕ µν ΑΒ τù ΒΤ ™στιν ‡σον· ™πί τε γ¦ρ τÁς αÙτÁς βάσεώς ε„σι τÁς ΖΚ κሠØπÕ τÕ αÙτÕ Ûψος· τÕ δ Γ∆ στερεÕν τù ∆Ψ ™στιν ‡σον· ™πί τε γ¦ρ πάλιν τÁς αÙτÁς βάσεώς ε„σι τÁς ΡΞ κሠØπÕ τÕ αÙτÕ Ûψος· κሠτÕ ΒΤ ¥ρα στερεÕν τù ∆Ψ στερεù ‡σον ™στίν. œστιν ¥ρα æς ¹ ΖΚ βάσις πρÕς τ¾ν ΞΡ βάσιν, οÛτως τÕ τοà ∆Ψ στε-

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So, let the (straight-lines) standing up, F E, BL, GA, KH, ON , DP , M C, and RQ, not be at right-angles to their bases. And let perpendiculars have been drawn to the planes through EH and N Q from points F , G, B, K, O, M , R, and D, and let them have joined the planes at (points) S, T , U , V , W , X, Y , and a (respectively). And let the solids F V and OY have been completed. In this case, also, I say that the solids AB and CD being equal, their bases are reciprocally proportional to their heights, and (so) as base EH is to base N Q, so the height of solid CD (is) to the height of solid AB. Since solid AB is equal to solid CD, but AB is equal to BT . For they are on the same base F K, and (have) the same height [Props. 11.29, 11.30]. And solid CD is equal is equal to DX. For, again, they are on the same base RO, and (have) the same height [Props. 11.29, 11.30]. Solid BT is thus also equal to solid DX. Thus, as base F K (is)

463

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

ρεοà Ûψος πρÕς τÕ τοà ΒΤ στερεοà Ûψος. ‡ση δ ¹ µν ΖΚ βάσις τÍ ΕΘ βάσει, ¹ δ ΞΡ βάσις τÍ ΝΠ βάσει· œστιν ¥ρα æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως τÕ τοà ∆Ψ στερεοà Ûψος πρÕς τÕ τοà ΒΤ στερεοà Ûψος. τ¦ δ' αÙτ¦ Ûψη ™στˆ τîν ∆Ψ, ΒΤ στερεîν κሠτîν ∆Γ, ΒΑ· œστιν ¥ρα æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως τÕ τοà ∆Γ στερεοà Ûψος πρÕς τÕ τοà ΑΒ στερεοà Ûψος. τîν ΑΒ, Γ∆ ¥ρα στερεîν παραλληλεπιπέδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν. Πάλιν δ¾ τîν ΑΒ, Γ∆ στερεîν παραλληλεπιπέδων ¢ντιπεπονθέτωσαν αƒ βάσεις το‹ς Ûψεσιν, κሠœστω æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως τÕ τοà Γ∆ στερεοà Ûψος πρÕς τÕ τοà ΑΒ στερεοà Ûψος· λέγω, Óτι ‡σον ™στˆ τÕ ΑΒ στερεÕν τù Γ∆ στερεù. Τîν γ¦ρ αÙτîν κατασκευασθέντων, ™πεί ™στιν æς ¹ ΕΘ βάσις πρÕς τ¾ν ΝΠ βάσιν, οÛτως τÕ τοà Γ∆ στερεοà Ûψος πρÕς τÕ τοà ΑΒ στερεοà Ûψος, ‡ση δ ¹ µν ΕΘ βάσις τÍ ΖΚ βάσει, ¹ δ ΝΠ τÍ ΞΡ, œστιν ¥ρα æς ¹ ΖΚ βάσις πρÕς τ¾ν ΞΡ βάσιν, οÛτως τÕ τοà Γ∆ στερεοà Ûψος πρÕς τÕ τοà ΑΒ στερεοà Ûψος. τ¦ δ' αÙτ¦ Ûψη ™στˆ τîν ΑΒ, Γ∆ στερεîν κሠτîν ΒΤ, ∆Ψ· œστιν ¥ρα æς ¹ ΖΚ βάσις πρÕς τ¾ν ΞΡ βάσιν, οÛτως τÕ τοà ∆Ψ στερεοà Ûψος πρÕς τÕ τοà ΒΤ στερεοà Ûψος. τîν ΒΤ, ∆Ψ ¥ρα στερεîν παραλληλεπιπέδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν· ‡σον ¥ρα ™στˆ τÕ ΒΤ στερεÕν τù ∆Ψ στερεù. ¢λλ¦ τÕ µν ΒΤ τù ΒΑ ‡σον ™στίν· ™πί τε γ¦ρ τÁς αÙτÁς βάσεως [ε„σι] τÁς ΖΚ κሠØπÕ τÕ αÙτÕ Ûψος. τÕ δ ∆Ψ στερεÕν τù ∆Γ στερεù ‡σον ™στίν. κሠτÕ ΑΒ ¥ρα στερεÕν τù Γ∆ στερεù ™στιν ‡σον· Óπερ œδει δε‹ξαι.

to base OR, so the height of solid DX (is) to the height of solid BT (see first part of proposition). And base F K (is) equal to base EH, and base OR to N Q. Thus, as base EH is to base N Q, so the height of solid DX (is) to the height of solid BT . And solids DX, BT are the same height as (solids) DC, BA (respectively). Thus, as base EH is to base N Q, so the height of solid DC (is) to the height of solid AB. Thus, the bases of the parallelepiped solids AB and CD are reciprocally proportional to their heights. So, again, let the bases of the parallelepiped solids AB and CD be reciprocally proportional to their heights, and (so) let base EH be to base N Q, as the height of solid CD (is) to the height of solid AB. I say that solid AB is equal to solid CD. For, with the same construction (as before), since as base EH is to base N Q, so the height of solid CD (is) to the height of solid AB, and base EH (is) equal to base F K, and N Q to OR, thus as base F K is to base OR, so the height of solid CD (is) to the height of solid AB. And solids AB, CD are the same height as (solids) BT , DX (respectively). Thus, as base F K is to base OR, so the height of solid DX (is) to the height of solid BT . Thus, the bases of the parallelepiped solids BT and DX are reciprocally proportional to their heights. Thus, solid BT is equal to solid DX (see first part of proposition). But, BT is equal to BA. For [they are] on the same base F K, and (have) the same height [Props. 11.29, 11.30]. And solid DX is equal to solid DC [Props. 11.29, 11.30]. Thus, solid AB is also equal to solid CD. (Which is) the very thing it was required to show.

†

This proposition assumes that (a) if two parallelepipeds are equal, and have equal bases, then their heights are equal, and (b) if the bases of two equal parallelepipeds are unequal, then that solid which has the lesser base has the greater height.

λε΄.

Proposition 35

'Ε¦ν ðσι δύο γωνίαι ™πίπεδοι ‡σαι, ™πˆ δ τîν κορυφîν αÙτîν µετέωροι εÙθε‹αι ™πισταθîσιν ‡σας γωνίας περιέχουσαι µετ¦ τîν ™ξ ¢ρχÁς εÙθειîν ˜κατέραν ˜κατέρv, ™πˆ δ τîν µετεώρων ληφθÍ τυχόντα σηµε‹α, κሠ¢π' αÙτîν ™πˆ τ¦ ™πίπεδα, ™ν οŒς ε„σιν αƒ ™ξ ¢ρχÁς γωνίαι, κάθετοι ¢χθîσιν, ¢πÕ δ τîν γενοµένων σηµείων ™ν το‹ς ™πιπέδοις ™πˆ τ¦ς ™ξ ¢ρχÁς γωνίας ™πιζευχθîσιν εÙθε‹αι, ‡σας γωνίας περιέξουσι µετ¦ τîν µετεώρων. ”Εστωσαν δύο γωνίαι εÙθύγραµµοι ‡σαι αƒ ØπÕ ΒΑΓ, Ε∆Ζ, ¢πÕ δ τîν Α, ∆ σηµείων µετέωροι εÙθε‹αι ™φεστάτωσαν αƒ ΑΗ, ∆Μ ‡σας γωνίας περιέχουσιν µετ¦ τîν ™ξ ¢ρχÁς εÙθειîν ˜κατέραν ˜κατέρv, τ¾ν µν ØπÕ Μ∆Ε τÍ ØπÕ ΗΑΒ, τ¾ν δ ØπÕ Μ∆Ζ τÍ ØπÕ ΗΑΓ, κሠε„λήφθω ™πˆ τîν ΑΗ, ∆Μ τυχόντα σηµε‹α τ¦ Η, Μ, καˆ

If there are two equal plane angles, and raised straight-lines are stood on the apexes of them, containing equal angles respectively with the original straight-lines (forming the angles), and random points are taken on the raised (straight-lines), and perpendiculars are drawn from them to the planes in which the original angles are, and straight-lines are joined from the points created in the planes to the (vertices of the) original angles, then they will enclose equal angles with the raised (straightlines). Let BAC and EDF be two equal rectilinear angles. And let the raised straight-lines AG and DM have been stood on points A and D, containing equal angles respectively with the original straight-lines. (That is) M DE

464

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

½χθωσαν ¢πÕ τîν Η, Μ σηµείων ™πˆ τ¦ δι¦ τîν ΒΑΓ, Ε∆Ζ ™πίπεδα κάθετοι αƒ ΗΛ, ΜΝ, κሠσυµβαλλέτωσαν το‹ς ™πιπέδοις κατ¦ τ¦ Λ, Ν, κሠ™πεζεύχθωσαν αƒ ΛΑ, Ν∆· λέγω, Óτι ‡ση ™στˆν ¹ ØπÕ ΗΑΛ γωνία τÍ ØπÕ Μ∆Ν γωνίv.

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(equal) to GAB, and M DF (to) GAC. And let the random points G and M have been taken on AG and DM (respectively). And let the GL and M N have been drawn from points G and M perpendicular to the planes through BAC and EDF (respectively). And let them have joined the planes at points L and N (respectively). And let LA and N D have been joined. I say that angle GAL is equal to angle M DN .

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Κείσθω τÍ ∆Μ ‡ση ¹ ΑΘ, κሠ½χθω δι¦ τοà Θ σηµείου τÍ ΗΛ παράλληλος ¹ ΘΚ. ¹ δ ΗΛ κάθετός ™στιν ™πˆ τÕ δι¦ τîν ΒΑΓ ™πίπεδον· κሠ¹ ΘΚ ¥ρα κάθετός ™στιν ™πˆ τÕ δι¦ τîν ΒΑΓ ™πίπεδον. ½χθωσαν ¢πÕ τîν Κ, Ν σηµείων ™πˆ τ¦ς ΑΓ, ∆Ζ, ΑΒ, ∆Ε εÙθείας κάθετοι αƒ ΚΓ, ΝΖ, ΚΒ, ΝΕ, κሠ™πεζεύχθωσαν αƒ ΘΓ, ΓΒ, ΜΖ, ΖΕ. ™πεˆ τÕ ¢πÕ τÁς ΘΑ ‡σον ™στˆ το‹ς ¢πÕ τîν ΘΚ, ΚΑ, τù δ ¢πÕ τÁς ΚΑ ‡σα ™στˆ τ¦ ¢πÕ τîν ΚΓ, ΓΑ, κሠτÕ ¢πÕ τÁς ΘΑ ¥ρα ‡σον ™στˆ το‹ς ¢πÕ τîν ΘΚ, ΚΓ, ΓΑ. το‹ς δ ¢πÕ τîν ΘΚ, ΚΓ ‡σον ™στˆ τÕ ¢πÕ τÁς ΘΓ· τÕ ¥ρα ¢πÕ τÁς ΘΑ ‡σον ™στˆ το‹ς ¢πÕ τîν ΘΓ, ΓΑ. Ñρθ¾ ¥ρα ™στˆν ¹ ØπÕ ΘΓΑ γωνία. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ ∆ΖΜ γωνία Ñρθή ™στιν. ‡ση ¥ρα ™στˆν ¹ ØπÕ ΑΓΘ γωνία τÍ ØπÕ ∆ΖΜ. œστι δ κሠ¹ ØπÕ ΘΑΓ τÍ ØπÕ Μ∆Ζ ‡ση. δύο δ¾ τρίγωνά ™στι τ¦ Μ∆Ζ, ΘΑΓ δύο γωνίας δυσˆ γωνίαις ‡σας œχοντα ˜κατέραν ˜κατέρv κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην τ¾ν Øποτείνουσαν ØπÕ µίαν τîν ‡σων γωνιîν τ¾ν ΘΑ τÍ Μ∆· κሠτ¦ς λοιπ¦ς ¥ρα πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξει ˜κατέραν ˜καρέρv. ‡ση ¥ρα ™στˆν ¹ ΑΓ τÍ ∆Ζ. еοίως δ¾ δείξοµεν, Óτι κሠ¹ ΑΒ τÍ ∆Ε ™στιν ‡ση. ™πεˆ οâν ‡ση ™στˆν ¹ µν ΑΓ τÍ ∆Ζ, ¹ δ ΑΒ τÍ ∆Ε, δύο δ¾ αƒ ΓΑ, ΑΒ δυσˆ τα‹ς Ζ∆, ∆Ε ‡σαι ε„σίν. ¢λλ¦ κሠγωνία ¹ ØπÕ ΓΑΒ γωνίv τÍ ØπÕ Ζ∆Ε ™στιν ‡ση· βάσις ¥ρα ¹ ΒΓ βάσει τÍ ΕΖ ‡ση ™στˆ κሠτÕ τρίγωνον τù τριγώνJ καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις· ‡ση ¥ρα ¹ ØπÕ ΑΓΒ γωνία τÍ ØπÕ ∆ΖΕ. œστι δ κሠÑρθ¾ ¹ ØπÕ ΑΓΚ ÑρθÍ τÍ ØπÕ ∆ΖΝ ‡ση· κሠλοιπ¾ ¥ρα ¹

Let AH be made equal to DM . And let HK have been drawn through point H parallel to GL. And GL is perpendicular to the plane through BAC. Thus, HK is also perpendicular to the plane through BAC [Prop. 11.8]. And let KC, N F , KB, and N E have been drawn from points K and N perpendicular to the straight-lines AC, DF , AB, and DE. And let HC, CB, M F , and F E have been joined. Since the (square) on HA is equal to the (sum of the squares) on HK and KA [Prop. 1.47], and the (sum of the squares) on KC and CA is equal to the (square) on KA [Prop. 1.47], thus the (square) on HA is equal to the (sum of the squares) on HK, KC, and CA. And the (square) on HC is equal to the (sum of the squares) on HK and KC [Prop. 1.47]. Thus, the (square) on HA is equal to the (sum of the squares) on HC and CA. Thus, angle HCA is a right-angle [Prop. 1.48]. So, for the same (reasons), angle DF M is also a right-angle. Thus, angle ACH is equal to (angle) DF M . And HAC is also equal to M DF . So, M DF and HAC are two triangles having two angles equal to two angles, respectively, and one side equal to one side— (namely), that subtending one of the equal angles —(that is), HA (equal) to M D. Thus, they will also have the remaining sides equal to the remaining sides, respectively [Prop. 1.26]. Thus, AC is equal to DF . So, similarly, we can show that AB is also equal to DE. Therefore, since AC is equal to DF , and AB to DE, so the two (straightlines) CA and AB are equal to the two (straight-lines)

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ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

ØπÕ ΒΓΚ λοιπÍ τÍ ØπÕ ΕΖΝ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ ΓΒΚ τÍ ØπÕ ΖΕΝ ™στιν ‡ση. δύο δ¾ τρίγωνά ™στι τ¦ ΒΓΚ, ΕΖΝ [τ¦ς] δύο γωνίας δυσˆ γωνίαις ‡σας œχοντα ˜κατέραν ˜κατέρv κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην τ¾ν πρÕς ταˆς ‡σαις γωνίαις τ¾ν ΒΓ τÍ ΕΖ· κሠτ¦ς λοιπ¦ς ¥ρα πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξουσιν. ‡ση ¥ρα ™στˆν ¹ ΓΚ τÍ ΖΝ. œστι δ κሠ¹ ΑΓ τÍ ∆Ζ ‡ση· δύο δ¾ αƒ ΑΓ, ΓΚ δυσˆ τα‹ς ∆Ζ, ΖΝ ‡σαι ε„σίν· κሠÑρθ¦ς γωνίας περιέχουσιν. βάσις ¥ρα ¹ ΑΚ βάσει τÍ ∆Ν ‡ση ™στίν. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΘ τÍ ∆Μ, ‡σον ™στˆ κሠτÕ ¢πÕ τÁς ΑΘ τù ¢πÕ τÁς ∆Μ. ¢λλ¦ τù µν ¢πÕ τÁς ΑΘ ‡σα ™στˆ τ¦ ¢πÕ τîν ΑΚ, ΚΘ· Ñρθ¾ γ¦ρ ¹ ØπÕ ΑΚΘ· τù δ ¢πÕ τÁς ∆Μ ‡σα τ¦ ¢πÕ τîν ∆Ν, ΝΜ· Ñρθ¾ γ¦ρ ¹ ØπÕ ∆ΝΜ· τ¦ ¥ρα ¢πÕ τîν ΑΚ, ΚΘ ‡σα ™στˆ το‹ς ¢πÕ τîν ∆Ν, ΝΜ, ïν τÕ ¢πÕ τÁς ΑΚ ‡σον ™στˆ τù ¢πÕ τÁς ∆Ν· λοιπÕν ¥ρα τÕ ¢πÕ τÁς ΚΘ ‡σον ™στˆ τù ¢πÕ τÁς ΝΜ· ‡ση ¥ρα ¹ ΘΚ τÍ ΜΝ. κሠ™πεˆ δύο αƒ ΘΑ, ΑΚ δυσˆ τα‹ς Μ∆, ∆Ν ‡σαι ε„σˆν ˜κατέρα ˜κατέρv, κሠβάσις ¹ ΘΚ βάσει τÍ ΜΝ ™δείχθη ‡ση, γωνία ¥ρα ¹ ØπÕ ΘΑΚ γωνίv τÍ ØπÕ Μ∆Ν ™στιν ‡ση. 'Ε¦ν ¥ρα ðσι δύο γωνίαι ™πίπεδοι ‡σαι κሠτ¦ ˜ξÁς τÁς προτάσεως [Óπερ œδει δε‹ξαι].

F D and DE (respectively). But, angle CAB is also equal to angle F DE. Thus, base BC is equal to base EF , and triangle (ACB) to triangle (DF E), and the remaining angles to the remaining angles (respectively) [Prop. 1.4]. Thus, angle ACB (is) equal to DF E. And the right-angle ACK is also equal to the right-angle DF N . Thus, the remainder BCK is equal to the remainder EF N . So, for the same (reasons), CBK is also equal to F EN . So, BCK and EF N are two triangles having two angles equal to two angles, respectively, and one side equal to one side—(namely), that by the equal angles—(that is), BC (equal) to EF . Thus, they will also have the remaining sides equal to the remaining sides (respectively) [Prop. 1.26]. Thus, CK is equal to F N . And AC (is) also equal to DF . So, the two (straight-lines) AC and CK are equal to the two (straight-lines) DF and F N (respectively). And they enclose right-angles. Thus, base AK is equal to base DN [Prop. 1.4]. And since AH is equal to DM , the (square) on AH is also equal to the (square) on DM . But, the the (sum of the squares) on AK and KH is equal to the (square) on AH. For angle AKH (is) a right-angle [Prop. 1.47]. And the (sum of the squares) on DN and N M (is) equal to the square on DM . For angle DN M (is) a right-angle [Prop. 1.47]. Thus, the (sum of the squares) on AK and KH is equal to the (sum of the squares) on DN and N M , of which the (square) on AK is equal to the (square) on DN . Thus, the remaining (square) on KH is equal to the (square) on N M . Thus, HK (is) equal to M N . And since the two (straight-lines) HA and AK are equal to the two (straight-lines) M D and DN , respectively, and base HK was shown (to be) equal to base M N , angle HAK is thus equal to angle M DN [Prop. 1.8]. Thus, if there are two equal plane angles, and so on of the proposition. [(Which is) the very thing it was required to show].

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι, ˜¦ν ðσι δύο γωνίαι ™πίπεδοι ‡σαι, ™πισταθîσι δ ™π' αÙτîν µετέωροι εÙθε‹αι ‡σαι ‡σας γωνίας περιέχουσαι µετ¦ τîν ™ξ ¢ρχÁς εÙθειîν ˜κατέραν ˜κατέρv, αƒ ¢π' αÙτîν κάθετοι ¢γόµεναι ™πˆ τ¦ ™πίπεδα, ™ν οŒς ε„σιν αƒ ™ξ ¢ρχÁς γωνίαι, ‡σαι ¢λλήλαις ε„σίν. Óπερ œδει δε‹ξαι.

So, it is clear, from this, that if there are two equal plane angles, and equal raised straight-lines are stood on them (at their apexes), containing equal angles respectively with the original straight-lines (forming the angles), then the perpendiculars drawn from (the raised ends of) them to the planes in which the original angles lie are equal to one another. (Which is) the very thing it was required to show.

λ$΄.

Proposition 36

'Ε¦ν τρε‹ς εÙθε‹αι ¢νάλογον ðσιν, τÕ ™κ τîν τριîν If three straight-lines are (continuously) proportional στερεÕν παραλληλεπίπεδον ‡σον ™στˆ τù ¢πÕ τÁς µέσης then the parallelepiped solid (formed) from the three 466

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

στερεù παραλληλεπιπέδJ „σοπλεύρJ µέν, „σογωνίJ δ (straight-lines) is equal to the equilateral parallelepiped τù προειρηµένJ. solid on the middle (straight-line which is) equiangular to the aforementioned (parallelepiped solid).

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”Εστωσαν τρε‹ς εÙθε‹αι ¢νάλογον αƒ Α, Β, Γ, æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Β πρÕς τ¾ν Γ· λέγω, Óτι τÕ ™κ τîν Α, Β, Γ στερεÕν ‡σον ™στˆ τù ¢πÕ τÁς Β στερεù „σοπλεύρJ µέν, „σογωνίJ δ τù προειρηµένJ. 'Εκκείσθω στερε¦ γωνία ¹ πρÕς τù Ε περιεχοµένη ØπÕ τîν ØπÕ ∆ΕΗ, ΗΕΖ, ΖΕ∆, κሠκείσθω τÍ µν Β ‡ση ˜κάστη τîν ∆Ε, ΗΕ, ΕΖ, κሠσυµπεπληρώσθω τÕ ΕΚ στερεÕν παραλληλεπίπεδον, τÍ δ Α ‡ση ¹ ΛΜ, κሠσυνεστάτω πρÕς τÍ ΛΜ εÙθείv κሠτù πρÕς αÙτÍ σηµείJ τù Λ τÍ πρÕς τù Ε στερε´ γωνίv ‡ση στερε¦ γωνία ¹ περειχοµένη ØπÕ τîν ΝΛΞ, ΞΛΜ, ΜΛΝ, κሠκείσθω τÍ µν Β ‡ση ¹ ΛΞ, τÍ δ Γ ‡ση ¹ ΛΝ. κሠ™πεί ™στιν æς ¹ Α πρÕς τ¾ν Β, οÛτως ¹ Β πρÕς τ¾ν Γ, ‡ση δ ¹ µν Α τÍ ΛΜ, ¹ δ Β ˜κατέρv τîν ΛΞ, Ε∆, ¹ δ Γ τÍ ΛΝ, œστιν ¥ρα æς ¹ ΛΜ πρÕς τ¾ν ΕΖ, οÛτως ¹ ∆Ε πρÕς τ¾ν ΛΝ. κሠπερˆ ‡σας γωνίας τ¦ς ØπÕ ΝΛΜ, ∆ΕΖ αƒ πλευρሠ¢ντιπεπόνθασιν· ‡σον ¥ρα ™στˆ τÕ ΜΝ παραλληλόγραµµον τù ∆Ζ παραλληλογραµάµµJ. κሠ™πεˆ δύο γωνίαι ™πίπεδοι εÙθύγραµµοι ‡σαι ε„σˆν αƒ ØπÕ ∆ΕΖ, ΝΛΜ, κሠ™π' αÙτîν µετέωροι εÙθε‹αι ™φεστ©σιν αƒ ΛΞ, ΕΗ ‡σαι τε ¢λλήλαις κሠ‡σας γωνίας περιέχουσαι µετ¦ τîν ™ξ ¢ρχÁς εÙθειîν ˜κατέραν ˜κατέρv, αƒ ¥ρα ¢πÕ τîν Η, Ξ σηµείων κάθετοι ¢γόµεναι ™πˆ τ¦ δι¦ τîν ΝΛΜ, ∆ΕΖ ™πίπεδα ‡σαι ¢λλήλαις ε„σίν· éστε τ¦ ΛΘ, ΕΚ στερε¦ ØπÕ τÕ αÙτÕ Ûψος ™στίν. τ¦ δ ™πˆ ‡σων βάσεων στερε¦ παραλληλεπίπεδα κሠØπÕ τÕ αÙτÕ Ûψος ‡σα ¢λλήλοις ™στίν· ‡σον ¥ρα ™στˆ τÕ ΘΛ στερεÕν τù ΕΚ στερεù. καί ™στι τÕ µν ΛΘ τÕ ™κ τîν Α, Β, Γ στερεόν, τÕ δ ΕΚ τÕ ¢πÕ τÁς Β στερεόν· τÕ ¥ρα ™κ τîν Α, Β, Γ στερεÕν παραλληλεπίπεδον ‡σον ™στˆ τù ¢πÕ τÁς Β στερεù „σοπλεύρJ µέν, „σογωνίJ δ τù προειρηµένJ· Óπερ œδει δε‹ξαι.

Let A, B, and C be three (continuously) proportional straight-lines, (such that) as A (is) to B, so B (is) to C. I say that the (parallelepiped) solid (formed) from A, B, and C is equal to the equilateral solid on B (which is) equiangular with the aforementioned (solid). Let the solid angle at E, contained by DEG, GEF , and F ED, be set out. And let DE, GE, and EF each be made equal to B. And let the parallelepiped solid EK have been completed. And (let) LM (be made) equal to A. And let the solid angle contained by N LO, OLM , and M LN have been constructed on the straightline LM , and at the point L on it, (so as to be) equal to the solid angle E [Prop. 11.23]. And let LO be made equal to B, and LN equal to C. And since as A (is) to B, so B (is) to C, and A (is) equal to LM , and B to each of LO and ED, and C to LN , thus as LM (is) to EF , so DE (is) to LN . And (so) the sides around the equal angles N LM and DEF are reciprocally proportional. Thus, parallelogram M N is equal to parallelogram DF [Prop. 6.14]. And since the two plane rectilinear angles DEF and N LM are equal, and the raised straight-lines stood on them (at their apexes), LO and EG, are equal to one another, and contain equal angles respectively with the original straight-lines (forming the angles), the perpendiculars drawn from points G and O to the planes through N LM and DEF (respectively) are thus equal to one another [Prop. 11.35 corr.]. Thus, the solids LH and EK (have) the same height. And parallelepiped solids on equal bases, and with the same height, are equal to one another [Prop. 11.31]. Thus, solid HL is equal to solid EK. And LH is the solid (formed) from A, B, and C, and EK the solid on B. Thus, the parallelepiped solid (formed) from A, B, and C is equal to the equilateral solid on B (which is) equiangular with the aforementioned (solid). (Which is) the very thing it was required to show.

467

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11 λζ΄.

Proposition 37†

'Ε¦ν τέσσαρες εÙθε‹αι ¢νάλογον ðσιν, κሠτ¦ ¢π' αÙτîν στερε¦ παραλληλεπίπεδα Óµοιά τε καˆ Ðµοίως ¢ναγραφόµενα ¢νάλογον œσται· κሠ™¦ν τ¦ ¢π' αÙτîν στερε¦ παραλληλεπίπεδα Óµοιά τε καˆ Ðµοίως ¢ναγραφόµενα ¢νάλογον Ï, κሠαÙταˆ αƒ εÙθε‹αι ¢νάλογον œσονται.

If four straight-lines are proportional then the similar, and similarly described, parallelepiped solids on them will also be proportional. And if the similar, and similarly described, parallelepiped solids on them are proportional then the straight-lines themselves will be proportional.

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”Εστωσαν τέσσαρες εÙθε‹αι ¢νάλογον αƒ ΑΒ, Γ∆, ΕΖ, ΗΘ, æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΗΘ, κሠ¢ναγεγράφθωσαν ¢πÕ τîν ΑΒ, Γ∆, ΕΖ, ΗΘ Óµοιά τε καˆ Ðµοίως κείµενα στερε¦ παραλληλεπίπεδα τ¦ ΚΑ, ΛΓ, ΜΕ, ΝΗ· λέγω, Óτι ™στˆν æς τÕ ΚΑ πρÕς τÕ ΛΓ, οÛτως τÕ ΜΕ πρÕς τÕ ΝΗ. 'Επεˆ γ¦ρ Óµοιόν ™στι τÕ ΚΑ στερεÕν παραλληλεπίπεδον τù ΛΓ, τÕ ΚΑ ¥ρα πρÕς τÕ ΛΓ τριπλασίονα λόγον œχει ½περ ¹ ΑΒ πρÕς τ¾ν Γ∆. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ΜΕ πρÕς τÕ ΝΗ τριπλασίονα λόγον œχει ½περ ¹ ΕΖ πρÕς τ¾ν ΗΘ. καί ™στιν æς ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΗΘ. κሠæς ¥ρα τÕ ΑΚ πρÕς τÕ ΛΓ, οÛτως τÕ ΜΕ πρÕς τÕ ΝΗ. 'Αλλ¦ δ¾ œστω æς τÕ ΑΚ στερεÕν πρÕς τÕ ΛΓ στερεόν, οÛτως τÕ ΜΕ στερεÕν πρÕς τÕ ΝΗ· λέγω, Óτι ™στˆν æς ¹ ΑΒ εÙθε‹α πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΗΘ. 'Επεˆ γ¦ρ πάλιν τÕ ΚΑ πρÕς τÕ ΛΓ τριπλασίονα λόγον œχει ½περ ¹ ΑΒ πρÕς τ¾ν Γ∆, œχει δ κሠτÕ ΜΕ πρÕς τÕ ΝΗ τριπλασίονα λόγον ½περ ¹ ΕΖ πρÕς τ¾ν ΗΘ, καί ™στιν æς τÕ ΚΑ πρÕς τÕ ΛΓ, οÛτως τÕ ΜΕ πρÕς τÕ ΝΗ, κሠæς ¥ρα ¹ ΑΒ πρÕς τ¾ν Γ∆, οÛτως ¹ ΕΖ πρÕς τ¾ν ΗΘ. 'Ε¦ν ¥ρα τέσσαρες εÙθε‹αι ¢νάλογον ðσι κሠτ¦ ˜ξÁς τÁς προτάσεως· Óπερ œδει δε‹ξαι. †

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Let AB, CD, EF , and GH, be four proportional straight-lines, (such that) as AB (is) to CD, so EF (is) to GH. And let the similar, and similarly laid out, parallelepiped solids KA, LC, M E and N G have been described on AB, CD, EF , and GH (respectively). I say that as KA is to LC, so M E (is) to N G. For since the parallelepiped solid KA is similar to LC, KA thus has to LC the cubed ratio that AB (has) to CD [Prop. 11.33]. So, for the same (reasons), M E also has to N G the cubed ratio that EF (has) to GH [Prop. 11.33]. And since as AB is to CD, so EF (is) to GH, thus, also, as AK (is) to LC, so M E (is) to N G. And so let solid AK be to solid LC, as solid M E (is) to N G. I say that as straight-line AB is to CD, so EF (is) to GH. For, again, since KA has to LC the cubed ratio that AB (has) to CD [Prop. 11.33], and M E also has to N G the cubed ratio that EF (has) to GH [Prop. 11.33], and as KA is to LC, so M E (is) to N G, thus, also, as AB (is) to CD, so EF (is) to GH. Thus, if four straight-lines are proportional, and so on of the proposition. (Which is) the very thing it was required to show.

This proposition assumes that if two ratios are equal then the cube of the former is also equal to the cube of the latter, and vice versa.

λη΄.

Proposition 38

'Ε¦ν κύβου τîν ¢πεναντίον ™πιπέδων αƒ πλευρሠδίχα τµηθîσιν, δι¦ δ τîν τοµîν ™πίπεδα ™κβληθÍ, ¹ κοιν¾

If the sides of the opposite planes of a cube are cut in half, and planes are produced through the pieces, then

468

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11

τοµ¾ τîν ™πιπέδων κሠ¹ τοà κύβου διάµετρος δίχα τέµνουσιν ¢λλήλας.

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the common section of the (latter) planes and the diameter of the cube cut one another in half.

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Κύβου γ¦ρ τοà ΑΖ τîν ¢πεναντίον ™πιπέδων τîν ΓΖ, ΑΘ αƒ πλευρሠδίχα τετµήσθωσαν κατ¦ τ¦ Κ, Λ, Μ, Ν, Ξ, Π, Ο, Ρ σηµε‹α, δι¦ δ τîν τοµîν ™πίπεδα ™κβεβλήσθω τ¦ ΚΝ, ΞΡ, κοιν¾ δ τοµ¾ τîν ™πιπέδων œστω ¹ ΥΣ, τοà δ ΑΖ κύβου διαγώνιος ¹ ∆Η. λέγω, Óτι ‡ση ™στˆν ¹ µν ΥΤ τÍ ΤΣ, ¹ δ ∆Τ τÍ ΤΗ. 'Επεζεύχθωσαν γ¦ρ αƒ ∆Υ, ΥΕ, ΒΣ, ΣΗ. κሠ™πεˆ παράλληλός ™στιν ¹ ∆Ξ τÍ ΟΕ, αƒ ™ναλλ¦ξ γωνίαι αƒ ØπÕ ∆ΞΥ, ΥΟΕ ‡σαι ¢λλήλαις ε„σίν. κሠ™πεˆ ‡ση ™στˆν ¹ µν ∆Ξ τÍ ΟΕ, ¹ δ ΞΥ τÍ ΥΟ, κሠγωνίας ‡σας περιέχουσιν, βάσις ¥ρα ¹ ∆Υ τÍ ΥΕ ™στιν ‡ση, κሠτÕ ∆ΞΥ τρίγωνον τù ΟΥΕ τριγώνJ ™στˆν ‡σον καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι· ‡ση ¥ρα ¹ ØπÕ ΞΥ∆ γωνία τÍ ØπÕ ΟΥΕ γωνίv. δι¦ δ¾ τοàτο εÙθε‹ά ™στιν ¹ ∆ΥΕ. δι¦ τ¦ αÙτ¦ δ¾ κሠΒΣΗ εÙθε‹ά ™στιν, κሠ‡ση ¹ ΒΣ τÍ ΣΗ. κሠ™πεˆ ¹ ΓΑ τÍ ∆Β ‡ση ™στˆ κሠπαράλληλος, ¢λλ¦ ¹ ΓΑ κሠτÍ ΕΗ ‡ση τέ ™στι κሠπαράλληλος, κሠ¹ ∆Β ¥ρα τÍ ΕΗ ‡ση τέ ™στι κሠπαράλληλος. κሠ™πιζευγνύουσιν αÙτ¦ς εÙθε‹αι αƒ ∆Ε, ΒΗ· παράλληλος ¥ρα ™στˆν ¹ ∆Ε τÍ ΒΗ. ‡ση ¥ρα ¹ µν ØπÕ Ε∆Τ γωνία τÍ ØπÕ ΒΗΤ· ™ναλλ¦ξ γάρ· ¹ δ ØπÕ ∆ΤΥ τÍ ØπÕ ΗΤΣ. δύο δ¾ τρίγωνά ™στι τ¦ ∆ΤΥ, ΗΤΣ τ¦ς δύο γωνίας τα‹ς δυσˆ γωνίαις ‡σας œχοντα κሠµίαν πλευρ¦ν µι´ πλευρ´ ‡σην τ¾ν Øποτείνουσαν ØπÕ µίαν τîν ‡σων γωνιîν τ¾ν ∆Υ τÍ ΗΣ· ¹µίσειαι γάρ ε„σι τîν ∆Ε, ΒΗ· κሠτ¦ς λοιπ¦ς πλευρ¦ς τα‹ς λοιπα‹ς πλευρα‹ς ‡σας ›ξει. ‡ση ¥ρα ¹ µν ∆Τ τÍ ΤΗ, ¹ δ ΥΤ τÍ ΤΣ. 'Ε¦ν ¥ρα κύβον τîν ¢πεναντίον ™πιπέδων αƒ πλευρሠδίχα τµηθîσιν, δι¦ δ τîν τοµîν ™πίπεδα ™κβληθÍ, ¹ κοιν¾ τοµ¾ τîν ™πιπέδων κሠ¹ τοà κύβον διάµετρος δίχα τέµνουσιν ¢λλήλας· Óπερ œδει δε‹ξαι.

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For let the opposite planes CF and AH of the cube AF have been cut in half at the points K, L, M , N , O, Q, P , and R. And let the planes KN and OR have been produced through the pieces. And let U S be the common section of the planes, and DG the diameter of cube AF . I say that U T is equal to T S, and DT to T G. For let DU , U E, BS, and SG have been joined. And since DO is parallel to P E, the alternate angles DOU and U P E are equal to one another [Prop. 1.29]. And since DO is equal to P E, and OU to U P , and they contain equal angles, base DU is thus equal to base U E, and triangle DOU is equal to triangle P U E, and the remaining angles equal to the remaining angles [Prop. 1.4]. Thus, angle OU D (is) equal to angle P U E. So, on account of this, DU E is a straight-line [Prop. 1.14]. So, for the same (reasons), BSG is also a straight-line, and BS equal to SG. And since CA is equal and parallel to DB, but CA is also equal and parallel to EG, DB is thus also equal and parallel to EG [Prop. 11.9]. And the straightlines DE and BG join them. DE is thus parallel to BG [Prop. 1.33]. Thus, angle EDT (is) equal to BGT . For (they are) alternate [Prop. 1.29]. And (angle) DT U (is equal) to GT S [Prop. 1.15]. So, DT U and GT S are two triangles having two angles equal to two angles, and one side equal to one side—(namely), that subtended by one of the equal angles—(that is), DU (equal) to GS. For they are halves of DE and BG (respectively). (Thus), they will also have the remaining sides equal to the remaining sides [Prop. 1.26]. Thus, DT (is) equal to T G, and U T to T S. Thus, if the sides of the opposite planes of a cube are cut in half, and planes are produced through the pieces, then the common section of the (latter) planes and the diameter of the cube cut one another in half. (Which is)

469

ΣΤΟΙΧΕΙΩΝ ια΄.

ELEMENTS BOOK 11 the very thing it was required to show.

λθ΄.

Proposition 39

'Ε¦ν Ï δύο πρίσµατα „σοϋψÁ, κሠτÕ µν œχÍ βάσιν If there are two equal height prisms, and one has a παραλληλόγραµµον, τÕ δ τρίγωνον, διπλάσιον δ Ï parallelogram, and the other a triangle, (as a) base, and τÕ παραλληλόγραµµον τοà τριγώνου, ‡σα œσται τ¦ the parallelogram is double the triangle, then the prisms πρίσµατα. will be equal.

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”Εστω δύο πρίσµατα „σοϋψÁ τ¦ ΑΒΓ∆ΕΖ, ΗΘΚΛΜΝ, κሠτÕ µν ™χέτω βάσιν τÕ ΑΖ παραλληλόγραµµον, τÕ δ τÕ ΗΘΚ τρίγωνον, διπλάσιον δ œστω τÕ ΑΖ παραλληλόγραµµον τοà ΗΘΚ τριγώνου· λέγω, Óτι ‡σον ™στˆ τÕ ΑΒΓ∆ΕΖ πρίσµα τù ΗΘΚΛΜΝ πρίσµατι. Συµπεπληρώσθω γ¦ρ τ¦ ΑΞ, ΗΟ στερεά. ™πεˆ διπλάσιόν ™στι τÕ ΑΖ παραλληλόγραµµον τοà ΗΘΚ τριγώνου, œστι δ κሠτÕ ΘΚ παραλληλόγραµµον διπλάσιον τοà ΗΘΚ τριγώνου, ‡σον ¥ρα ™στˆ τÕ ΑΖ παραλληλόγραµµον τù ΘΚ παραλληλογράµµJ. τ¦ δ ™πˆ ‡σων βάσεων Ôντα στερε¦ παραλληλεπίπεδα κሠØπÕ τÕ αÙτÕ Ûψος ‡σα ¢λλήλοις ™στˆν· ‡σον ¥ρα ™στˆ τÕ ΑΞ στερεÕν τù ΗΟ στερεù. καί ™στι τοà µν ΑΞ στερεοà ¼µισυ τÕ ΑΒΓ∆ΕΖ πρίσµα, τοà δ ΗΟ στερεοà ¼µισυ τÕ ΗΘΚΛΜΝ πρίσµα· ‡σον ¥ρα ™στˆ τÕ ΑΒΓ∆ΕΖ πρίσµα τù ΗΘΚΛΜΝ πρίσµατι. 'Ε¦ν ¥ρα Ï δύο πρίσµατα „σοϋψÁ, κሠτÕ µν œχÍ βάσιν παραλληλόγραµµον, τÕ δ τρίγωνον, διπάσιον δ Ï τÕ παραλληλόγραµµον τοà τριγώνου, ‡σα œστˆ τ¦ πρίσµατα· Óπερ œδει δε‹ξαι.

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Let ABCDEF and GHKLM N be two equal height prisms, and let the former have the parallelogram AF , and the latter the triangle GHK, as a base. And let parallelogram AF be twice triangle GHK. I say that prism ABCDEF is equal to prism GHKLM N . For let the solids AO and GP have been completed. Since parallelogram AF is double triangle GHK, and parallelogram HK is also double triangle GHK [Prop. 1.34], parallelogram AF is thus equal to parallelogram HK. And parallelepiped solids which are on equal bases, and (have) the same height, are equal to one another [Prop. 11.31]. Thus, solid AO is equal to solid GP . And prism ABCDEF is half of solid AO, and prism GHKLM N half of solid GP [Prop. 11.28]. Prism ABCDEF is thus equal to prism GHKLM N . Thus, if there are two equal height prisms, and one has a parallelogram, and the other a triangle, (as a) base, and the parallelogram is double the triangle, then the prisms are equal. (Which is) the very thing it was required to show.

470

ELEMENTS BOOK 12 Proportional stereometry†

† The novel feature of this book is the use of the so-called method of exhaustion (see Prop. 10.1), a precursor to integration which is generally attributed to Eudoxus of Cnidus.

471

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

α΄.

Proposition 1

Τ¦ ™ν το‹ς κύκλοις Óµοια πολύγωνα πρÕς ¥λληλά Similar polygons (inscribed) in circles are to one an™στιν æς τ¦ ¢πÕ τîν διαµέτρων τετράγωνα. other as the squares on the diameters (of the circles). A A

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”Εστωσαν κύκλοι οƒ ΑΒΓ, ΖΗΘ, κሠ™ν αÙτο‹ς Óµοια πολύγωνα œστω τ¦ ΑΒΓ∆Ε, ΖΗΘΚΛ, διάµετροι δ τîν κύκλων œστωσαν ΒΜ, ΗΝ· λέγω, Óτι ™στˆν æς τÕ ¢πÕ τÁς ΒΜ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΗΝ τετράγωνον, οÛτως τÕ ΑΒΓ∆Ε πολύγωνον πρÕς τÕ ΖΗΘΚΛ πολύγωνον. 'Επεζεύχθωσαν γ¦ρ αƒ ΒΕ, ΑΜ, ΗΛ, ΖΝ. κሠ™πεˆ Óµοιον τÕ ΑΒΓ∆Ε πολύγωνον τù ΖΗΘΚΛ πλουγώνJ, ‡ση ™στˆ κሠ¹ ØπÕ ΒΑΕ γωνία τÍ ØπÕ ΗΖΛ, καί ™στιν æς ¹ ΒΑ πρÕς τ¾ν ΑΕ, οÛτως ¹ ΗΖ πρÕς τ¾ν ΖΛ. δύο δ¾ τρίγωνά ™στι τ¦ ΒΑΕ, ΗΖΛ µίαν γωνίαν µι´ γωνίv ‡σην œχοντα τ¾ν ØπÕ ΒΑΕ τÍ ØπÕ ΗΖΛ, περˆ δ τ¦ς ‡σας γωνίας τ¦ς πλευρ¦ς ¢νάλογον· „σογώνιον ¥ρα ™στˆ τÕ ΑΒΕ τρίγωνον τù ΖΗΛ τριγώνJ. ‡ση ¥ρα ™στˆν ¹ ØπÕ ΑΕΒ γωνία τÍ ØπÕ ΖΛΗ. ¢λλ' ¹ µν ØπÕ ΑΕΒ τÍ ØπÕ ΑΜΒ ™στιν ‡ση· ™πˆ γ¦ρ τÁς αÙτÁς περιφερείας βεβήκασιν· ¹ δ ØπÕ ΖΛΗ τÍ ØπÕ ΖΝΗ· κሠ¹ ØπÕ ΑΜΒ ¥ρα τÍ ØπÕ ΖΝΗ ™στιν ‡ση. œστι δ κሠÑρθ¾ ¹ ØπÕ ΒΑΜ ÑρθÍ τÍ ØπÕ ΗΖΝ ‡ση· κሠ¹ λοιπ¾ ¥ρα τÍ λοιπÍ ™στιν ‡ση. „σογώνιον ¥ρα ™στˆ τÕ ΑΒΜ τρίγωνον τù ΖΗΝ τρίγωνJ. ¢νάλογον ¥ρα ™στˆν æς ¹ ΒΜ πρÕς τ¾ν ΗΝ, οÛτως ¹ ΒΑ πρÕς τ¾ν ΗΖ. ¢λλ¦ τοà µν τÁς ΒΜ πρÕς τ¾ν ΗΝ λόγον διπλασίων ™στˆν Ð τοà ¢πÕ τÁς ΒΜ τετραγώνου πρÕς τÕ ¢πÕ τÁς ΗΝ τετράγωνον, τοà δ τÁς ΒΑ πρÕς τ¾ν ΗΖ διπλασίων ™στˆν Ð τοà ΑΒΓ∆Ε πολυγώνου πρÕς τÕ ΖΗΘΚΛ πολύγωνον· κሠæς ¥ρα τÕ ¡πÕ τÁς ΒΜ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΗΝ τετράγωνον, οÛτως τÕ ΑΒΓ∆Ε πολύγωνον πρÕς τÕ ΖΗΘΚΛ πολύγωνον. Τ¦ ¥ρα ™ν το‹ς κύκλοις Óµοια πολύγωνα πρÕς ¥λληλά ™στιν æς τ¦ ¢πÕ τîν διαµέτρων τετράγωνα· Óπερ œδει δε‹ξαι.

H K D Let ABC and F GH be circles, and let ABCDE and F GHKL be similar polygons (inscribed) in them (respectively), and let BM and GN be the diameters of the circles (respectively). I say that as the square on BM is to the square on GN , so polygon ABCDE (is) to polygon F GHKL. For let BE, AM , GL, and F N have been joined. And since polygon ABCDE (is) similar to polygon F GHKL, angle BAE is also equal to (angle) GF L, and as BA is to AE, so GF (is) to F L [Def. 6.1]. So, BAE and GF L are two triangles having one angle equal to one angle, (namely), BAE (equal) to GF L, and the sides around the equal angles proportional. Triangle ABE is thus equiangular with triangle F GL [Prop. 6.6]. Thus, angle AEB is equal to (angle) F LG. But, AEB is equal to AM B, and F LG to F N G, for they stand on the same circumference [Prop. 3.27]. Thus, AM B is also equal to F N G. And the right-angle BAM is also equal to the right-angle GF N [Prop. 3.31]. Thus, the remaining (angle) is also equal to the remaining (angle) [Prop. 1.32]. Thus, triangle ABM is equiangular with triangle F GN . Thus, proportionally, as BM is to GN , so BA (is) to GF [Prop. 6.4]. But, the (ratio) of the square on BM to the square on GN is the square of the ratio of BM to GN , and the (ratio) of polygon ABCDE to polygon F GHKL is the square of the (ratio) of BA to GF [Prop. 6.20]. And, thus, as the square on BM (is) to the square on GN , so polygon ABCDE (is) to polygon F GHKL. Thus, similar polygons (inscribed) in circles are to one another as the squares on the diameters (of the circles). (Which is) the very thing it was required to show.

β΄.

Proposition 2

Οƒ κύκλοι πρÕς ¢λλήλους ε„σˆν æς τ¦ ¢πÕ τîν Circles are to one another as the squares on (their) διαµέτρων τετράγωνα. diameters.

472

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

”Εστωσαν κύκλοι οƒ ΑΒΓ∆, ΕΖΗΘ, διάµετροι δ Let ABCD and EF GH be circles, and [let] BD and αÙτîν [œστωσαν] αƒ Β∆, ΖΘ· λέγω, Óτι ™στˆν æς Ð ΑΒΓ∆ F H [be] their diameters. I say that as circle ABCD is to κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως τÕ ¢πÕ τÁς Β∆ circle EF GH, so the square on BD (is) to the square on τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΘ τετράγωνον. F H.

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Ε„ γ¦ρ µή ™στιν æς Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ, οÛτως τÕ ¢πÕ τÁς Β∆ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΘ, œσται æς τÕ ¢πÕ τÁς Β∆ πρÕς τÕ ¢πÕ τÁς ΖΘ, οÛτως Ð ΑΒΓ∆ κύκλος ½τοι πρÕς œλασσόν τι τοà ΕΖΗΘ κύκλου χωρίον À πρÕς µε‹ζον. œστω πρότερον πρÕς œλασσον τÕ Σ. και ™γγεγράφθω ε„ς τÕν ΕΖΗΘ κύκλον τετράγωνον τÕ ΕΖΗΘ. τÕ δ¾ ™γγεγραµµένον τετράγωνον µε‹ζόν ™στιν À τÕ ¼µισυ τοà ΕΖΗΘ κύκλου, ™πειδήπερ ™¦ν δι¦ τîν Ε, Ζ, Η, Θ σηµείων ™φαπτοµένας [εÙθείας] τοà κύκλου ¢γάγωµεν, τοà περιγραφοµένου περˆ τÕν κύκλον τετραγώνου ¼µισύ ™στι τÕ ΕΖΗΘ τετράγωνον, τοà δ περιγραφέντος τετραγώνου ™λάττων ™στˆν Ð κύκλος· éστε τÕ ΕΖΗΘ ™γγεγραµµένον τετράγωνον µε‹ζόν ™στι τοà ¹µίσεως τοà ΕΖΗΘ κύκλου. τετµήσθωσαν δίχα αƒ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι κατ¦ τ¦ Κ, Λ, Μ, Ν σηµε‹α, κሠ™πεζεύχθωσαν αƒ ΕΚ, ΚΖ, ΖΛ, ΛΗ, ΗΜ, ΜΘ, ΘΝ, ΝΕ· κሠ›καστον ¥ρα τîν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων µε‹ζόν ™στιν À τÕ ¼µισυ τοà καθ' ˜αυτÕ τµήµατος τοà κύκλου, ™πειδήπερ ™¦ν δι¦ τîν Κ, Λ, Μ, Ν σηµείων ™φαπτοµένας τοà κύκλου ¢γάγωµεν κሠ¢ναπληρώσωµεν τ¦ ™πˆ τîν ΕΖ, ΖΗ, ΗΘ, ΘΕ εÙθειîν παραλληλόγραµµα, ›καστον τîν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων ¼µισυ œσται τοà καθ' ˜αυτÕ παραλληλογράµµου, ¢λλ¦ τÕ καθ' ˜αυτÕ τµÁµα œλαττόν ™στι τοà παραλληλογράµµου· éστε ›καστον τîν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων µε‹ζόν ™στι τοà ¹µίσεως τοà καθ' ˜αυτÕ τµήµατος τοà κύκλου. τέµνοντες δ¾ τ¦ς Øπολει-

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For if the circle ABCD is not to the (circle) EF GH, as the square on BD (is) to the (square) on F H, then as the (square) on BD (is) to the (square) on F H, so circle ABCD will be to some area either less than, or greater than, circle EF GH. Let it, first of all, be (in that ratio) to (some) lesser (area), S. And let the square EF GH have been inscribed in circle EF GH [Prop. 4.6]. So the inscribed square is greater than half of circle EF GH, inasmuch as if we draw tangents to the circle through the points E, F , G, and H, then square EF GH is half of the square circumscribed about the circle [Prop. 1.47], and the circle is less than the circumscribed square. Hence, the inscribed square EF GH is greater than half of circle EF GH. Let the circumferences EF , F G, GH, and HE have been cut in half at points K, L, M , and N (respectively), and let EK, KF , F L, LG, GM , M H, HN , and N E have been joined. And, thus, each of the triangles EKF , F LG, GM H, and HN E is greater than half of the segment of the circle about it, inasmuch as if we draw tangents to the circle through points K, L, M , and N , and complete the parallelograms on the straight-lines EF , F G, GH, and HE, then each of the triangles EKF , F LG, GM H, and HN E will be half of the parallelogram about it, but the segment about it is less than the parallelogram. Hence, each of the triangles EKF , F LG, GM H, and HN E is greater than half of the segment of the circle about it. So, by cutting the circumferences remaining behind in half, and joining

473

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

ποµένας περιφερείας δίχα κሠ™πιζευγνύντες εÙθείας κሠτοàτο ¢εˆ ποιοàντες καταλείψοµέν τινα ¢ποτµήµατα τοà κύκλου, § œσται ™λάσσοντα τÁς ØπεροχÁς, Î Øπερέχει Ð ΕΖΗΘ κύκλος τοà Σ χωρίου. ™δείχθη γ¦ρ ™ν τù πρώτJ θεωρήµατι τοà δεκάτου βιβλίου, Óτι δύο µεγεθîν ¢νίσων ™κκειµένων, ™¦ν ¢πÕ τοà µείζονες ¢φαιρεθÍ µε‹ζον À τÕ ¼µισυ κሠτοà καταλειποµένου µε‹ζον À τÕ ¼µισυ, κሠτοàτο ¢εˆ γίγνηται, λειφθήσεταί τι µέγεθος, Ö œσται œλασσον τοà ™κκειµένου ™λάσσονος µεγέθους. λελείφθω οâν, κሠœστω τ¦ ™πˆ τîν ΕΚ, ΚΖ, ΖΛ, ΛΗ, ΗΜ, ΜΘ, ΘΝ, ΝΕ τµήµατα τοà ΕΖΗΘ κύκλου ™λάττονα τÁς ØπεροχÁς, Î Øπερέχει Ð ΕΖΗΘ κύκλος τοà Σ χωρίου. λοιπÕν ¥ρα τÕ ΕΚΖΛΗΜΘΝ πολύγωνον µε‹ζόν ™στι τοà Σ χωρίου. ™γγεγράφθω κሠε„ς τÕν ΑΒΓ∆ κύκλον τù ΕΚΖΛΗΜΘΝ πολυγώνJ Óµοιον πολύγωνον τÕ ΑΞΒΟΓΠ∆Ρ· œστιν ¥ρα æς τÕ ¢πÕ τÁς Β∆ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΘ τετράγωνον, οÛτως τÕ ΑΞΒΟΓΠ∆Ρ πολύγωνον πρÕς τÕ ΕΚΖΛΗΜΘΝ πολύγωνον. ¢λλ¦ κሠæς τÕ ¢πÕ τÁς Β∆ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΘ, οÛτως Ð ΑΒΓ∆ κύκλος πρÕς τÕ Σ χωρίον· κሠæς ¥ρα Ð ΑΒΓ∆ κύκλος πρÕς τÕ Σ χωρίον, οÛτως τÕ ΑΞΒΟΓΠ∆Ρ πολύγωνον πρÕς τÕ ΕΚΖΛΗΜΘΝ πολύγωνον· ™ναλλ¦ξ ¥ρα æς Ð ΑΒΓ∆ κύκλος πρÕς τÕ ™ν αÙτù πολύγωνον, οÛτως τÕ Σ χωρίον πρÕς τÕ ΕΚΖΛΗΜΘΝ πολύγωνον. µείζων δ Ð ΑΒΓ∆ κύκλος τοà ™ν αÙτù πολυγώνου· µείζων ¥ρα κሠτÕ Σ χωρίον τοà ΕΚΖΛΗΜΘΝ πολυγώνου. ¢λλ¦ κሠœλαττον· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ™στˆν æς τÕ ¢πÕ τÁς Β∆ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΘ, οÛτως Ð ΑΒΓ∆ κύκλος πρÕς œλασσόν τι τοà ΕΖΗΘ κύκλου χωρίον. еοίως δ¾ δείξοµεν, Óτι οÙδ æς τÕ ¢πÕ ΖΘ πρÕς τÕ ¢πÕ Β∆, οÛτως Ð ΕΖΗΘ κύκλος πρÕς œλασσόν τι τοà ΑΒΓ∆ κύκλου χωρίον. Λέγω δή, Óτι οÙδ æς τÕ ¢πÕ τÁς Β∆ πρÕς τÕ ¢πÕ τÁς ΖΘ, οÛτως Ð ΑΒΓ∆ κύκλος πρÕς µε‹ζόν τι τοà ΕΖΗΘ κύκλου χωρίον. Ε„ γ¦ρ δυνατόν, œστω πρÕς µε‹ζον τÕ Σ. ¢νάπαλιν ¥ρα [™στˆν] æς τÕ ¢πÕ τÁς ΖΘ τετράγωνον πρÕς τÕ ¢πÕ τÁς ∆Β, οÛτως τÕ Σ χωρίον πρÕς τÕν ΑΒΓ∆ κύκλον. ¢λλ' æς τÕ Σ χωρίον πρÕς τÕν ΑΒΓ∆ κύκλον, οÛτως Ð ΕΖΗΘ κύκλος πρÕς œλαττόν τι τοà ΑΒΓ∆ κύκλου χωρίον· κሠæς ¥ρα τÕ ¢πÕ τÁς ΖΘ πρÕς τÕ ¢πÕ τÁς Β∆, οÛτως Ð ΕΖΗΘ κύκλος πρÕς œλασσόν τι τοà ΑΒΓ∆ κύκλου χωρίον· Óπερ ¢δύνατον ™δείχθη. οÙκ ¥ρα ™στˆν æς τÕ ¢πÕ τÁς Β∆ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΘ, οÛτως Ð ΑΒΓ∆ κύκλος πρÕς µε‹ζόν τι τοà ΕΖΗΘ κύκλου χωρίον. ™δείχθη δέ, Óτι οÙδ πρÕς œλασσον· œστιν ¥ρα æς τÕ ¢πÕ τÁς Β∆ τετράγωνον πρÕς τÕ ¢πÕ τÁς ΖΘ, οÛτως Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον. Οƒ ¥ρα κύκλοι πρÕς ¢λλήλους ε„σˆν æς τ¦ ¢πÕ τîν

straight-lines, and doing this continually, we will (eventually) leave behind some segments of the circle whose (sum) will be less than the excess by which circle EF GH exceeds the area S. For we showed in the first theorem of the tenth book that if two unequal magnitudes are laid out, and if (a part) greater than a half is subtracted from the greater, and (if from) the remainder (a part) greater than a half (is subtracted), and this happens continually, then some magnitude will (eventually) be left which will be less than the lesser laid out magnitude [Prop. 10.1]. Therefore, let the (segments) have been left, and let the (sum of the) segments of the circle EF GH on EK, KF , F L, LG, GM , M H, HN , and N E be less than the excess by which circle EF GH exceeds area S. Thus, the remaining polygon EKF LGM HN is greater than area S. And let the polygon AOBP CQDR, similar to the polygon EKF LGM HN , have been inscribed in circle ABCD. Thus, as the square on BD is to the square on F H, so polygon AOBP CQDR (is) to polygon EKF LGM HN [Prop. 12.1]. But, also, as the square on BD (is) to the square on F H, so circle ABCD (is) to area S. And, thus, as circle ABCD (is) to area S, so polygon AOBP GQDR (is) to polygon EKF LGM HN [Prop. 5.11]. Thus, alternately, as circle ABCD (is) to the polygon (inscribed) within it, so area S (is) to polygon EKF LGM HN [Prop. 5.16]. And circle ABCD (is) greater than the polygon (inscribed) within it. Thus, area S is also greater than polygon EKF LGM HN . But, (it is) also less. The very thing is impossible. Thus, the square on BD is not to the (square) on F H, as circle ABCD (is) to some area less than circle EF GH. So, similarly, we can show that the (square) on F H (is) not to the (square) on BD as circle EF GH (is) to some area less than circle ABCD either. So, I say that neither (is) the (square) on BD to the (square) on F H, as circle ABCD (is) to some area greater than circle EF GH. For, if possible, let it be (in that ratio) to (some) greater (area), S. Thus, inversely, as the square on F H [is] to the (square) on DB, so area S (is) to circle ABCD [Prop. 5.7 corr.]. But, as area S (is) to circle ABCD, so circle EF GH (is) to some area less than circle ABCD (see lemma). And, thus, as the (square) on F H (is) to the (square) on BD, so circle EF GH (is) to some area less than circle ABCD [Prop. 5.11]. The very thing was shown (to be) impossible. Thus, as the square on BD is to the (square) on F H, so circle ABCD (is) not to some area greater than circle EF GH. And it was shown that neither (is it in that ratio) to (some) lesser (area). Thus, as the square on BD is to the (square) on F H, so circle ABCD (is) to circle EF GH.

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ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

διαµέτρων τετράγωνα· Óπερ œδει δε‹ξαι.

Thus, circles are to one another as the squares on (their) diameters. (Which is) the very thing it was required to show.

ΛÁµµα.

Lemma

Λέγω δή, Óτι τοà Σ χωρίου µείζονος Ôντος τοà ΕΖΗΘ κύκλου ™στˆν æς τÕ Σ χωρίον πρÕς τÕν ΑΒΓ∆ κύκλον, οÛτως Ð ΕΖΗΘ κύκλος πρÕς œλαττόν τι τοà ΑΒΓ∆ κύκλου χωρίον. Γεγονέτω γ¦ρ æς τÕ Σ χωρίον πρÕς τÕν ΑΒΓ∆ κύκλον, οÛτως Ð ΕΖΗΘ κύκλος πρÕς τÕ Τ χωρίον. λέγω, Óτι œλαττόν ™στι τÕ Τ χωρίον τοà ΑΒΓ∆ κύκλου. ™πεˆ γάρ ™στιν æς τÕ Σ χωρίον πρÕς τÕν ΑΒΓ∆ κύκλον, οÛτως Ð ΕΖΗΘ κύκλος πρÕς τÕ Τ χωρίον, ™ναλλάξ ™στιν æς τÕ Σ χωρίον πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως Ð ΑΒΓ∆ κύκλος πρÕς τÕ Τ χωρίον. µε‹ζον δ τÕ Σ χωρίον τοà ΕΖΗΘ κύκλου· µείζων ¥ρα καˆ Ð ΑΒΓ∆ κύκλος τοà Τ χωρίου. éστε ™στˆν æς τÕ Σ χωρίον πρÕς τÕν ΑΒΓ∆ κύκλον, οÛτως Ð ΕΖΗΘ κύκλος πρÕς œλαττόν τι τοà ΑΒΓ∆ κύκλου χωρίον· Óπερ œδει δε‹ξαι.

So, I say that, area S being greater than circle EF GH, as area S is to circle ABCD, so circle EF GH (is) to some area less than circle ABCD. For let it have been contrived that as area S (is) to circle ABCD, so circle EF GH (is) to area T . I say that area T is less than circle ABCD. For since as area S is to circle ABCD, so circle EF GH (is) to area T , alternately, as area S is to circle EF GH, so circle ABCD (is) to area T [Prop. 5.16]. And area S (is) greater than circle EF GH. Thus, circle ABCD (is) also greater than area T [Prop. 5.14]. Hence, as area S is to circle ABCD, so circle EF GH (is) to some area less than circle ABCD. (Which is) the very thing it was required to show.

γ΄.

Proposition 3

Π©σα πυρᵈς τρίγωνον œχουσα βάσιν διαιρε‹ται ε„ς δύο πυραµίδας ‡σας τε καˆ Ðµοίας ¢λλήλαις κሠ[еοίας] τÍ ÓλÍ τριγώνους ™χουσας βάσεις κሠε„ς δύο τρίσµατα ‡σα· κሠτ¦ δύο πρίσµατα µείζονά ™στιν À τÕ ¼µισυ τÁς Óλης πυραµίδος.

Any pyramid having a triangular base is divided into two pyramids having triangular bases (which are) equal, similar to one another, and [similar] to the whole, and into two equal prisms. And the (sum of the) two prisms is greater than half of the whole pyramid.

D

J

L G

H

A

D

E

L

K

H

K C

Z

G

B

A

”Εστω πυραµίς, Âς βάσις µέν ™στι τÕ ΑΒΓ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον· λέγω, Óτι ¹ ΑΒΓ∆ πυρᵈς διαιρε‹ται ε„ς δύο πυραµίδας ‡σας ¢λλήλαις τριγώνους βάσεις ™χούσας καˆ Ðµοίας τÍ ÓλÍ κሠε„ς δύο πρίσµατα ‡σα· κሠτ¦ δύο πρίσµατα µείζονά ™στιν À τÕ ¼µισυ τÁς Óλης πυραµίδος. Τετµήσθωσαν γ¦ρ αƒ ΑΒ, ΒΓ, ΓΑ, Α∆, ∆Β, ∆Γ δίχα κατ¦ τ¦ Ε, Ζ, Η, Θ, Κ, Λ σηµε‹α, κሠ™πεζεύχθωσαν αƒ ΘΕ, ΕΗ, ΗΘ, ΘΚ, ΚΛ, ΛΘ, ΚΖ, ΖΗ. ™πεˆ ‡ση

F E

B

Let there be a pyramid whose base is triangle ABC, and (whose) apex (is) point D. I say that pyramid ABCD is divided into two pyramids having triangular bases (which are) equal to one another, and similar to the whole, and into two equal prisms. And the (sum of the) two prisms is greater than half of the whole pyramid. For let AB, BC, CA, AD, DB, and DC have been cut in half at points E, F , G, H, K, and L (respectively). And let HE, EG, GH, HK, KL, LH, KF , and F G have

475

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

™στˆν ¹ µν ΑΕ τÍ ΕΒ, ¹ δ ΑΘ τÍ ∆Θ, παράλληλος ¥ρα ™στˆν ¹ ΕΘ τÍ ∆Β. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΘΚ τÍ ΑΒ παράλληλός ™στιν. παραλληλόγραµµον ¥ρα ™στˆ τÕ ΘΕΒΚ· ‡ση ¥ρα ™στˆν ¹ ΘΚ τÍ ΕΒ. ¢λλ¦ ¹ ΕΒ τÍ ΕΑ ™στιν ‡ση· κሠ¹ ΑΕ ¥ρα τÍ ΘΚ ™στιν ‡ση. œστι δ κሠ¹ ΑΘ τÍ Θ∆ ‡ση· δύο δ¾ αƒ ΕΑ, ΑΘ δυσˆ τα‹ς ΚΘ, Θ∆ ‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΕΑΘ γωνίv τÍ ØπÕ ΚΘ∆ ‡ση· βάσις ¥ρα ¹ ΕΘ βάσει τÍ Κ∆ ™στιν ‡ση. ‡σον ¥ρα καˆ Óµοιόν ™στι τÕ ΑΕΘ τρίγωνον τù ΘΚ∆ τριγώνJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ΑΘΗ τρίγωνον τù ΘΛ∆ τριγώνJ ‡σον τέ ™στι καˆ Óµοιον. κሠ™πεˆ δύο εÙθε‹αι ¡πτόµεναι ¢λλήλων αƒ ΕΘ, ΘΗ παρ¦ δύο εÙθείας ¡πτοµένας ¢λλήλων τ¦ς Κ∆, ∆Λ ε„σιν οÙκ ™ν τù αÙτù ™πιπέδJ οâσαι, ‡σας γωνίας περιέξουσιν. ‡ση ¥ρα ™στˆν ¹ ØπÕ ΕΘΗ γωνία τÍ ØπÕ Κ∆Λ γωνίv. κሠ™πεˆ δύο εÙθε‹αι αƒ ΕΘ, ΘΗ δυσˆ τα‹ς Κ∆, ∆Λ ‡σαι ε„σˆν ˜κατέρα εκατέρv, κሠγωνία ¹ ØπÕ ΕΘΗ γωνίv τÍ ØπÕ Κ∆Λ ™στιν ‡ση, βάσις ¥ρα ¹ ΕΗ βάσει τÍ ΚΛ [™στιν] ‡ση· ‡σον ¥ρα καˆ Óµοιόν ™στι τÕ ΕΘΗ τρίγωνον τù Κ∆Λ τριγώνJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ΑΕΗ τρίγωνον τù ΘΚΛ τριγώνJ ‡σον τε καˆ Óµοιόν ™στιν. ¹ ¥ρα πυραµίς, Âς βάσις µέν ™στι τÕ ΑΕΗ τρίγωνον, κορυφ¾ δ τÕ Θ σηµε‹ον, ‡ση καˆ Ðµοία ™στˆ πυραµίδι, Âς βάσις µέν ™στι τÕ ΘΚΛ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον. κሠ™πεˆ τριγώνου τοà Α∆Β παρ¦ µίαν τîν πλευρîν τ¾ν ΑΒ Ãκται ¹ ΘΚ, „σογώνιόν ™στι τÕ Α∆Β τρίγωνον τù ∆ΘΚ τριγώνJ, κሠτ¦ς πλευρ¦ς ¢νάλογον œχουσιν· Óµοιον ¥ρα ™στˆ τÕ Α∆Β τρίγωνον τù ∆ΘΚ τριγώνJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ µν ∆ΒΓ τρίγωνον τù ∆ΚΛ τριγώνJ Óµοιόν ™στιν, τÕ δ Α∆Γ τù ∆ΛΘ. κሠ™πεˆ δύο εÙθε‹αι ¡πτόµεναι ¢λλήλων αƒ ΒΑ, ΑΓ παρ¦ δύο εÙθείας ¡πτοµένας ¢λλήλων τ¦ς ΚΘ, ΘΛ ε„σιν οÙκ ™ν τù αÙτù ™πιπέδJ, ‡σας γωνίας περιέξουσιν. ‡ση ¥ρα ™στˆν ¹ ØπÕ ΒΑΓ γωνία τÍ ØπÕ ΚΘΛ. καί ™στιν æς ¹ ΒΑ πρÕς τ¾ν ΑΓ, οÛτως ¹ ΚΘ πρÕς τ¾ν ΘΛ· Óµοιον ¥ρα ™στˆ τÕ ΑΒΓ τρίγωνον τù ΘΚΛ τριγώνJ. κሠπυρᵈς ¥ρα, Âς βάσις µέν ™στι τÕ ΑΒΓ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον, еοία ™στˆ πυραµίδι, Âς βάσις µέν ™στι τÕ ΘΚΛ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον. ¢λλ¦ πυραµίς, Âς βάσις µέν [™στι] τÕ ΘΚΛ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον, еοία ™δείχθη πυραµίδι, Âς βάσις µέν ™στι τÕ ΑΕΗ τρίγωνον, κορυφ¾ δ τÕ Θ σηµε‹ον. ˜κατέρα ¥ρα τîν ΑΕΗΘ, ΘΚΛ∆ πυραµίδων еοία ™στˆ τÍ ÓλV τÍ ΑΒΓ∆ πυραµίδι. Κሠ™πεˆ ‡ση ™στˆν ¹ ΒΖ τÍ ΖΓ, διπλάσιόν ™στι τÕ ΕΒΖΗ παραλληλόγραµµον τοà ΗΖΓ τριγώνου. κሠ™πεˆ, ™¦ν Ï δύο πρίσµατα „σοϋψÁ, κሠτÕ µν œχV βάσιν παραλληλόγραµµον, τÕ δ τρίγωνον, διπλάσιον δ Ï τÕ παραλληλόγραµµον τοà τριγώνου, ‡σα ™στˆ τ¦ πρίσµατα, ‡σον ¥ρα ™στˆ τÕ πρίσµα τÕ περιεχόµενον ØπÕ δύο µν τριγώνων τîν ΒΚΖ, ΕΘΗ, τριîν δ παραλλη-

been joined. Since AE is equal to EB, and AH to DH, EH is thus parallel to DB [Prop. 6.2]. So, for the same (reasons), HK is also parallel to AB. Thus, HEBK is a parallelogram. Thus, HK is equal to EB [Prop. 1.34]. But, EB is equal to EA. Thus, AE is also equal to HK. And AH is also equal to HD. So the two (straight-lines) EA and AH are equal to the two (straight-lines) KH and HD, respectively. And angle EAH (is) equal to angle KHD [Prop. 1.29]. Thus, base EH is equal to base KD [Prop. 1.4]. Thus, triangle AEH is equal and similar to triangle HKD [Prop. 1.4]. So, for the same (reasons), triangle AHG is also equal and similar to triangle HLD. And since EH and HG are two straight-lines joining one another (which are respectively) parallel to two straight-lines joining one another, KD and DL, not being in the same plane, they will contain equal angles [Prop. 11.10]. Thus, angle EHG is equal to angle KDL. And since the two straight-lines EH and HG are equal to the two straight-lines KD and DL, respectively, and angle EHG is equal to angle KDL, base EG [is] thus equal to base KL [Prop. 1.4]. Thus, triangle EHG is equal and similar to triangle KDL. So, for the same (reasons), triangle AEG is also equal and similar to triangle HKL. Thus, the pyramid whose base is triangle AEG, and apex the point H, is equal and similar to the pyramid whose base is triangle HKL, and apex the point D [Def. 11.10]. And since HK has been drawn parallel to one of the sides, AB, of triangle ADB, triangle ADB is equiangular to triangle DHK [Prop. 1.29], and they have proportional sides. Thus, triangle ADB is similar to triangle DHK [Def. 6.1]. So, for the same (reasons), triangle DBC is also similar to triangle DKL, and ADC to DLH. And since two straight-lines joining one another, BA and AC, are parallel to two straight-lines joining one another, KH and HL, not in the same plane, they will contain equal angles [Prop. 11.10]. Thus, angle BAC is equal to (angle) KHL. And as BA is to AC, so KH (is) to HL. Thus, triangle ABC is similar to triangle HKL [Prop. 6.6]. And, thus, the pyramid whose base is triangle ABC, and apex the point D, is similar to the pyramid whose base is triangle HKL, and apex the point D [Def. 11.9]. But, the pyramid whose base [is] triangle HKL, and apex the point D, was shown (to be) similar to the pyramid whose base is triangle AEG, and apex the point H. Thus, each of the pyramids AEGH and HKLD is similar to the whole pyramid ABCD. And since BF is equal to F C, parallelogram EBF G is double triangle GF C [Prop. 1.41]. And since, if two prisms (have) equal heights, and the former has a parallelogram as a base, and the latter a triangle, and the parallelogram (is) double the triangle, then the prisms

476

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

λογράµµων τîν ΕΒΖΗ, ΕΒΚΘ, ΘΚΖΗ τù πρισµατι τù περιεχοµένJ ØπÕ δύο µν τριγώνων τîν ΗΖΓ, ΘΚΛ, τριîν δ παραλληλογράµµων τîν ΚΖΓΛ, ΛΓΗΘ, ΘΚΖΗ. κሠφανερόν, Óτι ˜κάτρον τîν πρισµάτων, οá τε βάσις τÕ ΕΒΖΗ παραλληλόγραµµον, ¢πεναντίον δ ¹ ΘΚ εÙθε‹α, κሠοá βάσις τÕ ΗΖΓ τρίγωνον, ¢πεναντίον δ τÕ ΘΚΛ τρίγωνον, µε‹ζόν ™στιν ˜κατέρας τîν πυραµίδων, ïν βάσεις µν τ¦ ΑΕΗ, ΘΚΛ τρίγωνα, κορυφαˆ, δ τ¦ Θ, ∆ σηµε‹α, ™πειδήπερ [καί] ™¦ν ™πιζεύξωµεν τ¦ς ΕΖ, ΕΚ εÙθείας, τÕ µν πρίσµα, οá βάσις τÕ ΕΒΖΗ παραλληλόγραµµον, ¢πεναντίον δ ¹ ΘΚ εÙθε‹α, µε‹ζόν ™στι τÁς πυραµίδος, Âς βάσις τÕ ΕΒΖ τρίγωνον, κορυφ¾ δ τÕ Κ σηµε‹ον. ¢λλ' ¹ πυραµίς, Âς βάσις τÕ ΕΒΖ τρίγωνον, κορυφ¾ δ τÕ Κ σηµε‹ον, ‡ση ™στˆ πυραµίδι, Âς βάσις τÕ ΑΕΗ τρίγωνον, κορυφ¾ δ τÕ Θ σηµε‹ον· ØπÕ γ¦ρ ‡σων καˆ Ðµοίων ™πιπέδων περιέχονται. éστε κሠτÕ πρίσµα, οá βάσις µν τÕ ΕΒΖΗ παραλληλόγραµµον, ¢πεναντίον δ ¹ ΘΚ εÙθε‹α, µε‹ζόν ™στι πυραµίδος, Âς βάσις µν τÕ ΑΕΗ τρίγωνον, κορυφ¾ δ τÕ Θ σηµε‹ον. ‡σον δ τÕ µν πρίσµα, οá βάσις τÕ ΕΒΖΗ παραλληλόγραµµον, ¢πεναντίον δ ¹ ΘΚ εÙθε‹α, τù πρίσµατι, οá βάσις µν τÕ ΗΖΓ τρίγωνον, ¢πεναντίον δ τÕ ΘΚΛ τρίγωνον· ¹ δ πυραµίς, Âς βάσις τÕ ΑΕΗ τρίγωνον, κορυφ¾ δ τÕ Θ σηµε‹ον, ‡ση ™στˆ πυραµίδι, Âς βάσις τÕ ΘΚΛ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον. τ¦ ¥ρα ε„ρηµένα δύο πρίσµατα µείζονά ™στι τîν ε„ρηµένων δύο πυραµίδων, ïν βάσεις µν τ¦ ΑΕΗ, ΘΚΛ τρίγωνα, κορυφαˆ δ τ¦ Θ, ∆ σηµε‹α. `Η ¥ρα Óλη πυραµίς, Âς βάσις τÕ ΑΒΓ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον, διÇρηται ε‡ς τε δύο πυραµίδας ‡σας ¢λλήλαις [καˆ Ðµοίας τÍ ÓλV] κሠε„ς δύο πρίσµατα ‡σα, κሠτ¦ δύο πρίσµατα µείζονά ™στιν À τÕ ¼µισυ τÁς Óλης πυραµίδος· Óπερ œδει δε‹ξαι.

are equal [Prop. 11.39], the prism contained by the two triangles BKF and EHG, and the three parallelograms EBF G, EBKH, and HKF G, is thus equal to the prism contained by the two triangles GF C and HKL, and the three parallelograms KF CL, LCGH, and HKF G. And (it is) clear that each of the prisms whose base (is) parallelogram EBF G, and opposite (side) straight-line HK, and whose base (is) triangle GF C, and opposite (plane) triangle HKL, is greater than each of the pyramids whose bases are triangles AEG and HKL, and apexes the points H and D (respectively), inasmuch as, if we [also] join the straight-lines EF and EK, then the prism whose base (is) parallelogram EBF G, and opposite (side) straight-line HK, is greater than the pyramid whose base (is) triangle EBF , and apex the point K. But the pyramid whose base (is) triangle EBF , and apex the point K, is equal to the pyramid whose base is triangle AEG, and apex point H. For they are contained by equal and similar planes. And, hence, the prism whose base (is) parallelogram EBF G, and opposite (side) straightline HK, is greater than the pyramid whose base (is) triangle AEG, and apex the point H. And the prism whose base is parallelogram EBF G, and opposite (side) straight-line HK, (is) equal to the prism whose base (is) triangle GF C, and opposite (plane) triangle HKL. And the pyramid whose base (is) triangle AEG, and apex the point H, is equal to the pyramid whose base (is) triangle HKL, and apex the point D. Thus, the (sum of the) aforementioned two prisms is greater than the (sum of the) aforementioned two pyramids, whose bases (are) triangles AEG and HKL, and apexes the points H and D (respectively). Thus, the whole pyramid, whose base (is) triangle ABC, and apex the point D, has been divided into two pyramids (which are) equal to one another [and similar to the whole], and into two equal prisms. And the (sum of the) two prisms is greater than half of the whole pyramid. (Which is) the very thing it was required to show.

δ΄.

Proposition 4

'Ε¦ν ðσι δύο πυραµίδες ØπÕ τÕ αÙτÕ Ûψος τριγώνους œχουσαι βάσεις, διαιρεθÍ δ ˜κατέρα αÙτîν ε‡ς τε δύο πυραµίδας ‡σας ¢λλήλαις καˆ Ðµοίας τÍ ÓλV κሠε„ς δύο πρίσµατα ‡σα, œσται æς ¹ τÁς µι©ς πυραµίδος βάσις πρÕς τ¾ν τÁς ˜τέρας πυραµίδος βάσιν, οÛτως τ¦ ™ν τÍ µι´ πυραµίδι πρίσµατα πάντα πρÕς τ¦ ™ν τÍ ˜τάρv πυραµίδι πρίσµατα πάντα „σοπληθÍ. ”Εστωσαν δύο πυραµίδες ØπÕ τÕ αÙτÕ Ûψος τριγώνους œχουσαι βάσεις τ¦ς ΑΒΓ, ∆ΕΖ, κορυφ¦ς δ τ¦ Η, Θ σηµε‹α, κሠδιVρήσθω ˜κατέρα αÙτîν ε‡ς τε δύο πυ-

If there are two pyramids with the same height, having trianglular bases, and each of them is divided into two pyramids equal to one another, and similar to the whole, and into two equal prisms, then as the base of one pyramid (is) to the base of the other pyramid, so (the sum of) all the prisms in one pyramid will be to (the sum of all) the equal number of prisms in the other pyramid. Let there be two pyramids with the same height, having the triangular bases ABC and DEF , (with) apexes the points G and H (respectively). And let each of them

477

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

ραµίδας ‡σας ¢λλήλαις καˆ Ðµοίας τÍ ÓλV κሠε„ς δύο πρίσµατα ‡σα· λέγω, Óτι ™στˆν æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τ¦ ™ν τÍ ΑΒΓΗ πυραµίδι πρίσµατα πάντα πρÕς τ¦ ™ν τÍ ∆ΕΖΘ πυραµίδι πρίσµατα „σοπληθÁ.

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have been divided into two pyramids equal to one another, and similar to the whole, and into two equal prisms [Prop. 12.3]. I say that as base ABC is to base DEF , so (the sum of) all the prisms in pyramid ABCG (is) to (the sum of) all the equal number of prisms in pyramid DEF H.

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'Επεˆ γ¦ρ ‡ση ™στˆν ¹ µν ΒΞ τÍ ΞΓ, ¹ δ ΑΛ τÍ ΛΓ, παράλληλος ¥ρα ™στˆν ¹ ΛΞ τÍ ΑΒ καˆ Óµοιον τÕ ΑΒΓ τρίγωνον τù ΛΞΓ τριγώνJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ ∆ΕΖ τρίγωνον τù ΡΦΖ τριγώνJ Óµοιόν ™στιν. κሠ™πεˆ διπλασίων ™στˆν ¹ µν ΒΓ τÁς ΓΞ, ¹ δ ΕΖ τÁς ΖΦ, œστιν ¥ρα æς ¹ ΒΓ πρÕς τ¾ν ΓΞ, οÛτως ¹ ΕΖ πρÕς τ¾ν ΖΦ. κሠ¢ναγέγραπται ¢πÕ µν τîν ΒΓ, ΓΞ Óµοιά τε καˆ Ðµοίως κείµενα εÙθύγραµµα τ¦ ΑΒΓ, ΛΞΓ, ¢πÕ δ τîν ΕΖ, ΖΦ Óµοιά τε καˆ Ðµοίως κείµενα [εÙθύγραµµα] τ¦ ∆ΕΖ, ΡΦΖ· œστιν ¥ρα æς τÕ ΑΒΓ τρίγωνον πρÕς τÕ ΛΞΓ τρίγωνον, οÛτως τÕ ∆ΕΖ τρίγωνον πρÕς τÕ ΡΦΖ τρίγωνον· ™ναλλ¦ξ ¥ρα ™στˆν æς τÕ ΑΒΓ τρίγωνον πρÕς τÕ ∆ΕΖ [τρίγωνον], οÛτως τÕ ΛΞΓ [τρίγωνον] πρÕς τÕ ΡΦΖ τρίγωνον. ¢λλ' æς τÕ ΛΞΓ τρίγωνον πρÕς τÕ ΡΦΖ τρίγωνον, οÛτως τÕ πρίσµα, οá βάσις µέν [™στι] τÕ ΛΞΓ τρίγωνον, ¢πεναντίον δ τÕ ΟΜΝ, πρÕς τÕ πρίσµα, οá βάσις µν τÕ ΡΦΖ τρίγωνον, ¢πεναντίον δ τÕ ΣΤΥ· κሠæς ¥ρα τÕ ΑΒΓ τρίγωνον πρÕς τÕ ∆ΕΖ τρίγωνον, οÛτως τÕ πρίσµα, οá βάσις µν τÕ ΛΞΓ τρίγωνον, ¢πεναντίον δ τÕ ΟΜΝ, πρÕς τÕ πρίσµα, οá βάσις µν τÕ ΡΦΖ τρίγωνον, ¢πεναντίον δ τÕ ΣΤΥ. æς δ τ¦ ε„ρηµένα πρίσµατα πρÕς ¥λληλα, οÛτως τÕ πρίσµα, οá βάσις µν τÕ ΚΒΞΛ παραλληλόγραµµον, ¢πεναντίον δ ¹ ΟΜ εÙθε‹α, πρÕς τÕ πρίσµα, οá βάσις µν τÕ ΠΕΦΡ παραλληλόγραµµον, ¢πεναντίον δ ¹ ΣΤ εÙθε‹α. κሠτ¦ δύο ¥ρα πρίσµατα, οá τε βάσις µν τÕ ΚΒΞΛ παραλληλόγραµµον, ¢πεναντίον δ ¹ ΟΜ, κሠοá βάσις µν τÕ ΛΞΓ, ¢πεναντίον δ τÕ ΟΜΝ, πρÕς τ¦ πρίσµατα, οá τε βάσις µν τÕ ΠΕΦΡ, ¢πεναντίον δ ¹ ΣΤ εÙθε‹α, κሠοá βάσις µν τÕ ΡΦΖ τρίγωνον, ¢πεναντίον δ τÕ ΣΤΥ. κሠæς ¥ρα ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τ¦ ε„ρηµένα δύο πρίσµατα πρÕς τ¦ ε„ρηµένα δύο

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For since BO is equal to OC, and AL to LC, LO is thus parallel to AB, and triangle ABC similar to triangle LOC [Prop. 12.3]. So, for the same (reasons), triangle DEF is also similar to triangle RV F . And since BC is double CO, and EF (double) F V , thus as BC (is) to CO, so EF (is) to F V . And the similar, and similarly laid out, rectilinear (figures) ABC and LOC have been described on BC and CO (respectively), and the similar, and similarly laid out, [rectilinear] (figures) DEF and RV F on EF and F V (respectively). Thus, as triangle ABC is to triangle LOC, so triangle DEF (is) to triangle RV F [Prop. 6.22]. Thus, alternately, as triangle ABC is to [triangle] DEF , so [triangle] LOC (is) to triangle RV F [Prop. 5.16]. But, as triangle LOC (is) to triangle RV F , so the prism whose base [is] triangle LOC, and opposite (plane) P M N , (is) to the prism whose base (is) triangle RV F , and opposite (plane) ST U (see lemma). And, thus, as triangle ABC (is) to triangle DEF , so the prism whose base (is) triangle LOC, and opposite (plane) P M N , (is) to the prism whose base (is) triangle RV F , and opposite (plane) ST U . And as the aforementioned prisms (are) to one another, so the prism whose base (is) parallelogram KBOL, and opposite (side) straight-line P M , (is) to the prism whose base (is) parallelogram QEV R, and opposite (side) straightline ST [Props. 11.39, 12.3]. Thus, also, (is) the (sum of the) two prisms—that whose base (is) parallelogram KBOL, and opposite (side) P M , and that whose base (is) LOC, and opposite (plane) P M N —to (the sum of) the (two) prisms—that whose base (is) QEV R, and opposite (side) straight-line ST , and that whose base (is) triangle RV F , and opposite (plane) ST U [Prop. 5.12].

478

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

πρίσµατα. Καˆ Ðµοίως, ™¦ν διαιρεθîσιν αƒ ΟΜΝΗ, ΣΤΥΘ πυραµίδες ε‡ς τε δύο πρίσµατα κሠδύο πυραµίδας, œσται æς ¹ ΟΜΝ βάσις πρÕς τ¾ν ΣΤΥ βάσιν, οÛτως τ¦ ™ν τÍ ΟΜΝΗ πυραµίδι δύο πρίσµατα πρÕς τ¦ ™ν τÍ ΣΤΥΘ πυραµίδι δύο πρίσµατα. ¢λλ' æς ¹ ΟΜΝ βάσις πρÕς τ¾ν ΣΤΥ βάσιν, οÛτως ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν· ‡σον γ¦ρ ˜κάτερον τîν ΟΜΝ, ΣΤΥ τριγώνων ˜κατέρJ τîν ΛΞΓ, ΡΦΖ. κሠæς ¥ρα ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τ¦ τέσσαρα πρίσµατα πρÕς τ¦ τέσσαρα πρίσµατα. еοίως δ κ¨ν τ¦ς Øπολειποµένας πυραµίδας διέλωµεν ε‡ς τε δύο πυραµίδας κሠε„ς δύο πρίσµατα, œσται æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τ¦ ™ν τÍ ΑΒΓΗ πυραµίδι πρίσµατα πάντα πρÕς τ¦ ™ν τÍ ∆ΕΖΘ πυραµίδι πρίσµατα πάντα „σοπληθÁ· Óπερ œδει δε‹ξαι.

And, thus, as base ABC (is) to base DEF , so the (sum of the first) aforementioned two prisms (is) to the (sum of the second) aforementioned two prisms. And, similarly, if pyramids P M N G and ST U H are divided into two prisms, and two pyramids, as base P M N (is) to base ST U , so (the sum of) the two prisms in pyramid P M N G will be to (the sum of) the two prisms in pyramid ST U H. But, as base P M N (is) to base ST U , so base ABC (is) to base DEF . For the triangles P M N and ST U (are) equal to LOC and RV F , respectively. And, thus, as base ABC (is) to base DEF , so (the sum of) the four prisms (is) to (the sum of) the four prisms [Prop. 5.12]. So, similarly, even if we divide the pyramids left behind into two pyramids and into two prisms, as base ABC (is) to base DEF , so (the sum of) all the prisms in pyramid ABCD will be to (the sum of) all the equal number of prisms in pyramid DEF H. (Which is) the very thing it was required to show.

ΛÁµµα.

Lemma

“Οτι δέ ™στιν æς τÕ ΛΞΓ τρίγωνον πρÕς τÕ ΡΦΖ τρίγωνον, οÛτως τÕ πρίσµα, οá βάσις τÕ ΛΞΓ τρίγωνον, ¢πεναντίον δ τÕ ΟΜΝ, πρÕς τÕ πρίσµα, οá βάσις µν τÕ ΡΦΖ [τρίγωνον], ¢πεναντίον δ τÕ ΣΤΥ, οÛτω δεικτέον. 'Επˆ γ¦ρ τÁς αÙτÁς καταγραφÁς νενοήσθωσαν ¢πÕ τîν Η, Θ κάθετοι ™πˆ τ¦ ΑΒΓ, ∆ΕΖ ™πίπεδα, ‡σαι δηλαδ¾ τυγχάνουσαι δι¦ τÕ „σοϋψε‹ς Øποκε‹σθαι τ¦ς πυραµίδας. κሠ™πεˆ δύο εÙθε‹αι ¼ τε ΗΓ κሠ¹ ¢πÕ τοà Η κάθετος ØπÕ παραλλήλων ™πιπέδων τîν ΑΒΓ, ΟΜΝ τέµνονται, ε„ς τοÝς αÙτοÝς λόγους τµηθήσονται. κሠτέτµηται ¹ ΗΓ δίχα ØπÕ τοà ΟΜΝ ™πιπέδου κατ¦ τÕ Ν· κሠ¹ ¢πÕ τοà Η ¥ρα κάθετος ™πˆ τÕ ΑΒΓ ™πίπεδον δίχα τµηθήσεται ØπÕ τοà ΟΜΝ ™πιπέδου. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ¢πÕ τοà Θ κάθετος ™πˆ τÕ ∆ΕΖ ™πίπεδον δίχα τµηθήσεται ØπÕ τοà ΣΤΥ ™πιπέδου. καί ε„σιν ‡σαι αƒ ¢πÕ τîν Η, Θ κάθετοι ™πˆ τ¦ ΑΒΓ, ∆ΕΖ ™πίπεδα· ‡σαι ¥ρα καˆ αƒ ¢πÕ τîν ΟΜΝ, ΣΤΥ τριγώνων ™πˆ τ¦ ΑΒΓ, ∆ΕΖ κάθετοι. „σοϋψÁ ¥ρα [™στˆ] τ¦ πρίσµατα ïν βάσεις µέν ε„σι τ¦ ΛΞΓ, ΡΦΖ τρίγωνα, ¢πεναντίον δ τ¦ ΟΜΝ, ΣΤΥ. éστε κሠτ¦ στερε¦ παραλληλεπίπεδα τ¦ ¢πÕ τîν ε„ρηµένων πρισµάτων ¢ναγραφόµενα „σοϋψÁ κሠπρÕς ¥λληλά [ε„σιν] æς αƒ βάσεις· κሠτ¦ ¹µίση ¥ρα ™στˆν æς ¹ ΛΞΓ βάσις πρÕς τ¾ν ΡΦΖ βάσιν, οÛτως τ¦ ε„ρηµένα πρίσµατα πρÕς ¥λληλα· Óπερ œδει δε‹ξαι.

And one may show as follows that as triangle LOC is to triangle RV F , so the prism whose base (is) triangle LOC, and opposite (plane) P M N , (is) to the prism whose base (is) [triangle] RV F , and opposite (plane) ST U . For, in the same figure, let perpendiculars have been conceived (drawn) from (points) G and H to the planes ABC and DEF (respectively). These clearly turn out to be equal, on account of the pyramids being assumed (to be) of equal height. And since two straight-lines, GC and the perpendicular from G are cut by the parallel planes ABC and P M N , they will be cut in the same ratios [Prop. 11.17]. And GC was cut in half by the plane P M N at N . Thus, the perpendicular from G to the plane ABC will also be cut in half by the plane P M N . So, for the same (reasons), the perpendicular from H to the plane DEF will also be cut in half by the plane ST U . And the perpendiculars from G and H to the planes ABC and DEF (respectively) are equal. Thus, the perpendiculars from the triangles P M N and ST U to ABC and DEF (respectively, are) also equal. Thus, the prisms whose bases are triangles LOC and RV F , and opposite (sides) P M N and ST U (respectively), [are] of equal height. And, hence, the parallelepiped solids described on the aforementioned prisms [are] of equal height and to one another as their bases [Prop. 11.32]. Likewise, the halves (of the solids) [Prop. 11.28]. Thus, as base LOC is to base RV F , so the aforementioned prisms (are) to one another. (Which is) the very thing it was required to

479

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12 show.

ε΄.

Proposition 5

Αƒ ØπÕ τÕ αÙτÕ Ûψος οâσαι πυραµίδες κሠτριγώνους œχουσαι βάσεις πρÕς ¢λλήλας ε„σˆν æς αƒ βάσεις.

Pyramids which are of the same height, and have triangular bases, are to one another as their bases.

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D

L K B

”Εστωσαν ØπÕ τÕ αÙτÕ Ûψος πυραµίδες, ïν βάσεις µν τ¦ ΑΒΓ, ∆ΕΖ τρίγωνα, κορυφαˆ δ τ¦ Η, Θ σηµε‹α· λέγω, Óτι ™στˆν æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως ¹ ΑΒΓΗ πυρᵈς πρÕς τ¾ν ∆ΕΖΘ πυραµίδα. Ε„ γ¦ρ µή ™στιν æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως ¹ ΑΒΓΗ πυρᵈς πρÕς τ¾ν ∆ΕΖΘ πυραµίδα, œσται æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως ¹ ΑΒΓΗ πυρᵈς ½τοι πρÕς œλασσόν τι τÁς ∆ΕΖΘ πυραµίδος στερεÕν À πρÕς µε‹ζον. œστω πρότερον πρÕς œλασσον τÕ Χ, κሠδιVρ¾σθω ¹ ∆ΕΖΘ πυρᵈς ε‡ς τε δύο πυραµίδας ‡σας ¢λλήλαις καˆ Ðµοίας τÍ ÓλV κሠε„ς δύο πρίσµατα ‡σα· τ¦ δ¾ δύο πρίσαµτα µείζονά ™στιν À τÕ ¼µισυ τÁς Óλης πυραµίδος. κሠπάλιν αƒ ™κ τÁς διαιρέσεως γινόµεναι πυραµίδες еοίως διVρήσθωσαν, κሠτοàτο ¢εˆ γινέσθω, ›ως οá λειφθîσί τινες πυταµίδες ¢πÕ τÁς ∆ΕΖΘ πυραµίδος, α† ε„σιν ™λάττονες τÁς ØπεροχÁς, Î Øπερέχει ¹ ∆ΕΖΘ πυρᵈς τοà Χ στερεοà. λελείφθωσαν κሠœστωσαν λόγου ›νεκεν αƒ ∆ΠΡΣ, ΣΤΥΘ· λοιπ¦ ¥ρα τ¦ ™ν τÍ ∆ΕΖΘ πυραµίδι πρίσµατα µείζονά ™στι τοà Χ στερεοà. διVρήσθω κሠ¹ ΑΒΓΗ πυρᵈς еοίως κሠ„σοπληθîς τÍ ∆ΕΖΘ πυραµίδι· œστιν ¥ρα æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τ¦ ™ν τÍ ΑΒΓΗ πυραµίδι πρίσµατα πρÕς τ¦ ™ν τÍ ∆ΕΖΘ πυραµίδι πρίσµατα, ¢λλ¦ κሠæς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως ¹ ΑΒΓΗ πυρᵈς πρÕς τÕ Χ στερεόν· κሠæς ¥ρα ¹ ΑΒΓΗ πυρᵈς πρÕς τÕ Χ στερεόν, οÛτως τ¦ ™ν τÍ ΑΒΓΗ πυραµίδι πρίσµατα πρÕς τ¦ ™ν τÍ ∆ΕΖΘ πυραµίδι πρίσµατα· ™ναλλ¦ξ ¥ρα æς ¹ ΑΒΓΗ πυρᵈς πρÕς τ¦ ™ν αÙτÍ πρίσµατα, οÛτως τÕ Χ στερεÕν πρÕς τ¦ ™ν τÍ ∆ΕΖΘ πυραµίδι πρίσµατα. µε‹ζων δ ¹ ΑΒΓΗ πυρᵈς τîν ™ν αÙτÍ πρισµάτων· µε‹ζον ¥ρα κሠτÕ Χ στερεÕν τîν ™ν τÍ ∆ΕΖΘ πυραµίδι πρισµάτων. ¢λλ¦ κሠœλαττον· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ™στˆν æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως ¹ ΑΒΓΗ πυρᵈς πρÕς œλασσόν τι τÁς ∆ΕΖΘ πυραµίδος στερεόν. еοίως δ¾ δειχθήσεται,

W

T

O

C

F

R Q

V E

Let there be pyramids of the same height whose bases (are) the triangles ABC and DEF , and apexes the points G and H (respectively). I say that as base ABC is to base DEF , so pyramid ABCG (is) to pyramid DEF H. For if base ABC is not to base DEF , as pyramid ABCG (is) to pyramid DEF H, then base ABC (is) to base DEF , as pyramid ABCG will be to some solid either less than, or greater than, pyramid DEF H. Let it, first of all, be (in this ratio) to (some) lesser (solid), W . And let pyramid DEF H have been divided into two pyramids equal to one another, and similar to the whole, and into two equal prisms. So, the (sum of the) two prisms is greater than half of the whole pyramid [Prop. 12.3]. And, again, let the pyramids generated by the division have been similarly divided, and let this be done continually until some pyramids are left from pyramid DEF H which (when added together) are less than the excess by which pyramid DEF H exceeds the solid W [Prop. 10.1]. Let them have been left, and, for the sake of argument, let them be DQRS and ST U H. Thus, the (sum of the) remaining prisms within pyramid DEF H is greater than solid W . Let pyramid ABCG also have been divided similarly, and a similar number of times, as pyramid DEF H. Thus, as base ABC is to base DEF , so the (sum of the) prisms within pyramid ABCG (is) to the (sum of the) prisms within pyramid DEF H [Prop. 12.4]. But, also, as base ABC (is) to base DEF , so pyramid ABCG (is) to solid W . And, thus, as pyramid ABCG (is) to solid W , so the (sum of the) prisms within pyramid ABCG (is) to the (sum of the) prisms within pyramid DEF H [Prop. 5.11]. Thus, alternately, as pyramid ABCG (is) to the (sum of the) prisms within it, so solid W (is) to the (sum of the) prisms within pyramid DEF H [Prop. 5.16]. And pyramid ABCG (is) greater than the (sum of the) prisms within it. Thus, solid W (is) also greater than the (sum of the) prisms within pyramid DEF H [Prop. 5.14].

480

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

Óτι οÙδ æς ¹ ∆ΕΖ βάσις πρÕς τ¾ν ΑΒΓ βάσιν, οÛτως ¹ ∆ΕΖΘ πυρᵈς πρÕς œλαττόν τι τÁς ΑΒΓΗ πυραµίδος στερεόν. Λέγω δή, Óτι οÙκ œστιν οÙδ æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως ¹ ΑΒΓΗ πυρᵈς πρÕς µε‹ζόν τι τÁς ∆ΕΖΘ πυραµίδος στερεόν. Ε„ γ¦ρ δυνατόν, œστω πρÕς µε‹ζον τÕ Χ· ¢νάπαλιν ¥ρα ™στˆν æς ¹ ∆ΕΖ βάσις πρÕς τ¾ν ΑΒΓ βάσιν, οÛτως τÕ Χ στερεÕν πρÕς τ¾ν ΑΒΓΗ πυραµίδα. æς δ τÕ Χ στερεÕν πρÕς τ¾ν ΑΒΓΗ πυραµίδα, οÛτως ¹ ∆ΕΖΘ πυρᵈς πρÕς œλασσόν τι τÁς ΑΒΓΗ πυραµίδος, æς œµπροσθεν ™δείχθη· κሠæς ¥ρα ¹ ∆ΕΖ βάσις πρÕς τ¾ν ΑΒΓ βάσιν, οÛτως ¹ ∆ΕΖΘ πυρᵈς πρÕς œλασσόν τι τÁς ΑΒΓΗ πυραµίδος· Óπερ ¥τοπον ™δείχθη. οÙκ ¥ρα ™στˆν æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως ¹ ΑΒΓΗ πυρᵈς πρÕς µε‹ζόν τι τÁς ∆ΕΖΘ πυραµίδος στερεόν. ™δείχθη δέ, Óτι οÙδ πρÕς œλασσον. œστιν ¥ρα æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως ¹ ΑΒΓΗ πυρᵈς πρÕς τ¾ν ∆ΕΖΘ πυραµίδα· Óπερ œδει δε‹ξαι.

But, (it is) also less. This very thing is impossible. Thus, as base ABC is to base DEF , so pyramid ABCG (is) not to some solid less than pyramid DEF H. So, similarly, we can show that base DEF is not to base ABC, as pyramid DEF H (is) to some solid less than pyramid ABCG either. So, I say that base ABC is not to base DEF , as pyramid ABCG (is) to some solid greater than pyramid DEF H either. For, if possible, let it be (in this ratio) to some greater (solid), W . Thus, inversely, as base DEF (is) to base ABC, so solid W (is) to pyramid ABCG [Prop. 5.7. corr.]. And as solid W (is) to pyramid ABCG, so pyramid DEF H (is) to some (solid) less than pyramid ABCG, as shown before [Prop. 12.2 lem.]. And, thus, as base DEF (is) to base ABC, so pyramid DEF H (is) to some (solid) less than pyramid ABCG [Prop. 5.11]. The very thing was shown (to be) absurd. Thus, base ABC is not to base DEF , as pyramid ABCG (is) to some solid greater than pyramid DEF H. And, it was shown that neither (is it in this ratio) to a lesser (solid). Thus, as base ABC is to base DEF , so pyramid ABCG (is) to pyramid DEF H. (Which is) the very thing it was required to show.

$΄.

Proposition 6

Αƒ ÙπÕ τÕ αÙτÕ Ûψος οâσαι πυραµίδες κሠπολυγώνους œχουσαι βάσεις πρÕς ¢λλήλας ε„σˆν æς αƒ βάσεις.

Pyramids which are of the same height, and have polygonal bases, are to one another as their bases.

N

M

G D

B A

E

K

M

J

C

H

L

N

K

H

B

D

G

L A

Z

E

”Εστωσαν ØπÕ τÕ αÙτÕ Ûψος πυραµίδες, ïν [αƒ] βάσεις µν τ¦ ΑΒΓ∆Ε, ΖΗΘΚΛ πολύγωνα, κορυφαˆ δ τ¦ Μ, Ν σηµε‹α· λέγω, Óτι ™στˆν æς ¹ ΑΒΓ∆Ε βάσις πρÕς τ¾ν ΖΗΘΚΛ βάσιν, οÛτως ¹ ΑΒΓ∆ΕΜ πυρᵈς πρÕς τ¾ν ΖΗΘΚΛΝ πυραµίδα. 'Επεζεύχθωσαν γ¦ρ αƒ ΑΓ, Α∆, ΖΘ, ΖΚ. ™πεˆ οâν δύο πυραµίδες ε„σˆν αƒ ΑΒΓΜ, ΑΓ∆Μ τριγώνους œχουσαι βάσεις κሠÛψος ‡σον, πρÕς ¢λλήλας ε„σˆν æς αƒ βάσεις· œστιν ¥ρα æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ΑΓ∆

F

Let there be pyramids of the same height whose bases (are) the polygons ABCDE and F GHKL, and apexes the points M and N (respectively). I say that as base ABCDE is to base F GHKL, so pyramid ABCDEM (is) to pyramid F GHKLN . For let AC, AD, F H, and F K have been joined. Therefore, since ABCM and ACDM are two pyramids having triangular bases and equal height, they are to one another as their bases [Prop. 12.5]. Thus, as base

481

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

βάσιν, οÛτως ¹ ΑΒΓΜ πυρᵈς πρÕς τ¾ν ΑΓ∆Μ πυραµίδα. κሠσυνθέντι æς ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΑΓ∆ βάσιν, οÛτως ¹ ΑΒΓ∆Μ πυρᵈς πρÕς τ¾ν ΑΓ∆Μ πυραµίδα. ¢λλ¦ κሠæς ¹ ΑΓ∆ βάσις πρÕς τ¾ν Α∆Ε βάσιν, οÛτως ¹ ΑΓ∆Μ πυρᵈς πρÕς τ¾ν Α∆ΕΜ πυραµίδα. δι' ‡σου ¥ρα æς ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν Α∆Ε βάσιν, οÛτως ¹ ΑΒΓ∆Μ πυρᵈς πρÕς τ¾ν Α∆ΕΜ πυραµίδα. κሠσυνθέντι πάλιν, æς ¹ ΑΒΓ∆Ε βάσις πρÕς τ¾ν Α∆Ε βάσιν, οÛτως ¹ ΑΒΓ∆ΕΜ πυρᵈς πρÕς τ¾ν Α∆ΕΜ πυραµίδα. еοίως δ¾ δειχθήσεται, Óτι κሠæς ¹ ΖΗΘΚΛ βάσις πρÕς τ¾ν ΖΗΘ βάσιν, οÛτως κሠ¹ ΖΗΘΚΛΝ πυρᵈς πρÕς τ¾ν ΖΗΘΝ πυραµίδα. κሠ™πεˆ δύο πυραµίδες εƒσˆν αƒ Α∆ΕΜ, ΖΗΘΝ τριγώνους œχουσαι βάσεις κሠÛψος ‡σον, œστιν ¥ρα æς ¹ Α∆Ε βάσις πρÕς τ¾ν ΖΗΘ βάσιν, οÛτως ¹ Α∆ΕΜ πυρᵈς πρÕς τ¾ν ΖΗΘΝ πυραµίδα. ¢λλ' æς ¹ Α∆Ε βάσις πρÕς τ¾ν ΑΒΓ∆Ε βάσιν, οÛτως Ãν ¹ Α∆ΕΜ πυρᵈς πρÕς τ¾ν ΑΒΓ∆ΕΜ πυραµίδα. κሠδι' ‡σου ¥ρα æς ¹ ΑΒΓ∆Ε βάσις πρÕς τ¾ν ΖΗΘ βάσιν, οÛτως ¹ ΑΒΓ∆ΕΜ πυρᵈς πρÕς τ¾ν ΖΗΘΝ πυραµίδα. ¢λλ¦ µ¾ν κሠæς ¹ ΖΗΘ βάσις πρÕς τ¾ν ΖΗΘΚΛ βάσιν, οÛτως Ãν κሠ¹ ΖΗΘΝ πυρᵈς πρÕς τ¾ν ΖΗΘΚΛΝ πυραµίδα, κሠδι' ‡σου ¥ρα æς ¹ ΑΒΓ∆Ε βάσις πρÕς τ¾ν ΖΗΘΚΛ βάσιν, οÛτως ¹ ΑΒΓ∆ΕΜ πυρᵈς πρÕς τ¾ν ΖΗΘΚΛΝ πυραµίδα· Óπερ œδει δε‹ξαι.

ABC is to base ACD, so pyramid ABCM (is) to pyramid ACDM . And, via composition, as base ABCD (is) to base ACD, so pyramid ABCDM (is) to pyramid ACDM [Prop. 5.18]. But, as base ACD (is) to base ADE, so pyramid ACDM (is) also to pyramid ADEM [Prop. 12.5]. Thus, via equality, as base ABCD (is) to base ADE, so pyramid ABCDM (is) to pyramid ADEM [Prop. 5.22]. And, again, via composition, as base ABCDE (is) to base ADE, so pyramid ABCDEM (is) to pyramid ADEM [Prop. 5.18]. So, similarly, it can also be shown that as base F GHKL (is) to base F GH, so pyramid F GHKLN (is) also to pyramid F GHN . And since ADEM and F GHN are two pyramids having triangular bases and equal height, thus as base ADE (is) to base F GH, so pyramid ADEM (is) to pyramid F GHN [Prop. 12.5]. But, as base ADE (is) to base ABCDE, so pyramid ADEM (was) to pyramid ABCDEM . Thus, via equality, as base ABCDE (is) to base F GH, so pyramid ABCDEM (is) also to pyramid F GHN [Prop. 5.22]. But, furthermore, as base F GH (was) to base F GHKL, so pyramid F GHN (is) also to pyramid F GHKLN . Thus, via equality, as base ABCDE (is) to base F GHKL, so pyramid ABCDEM (is) also to pyramid F GHKLN [Prop. 5.22]. (Which is) the very thing it was required to show.

ζ΄.

Proposition 7

Π©ν πρίσµα τρίγωνον œχον βάσιν διαιρε‹ται ε„ς τρε‹ς Any prism having a triangular base is divided into πυραµίδας ‡σας ¢λλήλαις τριγώνους βάσεις ™χούσας. three pyramids having triangular bases (which are) equal to one another.

Z

E

F

D

E

D

G B

C

A

B

A

”Εστω πρίσµα, οá βάσις µν τÕ ΑΒΓ τρίγωνον, ¢πεLet there be a prism whose base (is) triangle ABC, ναντίον δ τÕ ∆ΕΖ· λέγω, Óτι τÕ ΑΒΓ∆ΕΖ πρίσµα and opposite (plane) DEF . I say that prism ABCDEF διαιρε‹ται ε„ς τρε‹ς πυραµίδας ‡σας ¢λλήλαις τριγώνους is divided into three pyramids having triangular bases ™χούσας βάσεις. (which are) equal to one another. 'Επεζεύχθωσαν γ¦ρ αƒ Β∆, ΕΓ, Γ∆. ™πεˆ παραλFor let BD, EC, and CD have been joined. Since 482

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

ληλόγραµµόν ™στι τÕ ΑΒΕ∆, διάµετρος δ αÙτÕà ™στιν ¹ Β∆, ‡σον ¥ρα ™στι τÕ ΑΒ∆ τρίγωνον τù ΕΒ∆ τρίγωνJ· κሠ¹ πυρᵈς ¥ρα, Âς βάσις µν τÕ ΑΒ∆ τρίγωνον, κορυφ¾ δ τÕ Γ σηµε‹ον, ‡ση ™στˆ πυραµίδι, Âς βάσις µέν ™στι τÕ ∆ΕΒ τρίγωνον, κορυφ¾ δ τÕ Γ σηµε‹ον. ¢λλ¦ ¹ πυραµίς, Âς βάσις µέν ™στι τÕ ∆ΕΒ τρίγωνον, κορυφ¾ δ τÕ Γ σηµε‹ον, ¹ αÙτή ™στι πυραµίδι, Âς βάσις µέν ™στι τÕ ΕΒΓ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον· ØπÕ γ¦ρ τîν αÙτîν ™πιπέδων περιέχεται. κሠπυρᵈς ¥ρα, Âς βάσις µέν ™στι τÕ ΑΒ∆ τρίγωνον, κορυφ¾ δ τÕ Γ σηµε‹ον, ‡ση ™στˆ πυραµίδι, Âς βάσις µέν ™στι τÕ ΕΒΓ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον. πάλιν, ™πεˆ παραλληλόγραµµόν ™στι τÕ ΖΓΒΕ, διάµετρος δέ ™στιν αÙτοà ¹ ΓΕ, ‡σον ™στˆ τÕ ΓΕΖ τρίγωνον τù ΓΒΕ τριγώνJ. κሠπυρᵈς ¥ρα, Âς βάσις µέν ™στι τÕ ΒΓΕ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον, ‡ση ™στˆ πυραµίδι, Âς βάσις µέν ™στι τÕ ΕΓΖ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον. ¹ δ πυραµίς, Âς βάσις µέν ™στι τÕ ΒΓΕ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον, ‡ση ™δείχθη πυραµίδι, Âς βάσις µέν ™στι τÕ ΑΒ∆ τρίγωνον, κορυφ¾ δ τÕ Γ σηµε‹ον· κሠπυρᵈς ¥ρα, Âς βάσις µέν ™στι τÕ ΓΕΖ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον, ‡ση ™στˆ πυραµίδι, Âς βάσις µέν [™στι] τÕ ΑΒ∆ τρίγωνον, κορυφ¾ δ τÕ Γ σηµε‹ον· διÇρηται ¥ρα τÕ ΑΒΓ∆ΕΖ πρίσµα ε„ς τρε‹ς πυραµίδας ‡σας ¢λλήλαις τριγώνους ™χούσας βάσεις. Κሠ™πεˆ πυραµίς, Âς βάσις µέν ™στι τÕ ΑΒ∆ τρίγωνον, κορυφ¾ δ τÕ Γ σηµε‹ον, ¹ αÙτή ™στι πυραµίδι, Âς βάσις τÕ ΓΑΒ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον· ØπÕ γ¦ρ τîν αÙτîν ™πιπέδων περιέχονται· ¹ δ πυραµίς, Âς βάσις τÕ ΑΒ∆ τρίγωνον, κορυφ¾ δ τÕ Γ σηµε‹ον, τρίτον ™δείχθη τοà πρίσµατος, οá βάσις τÕ ΑΒΓ τρίγωνον, ¢πεναντίον δ τÕ ∆ΕΖ, κሠ¹ πυρᵈς ¥ρα, Âς βάσις τÕ ΑΒΓ τρίγωνον, κορυφ¾ δ τÕ ∆ σηµε‹ον, τρίτον ™στˆ τοà πρίσµατος τοà œχοντος βάσις τ¾ν αÙτ¾ν τÕ ΑΒΓ τρίγωνον, ¢πεναντίον δ τÕ ∆ΕΖ.

ABED is a parallelogram, and BD is its diagonal, triangle ABD is thus equal to triangle EBD [Prop. 1.34]. And, thus, the pyramid whose base (is) triangle ABD, and apex the point C, is equal to the pyramid whose base is triangle DEB, and apex the point C [Prop. 12.5]. But, the pyramid whose base is triangle DEB, and apex the point C, is the same as the pyramid whose base is triangle EBC, and apex the point D. For they are contained by the same planes. And, thus, the pyramid whose base is ABD, and apex the point C, is equal to the pyramid whose base is EBC and apex the point D. Again, since F CBE is a parallelogram, and CE is its diagonal, triangle CEF is equal to triangle CBE [Prop. 1.34]. And, thus, the pyramid whose base is triangle BCE, and apex the point D, is equal to the pyramid whose base is triangle ECF , and apex the point D [Prop. 12.5]. And the pyramid whose base is triangle BCE, and apex the point D, was shown (to be) equal to the pyramid whose base is triangle ABD, and apex the point C. Thus, the pyramid whose base is triangle CEF , and apex the point D, is also equal to the pyramid whose base [is] triangle ABD, and apex the point C. Thus, the prism ABCDEF has been divided into three pyramids having triangular bases (which are) equal to one another. And since the pyramid whose base is triangle ABD, and apex the point C, is the same as the pyramid whose base is triangle CAB, and apex the point D. For they are contained by the same planes. And the pyramid whose base (is) triangle ABD, and apex the point C, was shown (to be) a third of the prism whose base is triangle ABC, and opposite (plane) DEF , the pyramid whose base is triangle ABC, and apex the point D, is thus also a third of the pyramid having the same base, triangle ABC, and opposite (plane) DEF .

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι π©σα πυρᵈς τρίτον And, from this, (it is) clear that any pyramid is the µέρος ™στˆ τοà πρίσµατος τοà τ¾ν αÙτ¾ν βάσιν œχοντος third part of the prism having the same base as it, and an αÙτÍ κሠÛψος ‡σον· Óπερ œδει δε‹ξαι. equal height. (Which is) the very thing it was required to show.

η΄.

Proposition 8

Αƒ Óµοιαι πυραµίδες κሠτριγώνους œχουσαι βάσεις ™ν τριπλασίονι λόγJ ε„σˆ τîν еολόγων πλευρîν. ”Εστωσαν Óµοιαι καˆ Ðµοίως κείµεναι πυραµίδες, ïν βάσεις µέν ε„σι τ¦ ΑΒΓ, ∆ΕΖ τρίγωνα, κορυφαˆ δ τ¦ Η, Θ σηµε‹α· λέγω, Óτι ¹ ΑΒΓΗ πυρᵈς πρÕς τ¾ν ∆ΕΖΘ πυραµίδα τριπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ.

Similar pyramids which also have triangular bases are in the cubed ratio of their corresponding sides. Let there be similar, and similarly laid out, pyramids whose bases are triangles ABC and DEF , and apexes the points G and H (respectively). I say that pyramid ABCG has to pyramid DEF H the cubed ratio of that BC (has) to EF .

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Συµπεπληρώσθω γ¦ρ τ¦ ΒΗΜΛ, ΕΘΠΟ στερε¦ παραλληλεπίπεδα. κሠ™πεˆ Ðµοία ™στˆν ¹ ΑΒΓΗ πυρᵈς τÍ ∆ΕΖΘ πυραµίδι, †ση ¥ρα ™στˆν ¹ µν ØπÕ ΑΒΓ γωνία τÍ ØπÕ ∆ΕΖ γωνίv, ¹ δ ØπÕ ΗΒΓ τÍ ØπÕ ΘΕΖ, ¹ δ ØπÕ ΑΒΗ τÍ ØπÕ ∆ΕΘ, καί ™στιν æς ¹ ΑΒ πρÕς τ¾ν ∆Ε, οÛτως ¹ ΒΓ πρÕς τ¾ν ΕΖ, κሠ¹ ΒΗ πρÕς τ¾ν ΕΘ. κሠ™πεί ™στιν æς ¹ ΑΒ πρÕς τ¾ν ∆Ε, οÛτως ¹ ΒΓ πρÕς τ¾ν ΕΖ, κሠπερˆ ‡σας γωνίας αƒ πλευρሠ¢νάλογόν ε„σιν, Óµοιον ¥ρα ™στˆ τÕ ΒΜ παραλληλόγραµµον τù ΕΠ παραλληλογράµµJ. δι¦ τ¦ αÙτ¦ δ¾ κሠτÕ µν ΒΝ τù ΕΡ Óµοιόν ™στι, τÕ δ ΒΚ τù ΕΞ· τ¦ τρία ¥ρα τ¦ ΜΒ, ΒΚ, ΒΝ τρισˆ το‹ς ΕΠ, ΕΞ, ΕΡ Óµοιά ™στιν. ¢λλ¦ τ¦ µν τρία τ¦ ΜΒ, ΒΚ, ΒΝ τρισˆ το‹ς ¢πεναντίον ‡σα τε καˆ Óµοιά ™στιν, τ¦ δ τρία τ¦ ΕΠ, ΕΞ, ΕΡ τρισˆ το‹ς ¢πεναντίον ‡σα τε καˆ Óµοιά ™στιν. τ¦ ΒΗΜΛ, ΕΘΠΟ ¥ρα στερε¦ ØπÕ Ðµοίων ™πιπέδων ‡σων τÕ πλÁθος περιέχεται. Óµοιον ¥ρα ™στˆ τÕ ΒΗΜΛ στερεÕν τù ΕΘΠΟ στερεù. τ¦ δ Óµοια στερε¦ παραλληλεπίπεδα ™ν τριπλασίονι λόγJ ™στˆ τîν еολόγων πλευρîν. τÕ ΒΗΜΛ ¥ρα στερεÕν πρÕς τÕ ΕΘΠΟ στερεÕν τριπλασίονα λόγον œχει ½περ ¹ еόλογος πλευρ¦ ¹ ΒΓ πρÕς τ¾ν еόλογον πλευρ¦ν τ¾ν ΕΖ. æς δ τÕ ΒΗΜΛ στερεÕν πρÕς τÕ ΕΘΠΟ στερεόν, οÛτως ¹ ΑΒΓΗ πυρᵈς πρÕς τ¾ν ∆ΕΖΘ πυραµίδα, ™πειδήπερ ¹ πυρᵈς ›κτον µέρος ™στˆ τοà στερεοà δι¦ τÕ κሠτÕ πρίσµα ¼µισυ ×ν τοà στερεοà παραλληλεπιπέδου τριπλάσιον εναι τÁς πυραµίδος. κሠ¹ ΑΒΓΗ ¥ρα πυρᵈς πρÕς τ¾ν ∆ΕΖΘ πυραµίδα τριπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ· Óπερ œδει δε‹ξαι.

For let the parallelepiped solids BGM L and EHQP have been completed. And since pyramid ABCG is similar to pyramid DEF H, angle ABC is thus equal to angle DEF , and GBC to HEF , and ABG to DEH. And as AB is to DE, so BC (is) to EF , and BG to EH [Def. 11.9]. And since as AB is to DE, so BC (is) to EF , and (so) the sides around equal angles are proportional, parallelogram BM is thus similar to paralleleogram EQ. So, for the same (reasons), BN is also similar to ER, and BK to EO. Thus, the three (parallelograms) M B, BK, and BN are similar to the three (parallelograms) EQ, EO, ER (respectively). But, the three (parallelograms) M B, BK, and BN are (both) equal and similar to the three opposite (parallelograms), and the three (parallelograms) EQ, EO, and ER are (both) equal and similar to the three opposite (parallelograms) [Prop. 11.24]. Thus, the solids BGM L and EHQP are contained by equal numbers of similar (and similarly laid out) planes. Thus, solid BGM L is similar to solid EHQP [Def. 11.9]. And similar parallelepiped solids are in the cubed ratio of corresponding sides [Prop. 11.33]. Thus, solid BGM L has to solid EHQP the cubed ratio that the corresponding side BC (has) to the corresponding side EF . And as solid BGM L (is) to solid EHQP , so pyramid ABCG (is) to pyramid DEF H, inasmuch as the pyramid is the sixth part of the solid, on account of the prism, being half of the parallelepiped solid [Prop. 11.28], also being three times the pyramid [Prop. 12.7]. Thus, pyramid ABCG also has to pyramid DEF H the cubed ratio that BC (has) to EF . (Which is) the very thing it was required to show.

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι καˆ αƒ πολυγώνους œχουσαι βάσεις Óµοιαι πυραµίδες πρÕς ¢λλήλας ™ν τριπλασίονι λόγJ ε„σˆ τîν еολόγων πλευρîν. διαιρεθεισîν γ¦ρ αÙτîν ε„ς τ¦ς ™ν αÙτα‹ς πυραµίδας τριγώνους βάσεις ™χούσας τù κሠτ¦ Óµοια πολύγωνα τîν βάσεων ε„ς Óµοια τρίγωνα διαιρε‹σθαι κሠ‡σα τù πλήθει καˆ

So, from this, (it is) also clear that similar pyramids having polygonal bases (are) to one another as the cubed ratio of corresponding sides. For, dividing them into the pyramids (contained) within them which have triangular bases, with the similar polygons of the bases also being divided into similar triangles (which are) both equal in

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еόλογα το‹ς Óλοις œσται æς [¹] ™ν τÍ ˜τέρv µία πυρᵈς τρίγωνον œχουσα βάσιν πρÕς τ¾ν ™ν τÍ ˜τέρv µίαν πυραµίδα τρίγωνον œχουσαν βάσιν, οÛτως κሠ¤πασαι αƒ ™ν τÍ ˜τέρv πυραµίδι πυραµίδες τριγώνους œχουσαι βάσεις πρÕς τ¦ς ™ν τÍ ˜τέρv πυραµίδι πυραµίδας τριγώνους βάσεις ™χούσας, τουτέστιν αÙτ¾ ¹ πολύγωνον βάσιν œχουσα πυρᵈς πρÕς τ¾ν πολύγωνον βάσιν œχουσα πυραµίδα. ¹ δ τρίγωνον βάσιν œχουσα πυρᵈς πρÕς τ¾ν τρίγωνον βάσιν œχουσαν ™ν τριπλασίονι λόγJ ™στˆ τîν еολόγον πλευρîν· κሠ¹ πολύγωνον ¥ρα βάσιν œχουσα πρÕς τ¾ν еοίαν βάσιν œχουσαν τριπλασίονα λόγον œχει ½περ ¹ πλευρ¦ πρÕς τ¾ν πλευράν.

number, and corresponding, to the wholes [Prop. 6.20]. As one pyramid having a triangular base in the former (pyramid having a polygonal base is) to one pyramid having a triangular base in the latter (pyramid having a polygonal base), so (the sum of) all the pyramids having triangular bases in the former pyramid will also be to (the sum of) all the pyramids having triangular bases in the latter pyramid [Prop. 5.12]—that is to say, the (former) pyramid itself having a polygonal base to the (latter) pyramid having a polygonal base. And a pyramid having a triangular base is to a (pyramid) having a triangular base in the cubed ratio of corresponding sides [Prop. 12.8]. Thus, a (pyramid) having a polygonal base has to to a (pyramid) having a similar base the cubed ratio of a (corresponding) side to a (corresponding) side.

θ΄.

Proposition 9

Τîν ‡σων πυραµίδων κሠτριγώνους βάσεις ™χουσîν ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν· κሠïν πυραµίδων τριγώνους βάσεις ™χουσîν ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν, ‡σαι ε„σˆν ™κε‹ναι.

The bases of equal pyramids which also have triangular bases are reciprocally proportional to their heights. And those pyramids which have triangular bases whose bases are reciprocally proportional to their heights are equal.

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”Εστωσαν γ¦ρ ‡σαι πυραµίδες τριγώνους βάσεις œχουσαι τ¦ς ΑΒΓ, ∆ΕΖ, κορυφ¦ς δ τ¦ Η, Θ σηµε‹α· λέγω, Óτι τîν ΑΒΓΗ, ∆ΕΖΘ πυραµίδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν, καί ™στιν æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τÕ τÁς ∆ΕΖΘ πυραµίδος Ûψος πρÕς τÕ τÁς ΑΒΓΗ πυραµίδος Ûψος. Συµπεπληρώσθω γ¦ρ τ¦ ΒΗΜΛ, ΕΘΠΟ στερε¦ παραλληλεπίπεδα. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΒΓΗ πυρᵈς τÍ ∆ΕΖΘ πυραµίδι, καί ™στι τÁς µν ΑΒΓΗ πυραµίδος ˜ξαπλάσιον τÕ ΒΗΜΛ στερεόν, τÁς δ ∆ΕΖΘ πυραµίδος ˜ξαπλάσιον τÕ ΕΘΠΟ στερεόν, ‡σον ¥ρα ™στˆ τÕ ΒΗΜΛ στερεÕν τù ΕΘΠΟ στερεù. τîν δ ‡σων στερεîν παραλληλεπιπώδων ¢ντιπεπόνθασιν αƒ βάσεις

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For let there be (two) equal pyramids having the triangular bases ABC and DEF , and apexes the points G and H (respectively). I say that the bases of the pyramids ABCG and DEF H are reciprocally proportional to their heights, and (so) that as base ABC is to base DEF , so the height of pyramid DEF H (is) to the height of pyramid ABCG. For let the parallelepiped solids BGM L and EHQP have been completed. And since pyramid ABCG is equal to pyramid DEF H, and solid BGM L is six times pyramid ABCG (see previous proposition), and solid EHQP (is) six times pyramid DEF H, solid BGM L is thus equal to solid EHQP . And the bases of equal par-

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το‹ς Ûψεσιν· œστιν ¥ρα æς ¹ ΒΜ βάσις πρÕς τ¾ν ΕΠ βάσιν, οÛτως τÕ τοà ΕΘΠΟ στερεοà Ûψος πρÕς τÕ τοà ΒΗΜΛ στερεοà Ûψος. ¢λλ' æς ¹ ΒΜ βάσις πρÕς τ¾ν ΕΠ, οÛτως τÕ ΑΒΓ τρίγωνον πρÕς τÕ ∆ΕΖ τρίγωνον. κሠæς ¥ρα τÕ ΑΒΓ τρίγωνον πρÕς τÕ ∆ΕΖ τρίγωνον, οÛτως τÕ τοà ΕΘΠΟ στερεοà Ûψος πρÕς τÕ τοà ΒΗΜΛ στερεοà Ûψος. ¢λλ¦ τÕ µν τοà ΕΘΠΟ στερεοà Ûψος τÕ αÙτÕ ™στι τù τÁς ∆ΕΖΘ πυραµίδος Ûψει, τÕ δ τοà ΒΗΜΛ στερεοà Ûψος τÕ αÙτό ™στι τù τÁς ΑΒΓΗ πυραµίδος Ûψει· œστιν ¥ρα æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τÕ τÁς ∆ΕΖΘ πυραµίδος Ûψος πρÕς τÕ τÁς ΑΒΓΗ πυραµίδος Ûψος. τîν ΑΒΓΗ, ∆ΕΖΘ ¥ρα πυραµίδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν. 'Αλλ¦ δ¾ τîν ΑΒΓΗ, ∆ΕΖΘ πυραµίδων ¢ντιπεπονθέτωσαν αƒ βάσεις το‹ς Ûψεσιν, κሠœστω æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τÕ τÁς ∆ΕΖΘ πυραµίδος Ûψος πρÕς τÕ τÁς ΑΒΓΗ πυραµίδος Ûψος· λέγω, Óτι ‡ση ™στˆν ¹ ΑΒΓΗ πυρᵈς τÍ ∆ΕΖΘ πυραµίδι. Τîν γ¦ρ αÙτîν κατασκευασθέντων, ™πεί ™στιν æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τÕ τÁς ∆ΕΖΘ πυραµίδος Ûψος πρÕς τÕ τÁς ΑΒΓΗ πυραµίδος Ûψος, ¢λλ' æς ¹ ΑΒΓ βάσις πρÕς τ¾ν ∆ΕΖ βάσιν, οÛτως τÕ ΒΜ παραλληλόγραµµον πρÕς τÕ ΕΠ παραλληλόγραµµον, κሠæς ¥ρα τÕ ΒΜ παραλληλόγραµµον πρÕς τÕ ΕΠ παραλληλόγραµµον, οÛτως τÕ τÁς ∆ΕΖΘ πυραµίδος Ûψος πρÕς τÕ τÁς ΑΒΓΗ πυραµίδος Ûψος. ¢λλ¦ τÕ [µν] τÁς ∆ΕΖΘ πυραµίδος Ûψος τÕ αÙτό ™στι τù τοà ΕΘΠΟ παραλληλεπιπέδου Ûψει, τÕ δ τÁς ΑΒΓΗ πυραµίδος Ûψος τÕ αÙτό ™στι τù τοà ΒΗΜΛ παραλληλεπιπέδου Ûψει· œστιν ¥ρα æς ¹ ΒΜ βάσις πρÕς τÁν ΕΠ βάσιν, οÛτως τÕ τοà ΕΘΠΟ παραλληλεπιπέδου Ûψος πρÕς τÕ τοà ΒΗΜΛ παραλληλεπιπέδου Ûψος. ïν δ στερεîν παραλληλεπιπέδων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν, ‡σα ™στˆν ™κε‹να· ‡σον ¥ρα ™στˆ τÕ ΒΗΜΛ στερεÕν παραλληλεπίπεδον τù ΕΘΠΟ στερεù παραλληλεπιπέδJ. καί ™στι τοà µν ΒΗΜΛ ›κτον µέρος ¹ ΑΒΓΗ πυραµίς, τοà δ ΕΘΠΟ παραλληλεπιπέδου ›κτον µέρος ¹ ∆ΕΖΘ πυραµίς· ‡ση ¥ρα ¹ ΑΒΓΗ πυρᵈς τÍ ∆ΕΖΘ πυραµίδι. Τîν ¥ρα ‡σων πυραµίδων κሠτριγώνους βάσεις ™χουσîν ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν· κሠïν πυραµίδων τριγώνους βάσεις ™χουσîν ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν, ‡σαι ε„σˆν ™κε‹ναι· Óπερ œδει δε‹ξαι.

allelepiped solids are reciprocally proportional to their heights [Prop. 11.34]. Thus, as base BM is to base EQ, so the height of solid EHQP (is) to the height of solid BGM L. But, as base BM (is) to base EQ, so triangle ABC (is) to triangle DEF [Prop. 1.34]. And, thus, as triangle ABC (is) to triangle DEF , so the height of solid EHQP (is) to the height of solid BGM L [Prop. 5.11]. But, the height of solid EHQP is the same as the height of pyramid DEF H, and the height of solid BGM L is the same as the height of pyramid ABCG. Thus, as base ABC is to base DEF , so the height of pyramid DEF H (is) to the height of pyramid ABCG. Thus, the bases of pyramids ABCG and DEF H are reciprocally proportional to their heights. And so, let the bases of pyramids ABCG and DEF H be reciprocally proportional to their heights, and (thus) let base ABC be to base DEF , as the height of pyramid DEF H (is) to the height of pyramid ABCG. I say that pyramid ABCG is equal to pyramid DEF H. For, with the same construction, since as base ABC is to base DEF , so the height of pyramid DEF H (is) to the height of pyramid ABCG, but as base ABC (is) to base DEF , so parallelogram BM (is) to parallelogram EQ [Prop. 1.34], thus as parallelogram BM (is) to parallelogram EQ, so the height of pyramid DEF H (is) also to the height of pyramid ABCG [Prop. 5.11]. But, the height of pyramid DEF H is the same as the height of parallelepiped EHQP , and the height of pyramid ABCG is the same as the height of parallelepiped BGM L. Thus, as base BM is to base EQ, so the height of parallelepiped EHQP (is) to the height of parallelepiped BGM L. And those parallelepiped solids whose bases are reciprocally proportional to their heights are equal [Prop. 11.34]. Thus, the parallelepiped solid BGM L is equal to the parallelepiped solid EHQP . And pyramid ABCG is a sixth part of BGM L, and pyramid DEF H a sixth part of parallelepiped EHQP . Thus, pyramid ABCG is equal to pyramid DEF H. Thus, the bases of equal pyramids which also have triangular bases are reciprocally proportional to their heights. And those pyramids having triangular bases whose bases are reciprocally proportional to their heights are equal. (Which is) the very thing it was required to show.

ι΄.

Proposition 10

Π©ς κîνος κυλίνδρου τρίτον µέρος ™στˆ τοà τ¾ν Every cone is the third part of the cylinder which has αÙτ¾ν βάσιν œχοντος αÙτù κሠÛψος ‡σον. the same base as it, and an equal height. 'Εχέτω γ¦ρ κîνος κυλίνδρù βάσιν τε τ¾ν αÙτ¾ν τÕν For let there be a cone (with) the same base as a cylinΑΒΓ∆ κύκλον κሠÛψος ‡σον· λέγω, Óτι Ð κîνος τοà der, (namely) the circle ABCD, and an equal height. I

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κυλίνδρου τρίτον ™στˆ µέρος, τουτέστιν Óτι Ð κύλινδρος τοà κώνου τριπλασιων ™στίν.

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say that the cone is the third part of the cylinder—that is to say, that the cylinder is three times the cone.

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Ε„ γ¦ρ µή ™στιν Ð κύλινδρος τοà κώνου τριπλασίων, œσται Ð κύλινδρος τοà κώνου ½τοι µείζων À τριπλασίων À ™λάσσων À τριπλασίων. œστω πρότερον µείζων À τριπλασίων, κሠ™γγεγράφθω ε„ς τÕν ΑΒΓ∆ κύκλον τετράγωνον τÕ ΑΒΓ∆· τÕ δ¾ ΑΒΓ∆ τετράγωνον µείζόν ™στιν À τÕ ¼µισυ τοà ΑΒΓ∆ κύκλου. κሠ¢νεστάτω ¢πÕ τοà ΑΒΓ∆ τετραγώνου πρίσµα „σοϋψς τù κυλίνδρJ. τÕ δ¾ ¢νιστάµενον πρίσµα µε‹ζόν ™στιν À τÕ ¼µισυ τοà κυλίνδου, ™πειδήπερ κ¨ν περˆ τÕν ΑΒΓ∆ κύκλον τετράγωνον περιγράψωµεν, τÕ ™γγεγραµµένον ε„ς τÕν ΑΒΓ∆ κύκλον τετράγωνον ¼µισύ ™στι τοà περιγεγραµµένου· καί ™στι τ¦ ¢π' αÙτîν ¢νιστάµενα στερε¦ παραλληλεπίπεδα πρίσµατα „σοϋψÁ· τ¦ δ ØπÕ τÕ αÙτÕ Ûψος Ôντα στερε¦ παραλληλεπίπεδα πρÕς ¥λληλά ™στιν æς αƒ βάσεις· κሠτÕ ™πˆ τοà ΑΒΓ∆ ¥ρα τετραγώνου ¢νασταθν πρίσµα ¼µισύ ™στι τοà ¢νασταθέντος πρίσµατος ¢πÕ τοà περˆ τÕν ΑΒΓ∆ κύκλον περιγραφέντος τετραγώνου· καί ™στιν Ð κύλινδρος ™λάττων τοà πρίσµατος τοà ¢νατραθέντος ¢πÕ τοà περˆ τÕν ΑΒΓ∆ κύκλον περιγραφέντος τετραγώνου· τÕ ¥ρα πρίσµα τÕ ¢νασταθν ¢πÕ τοà ΑΒΓ∆ τετραγώνου „σοϋψς τù κυλίνδρJ µε‹ζόν ™στι τοà ¹µίσεως τοà κυλίνδρου. τετµήσθωσαν αƒ ΑΒ, ΒΓ, Γ∆, ∆Α περιφέρειαι δίχα κατ¦ τ¦ Ε, Ζ, Η, Θ σηµε‹α, κሠ™πεζεύχθωσαν αƒ ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, Η∆, ∆Θ, ΘΑ· κሠ›καστον ¥ρα τîν ΑΕΒ, ΒΖΓ, ΓΗ∆, ∆ΘΑ τριγώνων µειζόν ™στιν À τÕ ¼µισυ τοà καθ' ˜αυτÕ τηήµατος τοà ΑΒΓ∆ κύκλου, æς œµπροσθεν ™δείκνυµεν. ¢νεστάτω ™φ' ˜κάστου τîν ΑΕΒ, ΒΖΓ, ΓΗ∆, ∆ΘΑ τριγώνων πρίσµατα „σοϋψÁ τù κυλίνδρJ· κሠ›καστον ¥ρα τîν ¢νασταθέντων πρισµάτων µε‹ζόν ™στιν À τÕ ¼µισυ µέρος τοà καθ' ˜αυτÕ τµήµατος τοà κυλίνδρου, ™πει΄δηπερ ™¦ν δι¦ τîν Ε, Ζ, Η, Θ σηµείων παραλλήλους τα‹ς ΑΒ, ΒΓ, Γ∆, ∆Α ¢γάγωµεν, καˆ

For if the cylinder is not three times the cone then the cylinder will be either more than three times, or less than three times, (the cone). Let it, first of all, be more than three times (the cone). And let the square ABCD have been inscribed in circle ABCD [Prop. 4.6]. So, square ABCD is more than half of circle ABCD [Prop. 12.2]. And let a prism of equal height to the cylinder have been set up on square ABCD. So, the prism set up is more than half of the cylinder, inasmuch as if we also circumscribe a square around circle ABCD [Prop. 4.7] then the square inscribed in circle ABCD is half of the circumscribed (square). And the prisms set up on them are parallelepiped solids of equal height.† And parallelepiped solids having the same height are to one another as their bases [Prop. 11.32]. And, thus, the prism set up on square ABCD is half of the prism set up on the square circumscribed about circle ABCD. And the cylinder is less than the prism set up on the square circumscribed about circle ABCD. Thus, the prism set up on square ABCD of the same height as the cylinder is more than half of the cylinder. Let the circumferences AB, BC, CD, and DA have been cut in half at points E, F , G, and H. And let AE, EB, BF , F C, CG, GD, DH, and HA have been joined. And thus each of the triangles AEB, BF C, CGD, and DHA is more than half of the segment of circle ABCD about it, as was shown previously [Prop. 12.2]. Let prisms of equal height to the cylinder have been set up on each of the triangles AEB, BF C, CGD, and DHA. And each of the prisms set up is greater than the half part of the segment of the cylinder about it—inasmuch as if we draw (straight-lines) parallel to AB, BC, CD, and DA through points E, F , G, and H (respectively), and complete the parallelograms on AB,

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συµπληρώσωµεν τ¦ ™πˆ τîν ΑΒ, ΒΓ, Γ∆, ∆Α παραλληλόγραµµα, κሠ¢π' αÙτîν ¢ναστήσωµεν στερε¦ παραλληλεπίπεδα „σοϋψÁ τù κυλίνδρJ, ˜κάσου τîν ¢νασταθέντων ¹µίση ™στˆ τ¦ πρίσµατα τ¦ ™πˆ τîν ΑΕΒ, ΒΖΓ, ΓΗ∆, ∆ΘΑ τριγώνων· καί ™στι τ¦ τοà κυλίνδρου τµήµατα ™λάττονα τîν ¢νασταθέντων στερεîν παραλληλεπιπέδων· éστε κሠτ¦ ™πˆ τîν ΑΕΒ, ΒΖΓ, ΓΗ∆, ∆ΘΑ τριγώνων πρίσµατα µείζονά ™στιν À τÕ ¼µισυ τîν καθ' ˜αυτ¦ τοà κυλίνδρου τµηµάτων. τέµνοντες δ¾ τ¦ς Øπολειποµένας περιφερείας δίχα κሠ™πιζευγνύντες εÙθείας κሠ¢νιστάντες ™φ' ˜κάσου τîν τριγώνων πρίσµατα „σοϋψÁ τù κυλίνδρJ κሠτοàτο ¢εˆ ποιοàντες καταλείψοµέν τινα ¢ποτµήµατα τοà κυλίνδρου, § œσται ™λάττονα τÁς ØπεροχÁς, Î Øπερέχει Ð κυλίνδρος τοà τριπλασίου τοà κώνου. λελείφθω, κሠœστω τ¦ ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, Η∆, ∆Θ, ΘΑ· λοιπÕν ¥ρα τÕ πρίσµα, οá βάσις µν τÕ ΑΕΒΖΓΗ∆Θ πολύγωνον, Ûψος δ τÕ αÙτÕ τù κυλίνδρù, µε‹ζόν ™στˆν À τριπλάσιον τοà κώνου. ¢λλ¦ τÕ πρίσµα, οá βάσις µν ™στˆ τÕ ΑΕΒΖΓΗ∆Θ πολύγωνον, Ûψος δ τÕ αÙτÕ τù κυλίνδρJ, τριπλάσιόν ™στι τÁς πυραµίδος, Âς βάσις µέν ™στι τÕ ΑΕΒΖΓΗ∆Θ πολύγωνον, κορυφ¾ δ ¹ αÙτ¾ τù κώνJ· κሠ¹ πυρᵈς ¥ρα, Âς βάσις µέν [™στι] τÕ ΑΕΒΖΓΗ∆Θ πολύγωνον, κορυφ¾ δ ¹ αÙτ¾ τù κώνJ, µείζων ™στˆ τοà κώνου τοà βάσιν œχοντες τÕν ΑΒΓ∆ κύκλον. ¢λλ¦ κሠ™λάττων· ™µπεριέχεται γ¦ρ Øπ' αÙτοà· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα ™στˆν Ð κύλινδρος τοà κώνου µε‹ζων À τριπλάσιος. Λέγω δή, Óτι οÙδ ™λάττων ™στˆν À τριπλάσιος Ð κύλινδρος τοà κώνου. Ε„ γ¦ρ δυνατόν, œστω ™λάττων À τριπλάσιος Ð κύλινδρος τοà κώνου· ¢νάπαλιν ¥ρα Ð κîνος τοà κυλίνδρου µε‹ζων ™στˆν À τρίτον µέρος. ™γγεγράφθω δ¾ ε„ς τÕν ΑΒΓ∆ κύκλον τετράγωνον τÕ ΑΒΓ∆· τÕ ΑΒΓ∆ ¥ρα τετράγωνον µε‹ζόν ™στιν À τÕ ¼µισυ τοà ΑΒΓ∆ κύκλου. κሠ¢νεστάτω ¢πÕ τοà ΑΒΓ∆ τετραγώνου πυρᵈς τ¾ν αÙτ¾ν κορυφ¾ν œχουσα τù κώνJ· º ¥ρα ¢νασταθε‹σα πυρᵈς µείζων ™στˆν À τÕ ¼µισυ µέρος τοà κώνου, ™πειδήπερ, æς ›µπροσθεν ™δείκνυµεν, Óτι ™¦ν περˆ τÕν κύκλον τετράγωνον περιγράψωµεν, œσται τÕ ΑΒΓ∆ τετράγωνον ¼µισυ τοà περˆ τÕν κύκλον περιγεγραµµένου τετραγώνου· κሠ™¦ν ¢πÕ τîν τετραγώνων στερε¦ παραλληλεπίπεδα ¢ναστήσωµεν „σοϋψÁ τù κώνJ, ¨ κሠκαλε‹ται πρίσµατα, œσται τÕ ¢νασταθν ¢πÕ τοà ΑΒΓ∆ τετραγώνου ¼µισυ τοà ¢νασταθέντος ¢πÕ τοà περˆ τÕν κύκλον περιγραφέντος τετραγώνου· πρÕς ¥λληλα γάρ ε„σιν æς αƒ βάσεις. éστε κሠτ¦ τρίτα· κሠπυρᵈς ¥ρα, Âς βάσις τÕ ΑΒΓ∆ τετράγωνον, ¼µισύ ™στι τÁς πυραµίδος τÁς ¢νασταθείσης ¢πÕ τοà περˆ τÕν κύκλον περιγραφέντος τετραγώνου. καί ™στι µείζων ¹ πυρᵈς ¹ ¢νασταθε‹σα ¢πÕ τοà περˆ τÕν κύκλον τετραγώνου τοà κώνου· ™µπεριέχει γ¦ρ αÙτόν. ¹ ¥ρα

BC, CD, and DA, and set up parallelepiped solids of equal height to the cylinder on them, then the prisms on triangles AEB, BF C, CGD, and DHA are each half of the set up (parallelepipeds). And the segments of the cylinder are less than the set up parallelepiped solids. Hence, the prisms on triangles AEB, BF C, CGD, and DHA are also greater than half of the segments of the cylinder about them. So (if) the remaining circumferences are cut in half, and straight-lines are joined, and prisms of equal height to the cylinder are set up on each of the triangles, and this is done continually, then we will (eventually) leave some segments of the cylinder whose (sum) is less than the excess by which the cylinder exceeds three times the cone [Prop. 10.1]. Let them have been left, and let them be AE, EB, BF , F C, CG, GD, DH, and HA. Thus, the remaining prism whose base (is) polygon AEBF CGDH, and height the same as the cylinder, is greater than three times the cone. But, the prism whose base is polygon AEBF CGDH, and height the same as the cylinder, is three times the pyramid whose base is polygon AEBF CGDH, and apex the same as the cone [Prop. 12.7 corr.]. And thus the pyramid whose base [is] polygon AEBF CGDH, and apex the same as the cone, is greater than the cone having (as) base circle ABCD. But (it is) also less. For it is encompassed by it. The very thing (is) impossible. Thus, the cylinder is not more than three times the cone. So, I say that neither (is) the cylinder less than three times the cone. For, if possible, let the cylinder be less than three times the cone. Thus, inversely, the cone is greater than the third part of the cylinder. So, let the square ABCD have been inscribed in circle ABCD [Prop. 4.6]. Thus, square ABCD is greater than half of circle ABCD. And let a pyramid having the same apex as the cone have been set up on square ABCD. Thus, the pyramid set up is greater than the half part of the cone, inasmuch as we showed previously that if we circumscribe a square about the circle [Prop. 4.7], then the square ABCD will be half of the square circumscribed about the circle [Prop. 12.2]. And if we set up on the squares parallelepiped solids—which are also called prisms—of the same height as the cone, then the (prism) set up on square ABCD will be half of the (prism) set up on the square circumscribed about the circle. For they are to one another as their bases [Prop. 11.32]. Hence, (the same) also (goes for) the thirds. Thus, the pyramid whose base is square ABCD is half of the pyramid set up on the square circumscribed about the circle [Prop. 12.7 corr.]. And the pyramid set up on the square circumscribed about the circle is greater than the cone. For it encompasses it. Thus, the pyramid

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πυρᵈς, Âς βάσις τÕ ΑΒΓ∆ τετράγωνον, κορυφ¾ δ ¹ αÙτ¾ τù κώνJ, µείζων ™στˆν À τÕ ¼µισυ τοà κώνου. τετµήσθωσαν αƒ ΑΒ, ΒΓ, Γ∆, ∆Α περιφέρειαι δίχα κατ¦ τ¦ Ε, Ζ, Η, Θ σηµε‹α, κሠ™πεζεύχθωσαν αƒ ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, Η∆, ∆Θ, ΘΑ· κሠ›καστον ¥ρα τîν ΑΕΒ, ΒΖΓ, ΓΗ∆, ∆ΘΑ τριγώνων µε‹ζόν ™στιν À τÕ ¼µισυ µέρος του καθ' ˜αυτÕ τµήµατος τοà ΑΒΓ∆ κύκλου. κሠ¢νεστάτωσαν ™φ' ˜κάστου τîν ΑΕΒ, ΒΖΓ, ΓΗ∆, ∆ΘΑ τριγώνων πυραµίδες τ¾ν αÙτ¾ν κορυφ¾ν œχουσαι τù κώνJ· κሠ˜κάστη ¥ρα τîν ¢νασταθεισîν πυραµίδων κατ¦ τÕν αÙτÕν τρόπον µείζων ™στˆν À τÕ ¼µισυ µέρος τοà καθ' ˜αυτ¾ν τµήµατος τοà κώνου. τέµνοντες δ¾ τ¦ς Øπολειποµένας περιφερείας δίχα κሠ™πιζευγνύντες εÙθείας κሠ¢νιστάντες ™φ' ˜κάστου τîν τριγώνων πυραµίδα τ¾ν αÙτ¾ν κορυφ¾ν œχουσαν τù κώνJ κሠτοàτο ¢εˆ ποιοàτες καταλείψοµέν τινα ¢ποτµήµατα τοà κώνου, § œσται ™λάττονα τÁς ØπεροχÁς, Î Øπερέχει Ð κîνος τοà τρίτου µέρους τοà κυλίνδρου. λελείφθω, κሠœστω τ¦ ™πˆ τîν ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, Η∆, ∆Θ, ΘΑ· λοιπ¾ ¥ρα ¹ πυραµίς, Âς βάσις µέν ™στι τÕ ΑΕΒΖΓΗ∆Θ πολύγωνον, κορυφ¾ δ ¹ αÙτ¾ τù κώνJ, µείζων ™στˆν À τρίτον µέρος τοà κυλίνδρου. ¢λλ' ¹ πυραµίς, Âς βάσις µέν ™στι τÕ ΑΕΒΖΓΗ∆Θ πολύγωνον, κορυφ¾ δ ¹ αυτ¾ τù κώνJ, τρίτον ™στˆ µέρος τοà πρίσµατος, οá βάσις µέν ™στι τÕ ΑΕΒΖΓΗ∆Θ πολύγωνον, Ûψος δ τÕ αÙτÕ τù κυλίνδρJ· τÕ ¥ρα πρίσµα, οá βάσις µέν ™στι τÕ ΑΕΒΖΓΗ∆Θ πολύγωνον, Ûψος δ τÕ αÙτÕ τù κυλίνδρJ, µε‹ζόν ™στι τοà κυλίνδρου, οá βάσις ™στˆν Ð ΑΒΓ∆ κύκλος. ¢λλ¦ κሠœλαττον· ™µπεριέχεται γ¦ρ Øπ' αÙτοà· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα Ð κύλινδρος τοà κώνου ™λάττων ™στˆν À τριπλάσιος. ™δείχθη δέ, Óτι οÙδ µείζων À τριπλάσιος· τριπλάσιος ¥ρα Ð κύλινδρος τοà κώνου· éστε Ð κîνος τρίτον ™στˆ µέρος τοà κυλίνδρου. Π©ς ¥ρα κîνος κυλίνδρου τρίτον µέρος ™στˆ τοà τ¾ν αÙτ¾ν βάσιν œχοντος αÙτù κሠÛψος ‡σον· Óπερ œδει δε‹ξαι.

†

whose base is square ABCD, and apex the same as the cone, is greater than half of the cone. Let the circumferences AB, BC, CD, and DA have been cut in half at points E, F , G, and H (respectively). And let AE, EB, BF , F C, CG, GD, DH, and HA have been joined. And, thus, each of the triangles AEB, BF C, CGF , and DHA is greater than the half part of the segment of circle ABCD about it [Prop. 12.2]. And let pyramids having the same apex as the cone have been set up on each of the triangles AEB, BF C, CGF , and DHA. And, thus, in the same way, each of the pyramids set up is more than the half part of the segment of the cone about it. So, (if) the remaining circumferences are cut in half, and straightlines are joined, and pyramids having the same apex as the cone are set up on each of the triangles, and this is done continually, then we will (eventually) leave some segments of the cone whose (sum) is less than the excess by which the cone exceeds the third part of the cylinder [Prop. 10.1]. Let them have been left, and let them be the (segments) on AE, EB, BF , F C, CG, GD, DH, and HA. Thus, the remaining pyramid whose base is polygon AEBF CGDH, and apex the same as the cone, is greater than the third part of the cylinder. But, the pyramid whose base is polygon AEBF CGDH, and apex the same as the cone, is the third part of the prism whose base is polygon AEBF CGDH, and height the same as the cylinder [Prop. 12.7 corr.]. Thus, the prism whose base is polygon AEBF CGDH, and height the same as the cylinder, is greater than the cylinder whose base is circle ABCD. But, (it is) also less. For it is encompassed by it. The very thing is impossible. Thus, the cylinder is not less than three times the cone. And it was shown that neither (is it) greater than three times (the cone). Thus, the cylinder (is) three times the cone. Hence, the cone is the third part of the cylinder. Thus, every cone is the third part of the cylinder which has the same base as it, and an equal height. (Which is) the very thing it was required to show.

The Greek text mistakenly inverts “prisms” and “parallelepiped solids”.

ια΄.

Proposition 11

Οƒ Øπο τÕ αÙτÕ Ûψος Ôντες κîνοι κሠκύλινδροι πρÕς ¢λλήλους ε„σˆν æς αƒ βάσεις. ”Εστωσαν ØπÕ τÕ αÙτÕ Ûψος κîνοι κሠκύλινδροι, ïν βάσεις µν [ε„σιν] οƒ ΑΒΓ∆, ΕΖΗΘ κύκλοι, ¥ξονες δ οƒ ΚΛ, ΜΝ, διάµετροι δ τîν βάσεων αƒ ΑΓ, ΕΗ· λέγω, Óτι ™στˆν æς Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως Ð ΑΛ κîνος πρÕς τÕν ΕΝ κîνον.

Cones and cylinders having the same height are to one another as their bases. Let there be cones and cylinders of the same height whose bases [are] the circles ABCD and EF GH, axes KL and M N , and diameters of the bases AC and EG (respectively). I say that as circle ABCD is to circle EF GH, so cone AL (is) to cone EN .

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Ε„ γ¦ρ µή, œσται æς Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως Ð ΑΛ κîνος ½τοι πρÕς œλασσόν τι τοà ΕΝ κώνου στερεÕν À πρÕς µε‹ζον. œστω πρότερον πρÕς œλασσον τÕ Ξ, κሠú œλασσόν ™στι τÕ Ξ στερεÕν τοà ΕΝ κώνου, ™κείνJ ‡σον œστω τÕ Ψ στερεόν· Ð ΕΝ κîνος ¥ρα ‡σος ™στˆ το‹ς Ξ, Ψ στερεο‹ς. ™γγεγράφθω ε„ς τÕν ΕΖΗΘ κύκλον τετράγωνον τÕ ΕΖΗΘ· τÕ ¥ρα τετράγωνον µε‹ζόν ™στιν À τÕ ¼µισυ τοà κύκλου. ¢νεστάτω ¢πÕ τοà ΕΖΗΘ τετραγώνου πυρᵈς „σοϋψ¾ς τù κώνJ· ¹ ¥ρα ¢νασταθε‹σα πυρᵈς µείζων ™στˆν À τÕ ¼µισυ τοà κώνου, ™πειδήπερ ™¦ν περιγράψωµεν περˆ τÕν κύκλον τετράγωνον, κሠ¢π' αÙτοà ¢ναστήσωµεν πυραµίδα „σοϋψÁ τù κώνJ, ¹ ™γγραφε‹σα πυρᵈς ¼µισύ ™στι τÁς περιγραφείσης· πρÕς ¢λλήλας γάρ ε„σιν æς αƒ βάσεις· ™λάττων δ Ð κîνος τÁς περιγραφείσης πυραµίδος. τετµήσθωσαν αƒ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι δίχα κατ¦ τ¦ Ο, Π, Ρ, Σ σηµε‹α, κሠ™πεζεύχθωσαν αƒ ΘΟ, ΟΕ, ΕΠ, ΠΖ, ΖΡ, ΡΗ, ΗΣ, ΣΘ. ›καστον ¥ρα τîν ΘΟΕ, ΕΠΖ, ΖΡΗ, ΗΖΘ τριγώνων µε‹ζόν ™στιν À τÕ ¼µισυ τοà καθ' ˜αυτÕ τµήµατος τοà κύκλου. ¢νεστάτω ™φ' ˜κάστου τîν ΘΟΕ, ΕΠΖ, ΖΡΗ, ΗΣΘ τριγώνων πυρᵈς „σοϋψ¾ς τù κώνJ· κሠ˜κάστη ¥ρα τîν ¢νασταθεισîν πυραµίδων µείζων ™στˆν À τÕ ¼µισυ τοà καθ' ˜αυτ¾ν τµήµατος τοà κώνου. τέµνοντες δ¾ τ¦ς Øπολειποµένας περιφερείας δίχα κሠ™πιζευγνύντες εÙθείας κሠ¢νιστάντες ™πˆ ˜κάστου τîν τριγώνων πυραµίδας „σοϋψε‹ς τù κώνJ κሠ¢εˆ τοàτο ποιοàντες καταλείψοµέν τινα ¢ποτµήµατα τοà κώνου, § œσται ™λάσσονα τοà Ψ στερεοà. λελείφθω, κሠœστω τ¦ ™πˆ τîν ΘΟΕ, ΕΠΖ, ΖΡΗ, ΗΣΘ· λοιπ¾ ¥ρα ¹ πυραµίς, Âς βάσις τÕ ΘΟΕΠΖΡΗΣ πολύγωνον, Ûψος δ τÕ αÙτÕ τù κώνJ, µείζων ™στˆ τοà Ξ στερεοà. ™γγεγράφθω κሠε„ς τÕν ΑΒΓ∆ κύκλον τù ΘΟΕΠΖΡΗΣ πολυγώνJ Óµοιόν τε καˆ Ðµοίως κείµενον πολύγωνον τÕ ∆ΤΑΥΒΦΓΧ, κሠ¢νεστάτω ™π' αÙτοà πυρᵈς „σοϋψ¾ς τù ΑΛ κώνJ. ™πεˆ οâν ™στιν æς τÕ ¢πÕ τÁς ΑΓ πρÕς τÕ ¢πÕ τÁς ΕΗ, οÛτως τÕ ∆ΤΑΥΒΦΓΧ πολύγωνον πρÕς τÕ ΘΟΕΠΖΡΗΣ πολύγωνον, æς δ τÕ ¢πÕ τÁς ΑΓ πρÕς τÕ ¢πÕ τÁς ΕΗ, οÛτως Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, κሠæς ¥ρα Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως τÕ ∆ΤΑΥΒΦΓΧ πολύγωνον πρÕς τÕ ΘΟΕΠΖΡΗΣ πολύγωνον. æς δ Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως Ð

C

K

E

N

S

P

G

M

X

O

R

Q F

For if not, then as circle ABCD (is) to circle EF GH, so cone AL will be to some solid either less than, or greater than, cone EN . Let it, first of all, be (in this ratio) to (some) lesser (solid), O. And let solid X be equal to that (magnitude) by which solid O is less than cone EN . Thus, cone EN is equal to (the sum of) solids O and X. Let the square EF GH have been inscribed in circle EF GH [Prop. 4.6]. Thus, the square is greater than half of the circle [Prop. 12.2]. Let a pyramid of the same height as the cone have been set up on square EF GH. Thus, the pyramid set up is greater than half of the cone, inasmuch as, if we circumscribe a square about the circle [Prop. 4.7], and set up a pyramid of the same height as the cone on it, then the inscribed pyramid is half of the circumscribed pyramid. For they are to one another as their bases [Prop. 12.6]. And the cone (is) less than the circumscribed pyramid. Let the circumferences EF , F G, GH, and HE have been cut in half at points P , Q, R, and S. And let HP , P E, EQ, QF , F R, RG, GS, and SH have been joined. Thus, each of the triangles HP E, EQF , F RG, and GSH is greater than half of the segment of the circle about it [Prop. 12.2]. Let pyramids of the same height as the cone have been set up on each of the triangles HP E, EQF , F RG, and GSH. And, thus, each of the pyramids set up is greater than half of the segment of the cone about it [Prop. 12.10]. So, (if) the remaining circumferences are cut in half, and straight-lines are joined, and pyramids of equal height to the cone are set up on each of the triangles, and this is done continually, then we will (eventually) leave some segments of the cone (the sum of) which is less than solid X [Prop. 10.1]. Let them have been left, and let them be the (segments) on HP E, EQF , F RG, and GSH. Thus, the remaining pyramid whose base is polygon HP EQF RGS, and height the same as the cone, is greater than solid O [Prop. 6.18]. And let the polygon DT AU BV CW , similar, and similarly laid out, to polygon HP EQF RHS, have been inscribed in circle ABCD. And let a pyramid of the same height as cone AL have been set up on it. Therefore, since as the (square) on AC is to the (square) on EG, so polygon DT AU BV CW (is) to polygon HP EQF RGS [Prop. 12.1], and as the (square) on AC (is) to the (square) on EG, so circle ABCD (is)

490

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

ΑΛ κîνος πρÕς τÕ Ξ στερεόν, æς δ τÕ ∆ΤΑΥΒΦΓΧ πολύγωνον πρÕς τÕ ΘΟΕΠΖΡΗΣ πολύγωνον, οÛτως ¹ πυραµίς, Âς βάσις µν τÕ ∆ΤΑΥΒΦΓΧ πολύγωνον, κορυφ¾ δ τÕ Λ σηµε‹ον, πρÕς τ¾ν πυραµίδα, Âς βάσις µν τÕ ΘΟΕΠΖΡΗΣ πολύγωνον, κορυφ¾ δ τÕ Ν σηµε‹ον. κሠæς ¥ρα Ð ΑΛ κîνος πρÕς τÕ Ξ στερεόν, οÛτως ¹ πυραµίς, Âς βάσις µν τÕ ∆ΤΑΥΒΦΓΧ πολύγωνον, κορυφ¾ δ τÕ Λ σηµε‹ον, πρÕς τ¾ν πυραµίδα, Âς βάσις µν τÕ ΘΟΕΠΖΡΗΣ πολύγωνον, κορυφ¾ δ τÕ Ν σηµε‹ον· ™ναλλ¦ξ ¥ρα ™στˆν æς Ð ΑΛ κîνος πρÕς τ¾ν ™ν αÙτù πυραµίδα, οÛτως τÕ Ξ στερεÕν πρÕς τ¾ν ™ν τù ΕΝ κώνJ πυραµίδα. µείζων δ Ð ΑΛ κîνος τÁς ™ν αÙτù πυραµίδος· µε‹ζον ¥ρα κሠτÕ Ξ στερεÕν τ¾ς ™ν τù ΕΝ κώνJ πυραµίδος. ¢λλ¦ κሠœλασσον· Óπερ ¥τοπον. οÙκ ¥ρα ™στˆν æς Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως Ð ΑΛ κîνος πρÕς œλασσόν τι τοà ΕΝ κώνου στερεόν. еοίως δ δείξοµεν, Óτι οÙδέ ™στιν æς Ð ΕΖΗΘ κύκλος πρÕς τÕν ΑΒΓ∆ κύκλον, οÛτως Ð ΕΝ κîνος πρÕς œλασσόν τι τοà ΑΛ κώνου στερεόν. Λέγω δή, Óτι οÙδέ ™στιν æς Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως Ð ΑΛ κîνος πρÕς µε‹ζόν τι τοà ΕΝ κώνου στερεόν. Ε„ γ¦ρ δυνατόν, ›στω πρÕς µε‹ζον τÕ Ξ· ¢νάπαλιν ¥ρα ™στˆν æς Ð ΕΖΗΘ κύκλος πρÕς τÕν ΑΒΓ∆ κύκλον, οÛτως τÕ Ξ στερεÕν πρÕς τÕν ΑΛ κîνον. ¢λλ' æς τÕ Ξ στερεÕν πρÕς τÕν ΑΛ κîνον, οÛτως Ð ΕΝ κîνος πρÕς œλασσόν τι τοà ΑΛ κώνου στερεόν· κሠæς ¥ρα Ð ΕΖΗΘ κύκλος πρÕς τÕν ΑΒΓ∆ κύκλον, οÛτως Ð ΕΝ κîνος πρÕς œλασσόν τι τοà ΑΛ κώνου στερεόν· Óπερ ¢δύνατον ™δείχθη. οÙκ ¥ρα ™στˆν æς Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως Ð ΑΛ κîνος πρÕς µε‹ζόν τι τοà ΕΝ κώνου στερεόν. ™δείχθη δέ, Óτι οÙδ πρÕς œλασσον· œστιν ¥ρα æς Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως Ð ΑΛ κîνος πρÕς τÕν ΕΝ κîνον. 'Αλλ' æς Ð κîνος πρÕς τÕν κîνον, Ð κύλινδρος πρÕς τÕν κύλινδρον· τριπλασίων γ¦ρ ˜κάτερος ˜κατέρου. κሠæς ¥ρα Ð ΑΒΓ∆ κύκλος πρÕς τÕν ΕΖΗΘ κύκλον, οÛτως οƒ ™π' αÙτîν „σοϋψε‹ς. Οƒ ¥ρα ØπÕ τÕ αÙτÕ Ûψος Ôντες κîνοι κሠκύλινδροι πρÕς ¢λλήλους ε„σˆν æς αƒ βάσεις· Óπερ œδει δε‹ξαι.

to circle EF GH [Prop. 12.2], thus as circle ABCD (is) to circle EF GH, so polygon DT AU BV CW also (is) to polygon HP EQF RGS. And as circle ABCD (is) to circle EF GH, so cone AL (is) to solid O. And as polygon DT AU BV CW (is) to polygon HP EQF RGS, so the pyramid whose base is polygon DT AU BV CW , and apex the point L, (is) to the pyramid whose base is polygon HP EQF RG, and base the point N [Prop. 12.6]. And, thus, as cone AL (is) to solid O, so the pyramid whose base is DT AU BV CW , and apex the point L, (is) to the pyramid whose base is polygon HP EQF RG, and apex the point N [Prop. 5.11]. Thus, alternately, as cone AL is to the pyramid within it, so solid O (is) to the pyramid within cone EN [Prop. 5.16]. But, cone AL (is) greater than the pyramid within it. Thus, solid O (is) also greater than the pyramid within cone EN [Prop. 5.14]. But, (it is) also less. The very thing (is) absurd. Thus, circle ABCD is not to circle EF GH, as cone AL (is) to some solid less than cone EN . So, similarly, we can show that neither is circle EF GH to circle ABCD, as cone EN (is) to some solid less than cone AL. So, I say that neither is circle ABCD to circle EF GH, as cone AL (is) to some solid greater than cone EN . For, if possible, let it be (in this ratio) to (some) greater (solid), O. Thus, inversely, as circle EF GH is to circle ABCD, so solid O (is) to cone AL [Prop. 5.7 corr.]. But, as solid O (is) to cone AL, so cone EN (is) to some solid less than cone AL [Prop. 12.2 lem.]. And, thus, as circle EF GH (is) to circle ABCD, so cone EN (is) to some solid less than cone AL. The very thing was shown (to be) impossible. Thus, circle ABCD is not to circle EF GH, as cone AL (is) to some solid greater than cone EN . And, it was shown that neither (is it in this ratio) to (some) lesser (solid). Thus, as circle ABCD is to circle EF GH, so cone AL (is) to cone EN . But, as the cone (is) to the cone, (so) the cylinder (is) to the cylinder. For each (is) three times each [Prop. 12.10]. Thus, circle ABCD (is) also to circle EF GH, as (the ratio of the cylinders) on them (having) the same height. Thus, cones and cylinders having the same height are to one another as their bases. (Which is) the very thing it was required to show.

ιβ΄.

Proposition 12

Οƒ Óµοιοι κîνοι κሠκύλινδροι πρÕς ¢λλήλους ™ν Similar cones and cylinders are to one another in the τριπλασίονι λόγJ ε„σˆ τîν ™ν τα‹ς βάσεσι διαµέτρων. cubed ratio of the diameters of their bases. ”Εστωσαν Óµοιοι κîνοι κሠκύλινδροι, ïν βάσεις µν Let there be similar cones and cylinders of which the οƒ ΑΒΓ∆, ΕΖΗΘ κύκλοι, διάµετροι δ τîν βάσεων αƒ bases (are) the circles ABCD and EF GH, the diameters Β∆, ΖΘ, ¥ξονες δ τîν κώνων κሠκυλίνδρων οƒ ΚΛ, of the bases (are) BD and F H, and the axes of the cones

491

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

ΜΝ· λέγω, Óτι Ð κîνος, οá βάσις µέν [™στιν] Ð ΑΒΓ∆ κύκλος, κορυφ¾ δ τÕ Λ σηµε‹ον, πρÕς τÕν κîνον, οá βάσις µέν [™στιν] Ð ΕΖΗΘ κύκλος, κορυφ¾ δ τÕ Ν σηµε‹ον, τριπλασίονα λόγον œχει ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ.

N

L T B

A

Q

G

F

E

N L E

O

A

Z

M R

H

X

B

P

D

K V

U C

Ε„ γ¦ρ µ¾ œχει Ð ΑΒΓ∆Λ κîνος πρÕς τÕν ΕΖΗΘΝ κîνον πριπλασίονα λόγον ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ, ›ξει Ð ΑΒΓ∆Λ κîνος À πρÕς œλασσόν τι τοà ΕΖΗΘΝ κώνου στερεÕν τριπλασίονα λόγον À πρÕς µε‹ζον. ™χέτω πρότερον πρÕς œλασσον τÕ Ξ, κሠ™γγεγράφθω ε„ς τÕν ΕΖΗΘ κύκλον τετράγωνον τÕ ΕΖΗΘ· τÕ ¥ρα ΕΖΗΘ τετράγωνον µε‹ζόν ™στιν À τÕ ¼µισυ τοà ΕΖΗΘ κύκλου. κሠ¢νεστάτω ™πˆ τοà ΕΖΗΘ τετραγώνου πυρᵈς τ¾ν αÙτ¾ν κορυφ¾ν œχουσα τù κώνJ· ¹ ¥ρα ¢νασταθε‹σα πυρᵈς µείζων ™στˆν À τÕ ¼µισυ µέρος τοà κώνου. τετµήσθωσαν δ¾ αƒ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι δίχα κατ¦ τ¦ Ο, Π, Ρ, Σ σηµε‹α, κሠ™πεζεύχθωσαν αƒ ΕΟ, ΟΖ, ΖΠ, ΠΗ, ΗΡ, ΡΘ, ΘΣ, ΣΕ. κሠ›καστον ¥ρα τîν ΕΟΖ, ΖΠΗ, ΗΡΘ, ΘΣΕ τριγώνων µε‹ζόν ™στιν À τÕ ¼µισυ µέρος τοà καθ' ˜αυτÕ τµήµατος τοà ΕΖΗΘ κύκλου. κሠ¢νεστάτω ™φ' ˜κάστου τîν ΕΟΖ, ΖΠΗ, ΗΡΘ, ΘΣΕ τριγώνων πυρᵈς τ¾ν αÙτ¾ν κορυφ¾ν œχουσα τù κώνJ· κሠ˜κάστη ¥ρα τîν ¢νασταθεισîν πυραµίδων µείζων ™στˆν À τÕ ¼µισυ µέρος τοà καθ' ˜αυτ¾ν τµήµατος τοà κώνου. τέµνοντες δ¾ τ¦ς Øπολειποµένας περιφερείας δίχα κሠ™πιζευγνύντες εÙθείας κሠ¢νιστάντες ™φ' ˜κάστου τîν τριγώνων πυραµίδας τ¾ν αÙτ¾ν κορυφ¾ν ™χούσας τù κώνJ κሠτοàτο ¢εˆ ποιοàντες καταλείψοµέν τινα ¢ποτµήµατα τοà κώνου, § œσται ™λάσσονα τÁς ØπεροχÁς, Î Øπερέχει Ð ΕΖΗΘΝ κîνος τοà Ξ στερεοà. λελείφθω, κሠœστω τ¦ ™πˆ τîν ΕΟ, ΟΖ, ΖΠ, ΠΗ, ΗΡ, ΡΘ, ΘΣ, ΣΕ· λοιπ¾ ¥ρα ¹ πυραµίς, Âς βάσις µέν ™στι τÕ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφ¾ δ τÕ Ν σηµε‹ον, µείζων ™στˆ τοà Ξ στερεοà. ™γγεγράφθω κሠε„ς τÕν ΑΒΓ∆ κύκλον τù ΕΟΖΠΗΡΘΣ πολυγώνJ Óµοιόν τε καˆ Ðµοίως κείµενον πολύγωνον τÕ ΑΤΒΥΓΦ∆Χ, κሠ¢νεστάτω ™πˆ τοà ΑΤΒΥΓΦ∆Χ πολυγώνου πυρᵈς τ¾ν αÙτ¾ν κορυφ¾ν œχουσα τù κώνJ, κሠτîν µν περιεχόντων τ¾ν πυραµίδα, Âς βάσις µέν ™στι τÕ ΑΤΒΥΓΦ∆Χ πολύγωνον, κορυφ¾ δ τÕ Λ σηµε‹ον,

S

P

W

T

D J

K U

S

and cylinders (are) KL and M N (respectively). I say that the cone whose base [is] circle ABCD, and apex the point L, has to the cone whose base [is] circle EF GH, and apex the point N , the cubed ratio that BD (has) to F H.

H

F

M

O

Q

R G

For if cone ABCDL does not have to cone EF GHN the cubed ratio that BD (has) to F H then cone ABCDL will have the cubed ratio to some solid either less than, or greater than, cone EF GHN . Let it, first of all, have (such a ratio) to (some) lesser (solid), O. And let the square EF GH have been inscribed in circle EF GH [Prop. 4.6]. Thus, square EF GH is greater than half of circle EF GH [Prop. 12.2]. And let a pyramid having the same apex as the cone have been set up on square EF GH. Thus, the pyramid set up is greater than the half part of the cone [Prop. 12.10]. So, let the circumferences EF , F G, GH, and HE have been cut in half at points P , Q, R, and S (respectively). And let EP , P F , F Q, QG, GR, RH, HS, and SE have been joined. And, thus, each of the triangles EP F , F QG, GRH, and HSE is greater than the half part of the segment of circle EF GH about it [Prop. 12.2]. And let a pyramid having the same apex as the cone have been set up on each of the triangles EP F , F QG, GRH, and HSE. And thus each of the pyramids set up is greater than the half part of the segment of the cone about it [Prop. 12.10]. So, (if) the the remaining circumferences are cut in half, and straight-lines are joined, and pyramids having the same apex as the cone are set up on each of the triangles, and this is done continually, then we will (eventually) leave some segments of the cone whose (sum) is less than the excess by which cone EF GHN exceeds solid O [Prop. 10.1]. Let them have been left, and let them be the (segments) on EP , P F , F Q, QG, GR, RH, HS, and SE. Thus, the remaining pyramid whose base is polygon EP F QGRHS, and apex the point N , is greater than solid O. And let the polygon AT BU CV DW , similar, and similarly laid out, to polygon EP F QGRHS, have been inscribed in circle ABCD [Prop. 6.18]. And let a pyramid having the same apex as the cone have been set up on polygon AT BU CV DW . And let LBT be one

492

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

žν τρίγωνον œστω τÕ ΛΒΤ, τîν δ περειχόντων τ¾ν πυραµίδα, Âς βάσις µέν ™στι τÕ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφ¾ δ τÕ Ν σηµε‹ον, žν τρίγωνον œστω τÕ ΝΖΟ, κሠ™πεζεύχθωσαν αƒ ΚΤ, ΜΟ. κሠ™πεˆ Óµοιός ™στιν Ð ΑΒΓ∆Λ κîνος τù ΕΖΗΘΝ κώνJ, œστιν ¥ρα æς ¹ Β∆ πρÕς τ¾ν ΖΘ, οÛτως Ð ΚΛ ¥ξων πρÕς τÕν ΜΝ ¥ξονα. æς δ ¹ Β∆ πρÕς τ¾ν ΖΘ, οÛτως ¹ ΒΚ πρÕς τ¾ν ΖΜ· κሠæς ¥ρα ¹ ΒΚ πρÕς τ¾ν ΖΜ, οÛτως ¹ ΚΛ πρÕς τ¾ν ΜΝ. κሠ™ναλλ¦ξ æς ¹ ΒΚ πρÕς τ¾ν ΚΛ, οÛτως ¹ ΖΜ πρÕς τ¾ν ΜΝ. κሠπερˆ ‡σας γωνίας τ¦ς ØπÕ ΒΚΛ, ΖΜΝ αƒ πλευρሠ¢νάλογόν ε„σιν· Óµοιον ¥ρα ™στˆ τÕ ΒΚΛ τρίγωνον τù ΖΜΝ τριγώνJ. πάλιν, ™πεί ™στιν æς ¹ ΒΚ πρÕς τ¾ν ΚΤ, οÛτως ¹ ΖΜ πρÕς τ¾ν ΜΟ, κሠπερˆ ‡σας γωνίας τ¦ς ØπÕ ΒΚΤ, ΖΜΟ, ™πειδήπερ, Ö µέρος ™στˆν ¹ ØπÕ ΒΚΤ γωνία τîν πρÕς τù Κ κέντρJ τεσσάρων Ñρθîν, τÕ αÙτÕ µέρος ™στˆ κሠ¹ ØπÕ ΖΜΟ γωνία τîν πρÕς τù Μ κέντρJ τεσσάρων Ñρθîν· ™πεˆ οâν περˆ ‡σας γωνίας αƒ πλευρሠ¢νάλογόν ε„σιν, Óµοιον ¥ρα ™στι τÕ ΒΚΤ τρίγωνον τù ΖΜΟ τριγώνJ. πάλιν, ™πεˆ ™δείχθη æς ¹ ΒΚ πρÕς τ¾ν ΚΛ, οÛτως ¹ ΖΜ πρÕς τ¾ν ΜΝ, ‡ση δ ¹ µν ΒΚ τÍ ΚΤ, ¹ δ ΖΜ τÍ ΟΜ, œστιν ¥ρα æς ¹ ΤΚ πρÕς τ¾ν ΚΛ, οÛτως ¹ ΟΜ πρÕς τ¾ν ΜΝ. κሠπερˆ ‡σας γωνίας τ¦ς ØπÕ ΤΚΛ, ΟΜΝ· Ñρθሠγάρ· αƒ πλευρሠ¢νάλογόν ε„σιν· Óµοιον ¥ρα ™στˆ τÕ ΛΚΤ τρίγωνον τù ΝΜΟ τριγώνJ. κሠ™πεˆ δι¦ τ¾ν еοιότητα τîν ΛΚΒ, ΝΜΖ τριγώνων ™στˆν æς ¹ ΛΒ πρÕς τ¾ν ΒΚ, οÛτως ¹ ΝΖ πρÕς τ¾ν ΖΜ, δι¦ δ τ¾ν еοιότητα τîν ΒΚΤ, ΖΜΟ τριγώνων ™στˆν æς ¹ ΚΒ πρÕς τ¾ν ΒΤ, οÛτως ¹ ΜΖ πρÕς τ¾ν ΖΟ, δι' ‡σου ¥ρα æς ¹ ΛΒ πρÕς τ¾ν ΒΤ, οÛτως ¹ ΝΖ πρÕς τ¾ν ΖΟ. πάλιν, ™πεˆ δι¦ τ¾ν οµοιότητα τîν ΛΤΚ, ΝΟΜ τριγώνων ™στˆν æς ¹ ΛΤ πρÕς τ¾ν ΤΚ, οÛτως ¹ ΝΟ πρÕς τ¾ν ΟΜ, δι¦ δ τ¾ν еοιότητα τîν ΤΚΒ, ΟΜΖ τριγώνων ™στˆν æς ¹ ΚΤ πρÕς τ¾ν ΤΒ, οÛτως ¹ ΜΟ πρÕς τ¾ν ΟΖ, δι' ‡σου ¥ρα æς ¹ ΛΤ πρÕς τ¾ν ΤΒ, οÛτως ¹ ΝΟ πρÕς τ¾ν ΟΖ. ™δείχθη δ κሠæς ¹ ΤΒ πρÕς τ¾ν ΒΛ, οÛτως ¹ ΟΖ πρÕς τ¾ν ΖΝ. δι' ‡σου ¥ρα æς ¹ ΤΛ πρÕς τ¾ν ΛΒ, οÛτως ¹ ΟΝ πρÕς τ¾ν ΝΖ. τîν ΛΤΒ, ΝΟΖ ¥ρα τριγώνων ¢νάλογόν ε„σιν αƒ πλευραί· „σογώνια ¥ρα ™στˆ τ¦ ΛΤΒ, ΝΟΖ τρίγωνα· éστε καˆ Óµοια. κሠπυρᵈς ¥ρα, Âς βάσις µν τÕ ΒΚΤ τρίγωνον, κορυφ¾ δ τÕ Λ σηµε‹ον, еοία ™στˆ πυραµίδι, Âς βάσις µν τÕ ΖΜΟ τρίγωνον, κορυφ¾ δ τÕ Ν σηµε‹ον· ØπÕ γ¦ρ Óµοίων ™πιπέδων περιέχονται ‡σων τÕ πλÁθος. αƒ δ Óµοιαι πυραµίδες κሠτριγώνους œχουσαι βάσεις ™ν τριπλασίονι λόγJ ε„σˆ τîν еολόγων πλευρîν. ¹ ¥ρα ΒΚΤΛ πυρᵈς πρÕς τ¾ν ΖΜΟΝ πυραµίδα τριπλασίονα λόγον œχει ½περ ¹ ΒΚ πρÕς τ¾ν ΖΜ. еοίως δ¾ ™πιζευγνύντες ¢πÕ τîν Α, Χ, ∆, Φ, Γ, Υ ™πˆ τÕ Κ εÙθείας κሠ¢πÕ τîν Ε, Σ, Θ, Ρ, Η, Π ™πˆ τÕ Μ κሠ¢νιστάντες ™φ' ˜κάστου τîν τριγώνων πυραµίδας τ¾ν αÙτ¾ν κορυφ¾ν ™χούσας

triangle which containes the pyramid whose base is polygon AT BU CV DW , and apex the point L. And let N F P be one triangle which contains the pyramid whose base is triangle EP F QGRHS, and apex the point N . And let KT and M P have been joined. And since cone ABCDL is similar to cone EF GHN , thus as BD is to F H, so axis KL (is) to axis M N [Def. 11.24]. And as BD (is) to F H, so BK (is) to F M . And, thus, as BK (is) to F M , so KL (is) to M N . And, alternately, as BK (is) to KL, so F M (is) to M N [Prop. 5.16]. And the sides around the equal angles BKL and F M N are proportional. Thus, triangle BKL is similar to triangle F M N [Prop. 6.6]. Again, since as BK (is) to KT , so F M (is) to M P , and (they are) about the equal angles BKT and F M P , inasmuch as whatever part angle BKT is of the four rightangles at center K, angle F M P is also the same part of the four right-angles at center M . Therefore, since the sides about equal angles are proportional, triangle BKT is thus similar to traingle F M P [Prop. 6.6]. Again, since it was shown that as BK (is) to KL, so F M (is) to M N , and BK (is) equal to KT , and F M to P M , thus as T K (is) to KL, so P M (is) to M N . And the sides about the equal angles T KL and P M N —for (they are both) rightangles—are proportional. Thus, triangle LKT (is) similar to triangle N M P [Prop. 6.6]. And since, on account of the similarity of triangles LKB and N M F , as LB (is) to BK, so N F (is) to F M , and, on account of the similarity of triangles BKT and F M P , as KB (is) to BT , so M F (is) to F P [Def. 6.1], thus, via equality, as LB (is) to BT , so N F (is) to F P [Prop. 5.22]. Again, since, on account of the similarity of triangles LT K and N P M , as LT (is) to T K, so N P (is) to P M , and, on account of the similarity of triangles T KB and P M F , as KT (is) to T B, so M P (is) to P F , thus, via equality, as LT (is) to T B, so N P (is) to P F [Prop. 5.22]. And it was shown that as T B (is) to BL, so P F (is) to F N . Thus, via equality, as T L (is) to LB, so P N (is) to N F [Prop. 5.22]. Thus, the sides of triangles LT B and N P F are proportional. Thus, triangles LT B and N P F are equiangular [Prop. 6.5]. And, hence, (they are) similar [Def. 6.1]. And, thus, the pyramid whose base is triangle BKT , and apex the point L, is similar to the pyramid whose base is triangle F M P , and apex the point N . For they are contained by equal numbers of similar planes [Def. 11.9]. And similar pyramids which also have triangular bases are in the cubed ratio of corresponding sides [Prop. 12.8]. Thus, pyramid BKT L has to pyramid F M P N the cubed ratio that BK (has) to F M . So, similarly, joining straight-lines from (points) A, W , D, V , C, and U to (center) K, and from (points) E, S, H, R, G, and Q to (center) M , and setting up pyramids having the same apexes as the cones

493

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

το‹ς κώνοις δείξοµεν, Óτι κሠ˜κάστη τîν еοταγîν πυραµίδων πρÕς ˜κάστην еοταγÁ πυραµίδα τριπλασίονα λόγον ›ξει ½περ ¹ ΒΚ Ðµόλογος πλευρ¦ πρÕς τ¾ν ΖΜ Ðµόλογον πλευράν, τουτέστιν ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ. κሠæς žν τîν ¹γουµένων πρÕς žν τîν ˜ποµένων, οÛτως ¤παντα τ¦ ¹γούµενα πρÕς ¤παντα τ¦ ˜πόµενα· œστιν ¥ρα κሠæς ¹ ΒΚΤΛ πυρᵈς πρÕς τ¾ν ΖΜΟΝ πυραµίδα, οÛτως ¹ Óλη πυραµίς, Âς βάσις τÕ ΑΤΒΥΓΦ∆Χ πολύγωνον, κορυφ¾ δ τÕ Λ σηµε‹ον, πρÕς τ¾ν Óλην πυραµίδα, Âς βάσις µν τÕ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφ¾ δ τÕ Ν σηµε‹ον· éστε κሠπυραµίς, Âς βάσις µν τÕ ΑΤΒΥΓΦ∆Χ, κορυφ¾ δ τÕ Λ, πρÕς τ¾ν πυραµίδα, Âς βάσις [µν] τÕ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφ¾ δ τÕ Ν σηµε‹ον, τριπλασίονα λόγον œχει ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ. Øπόκειται δ καˆ Ð κîνος, οá βάσις [µν] Ð ΑΒΓ∆ κύκλος, κορυφ¾ δ τÕ Λ σηµε‹ον, πρÕς τÕ Ξ στερεÕν τριπλασίονα λόγον œχων ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ· œστιν ¥ρα æς Ð κîνος, οá βάσις µέν ™στιν Ð ΑΒΓ∆ κύκλος, κορυφ¾ δ τÕ Λ, πρÕς τÕ Ξ στερεόν, οÛτως ¹ πυραµίς, Âς βάσις µν τÕ ΑΤΒΥΓΦ∆Χ [πολύγωνον], κορυφ¾ δ τÕ Λ, πρÕς τ¾ν πυραµίδα, Âς βάσις µέν ™στι τÕ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφ¾ δ τÕ Ν· ™ναλλ¦ξ ¥ρα, æς Ð κîνος, οá βάσις µν Ð ΑΒΓ∆ κύκλος, κορυφ¾ δ τÕ Λ, πρÕς τ¾ν ™ν αÙτù πυραµίδα, Âς βάσις µν τÕ ΑΤΒΥΓΦ∆Χ πολύγωνον, κορυφ¾ δ τÕ Λ, οÛτως τÕ Ξ [στερεÕν] πρÕς τ¾ν πυραµίδα, Âς βάσις µέν ™στι τÕ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφ¾ δ τÕ Ν. µείζων δ Ð ε„ρηµένος κîνος τÁς ™ν αÙτù πυραµίδος· ™µπεριέχει γ¦ρ αÙτ¾ν. µε‹ζον ¥ρα κሠτÕ Ξ στερεÕν τÁς πυραµίδος, Âς βάσις µέν ™στι τÕ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφ¾ δ τÕ Ν. ¢λλ¦ κሠœλαττον· Óπερ ™στˆν ¢δύνατον. οÙκ ¥ρα Ð κîνος, οá βάσις Ð ΑΒΓ∆ κύκλος, κορυφ¾ δ τÕ Λ [σηµε‹ον], πρÕς œλαττόν τι τοà κώνου στερεόν, οá βάσις µν Ð ΕΖΗΘ κύκλος, κορυφ¾ δ τÕ Ν σηµε‹ον, τριπλασίονα λόγον œχει ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ. еοίως δ¾ δείξοµεν, Óτι οÙδ Ð ΕΖΗΘΝ κîνος πρÕς œλαττόν τι τοà ΑΒΓ∆Λ κώνου στερεÕν τριπλασίονα λόγον œχει ½περ ¹ ΖΘ πρÕς τ¾ν Β∆. Λέγω δή, Óτι οÙδ Ð ΑΒΓ∆Λ κîνος πρÕς µε‹ζόν τι τοà ΕΖΗΘΝ κώνου στερεÕν τριπλασίονα λόγον œχει ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ. Ε„ γ¦ρ δυνατόν, ™χέτω πρÕς µε‹ζον τÕ Ξ. ¢νάπαλιν ¥ρα τÕ Ξ στερεÕν πρÕς τÕν ΑΒΓ∆Λ κîνον τριπλασίονα λόγον œχει ½περ ¹ ΖΘ πρÕς τ¾ν Β∆. æς δ τÕ Ξ στερεÕν πρÕς τÕν ΑΒΓ∆Λ κîνον, οÛτως Ð ΕΖΗΘΝ κîνος πρÕς œλαττόν τι τοà ΑΒΓ∆Λ κώνου στερεόν. καˆ Ð ΕΖΗΘΝ ¥ρα κîνος πρÕς œλαττόν τι τοà ΑΒΓ∆Λ κώνου στερεÕν τριπλασίονα λόγον œχει ½περ ¹ ΖΘ πρÕς τ¾ν Β∆· Óπερ ¢δύνατον ™δείχθη. οÙκ ¥ρα Ð ΑΒΓ∆Λ κîνος πρÕς µε‹ζόν τι τοà ΕΖΗΘΝ κώνου στερεÕν τριπλασίονα λόγον œχει ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ. ™δείχθη

on each of the triangles (so formed), we can also show that each of the pyramids (on base ABCD taken) in order will have to each of the pyramids (on base EF GH taken) in order the cubed ratio that the corresponding side BK (has) to the corresponding side F M —that is to say, that BD (has) to F H. And (for two sets of proportional magnitudes) as one of the leading (magnitudes is) to one of the following, so (the sum of) all of the leading (magnitudes is) to (the sum of) all of the following (magnitudes) [Prop. 5.12]. And, thus, as pyramid BKT L (is) to pyramid F M P N , so the whole pyramid whose base is polygon AT BU CV DW , and apex the point L, (is) to the whole pyramid whose base is polygon EP F QGRHS, and apex the point N . And, hence, the pyramid whose base is polygon AT BU CV DW , and apex the point L, has to the pyramid whose base is polygon EP F QGRHS, and apex the point N , the cubed ratio that BD (has) to F H. And it was also assumed that the cone whose base is circle ABCD, and apex the point L, has to solid O the cubed ratio that BD (has) to F H. Thus, as the cone whose base is circle ABCD, and apex the point L, is to solid O, so the pyramid whose base (is) [polygon] AT BU CV DW , and apex the point L, (is) to the pyramid whose base is polygon EP F QGRHS, and apex the point N . Thus, alternately, as the cone whose base (is) circle ABCD, and apex the point L, (is) to the pyramid within it whose base (is) the polygon AT BU CV DW , and apex the point L, so the [solid] O (is) to the pyramid whose base is polygon EP F QGRHS, and apex the point N [Prop. 5.16]. And the aforementioned cone (is) greater than the pyramid within it. For it encompasses it. Thus, solid O (is) also greater than the pyramid whose base is polygon EP F QGRHS, and apex the point N . But, (it is) also less. The very thing is impossible. Thus, the cone whose base (is) circle ABCD, and apex the [point] L, does not have to some solid less than the cone whose base (is) circle EF GH, and apex the point N , the cubed ratio that BD (has) to EH. So, similarly, we can show that neither does cone EF GHN have to some solid less than cone ABCDL the cubed ratio that F H (has) to BD. So, I say that neither does cone ABCDL have to some solid greater than cone EF GHN the cubed ratio that BD (has) to F H. For, if possible, let it have (such a ratio) to a greater (solid), O. Thus, inversely, solid O has to cone ABCDL the cubed ratio that F H (has) to BD [Prop. 5.7 corr.]. And as solid O (is) to cone ABCDL, so cone EF GHN (is) to some solid less than cone ABCDL [12.2 lem.]. Thus, cone EF GHN also has to some solid less than cone ABCDL the cubed ratio that F H (has) to BD. The very thing was shown (to be) impossible. Thus, cone ABCDL

494

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

δέ, Óτι οÙδ πρÕς œλαττον. Ð ΑΒΓ∆Λ ¥ρα κîνος πρÕς τÕν ΕΖΗΘΝ κîνον τριπλασίονα λόγον œχει ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ. `Ως δ Ð κîνος πρÕς τÕν κîνον, Ð κύλινδρος πρÕς τÕν κύλινδρον· τριπλάσιος γ¦ρ Ð κύλινδρος τοà κώνου Ð ™πˆ τÁς αÙτÁς βάσεως τù κώνJ κሠ„σοϋψ¾ς αÙτù. καˆ Ð κύλινδρος ¥ρα πρÕς τÕν κύλινδρον τριπλασίονα λόγον œχει ½περ ¹ Β∆ πρÕς τ¾ν ΖΘ. Οƒ ¥ρα Óµοιοι κîνοι κሠκύλινδροι πρÕς ¢λλήλους ™ν τριπλασίονι λόγJ ε„σˆ τîν ™ν τα‹ς βάσεσι διαµέτρων· Óπερ œδει δε‹ξαι.

does not have to some solid greater than cone EF GHN the cubed ratio than BD (has) to F H. And it was shown that neither (does it have such a ratio) to a lesser (solid). Thus, cone ABCDL has to cone EF GHN the cubed ratio that BD (has) to F G. And as the cone (is) to the cone, so the cylinder (is) to the cylinder. For a cylinder is three times a cone on the same base as the cone, and of the same height as it [Prop. 12.10]. Thus, the cylinder also has to the cylinder the cubed ratio that BD (has) to F H. Thus, similar cones and cylinders are in the cubed ratio of the diameters of their bases. (Which is) the very thing it was required to show.

ιγ΄.

Proposition 13

'Ε¦ν κύλινδρος ™πιπέδJ τµηθÍ παραλλήλJ Ôντι το‹ς ¢πεναντίον ™πιπέδοις, œσται æς Ð κύλινδρος πρÕς τÕν κύλινδρον, οÛτως Ð ¥ξων πρÕς τÕν ¥ξονα.

If a cylinder is cut by a plane which is parallel to the opposite planes (of the cylinder) then as the cylinder (is) to the cylinder, so the axis will be to the axis. R A T P G C V

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Κύλινδρος γ¦ρ Ð Α∆ ™πιπέδJ τù ΗΘ τετµήσθω παραλλήλJ Ôντι το‹ς ¢πεναντίον ™πιπέδοις το‹ς ΑΒ, Γ∆, κሠσυβαλλέτω τù ¥ξονι τÕ ΗΘ ™πίπεδον κατ¦ τÕ Κ σηµε‹ον· λέγω, Óτι ™στˆν æς Ð ΒΗ κύλινδρος πρÕς τÕν Η∆ κύλινδρον, οÛτως Ð ΕΚ ¥ξων πρÕς τÕν ΚΖ ¥ξονα. 'Εκβεβλήσθω γ¦ρ Ð ΕΖ ¥ξων ™φ' ˜κάτερα τ¦ µέρη ™πˆ τ¦ Λ, Μ σηµε‹α, κሠ™κκείσθωσαν τù ΕΚ ¥ξονι ‡σοι Ðσοιδηποτοàν οƒ ΕΝ, ΝΛ, τù δ ΖΚ ‡σοι Ðσοιδηποτοàν οƒ ΖΞ, ΞΜ, κሠνοείσθω Ð ™πˆ τοà ΛΜ ¥ξονος κύλινδρος Ð ΟΧ, οá βάσεις οƒ ΟΠ, ΦΧ κύκλοι. κሠ™κβεβλήσθω δι¦ τîν Ν, Ξ σηµείων ™πίπεδα παράλληλα το‹ς ΑΒ, Γ∆ κሠτα‹ς βάσεσι τοà ΟΧ κυλίνδρου κሠποιείτωσαν τοÝς ΡΣ, ΤΥ κύκλους περˆ τ¦ Ν, Ξ κέντρα. κሠ™πεˆ οƒ ΛΝ, ΝΕ, ΕΚ ¥ξονες ‡σοι ε„σˆν ¢λλήλοις, οƒ ¥ρα ΠΡ, ΡΒ, ΒΗ κύλινδροι πρÕς ¢λλήλους ε„σˆν æς αƒ βάσεις. ‡σαι δέ ε„σιν αƒ βάσεις· ‡σοι ¥ρα καˆ οƒ ΠΡ, ΡΒ, ΒΗ κύλινδροι ¢λλήλοις. επεˆ οâν οƒ ΛΝ, ΝΕ, ΕΚ ¥ξονες ‡σοι ε„σˆν ¢λλήλοις, ε„σˆ δ καˆ οƒ ΠΡ, ΡΒ, ΒΗ κύλινδροι ‡σοι ¢λλήλοις, καί ™στιν ‡σον τÕ πλÁθος τù πλήθει, Ðσαπλασίων ¥ρα Ð ΚΛ ¥ξων τοà ΕΚ ¥ξονος, τοσαυταπλασίων œσται καˆ Ð ΠΗ κύλινδρος τοà ΗΒ κυλίνδρου. δι¦ τ¦ αÙτ¦ δ¾ κሠÐσαπλασίων ™στˆν Ð ΜΚ ¥ξων τοà ΚΖ ¥ξονος, τοσαυταπλασίων ™στˆ καˆ Ð ΧΗ κύλινδρος τοà Η∆ κυλίνδρου. καˆ ε„ µν ‡σος ™στˆν Ð ΚΛ ¥ξων τù ΚΜ ¥ξονι, ‡σος œσται καˆ Ð ΠΗ κύλινδρος τù

N

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M

Q S B H D U W For let the cylinder AD have been cut by the plane GH which is parallel to the opposite planes (of the cylinder), AB and CD. And let the plane GH have met the axis at point K. I say that as cylinder BG is to cylinder GD, so axis EK (is) to axis KF . For let axis EF have been produced in each direction to points L and M . And let any number whatsoever (of lengths), EN and N L, equal to axis EK, be set out (on the axis EL), and any number whatsoever (of lengths), F O and OM , equal to (axis) F K, (on the axis KM ). And let the cylinder P W , whose bases (are) the circles P Q and V W , have been conceived on axis LM . And let planes parallel to AB, CD, and the bases of cylinder P W , have been produced through points N and O, and let them have made the circles RS and T U around the centers N and O (respectively). And since axes LN , N E, and EK are equal to one another, the cylinders QR, RB, and BG are to one another as their bases [Prop. 12.11]. But the bases are equal. Thus, the cylinders QR, RB, and BG (are) also equal to one another. Therefore, since the axes LN , N E, and EK are equal to one another, and the cylinders QR, RB, and BG are also equal to one another, and the number (of the former) is equal to the number (of the latter), thus as many multiples as axis KL is of axis EK, so many multiples is cylinder QG also of

495

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

ΗΧ κυλίνδρJ, ε„ δ µείζων Ð ¥ξων τοà ¥ξονος, µείζων καˆ Ð κύλινδρος τοà κυλίνδρου, καˆ ε„ ™λάσσων, ™λάσσων. τεσσάρων δ¾ µεγεθîν Ôντων, ¢ξόνων µν τîν ΕΚ, ΚΖ, κυλίνδρων δ τîν ΒΗ, Η∆, ε‡ληπται „σάκις πολλαπλάσια, τοà µν ΕΚ ¥ξονος κሠτοà ΒΗ κυλίνδρου Ó τε ΛΚ ¢ξων καˆ Ð ΠΗ κύλινδρος, τοà δ ΚΖ ¥ξονες κሠτοà Η∆ κυλίνδρου Ó τε ΚΜ ¤ξων καˆ Ð ΗΧ κύλινδρος, κሠδέδεικται, Óτι ε„ Øπερέχει Ð ΚΛ ¥ξων τοà ΚΜ ¥ξονος, Øπερέχει καˆ Ð ΠΗ κύλινδρος τοà ΗΧ κυλίνδρου, καˆ ε„ ‡σος, ‡σος, καˆ ε„ ™λάσσων, ™λάσσων. œστιν ¥ρα æς Ð ΕΚ ¥ξων πρÕς τÕν ΚΖ ¥ξονα, οÛτως Ð ΒΗ κύλινδρος πρÕς τÕν Η∆ κύλινδρον· Óπερ œδει δε‹ξαι.

cylinder GB. And so, for the same (reasons), as many multiples as axis M K is of axis KF , so many multiples is cylinder W G also of cylinder GD. And if axis KL is equal to axis KM then cylinder QG will also be equal to cylinder GW , and if the axis (is) greater than the axis then the cylinder (will also be) greater than the cylinder, and if (the axis is) less then (the cylinder will also be) less. So, there are four magnitudes—the axes EK and KF , and the cylinders BG and GD—and equal multiples have been taken of axis EK and cylinder BG—(namely), axis LK and cylinder QG—and of axis KF and cylinder GD—(namely), axis KM and cylinder GW . And it has been shown that if axis KL exceeds axis KM then cylinder QG also exceeds cylinder GW , and if (the axes are) equal then (the cylinders are) equal, and if (KL is) less then (QG is) less. Thus, as axis EK is to axis KF , so cylinder BG (is) to cylinder GD [Def. 5.5]. (Which is) the very thing it was required to show.

ιδ΄.

Proposition 14

Οƒ ™πˆ ‡σων βάσεων Ôντες κîνοι κሠκύλινδροι πρÕς Cones and cylinders which are on equal bases are to αλλήλους ε„σˆν æς τ¦ Ûψη. one another as their heights.

E A

H

Z

F E

G J B

K

L

G

D

C A

N M

”Εστωσαν γ¦ρ ™πˆ ‡σων βάσεων τîν ΑΒ, Γ∆ κύκλων κύλινδροι οƒ ΕΒ, Ζ∆· λέγω, Óτι ™στˆν æς Ð ΕΒ κύλινδρος πρÕς τÕν Ζ∆ κύλινδρον, οÛτως Ð ΗΘ ¥ξων πρÕς τÕν ΚΛ ¥ξονα. 'Εκβεβλήσθω γ¦ρ Ð ΚΛ ¥ξων ™πˆ τÕ Ν σηµε‹ον, κሠκείσθω τù ΗΘ ¥ξονι ‡σος Ð ΛΝ, κሠπερˆ ¥ξονα τÕν ΛΝ κύλινδρος νενοήσθω Ð ΓΜ. ™πεˆ οâν οƒ ΕΒ, ΓΜ κύλινδροι ØπÕ τÕ αÙτÕ Ûψος ε„σίν, πρÕς ¢λλήλους ε„σˆν æς αƒ βάσεις. ‡σαι δέ ε„σίν αƒ βάσεις ¢λλήλαις· ‡σοι ¥ρα ε„σˆ καˆ οƒ ΕΒ, ΓΜ κύλινδροι. κሠ™πεˆ κύλινδρος Ð ΖΜ ™πιπέδJ τέτµηται τù Γ∆ παραλλήλJ Ôντι το‹ς ¢πεναντίον ™πιπέδοις, œστιν ¥ρα æς Ð ΓΜ κύλινδρος πρÕς τÕν Ζ∆ κύλινδρον, οÛτως Ð ΛΝ ¥ξων πρÕς τÕν ΚΛ ¥ξωνα. ‡σος δέ ™στιν Ð µν ΓΜ κύλινδρος τù ΕΒ κυλίνδρJ, Ð δ ΛΝ ¥ξων τù ΗΘ ¥ξονι· œστιν ¥ρα æς Ð ΕΒ κύνλινδρος πρÕς τÕν Ζ∆ κύλινδρον, οÛτως Ð ΗΘ ¥ξων πρÕς τÕν ΚΛ ¥ξονα. æς δ Ð ΕΒ κύλινδρος πρÕς τÕν Ζ∆ κύλινδρον,

K

H

L

D

N

M

B

For let EB and F D be cylinders on equal bases, (namely) the circles AB and CD (respectively). I say that as cylinder EB is to cylinder DF , so axis GH (is) to axis KL. For let the axis KL have been produced to point N . And let LN be made equal to axis GH. And let the cylinder CM have been conceived about axis LN . Therefore, since cylinders EB and CM have the same height they are to one another as their bases [Prop. 12.11]. And the bases are equal to one another. Thus, cylinders EB and CM are also equal to one another. And since cylinder F M has been cut by the plane CD which is parallel to its opposite planes, thus as cylinder CM is to cylinder F D, so axis LN (is) to axis KL [Prop. 12.13]. And cylinder CM is equal to cylinder EB, and axis LN to axis GH. Thus, as cylinder EB is to cylinder F D, so axis GH (is) to axis KL. And as cylinder EB (is) to cylinder F D, so

496

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

οØτως Ð ΑΒΗ κîνος πρÕς τÕν Γ∆Κ κîνον. κሠæς ¥ρα Ð ΗΘ ¥ξων πρÕς τÕν ΚΛ ¥ξονα, οÛτως Ð ΑΒΗ κîνος πρÕς τÕν Γ∆Κ κîνον καˆ Ð ΕΒ κύλινδρος πρÕς τÕν Ζ∆ κύλινδρον· Óπερ œδει δε‹ξαι.

cone ABG (is) to cone CDK [Prop. 12.10]. Thus, also, as axis GH (is) to axis KL, so cone ABG (is) to cone CDK, and cylinder EB to cylinder F D. (Which is) the very thing it was required to show.

ιε΄.

Proposition 15

Τîν ‡σων κώνων κሠκυλίνδρων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν· κሠïν κώνων κሠκυλίνδρων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν, ‡σοι ε„σˆν ™κε‹νοι.

The bases of equal cones and cylinders are reciprocally proportional to their heights. And, those cones and cylinders whose bases (are) reciprocally proportional to their heights are equal.

L D A K

X

B

G

O M R

H

S P Y

J

N E

L

Z

O

D A

M R

G

S

F

Q

N H

C

K B

”Εστωσαν ‡σοι κîνοι κሠκύλινδροι, ïν βάσεις µν οƒ ΑΒΓ∆, ΕΖΗΘ κύκλοι, διάµετροι δ αÙτîν αƒ ΑΓ, ΕΗ, ¤ξονες δ οƒ ΚΛ, ΜΝ, ο†τινες κሠÛψη ε„σˆ τîν κώνων À κυλίνδρων, κሠσυµπεπληρώσθωσαν οƒ ΑΞ, ΕΟ κύλινδροι. λέγω, Óτι τîν ΑΞ, ΕΟ κυλίνδρων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν, καί ™στιν æς ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΕΖΗΘ βάσιν, οÛτως τÕ ΜΝ Ûψος πρÕς τÕ ΚΛ Ûψος. ΤÕ γ¦ρ ΛΚ Ûψος τù ΜΝ Ûψει ½τοι ‡σον ™στˆν À οÜ. œστω πρότερον ‡σον. œστι δ καˆ Ð ΑΞ κύλινδρος τù ΕΟ κυλίνδρJ ‡σος. οƒ δ ØπÕ τÕ αÙτÕ Ûψος Ôντες κîνοι κሠκύλινδροι πρÕς ¢λλήλους ε„σˆν æς αƒ βάσεις· ‡ση ¥ρα κሠ¹ ΑΒΓ∆ βάσις τÍ ΕΖΗΘ βάσει. éστε κሠ¢ντιπέπονθεν, æς ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΕΖΗΘ βάσιν, οÛτως τÕ ΜΝ Ûψος πρÕς τÕ ΚΛ Ûψος. ¢λλ¦ δ¾ µ¾ œστω τÕ ΛΚ Ûψος τù ΜΝ ‡σον, ¢λλ' œστω µε‹ζον τÕ ΜΝ, κሠ¢φVρήσθω ¢πÕ τοà ΜΝ Ûψους τù ΚΛ ‡σον τÕ ΠΝ, κሠδι¦ τοà Π σηµείου τετµήσθω Ð ΕΟ κύλινδρος ™πιπέδJ τù ΤΥΣ παραλλήλJ το‹ς τîν ΕΖΗΘ, ΡΟ κύκλων ™πιπέδοις, κሠ¢πÕ βάσεως µν τοà ΕΖΗΘ κύκλου, Ûψους δ τοà ΝΠ κύλινδρος νενοήσθω Ð ΕΣ. καί ™πεˆ ‡σος ™στˆν Ð ΑΞ κύλινδρος τù ΕΟ κυλίνδρJ, œστιν ¥ρα æς Ð ΑΞ κύλινδρος πρÕς τÕν ΕΣ κυλίνδρον, οÛτως Ð ΕΟ κύλινδρος πρÕς τÕν ΕΣ κύλινδρον. ¢λλ' æς µν Ð ΑΞ κύλινδρος πρÕς τÕν ΕΣ κύλινδρον, οÛτως ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΕΖΗΘ· ØπÕ γ¦ρ τÕ αÙτÕ Ûψος ε„σˆν οƒ ΑΞ, ΕΣ κύλινδροι· æς δ Ð ΕΟ κύλινδρος πρÕς τÕν ΕΣ, οÛτως τÕ ΜΝ Ûψος πρÕς τÕ ΠΝ Ûψος· Ð γ¦ρ ΕΟ κύλινδρος ™πιπέδJ τέτµηται παραλλήλJ Ôντι το‹ς ¢πεναντίον ™πιπέδοις. œστιν ¥ρα κሠæς ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΕΖΗΘ βάσιν, οÛτως τÕ ΜΝ Ûψος πρÕς τÕ ΠΝ Ûψος. ‡σον δ τÕ ΠΝ Ûψος τù ΚΛ Ûψει· œστιν ¥ρα æς ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΕΖΗΘ βάσιν, οÛτως τÕ ΜΝ

P

X

E

Let there be equal cones and cylinders whose bases are the circles ABCD and EF GH, and the diameters of (the bases), AC and EG, and (whose) axes (are) KL and M N , which are also the heights of the cones and cylinders (respectively). And let the cylinders AO and EP have been completed. I say that the bases of cylinders AO and EP are reciprocally proportional to their heights, and (so) as base ABCD is to base EF GH, so height M N (is) to height KL. For height LK is either equal to height M N , or not. Let it, first of all, be equal. And cylinder AO is also equal to cylinder EP . And cones and cylinders having the same height are to one another as their bases [Prop. 12.11]. Thus, base ABCD (is) also equal to base EF GH. And, hence, reciprocally, as base ABCD (is) to base EF GH, so height M N (is) to height KL. And so, let height LK not be equal to M N , but let M N be greater. And let QN , equal to KL, have been cut off from height M N . And let the cylinder EP have been cut, through point Q, by the plane T U S (which is) parallel to the planes of the circles EF GH and RP . And let cylinder ES have been conceived, with base the circle EF GH, and height N Q. And since cylinder AO is equal to cylinder EP , thus, as cylinder AO (is) to cylinder ES, so cylinder EP (is) to cylinder ES [Prop. 5.7]. But, as cylinder AO (is) to cylinder ES, so base ABCD (is) to base EF GH. For cylinders AO and ES (have) the same height [Prop. 12.11]. And as cylinder EP (is) to (cylinder) ES, so height M N (is) to height QN . For cylinder EP has been cut by a plane which is parallel to its opposite planes [Prop. 12.13]. And, thus, as base ABCD is to base EF GH, so height M N (is) to height QN [Prop. 5.11]. And height QN (is) equal to height KL. Thus, as base ABCD is to base

497

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

Ûψος πρÕς τÕ ΚΛ Ûψος. τîν ¥ρα ΑΞ, ΕΟ κυλίνδρων ¢ντιπεπόνθασιν αƒ βάσεις το‹ς Ûψεσιν. 'Αλλ¦ δ¾ τîν ΑΞ, ΕΟ κυλίνδρων ¢ντιπεπονθέτωσαν αƒ βάσεις το‹ς Ûψεσιν, κሠœστω æς ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΕΖΗΘ βάσιν, οÛτως τÕ ΜΝ Ûψος πρÕς τÕ ΚΛ Ûψος· λέγω, Óτι ‡σος ™στˆν Ð ΑΞ κύλινδρος τù ΕΟ κυλίνδρJ. Τîν γ¦ρ αÙτîν κατασκευασθέντων ™πεί ™στιν æς ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΕΖΗΘ βάσιν, οÛτως τÕ ΜΝ Ûψος πρÕς τÕ ΚΛ Ûψος, ‡σον δ τÕ ΚΛ Ûψος τù ΠΝ Ûψει, œσται ¥ρα æς ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΕΖΗΘ βάσιν, οÛτως τÕ ΜΝ Ûψος πρÕς τÕ ΠΝ Ûψος. ¢λλ' æς µν ¹ ΑΒΓ∆ βάσις πρÕς τ¾ν ΕΖΗΘ βάσιν, οÛτως Ð ΑΞ κύλινδρος πρÕς τÕν ΕΣ κύλινδρον· ØπÕ γ¦ρ τÕ αÙτÕ Ûψος ε„σίν· æς δ τÕ ΜΝ Ûψος πρÕς τÕ ΠΝ [Ûψος], οÛτως Ð ΕΟ κύλινδρος πρÕς τÕν ΕΣ κύλινδρον· œστιν ¥ρα æς Ð ΑΞ κύλινδρος πρÕς τÕν ΕΣ κύλινδρον, οÛτως Ð ΕΟ κύλινδρος πρÕς τÕν ΕΣ. ‡σος ¥ρα Ð ΑΞ κύλινδρος τù ΕΟ κυλίνδρJ. æσαύτως δ κሠ™πˆ τîν κώνων· Óπερ œδει δε‹ξαι.

EF GH, so height M N (is) to height KL. Thus, the bases of cylinders AO and EP are reciprocally proportional to their heights. And, so, let the bases of cylinders AO and EP be reciprocally proportional to their heights, and (thus) let base ABCD be to base EF GH, as height M N (is) to height KL. I say that cylinder AO is equal to cylinder EP . For, with the same construction, since base ABCD is to base EF GH, as height M N (is) to height KL, and height KL (is) equal to height QN , thus, as base ABCD (is) to base EF GH, so height M N will be to height QN . But, as base ABCD (is) to base EF GH, so cylinder AO (is) to cylinder ES. For they are the same height [Prop. 12.11]. And as height M N (is) to [height] QN , so cylinder EP (is) to cylinder ES [Prop. 12.13]. Thus, as cylinder AO is to cylinder ES, so cylinder EP (is) to (cylinder) ES [Prop. 5.11]. Thus, cylinder AO (is) equal to cylinder EP [Prop. 5.9]. In the same manner, (the proposition can) also (be demonstrated) for the cones. (Which is) the very thing it was required to show.

ι$΄.

Proposition 16

∆ύο κύκλων περˆ τÕ αÙτÕ κέντρον Ôντων ε„ς τÕν There being two circles about the same center, to µείζονα κύκλον πολύγωνον „σόπλευρόν τε κሠ¢ρτιόπλευ- inscribe an equilateral and even-sided polygon in the ρον ™γγράψαι µ¾ ψαàον τοà ™λάσσονος κύκλου. greater circle, not touching the lesser circle.

J B

E

K Z

A

H

L

L

H MD G

A

B

N

E

K

G M

D

N F

”Εστωσαν οƒ δοθέντες δύο κύκλοι οƒ ΑΒΓ∆, ΕΖΗΘ περˆ τÕ αÙτÕ κέντρον τÕ Κ· δε‹ δ¾ ε„ς τÕν µείζονα κύκλον τÕν ΑΒΓ∆ πολύγωνον „σόπλευρόν τε κሠ¢ρτιόπλευρον ™γγράψαι µ¾ ψαàον τοà ΕΖΗΘ κύκλου. ”Ηχθω γ¦ρ δι¦ τοà Κ κέντρου εÙθε‹α ¹ ΒΚ∆, κሠ¢πÕ τοà Η σηµείου τÍ Β∆ εÙθείv πρÕς Ñρθ¦ς ½χθω ¹ ΗΑ κሠδιήχθω ™πˆ τÕ Γ· ¹ ΑΓ ¥ρα ™φάπτεται τοà ΕΖΗΘ κύκλου. τέµνοντες δ¾ τ¾ν ΒΑ∆ περιφέρειαν δίχα κሠτ¾ν ¹µίσειαν αÙτÁς δίχα κሠτοàτο ¢εˆ ποιοàντες καταλείψοµεν περιφέρειαν ™λάσσονα τÁς Α∆. λελείφθω, κሠœστω ¹ Λ∆, κሠ¢πÕ τοà Λ ™πˆ τ¾ν Β∆ κάθετος ½χθω ¹ ΛΜ κሠδιήχθω ™πˆ τÕ Ν, κሠ™πεζεύχθωσαν

C

Let ABCD and EF GH be the given two circles, about the same center, K. So, it is necessary to inscribe an equilateral and even-sided polygon in the greater circle ABCD, not touching circle EF GH. Let the straight-line BKD have been drawn through the center K. And let GA have been drawn, at rightangles to the straight-line BD, through point G, and let it have been drawn through to C. Thus, AC touches circle EF GH [Prop. 3.16 corr.]. So, (by) cutting circumference BAD in half, and the half of it in half, and doing this continually, we will (eventually) leave a circumference less than AD [Prop. 10.1]. Let it have been left, and let it be

498

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

αƒ Λ∆, ∆Ν· ‡ση ¥ρα ™στˆν ¹ Λ∆ τÍ ∆Ν. κሠ™πεˆ παράλληλός ™στιν ¹ ΛΝ τÍ ΑΓ, ¹ δ ΑΓ ™φάπτεται τοà ΕΖΗΘ κύκλου, ¹ ΛΝ ¥ρα οÙκ ™φάπτεται τοà ΕΖΗΘ κύκλου· πολλù ¥ρα αƒ Λ∆, ∆Ν οÙκ ™φάπτονται τοà ΕΖΗΘ κύκλου. ™¦ν δ¾ τÍ Λ∆ εÙθείv ‡σας κατ¦ τÕ συνεχς ™ναρµόσωµεν ε„ς τÕν ΑΒΓ∆ κύκλον, ™γγραφήσεται ε„ς τÕν ΑΒΓ∆ κύκλον πολύγωνον „σόπλευρόν τε κሠ¢ρτιόπλευρον µ¾ ψαàον τοà ™λάσσονος κύκλου τοà ΕΖΗΘ· Óπερ œδει ποιÁσαι.

†

LD. And let LM have been drawn, from L, perpendicular to BD, and let it have been drawn through to N . And let LD and DN have been joined. Thus, LD is equal to DN [Props. 3.3, 1.4]. And since LN is parallel to AC [Prop. 1.28], and AC touches circle EF GH, LN thus does not touch circle EF GH. Thus, even more so, LD and DN do not touch circle EF GH. And if we continuously insert (straight-lines) equal to straight-line LD into circle ABCD [Prop. 4.1], then an equilateral and evensided polygon, not touching the lesser circle EF GH, will have been inscribed in circle ABCD.† (Which is) the very thing it was required to do.

Note that the chord of the polygon, LN , does not touch the inner circle either.

ιζ΄.

Proposition 17

∆ύο σφαιρîν περˆ τÕ αÙτÕ κέντρον οÙσîν ε„ς τ¾ν µείζονα σφα‹ραν στερεÕν πολύεδρον ™γγράψαι µ¾ ψαàον τÁς ™λάσσονος σφαίρας κατ¦ τ¾ν ™πιφάνειαν.

There being two spheres about the same center, to inscribe a polyhedral solid in the greater sphere, not touching the lesser sphere on its surface.

E

M L K

S

O W

P

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Q

Y

B

T

F

U R

L

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K

D

A J

G

T

S

B

N

U R

Q

P

O

F W

X

H

E

M

Y

V

G

D

A H

N

C

Νενοήσθωσαν δύο σφα‹ραι περˆ τÕ αÙτÕ κέντρον τÕ Α· δε‹ δ¾ ε„ς τ¾ν µείζονα σφα‹ραν στερεÕν πολύεδρον ™γγράψαι µ¾ ψαàον τÁς ™λάσσονος σφαίρας κατ¦ τ¾ν ™πιφάνειαν. Τετµήσθωσαν αƒ σφα‹ραι ™πιπέδJ τινˆ δι¦ τοà κέντρου· œσονται δ¾ αƒ τοµαˆ κύκλοι, ™πειδήπερ µενούσης τÁς διαµέτρου κሠπεριφεροµένου τοà ¹µικυκλίου ™γιγνετο ¹ σφα‹ρα· éστε κሠκαθ' ο†ας ¨ν θέσεως ™πινοήσωµεν τÕ ¹µικύκλιον, τÕ δι' αÙτοà ™κβαλλόµενον ™πίπεδον ποιήσει ™πˆ τÁς ™πιφανείας τÁς σφαίρας κύκλον. κሠφανερόν, Óτι κሠµέγιστον, ™πειδήπερ ¹ διάµετρος

Let two spheres have been conceived about the same center, A. So, it is necessary to inscribe a polyhedral solid in the greater sphere, not touching the lesser sphere on its surface. Let the spheres have been cut by some plane through the center. So, the sections will be circles, inasmuch as a sphere is generated by the diameter remaining behind, and a semi-circle being carried around [Def. 11.14]. And, hence, whatever position we conceive (of for) the semicircle, the plane produced through it makes a circle on the surface of the sphere. And (it is) clear that (it is)

499

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

τÁς σφαίρας, ¼τις ™στˆ κሠτοà ¹µικυκλίου διάµετρος δηλαδ¾ κሠτοà κύκλου, µείζων ™στˆ πασîν τîν ε„ς τÕν κύκλον À τ¾ν σφα‹ραν διαγοµένων [εÙθειîν]. œστω οâν ™ν µν τÍ µείζονι σφαίρv κύκλος Ð ΒΓ∆Ε, ™ν δ τÍ ™λάσσονι σφαίρv κύκλος Ð ΖΗΘ, κሠ½χθωσαν αÙτîν δύο διάµετροι πρÕς Ñρθ¦ς ¢λλήλαις αƒ Β∆, ΓΕ, κሠδύο κύκλων περˆ τÕ αÙτÕ κέντρον Ôντων τîν ΒΓ∆Ε, ΖΗΘ ε„ς τÕν µείζονα κύκλον τÕν ΒΓ∆Ε πολύγωνον „σόπλευρον κሠ¢ρτιόπλευρον ™γγεγράφθω µ¾ ψαàον τοà ™λάσσονος κύκλου τοà ΖΗΘ, οá πλευρሠœστωσαν ™ν τù ΒΕ τεταρτηµορίJ αƒ ΒΚ, ΚΛ, ΛΜ, ΜΕ, κሠ™πιζευχθε‹σα ¹ ΚΑ διήχθω ™πˆ τÕ Ν, κሠ¢νεστάτω ¢πÕ τοà Α σηµείου τù τοà ΒΓ∆Ε κύκλου ™πιπέδJ πρÕς Ñρθ¦ς ¹ ΑΞ κሠσυµβαλλέτω τÍ ™πιφανείv τÁς σφαίρας κατ¦ τÕ Ξ, κሠδι¦ τÁς ΑΞ κሠ˜κατέρας τîν Β∆, ΚΝ ™πίπεδα ™κβεβλήσθω· ποιήσουσι δ¾ δι¦ τ¦ ε„ρηµένα ™πˆ τÁς ™πιφανείας τÁς σφαίρας µεγίστους κύκλους. ποιείτωσαν, ïν ¹µικύκλια œστω ™πˆ τîν Β∆, ΚΝ διαµέτρων τ¦ ΒΞ∆, ΚΞΝ. κሠ™πεˆ ¹ ΞΑ Ñρθή ™στι πρÕς τÕ τοà ΒΓ∆Ε κύκλου ™πίπεδον, κሠπάντα ¥ρα τ¦ δι¦ τÁς ΞΑ ™πίπεδά ™στιν Ñρθ¦ πρÕς τÕ τοà ΒΓ∆Ε κύκλου ™πίπεδον· éστε κሠτ¦ ΒΞ∆, ΚΞΝ ¹µικύκλια Ñρθά ™στι πρÕς τÕ τοà ΒΓ∆Ε κύκλου ™πίπεδον. κሠ™πεˆ ‡σα ™στˆ τ¦ ΒΕ∆, ΒΞ∆, ΚΞΝ ¹µικύκλια· ™πˆ γ¦ρ ‡σων ε„σˆ διαµέτρων τîν Β∆, ΚΝ· ‡σα ™στˆ κሠτ¦ ΒΕ, ΒΞ, ΚΞ τεταρτηµόρια ¢λλήλοις. Ôσαι ¥ρα ε„σˆν ™ν τù ΒΕ τεταρτηµορίJ πλευρሠτοà πολυγώνου, τοσαàταί ε„σι κሠ™ν το‹ς ΒΞ, ΚΞ τεταρτηµορίοις ‡σαι τα‹ς ΒΚ, ΚΛ, ΛΜ, ΜΕ εÙθείαις. ™γγεγράφθωσαν κሠœστωσαν αƒ ΒΟ, ΟΠ, ΠΡ, ΡΞ, ΚΣ, ΣΤ, ΤΥ, ΥΞ, κሠ™πεζεύχθωσαν αƒ ΣΟ, ΤΠ, ΥΡ, κሠ¢πÕ τîν Ο, Σ ™πˆ τÕ τοà ΒΓ∆Ε κύκλου ™πίπεδον κάθετοι ½χθωσαν· πεσοàνται δ¾ ™πˆ τ¦ς κοιν¦ς τﵦς τîν ™πιπέδων τ¦ς Β∆, ΚΝ, ™πειδήπερ κሠτ¦ τîν ΒΞ∆, ΚΞΝ ™πίπεδα Ñρθά ™στι πρÕς τÕ τοà ΒΓ∆Ε κύκλου ™πίπεδον. πιπτέτωσαν, κሠœστωσαν αƒ ΟΦ, ΣΧ, κሠ™πεζεύχθω ¹ ΧΦ. κሠ™πεˆ ™ν ‡σοις ¹µικυκλίοις τοˆς ΒΞ∆, ΚΞΝ ‡σαι ¢πειληµµέναι ε„σˆν αƒ ΒΟ, ΚΣ, κሠκάθετοι ºγµέναι ε„σˆν αƒ ΟΦ, ΣΧ, ‡ση [¥ρα] ™στˆν ¹ µν ΟΦ τÍ ΣΧ, ¹ δ ΒΦ τÍ ΚΧ. œστι δ κሠÓλη ¹ ΒΑ ÐλV τÍ ΚΑ ‡ση· κሠλοιπ¾ ¥ρα ¹ ΦΑ λοιπÍ τÍ ΧΑ ™στιν ‡ση· œστιν ¥ρα æς ¹ ΒΦ πρÕς τ¾ν ΦΑ, οÛτως ¹ ΚΧ πρÕς τ¾ν ΧΑ· παράλληλος ¥ρα ™στˆν ¹ ΧΦ τÍ ΚΒ. κሠ™πεˆ ˜κατέρα τîν ΟΦ, ΣΧ Ñρθή ™στι πρÕς τÕ τοà ΒΓ∆Ε κύκλου ™πίπεδον, παράλληλος ¥ρα ™στˆν ¹ ΟΦ τÍ ΣΧ. ™δείχθη δ αÙτÍ κሠ‡ση· καˆ αƒ ΧΦ, ΣΟ ¥ρα ‡σαι ε„σˆ κሠπαράλληλοι. κሠ™πεˆ παράλληλός ™στιν ¹ ΧΦ τÍ ΣΟ, ¢λλ¦ ¹ ΧΦ τÍ ΚΒ ™στι παράλληλος, κሠ¹ ΣΟ ¥ρα τÍ ΚΒ ™στι παράλληλος. κሠ™πιζευγνύουσιν αÙτ¦ς αƒ ΒΟ, ΚΣ· τÕ ΚΒΟΣ ¥ρα τετράπλευρον ™ν ˜νί ™στιν ™πιπέδJ, ™πειδήπερ, ™¦ν ðσι δύο εÙθε‹αι παράλληλοι, κሠ™φ' ˜κατέρας αÙτîν ληφθÍ τυχόντα σηµε‹α, ¹ ™πˆ

also a great (circle), inasmuch as the diameter of the sphere, which is also manifestly the diameter of the semicircle and the circle, is greater than of all the (other) [straight-lines] drawn across in the circle or the sphere [Prop. 3.15]. Therefore, let BCDE be the circle in the greater sphere, and F GH the circle in the lesser sphere. And let two diameters of them have been drawn at rightangles to one another, (namely), BD and CE. And there being two circles about the same center—(namely), BCDE and F GH—let an equilateral and even-sided polygon have been inscribed in the greater circle, BCDE, not touching the lesser circle, F GH [Prop. 12.16], of which let the sides in the quadrant BE be BK, KL, LM , and M E. And, KA being joined, let it have been drawn across to N . And let AO have been set up at point A, at right-angles to the plane of circle BCDE. And let it meet the surface of the sphere at O. And let planes have been produced through AO and each of BD and KN . So, according to the aforementioned (discussion), they will make great circles on the surface of the sphere. Let them make (great circles), of which let BOD and KON be semi-circles on the diameters BD and KN (respectively). And since OA is at right-angles to the plane of circle BCDE, thus all of the planes through OA are also at right-angles to the plane of circle BCDE [Prop. 11.18]. And, hence, the semi-circles BOD and KON are also at right-angles to the plane of circle BCDE. And since semi-circles BED, BOD, and KON are equal—for (they are) on the equal diameters BD and KN [Def. 3.1]— the quadrants BE, BO, and KO are also equal to one another. Thus, as many sides of the polygon as are in quadrant BE, so many are also in quadrants BO and KO equal to the straight-lines BK, KL, LM , and M E. Let them have been inscribed, and let them be BP , P Q, QR, RO, KS, ST , T U , and U O. And let SP , T Q, and U R have been joined. And let perpendiculars have been drawn from P and S to the plane of circle BCDE [Prop. 11.11]. So, they will fall on the common sections of the planes BD and KN (with BCDE), inasmuch as the planes of BOD and KON are also at right-angles to the plane of circle BCDE [Def. 11.4]. Let them have fallen, and let them be P V and SW . And let W V have been joined. And since BP and KS are equal (circumferences) having been cut off in the equal semi-circles BOD and KON [Def. 3.28], and P V and SW are perpendiculars having been drawn (from them), P V is [thus] equal to SW , and BV to KW [Props. 3.27, 1.26]. And the whole of BA is also equal to the whole of KA. And, thus, as BV is to V A, so KW (is) to W A. W V is thus parallel to KB [Prop. 6.2]. And since P V and SW are each at right-angles to the plane of circle BCDE, P V is

500

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

τ¦ σηµε‹α ™πιζευγνυµένη εÙθε‹α ™ν τù αÙτù ™πιπέδJ ™στˆ τα‹ς παραλλήλοις. δι¦ τ¦ αÙτ¦ δ¾ κሠ˜κάτερον τîν ΣΟΠΤ, ΤΠΡΥ τετραπλεύρων ™ν ˜νί ™στιν ™πιπέδJ. œστι δ κሠτÕ ΥΡΞ τρίγωνον ™ν ˜νˆ ˜πιπέδJ. ™¦ν δ¾ νοήσωµεν ¢πÕ τîν Ο, Σ, Π, Τ, Ρ, Υ σηµείων ™πˆ τÕ Α ™πιζευγνυµένας εÙθείας, συσταθήσεταί τι σχÁµα στερεÕν πολύεδρον µαταξÝ τîν ΒΞ, ΚΞ περιφερειîν ™κ πυραµίδων συγκείµενον, ïν βάσεις µν τ¦ ΚΒΟΣ, ΣΟΠΤ, ΤΠΡΥ τετράπλευρα κሠτÕ ΥΡΞ τρίγωνον, κορυφ¾ δ τÕ Α σηµε‹ον. ™¦ν δ κሠ™πˆ ˜κάστης τîν ΚΛ, ΛΜ, ΜΕ πλευρîν καθάπερ ™πˆ τÁς ΒΚ τ¦ αÙτ¦ κατασκευάσωµεν κሠœτι τîν λοιπîν τριîν τεταρτηµορίων, συσταθήσεταί τι σχÁµα πολύεδρον ™γγεγραµµένον ε„ς τ¾ν σφα‹ραν πυραµίσι περιεχόµενον, ïν βάσιες [µν] τ¦ ε„ρηµένα τετράπλευρα κሠτÕ ΥΡΞ τρίγωνον κሠτ¦ еοταγÁ αÙτο‹ς, κορυφ¾ δ τÕ Α σηµε‹ον. Λέγω Óτι τÕ ε„ρηµένον πολύεδρον οÙκ ™φάψεται τÁς ™λάσσονος σφαίρας κατ¦ τ¾ν ™πιφάνειαν, ™φ' Âς ™στιν Ð ΖΗΘ κύκλος. ”Ηχθω ¢πÕ τοà Α σηµείου ™πˆ τÕ τοà ΚΒΟΣ τετραπλεύρου ™πίπεδον κάθετος ¹ ΑΨ κሠσυµβαλλέτω τù ™πιπέδJ κατ¦ τÕ Ψ σηµε‹ον, κሠ™πεζεύχθωσαν αƒ ΨΒ, ΨΚ. κሠ™πεˆ ¹ ΑΨ Ñρθή ™στι πρÕς τÕ τοà ΚΒΟΣ τετραπλεύρου ™πίπεδον, κሠπρÕς πάσας ¥ρα τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù τοà τετραπλεύρου ™πιπέδJ Ñρθή ™στιν. ¹ ΑΨ ¥ρα Ñρθή ™στι πρÕς ˜κατέραν τîν ΒΨ, ΨΚ. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΒ τÍ ΑΚ, †σον ™στˆ κሠτÕ ¢πÕ τÁς ΑΒ τù ¢πÕ τÁς ΑΚ. καί ™στι τù µν ¢πÕ τÁς ΑΒ ‡σα τ¦ ¢πÕ τîν ΑΨ, ΨΒ· Ñρθ¾ γ¦ρ ¹ πρÕς τù Ψ· τù δ ¢πÕ τÁς ΑΚ ‡σα τ¦ ¢πÕ τîν ΑΨ, ΨΚ. τ¦ ¥ρα ¢πÕ τîν ΑΨ, ΨΒ ‡σα ™στˆ το‹ς ¢πÕ τîν ΑΨ, ΨΚ. κοινÕν ¢φVρήσθω τÕ ¢πÕ τÁς ΑΨ· λοιπÕν ¥ρα τÕ ¢πÕ τÁς ΒΨ λοιπù τù ¢πÕ τÁς ΨΚ ‡σον ™στίν· ‡ση ¥ρα ¹ ΒΨ τÍ ΨΚ. еοίως δ¾ δείξοµεν, Óτι καˆ αƒ ¢πÕ τοà Ψ ™πˆ τ¦ Ο, Σ ™πιζευγνύµεναι εÙθε‹αι ‡σαι ε„σˆν ˜κατέρv τîν ΒΨ, ΨΚ. Ð ¥ρα κέντρJ τù Ψ κሠδιαστήµατι ˜νˆ τîν ΨΒ, ΨΚ γραφόµενος κύκλος ¼ξει κሠδι¦ τîν Ο, Σ, κሠœσται ™ν κύκλJ τÕ ΚΒΟΣ τετράπλευρον. Κሠ™πεˆ µείζων ™στˆν ¹ ΚΒ τÁς ΧΦ, ‡ση δ ¹ ΧΦ τÍ ΣΟ, µείζων ¥ρα ¹ ΚΒ τÁς ΣΟ. ‡ση δ ¹ ΚΒ ˜κατέρv τîν ΚΣ, ΒΟ· κሠ˜κατέρα ¥ρα τîν ΚΣ, ΒΟ τÁς ΣΟ µείζων ™στίν. κሠ™πεˆ ™ν κύκλJ τετράπλευρόν ™στι τÕ ΚΒΟΣ, κሠ‡σαι αƒ ΚΒ, ΒΟ, ΚΣ, κሠ™λάττων ¹ ΟΣ, κሠ™κ τοà κέντρου τοà κύκλου ™στˆν ¹ ΒΨ, τÕ ¥ρα ¢πÕ τÁς ΚΒ τοà ¢πÕ τÁς ΒΨ µε‹ζόν ™στιν À διπλάσιον. ½χθω ¢πÕ τοà Κ ™πˆ τ¾ν ΒΦ κάθετος ¹ ΚΩ. κሠ™πεˆ ¹ Β∆ τÁς ∆Ω ™λάττων ™στˆν À διπλÁ, καί ™στιν æς ¹ Β∆ πρÕς τ¾ν ∆Ω, οÛτως τÕ ØπÕ τîν ∆Β, ΒΩ πρÕς τÕ ØπÕ [τîν] ∆Ω, ΩΒ, ¢ναγραφοµένου ¢πÕ τÁς ΒΩ τετραγώνου κሠσυµπληρουµένου τοà ™πˆ τÁς Ω∆ παραλληλογράµµου

thus parallel to SW [Prop. 11.6]. And it was also shown (to be) equal to it. And, thus, W V and SP are equal and parallel [Prop. 1.33]. And since W V is parallel to SP , but W V is parallel to KB, SP is thus also parallel to KB [Prop. 11.1]. And BP and KS join them. Thus, the quadrilateral KBP S is in one plane, inasmuch as if there are two parallel straight-lines, and a random point is taken on each of them, then the straight-line joining the points is in the same plane as the parallel (straightlines) [Prop. 11.7]. So, for the same (reasons), each of the quadrilaterals SP QT and T QRU is also in one plane. And triangle U RO is also in one plane [Prop. 11.2]. So, if we conceive straight-lines joining points P , S, Q, T , R, and U to A then some solid polyhedral figure will have been constructed between the circumferences BO and KO, being composed of pyramids whose bases (are) the quadrilaterals KBP S, SP QT , T QRU , and the triangle U RO, and apex the point A. And if we also make the same construction on each of the sides KL, LM , and M E, just as on BK, and, further, (repeat the construction) in the remaining three quadrants, then some polyhedral figure which has been inscribed in the sphere will have been constructed, being contained by pyramids whose bases (are) the aforementioned quadrilaterals, and triangle U RO, and the (quadrilaterals and triangles) ranged in the same rows as them, and apex the point A. So, I say that the aforementioned polyhedron will not touch the lesser sphere on the surface on which the circle F GH is (situated). Let the perpendicular (straight-line) AX have been drawn from point A to the plane KBP S, and let it meet the plane at point X [Prop. 11.11]. And let XB and XK have been joined. And since AX is at right-angles to the plane of quadrilateral KBP S, it is thus also at rightangles to all of the straight-lines joined to it which are also in the plane of the quadrilateral [Def. 11.3]. Thus, AX is at right-angles to each of BX and XK. And since AB is equal to AK, the (square) on AB is also equal to the (square) on AK. And the (sum of the squares) on AX and XB is equal to the (square) on AB. For the angle at X (is) a right-angle [Prop. 1.47]. And the (sum of the squares) on AX and XK is equal to the (square) on AK. Thus, the (sum of the squares) on AX and XB is equal to the (sum of the squares) on AX and XK. Let the (square) on AX have been subtracted from both. Thus, the remaining (square) on BX is equal to the remaining (square) on XK. Thus, BX (is) equal to XK. So, similarly, we can show that the straight-lines joined from X to P and S are equal to each of BX and XK. Thus, a circle drawn (in the plane of the quadrilateral) with center

501

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

κሠτÕ ØπÕ ∆Β, ΒΩ ¥ρα τοà ØπÕ ∆Ω, ΩΒ œλαττόν ™στιν À διπλάσιον. καί ™στι τÁς Κ∆ ™πιζευγνυµένης τÕ µν ØπÕ ∆Β, ΒΩ ‡σον τö ¢πÕ τÁς ΒΚ, τÕ δ ØπÕ τîν ∆Ω, ΩΒ ‡σον τù ¢πÕ τÁς ΚΩ· τÕ ¥ρα ¢πÕ τÁς ΚΒ τοà ¢πÕ τÁς ΚΩ œλασσόν ™στιν À διπλάσιον. ¢λλ¦ τÕ ¢πÕ τÁς ΚΒ τοà ¢πÕ τÁς ΒΨ µε‹ζόν ™στιν À διπλάσιον· µε‹ζον ¥ρα τÕ ¢πÕ τÁς ΚΩ τοà ¢πÕ τÁς ΒΨ. κሠ™πεˆ ‡ση ™στˆν ¹ ΒΑ τÍ ΚΑ, ‡σον ™στˆ τÕ ¢πÕ τÁς ΒΑ τù ¢πÕ τÁς ΑΚ. καί ™στι τù µν ¢πÕ τÁς ΒΑ ‡σα τ¦ ¢πÕ τîν ΒΨ, ΨΑ, τù δ ¢πÕ τÁς ΚΑ ‡σα τ¦ ¢πÕ τîν ΚΩ, ΩΑ· τ¦ ¥ρα ¢πÕ τîν ΒΨ, ΨΑ ‡σα ™στˆ το‹ς ¢πÕ τîν ΚΩ, ΩΑ, ïν τÕ ¢πÕ τÁς ΚΩ µε‹ζον τοà ¢πÕ τÁς ΒΨ· λοιπÕν ¥ρα τÕ ¢πÕ τÁς ΩΑ œλασσόν ™στι τοà ¢πÕ τÁς ΨΑ. µείζων ¥ρα ¹ ΑΨ τÁς ΑΩ· πολλù ¥ρα ¹ ΑΨ µείζων ™στˆ τ¾ς ΑΗ. καί ™στιν ¹ µν ΑΨ ™πˆ µίαν τοà πολυέδρου βάσιν, ¹ δ ΑΗ ™πˆ τ¾ν τÁς ™λάσσονος σφαίρας ™πιφάνειαν· éστε τÕ πολύεδρον οÙ ψαύσει τÁς ™λάσσονος σφαίρας κατ¦ τ¾ν ™πιφάνειαν. ∆ύο ¥ρα σφαιρîν περˆ τÕ αÙτÕ κέντρον οÙσîν ε„ς τ¾ν µείζονα σφα‹ραν στερεÕν πολύεδρον ™γγέγραπται µ¾ ψαàον τÁς ™λάσσονος σφαίρας κατ¦ τ¾ν ™πιφάνειαν· Óπερ œδει ποιÁσαι.

X, and radius one of XB or XK, will also pass through P and S, and the quadrilateral KBP S will be inside the circle. And since KB is greater than V W , and V W (is) equal to SP , KB (is) thus greater than SP . And KB (is) equal to each of KS and BP . Thus, KS and BP are each greater than SP . And since quadrilateral KBP S is in a circle, and KB, BP , and KS are equal (to one another), and P S (is) less (than them), and BX is the radius of the circle, the (square) on KB is thus greater than double the (square) on BX.† Let the perpendicular KY have been drawn from K to BV .‡ And since BD is less than double DY , and as BD is to DY , so the (rectangle contained) by DB and BY (is) to the (rectangle contained) by DY and Y B—a square being described on BY , and a (rectangular) parallelogram (with short side equal to BY ) completed on Y D—the (rectangle contained) by DB and BY is thus also less than double the (rectangle contained) by DY and Y B. And, KD being joined, the (rectangle contained) by DB and BY is equal to the (square) on BK, and the (rectangle contained) by DY and Y B equal to the (square) on KY [Props. 3.31, 6.8 corr.]. Thus, the (square) on KB is less than double the (square) on KY . But, the (square) on KB is greater than double the (square) on BX. Thus, the (square) on KY (is) greater than the (square) on BX. And since BA is equal to KA, the (square) on BA is equal to the (square) on AK. And the (sum of the squares) on BX and XA is equal to the (square) on BA, and the (sum of the squares) on KY and Y A (is) equal to the (square) on KA [Prop. 1.47]. Thus, the (sum of the squares) on BX and XA is equal to the (sum of the squares) on KY and Y A, of which the (square) on KY (is) greater than the (square) on BX. Thus, the remaining (square) on Y A is less than the (square) on XA. Thus, AX (is) greater than AY . Thus, AX is much greater than AG.§ And AX is on one of the bases of the polyhedron, and AG (is) on the surface of the lesser sphere. Hence, the polyhedron will not touch the lesser sphere on its surface. Thus, there being two spheres about the same center, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere on its surface. (Which is) the very thing it was required to do. √

†

Since KB, BP , and KS are greater than the sides of an inscribed square, which are each of length

‡

Note that points Y and V are actually identical.

§

This conclusion depends on the fact that the chord of the polygon in proposition 12.16 does not touch the inner circle.

Πόρισµα.

2 BX.

Corollary

'Ε¦ν δ κሠε„ς ˜τάραν σφα‹ραν τù ™ν τÍ ΒΓ∆Ε

502

And, also, if a similar polyhedral solid to that in

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

σφαίρv στερεù πολυέδρJ Óµοιον στερεÕν πολύεδρον ™γγραφÍ, τÕ ™ν τÍ ΒΓ∆Ε σφαίρv στερεÕν πολύεδρον πρÕς τÕ ™ν τÍ ˜τέρv σφαίρv στερεÕν πολύεδρον τριπλασίονα λόγον œχει, ½περ ¹ τÁς ΒΓ∆Ε σφαίρας διάµετρος πρÕς τ¾ν τÁς ˜τέρας σφαίρας διάµετρον. διαιρεθέντων γ¦ρ τîν στερεîν ε„ς τ¦ς еοιοπληθε‹ς καˆ Ðµοιοταγε‹ς πυραµίδας œσονται αƒ πυραµίδες Óµοιαι. αƒ δ Óµοιαι πυραµίδες πρÕς ¢λλήλας ™ν τριπλασίονι λόγJ ε„σˆ τîν еολόγων πλευρîν· ¹ ¥ρα πυραµίς, Âς βάσις µέν ™στι τÕ ΚΒΟΣ τετράπλευρον, κορυφ¾ δ τÕ Α σηµε‹ον, πρÕς τ¾ν ™ν τÍ ˜τέρv σφαίρv еοιοταγÁ πυραµίδα τριπλασίονα λόγον œχει, ½περ ¹ еόλογος πλευρ¦ πρÕς τ¾ν еόλογον πλευράν, τουτέστιν ½περ ¹ ΑΒ ™κ τοà κέντρου τÁς σφαίρας τÁς περˆ κέντρον τÕ Α πρÕς τ¾ν ™κ τοà κέντρου τÁς ˜τέρας σφαίρας. еοίως κሠ˜κάστη πυρᵈς τîν ™ν τÍ περˆ κέντρον τÕ Α σφαίρv πρÕς ˜κάστην еοταγÁ πυραµίδα τîν ™ν τÍ ˜τέρv σφαίρv τριπλασίονα λόγον ›ξει, ½περ ¹ ΑΒ πρÕς τ¾ν ™κ τοà κέντρου τÁς ˜τέρας σφαίρας. κሠæς žν τîν ¹γουµένων πρÕς žν τîν ˜ποµένων, οÛτως ¤παντα τ¦ ¹γούµενα πρÕς ¤παντα τ¦ ˜πόµενα· éστε Óλον τÕ ™ν τÍ περˆ κέντρον τÕ Α σφαίρv στερεÕν πολύεδρον πρÕς Óλον τÕ ™ν τÍ ˜τέρv [σφαίρv] στερεÕν πολύεδρον τριπλασίονα λόγον ›ξει, ½περ ¹ ΑΒ πρÕς τ¾ν ™κ τοà κέντρου τÁς ˜τέρας σφαίρας, τουτέστιν ½περ ¹ Β∆ διάµετρος πρÕς τ¾ν τÁς ˜τέρας σφαίρας διάµετρον· Óπερ œδει δε‹ξαι.

sphere BCDE is inscribed in another sphere then the polyhedral solid in sphere BCDE has to the polyhedral solid in the other sphere the cubed ratio that the diameter of sphere BCDE has to the diameter of the other sphere. For if the spheres are divided into similarly numbered, and similarly situated, pyramids, then the pyramids will be similar. And similar pyramids are in the cubed ratio of corresponding sides [Prop. 12.8 corr.]. Thus, the pyramid whose base is quadrilateral KBP S, and apex the point A, will have to the similarly situated pyramid in the other sphere the cubed ratio that a corresponding side (has) to a corresponding side. That is to say, that of radius AB of the sphere about center A to the radius of the other sphere. And, similarly, each pyramid in the sphere about center A will have to each similarly situated pyramid in the other sphere the cubed ratio that AB (has) to the radius of the other sphere. And as one of the leading (magnitudes is) to one of the following (in two sets of proportional magnitudes), so (the sum of) all the leading (magnitudes is) to (the sum of) all of the following (magnitudes) [Prop. 5.12]. Hence, the whole polyhedral solid in the sphere about center A will have to the whole polyhedral solid in the other [sphere] the cubed ratio that AB (has) to the radius of the other sphere. That is to say, that diameter BD (has) to the diameter of the other sphere. (Which is) the very thing it was required to show.

ιη΄.

Proposition 18

Αƒ σφα‹ραι πρÕς ¢λλήλας ™ν τριπλασίονι λόγJ ε„σˆ Spheres are to one another in the cubed ratio of their τîν „δίων διαµέτρων. respective diameters. A A

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Νενοήσθωσαν σφα‹ραι αƒ ΑΒΓ, ∆ΕΖ, διάµετροι δ Let the spheres ABC and DEF have been conceived, αÙτîν αƒ ΒΓ, ΕΖ· λέγω, Óτι ¹ ΑΒΓ σφα‹ρα πρÕς τ¾ν and (let) their diameters (be) BC and EF (respectively). ∆ΕΖ σφα‹ραν τριπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς I say that sphere ABC has to sphere DEF the cubed ratio τ¾ν ΕΖ. that BC (has) to EF . 503

ΣΤΟΙΧΕΙΩΝ ιβ΄.

ELEMENTS BOOK 12

Ε„ γ¦ρ µ¾ ¹ ΑΒΓ σφα‹ρα πρÕς τ¾ν ∆ΕΖ σφα‹ραν τριπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ, ›ξει ¥ρα ¹ ΑΒΓ σφα‹ρα πρÕς ™λάσσονά τινα τÁς ∆ΕΖ σφαίρας τριπλασίονα λόγον À πρÕς µείζονα ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ. ™χέτω πρότερον πρÕς ™λάσσονα τ¾ν ΗΘΚ, κሠνενοήσθω ¹ ∆ΕΖ τÍ ΗΘΚ περˆ τÕ αÙτÕ κέντρον, κሠ™γγεγράφθω ε„ς τ¾ν µείζονα σφα‹ραν τ¾ν ∆ΕΖ στερεÕν πολύεδρον µ¾ ψαàον τÁς ™λάσσονος σφαίρας τÁς ΗΘΚ κατ¦ τ¾ν ™πιφάνειαν, ™γγεγράφθω δ κሠε„ς τ¾ν ΑΒΓ σφα‹ραν τù ™ν τÍ ∆ΕΖ σφαίρv στερεù πολυέδρJ Óµοιον στερεÕν πολύεδρον· τÕ ¥ρα ™ν τÍ ΑΒΓ στερεÕν πολύεδρον πρÕς τÕ ™ν τÍ ∆ΕΖ στερεÕν πολύεδρον τριπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ. œχει δ κሠ¹ ΑΒΓ σφα‹ρα πρÕς τ¾ν ΗΘΚ σφα‹ραν τριπλασίονα λόγον ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ· œστιν ¥ρα æς ¹ ΑΒΓ σφα‹ρα πρÕς τ¾ν ΗΘΚ σφα‹ραν, οÛτως τÕ ™ν τÍ ΑΒΓ σφαίρv στερεÕν πολύεδρον πρÕς τÕ ™ν τÍ ∆ΕΖ σφαίρv στερεÕν πολύεδρον· ™ναλλ¦ξ [¥ρα] æς ¹ ΑΒΓ σφα‹ρα πρÕς τÕ ™ν αÙτÍ πολύεδρον, οÛτως ¹ ΗΘΚ σφα‹ρα πρÕς τÕ ™ν τÍ ∆ΕΖ σφαίρv στερεÕν πολύεδρον. µείζων δ ¹ ΑΒΓ σφα‹ρα τοà ™ν αÙτÍ πολυέδρου· µείζων ¥ρα κሠ¹ ΗΘΚ σφα‹ρα τοà ™ν τÍ ∆ΕΖ σφαίρv πολυέδρου. ¢λλ¦ κሠ™λάττων· ™µπεριέχεται γ¦ρ Øπ' αÙτοà. οÙκ ¥ρα ¹ ΑΒΓ σφα‹ρα πρÕς ™λάσσονα τÁς ∆ΕΖ σφαίρας τριπλασίονα λόγον œχει ½περ ¹ ΒΓ διάµετρος πρÕς τ¾ν ΕΖ. еοίως δ¾ δείξοµεν, Óτι οÙδ ¹ ∆ΕΖ σφα‹ρα πρÕς ™λάσσονα τÁς ΑΒΓ σφαίρας τριπλασίονα λόγον œχει ½περ ¹ ΕΖ πρÕς τ¾ν ΒΓ. Λέγω δή, Óτι οÙδ ¹ ΑΒΓ σφα‹ρα πρÕς µείζονά τινα τÁς ∆ΕΖ σφαίρας τριπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ. Ε„ γ¦ρ δυνατόν, ™χέτω πρÕς µείζονα τ¾ν ΛΜΝ· ¢νάπαλιν ¥ρα ¹ ΛΜΝ σφα‹ρα πρÕς τ¾ν ΑΒΓ σφα‹ραν τριπλασίονα λόγον œχει ½περ ¹ ΕΖ διάµετρος πρÕς τ¾ν ΒΓ διάµετρον. æς δ ¹ ΛΜΝ σφα‹ρα πρÕς τ¾ν ΑΒΓ σφα‹ραν, οÛτως ¹ ∆ΕΖ σφα‹ρα πρÕς ™λάσσονά τινα τÁς ΑΒΓ σφαίρας, ™πειδήπερ µείζων ™στˆν ¹ ΛΜΝ τÁς ∆ΕΖ, æς œµπροσθεν ™δείχθη. κሠ¹ ∆ΕΖ ¥ρα σφα‹ρα πρÕς ™λάσσονά τινα τÁς ΑΒΓ σφαίρας τριπλασίονα λόγον œχει ½περ ¹ ΕΖ πρÕς τ¾ν ΒΓ· Óπερ ¢δύνατον ™δείχθη. οÙκ ¥ρα ¹ ΑΒΓ σφα‹ρα πρÕς µείζονά τινα τÁς ∆ΕΖ σφαίρας τριπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ. ™δείχθη δέ, Óτι οÙδ πρÕς ™λάσσονα. ¹ ¥ρα ΑΒΓ σφα‹ρα πρÕς τ¾ν ∆ΕΖ σφα‹ραν τριπλασίονα λόγον œχει ½περ ¹ ΒΓ πρÕς τ¾ν ΕΖ· Óπερ œδει δε‹ξαι.

For if sphere ABC does not have to sphere DEF the cubed ratio that BC (has) to EF , sphere ABC will have to some (sphere) either less than, or greater than, sphere DEF the cubed ratio that BC (has) to EF . Let it, first of all, have (such a ratio) to a lesser (sphere), GHK. And let DEF have been conceived about the same center as GHK. And let a polyhedral solid have been inscribed in the greater sphere DEF , not touching the lesser sphere GHK on its surface [Prop. 12.17]. And let a polyhedral solid, similar to the polyhedral solid in sphere DEF , have also been inscribed in sphere ABC. Thus, the polyhedral solid in sphere ABC has to the polyhedral solid in sphere DEF the cubed ratio that BC (has) to EF [Prop. 12.17 corr.]. And sphere ABC also has to sphere GHK the cubed ratio that BC (has) to EF . Thus, as sphere ABC is to sphere GHK, so the polyhedral solid in sphere ABC (is) to the polyhedral solid is sphere DEF . [Thus], alternately, as sphere ABC (is) to the polygon within it, so sphere GHK (is) to the polyhedral solid within sphere DEF [Prop. 5.16]. And sphere ABC (is) greater than the polyhedron within it. Thus, sphere GHK (is) also greater than the polyhedron within sphere DEF [Prop. 5.14]. But, (it is) also less. For it is encompassed by it. Thus, sphere ABC does not have to (a sphere) less than sphere DEF the cubed ratio that diameter BC (has) to EF . So, similarly, we can show that sphere DEF does not have to (a sphere) less than sphere ABC the cubed ratio that EF (has) to BC either. So, I say that sphere ABC does not have to some (sphere) greater than sphere DEF the cubed ratio that BC (has) to EF either. For, if possible, let it have (the cubed ratio) to a greater (sphere), LM N . Thus, inversely, sphere LM N (has) to sphere ABC the cubed ratio that diameter EF (has) to diameter BC [Prop. 5.7 corr.]. And as sphere LM N (is) to sphere ABC, so sphere DEF (is) to some (sphere) less than sphere ABC, inasmuch as LM N is greater than DEF , as shown before [Prop. 12.2 lem.]. And, thus, sphere DEF has to some (sphere) less than sphere ABC the cubed ratio that EF (has) to BC. The very thing was shown (to be) impossible. Thus, sphere ABC does not have to some (sphere) greater than sphere DEF the cubed ratio that BC (has) to EF . And it was shown that neither (does it have such a ratio) to a lesser (sphere). Thus, sphere ABC has to sphere DEF the cubed ratio that BC (has) to EF . (Which is) the very thing it was required to show.

504

ELEMENTS BOOK 13 The Platonic solids†

† The five regular solids—the cube, tetrahedron (i.e., pyramid), octahedron, icosahedron, and dodecahedron—were problably discovered by the school of Pythagoras. They are generally termed “Platonic” solids because they feature prominently in Plato’s famous dialogue Timaeus. Many of the theorems contained in this book—particularly those which pertain to the last two solids—are ascribed to Theaetetus of Athens.

505

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

α΄.

Proposition 1

'Ε¦ν εÙθε‹α γραµµ¾ ¥κρον κሠµέσον λόγον τµηθÍ, τÕ µε‹ζον τµÁµα προσλαβÕν τ¾ν ¹µίσειαν τÁς Óλης πενταπλάσιον δύναται τοà ¢πÕ τÁς ¹µισείας τετραγώνου.

If a straight-line is cut in extreme and mean ratio then the square on the greater piece, added to half of the whole, is equal to five times the square on the half.

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ΕÙθε‹α γ¦ρ γραµµ¾ ¹ ΑΒ ¥κρον κሠµέσον λόγον τετµήσθω κατ¦ τÕ Γ σηµε‹ον, κሠœστω µε‹ζον τµÁµα τÕ ΑΓ, κሠ™κβεβλήσθω ™π' εÙθείας τÍ ΓΑ εÙθε‹α ¹ Α∆, κሠκείσθω τÁς ΑΒ ¹µίσεια ¹ Α∆· λέγω, Óτι πενταπλάσιόν ™στι τÕ ¢πÕ τÁς Γ∆ τοà ¢πÕ τÁς ∆Α. 'Αναγεγράφθωσαν γ¦ρ ¢πÕ τîν ΑΒ, ∆Γ τετράγωνα τ¦ ΑΕ, ∆Ζ, κሠκαταγεγράφθω ™ν τù ∆Ζ τÕ σχÁµα, κሠδιήχθω ¹ ΖΓ ™πˆ τÕ Η. κሠ™πεˆ ¹ ΑΒ ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Γ, τÕ ¥ρα ØπÕ τîν ΑΒΓ ‡σον ™στˆ τù ¢πÕ τÁς ΑΓ. καί ™στι τÕ µν ØπÕ τîν ΑΒΓ τÕ ΓΕ, τÕ δ ¢πÕ τÁς ΑΓ τÕ ΖΘ· ‡σον ¥ρα τÕ ΓΕ τù ΖΘ. κሠ™πεˆ διπλÁ ™στιν ¹ ΒΑ τÁς Α∆, ‡ση δ ¹ µν ΒΑ τÍ ΚΑ, ¹ δ Α∆ τÍ ΑΘ, διπλÁ ¥ρα κሠ¹ ΚΑ τÁς ΑΘ. æς δ ¹ ΚΑ πρÕς τ¾ν ΑΘ, οÛτως τÕ ΓΚ πρÕς τÕ ΓΘ· διπλάσιον ¥ρα τÕ ΓΚ τοà ΓΘ. ε„σˆ δ κሠτÕ ΛΘ, ΘΓ διπλάσια τοà ΓΘ. ‡σον ¥ρα τÕ ΚΓ το‹ς ΛΘ, ΘΓ. ™δείχθη δ κሠτÕ ΓΕ τù ΘΖ ‡σον· Óλον ¥ρα τÕ ΑΕ τετράγωνον ‡σον ™στˆ τù ΜΝΞ γνώµονι. κሠ™πεˆ διπλÁ ™στιν ¹ ΒΑ τÁς Α∆, τετραπλάσιόν ™στι τÕ ¢πÕ τÁς ΒΑ τοà ¢πÕ τÁς Α∆, τουτέστι τÕ ΑΕ τοà ∆Θ. ‡σον δ τÕ ΑΕ τù ΜΝΞ γνώµονι· καˆ Ð ΜΝΞ ¥ρα γνώµων τετραπλάσιός ™στι τοà ΑΟ· Óλον ¥ρα τÕ ∆Ζ πενταπλάσιόν ™στι τοà ΑΟ. καί ™στι τÕ µν ∆Ζ τÕ ¢πÕ τÁς ∆Γ, τÕ δ ΑΟ τÕ ¢πÕ τÁς ∆Α· τÕ ¥ρα ¢πÕ τÁς Γ∆ πενταπλάσιόν ™στι τοà ¢πÕ τÁς ∆Α. 'Ε¦ν ¥ρα εÙθε‹α ¥κρον κሠµέσον λόγον τµηθÍ, τÕ µε‹ζον τµÁµα προσλαβÕν τ¾ν ¹µίσειαν τÁς Óλης πενταπλάσιον δύναται τοà ¢πÕ τÁς ¹µισείας τετραγώνου· Óπερ œδει δε‹ξαι.

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For let the straight-line AB have been cut in extreme and mean ratio at point C, and let AC be the greater piece. And let the straight-line AD have been produced in a straight-line with CA. And let AD be made (equal to) half of AB. I say that the (square) on CD is five times the (square) on DA. For let the squares AE and DF have been described on AB and DC (respectively). And let the figure in DF have been drawn. And let F C have been drawn across to G. And since AB has been cut in extreme and mean ratio at C, the (rectangle contained) by ABC is thus equal to the (square) on AC [Def. 6.3, Prop. 6.17]. And CE is the (rectangle contained) by ABC, and F H the (square) on AC. Thus, CE (is) equal to F H. And since BA is double AD, and BA (is) equal to KA, and AD to AH, KA (is) thus also double AH. And as KA (is) to AH, so CK (is) to CH [Prop. 6.1]. Thus, CK (is) double CH. And LH plus HC is also double CH [Prop. 1.43]. Thus, KC (is) equal to LH plus HC. And CE was also shown (to be) equal to HF . Thus, the whole square AE is equal to the gnomon M N O. And since BA is double AD, the (square) on BA is four times the (square) on AD—that is to say, AE (is four times) DH. And AE (is) equal to gnomon M N O. And, thus, gnomon M N O is also four times AP . Thus, the whole of DF is five times AP . And DF is the (square) on DC, and AP the (square) on DA. Thus, the (square) on CD is five times the (square) on DA. Thus, if a straight-line is cut in extreme and mean ratio then the square on the greater piece, added to half of

506

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13 the whole, is equal to five times the square on the half. (Which is) the very thing it was required to show.

β΄.

Proposition 2

'Ε¦ν εÙθε‹α γραµµ¾ τµήµατος ˜αυτÁς πενταπλάσιον δύνηται, τÁς διπλασίας τοà ε„ρηµένου τµήµατος ¥κρον κሠµέσον λόγον τεµνοµένης τÕ µε‹ζον τµÁµα τÕ λοιπÕν µέρος ™στˆ τÁς ™ξ ¢ρχÁς εÙθείας.

If the square on a straight-line is five times the (square) on a piece of it, and double the aforementioned piece is cut in extreme and mean ratio, then the greater piece is the remaining part of the original straight-line.

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ΕÙθε‹α γ¦ρ γραµµ¾ ¹ ΑΒ τµήµατος ˜αυτÁς τοà ΑΓ πενταπλάσιον δυνάσθω, τÁς δ ΑΓ διπλÁ œστω ¹ Γ∆. λέγω, Óτι τÁς Γ∆ ¥κρον κሠµέσον λόγον τεµνοµένος τÕ µε‹ζον τµÁµά ™στιν ¹ ΓΒ. 'Αναγεγράφθω γ¦ρ ¢φ' ˜κατέρας τîν ΑΒ, Γ∆ τετράγωνα τ¦ ΑΖ, ΓΗ, κሠκαταγεγράφθω ™ν τù ΑΖ τÕ σχÁµα, κሠδιήχθω ¹ ΒΕ. κሠ™πεˆ πενταπλάσιόν ™στι τÕ ¢πό τÁς ΒΑ τοà ¢πÕ τÁς ΑΓ, πενταπλάσιόν ™στι τÕ ΑΖ τοà ΑΘ. τετραπλάσιος ¥ρα Ð ΜΝΞ γνώµων τοà ΑΘ. κሠ™πεˆ διπλÁ ™στιν ¹ ∆Γ τÁς ΓΑ, τετραπλάσιον ¥ρα ™στˆ τÕ ¢πÕ ∆Γ τοà ¢πÕ ΓΑ, τουτέστι τÕ ΓΗ τοà ΑΘ. ™δείχθη δ καˆ Ð ΜΝΞ γνώµων τετραπλάσιος τοà ΑΘ· ‡σος ¥ρα Ð ΜΝΞ γνώµων τù ΓΗ. κሠ™πεˆ διπλÁ ™στιν ¹ ∆Γ τÁς ΓΑ, ‡ση δ ¹ µν ∆Γ τÍ ΓΚ, ¹ δ ΑΓ τÍ ΓΘ, [διπλÁ ¥ρα κሠ¹ ΚΓ τÁς ΓΘ], διπλάσιον ¥ρα κሠτÕ ΚΒ τοà ΒΘ. ε„σˆ δ κሠτ¦ ΛΘ, ΘΒ τοà ΘΒ διπλάσια· ‡σον ¥ρα τÕ ΚΒ το‹ς ΛΘ, ΘΒ. ™δείχθη δ κሠÓλος Ð ΜΝΞ γνώµων ÓλJ τù ΓΗ ‡σος· κሠλοιπÕν ¥ρα τÕ ΘΖ τù ΒΗ ™στιν ‡σον. καί ™στι τÕ µν ΒΗ τÕ ØπÕ τîν Γ∆Β· ‡ση γ¦ρ ¹ Γ∆ τÍ ∆Η· τÕ δ ΘΖ τÕ ¢πÕ τÁς ΓΒ· τÕ ¥ρα ØπÕ τîν Γ∆Β ‡σον ™στˆ τù ¢πÕ τÁς ΓΒ. œστιν ¥ρα æς ¹ ∆Γ πρÕς τ¾ν ΓΒ, οÛτως ¹ ΓΒ πρÕς τ¾ν Β∆. µείζων δ ¹ ∆Γ τÁς ΓΒ· µείζων ¥ρα κሠ¹ ΓΒ τÁς Β∆. τÁς Γ∆ ¥ρα εÙθείας ¥κρον κሠµέσον λόγον τεµνοµένης τÕ µε‹ζον τµÁµά ™στιν ¹ ΓΒ. 'Ε¦ν ¥ρα εÙθε‹α γραµµ¾ τµήµατος ˜αυτÁς πενταπλάσιον δύνηται, τÁς διπλασίας τοà ε„ρηµένου τµήµ-

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D

G

For let the square on the straight-line AB be five times the (square) on the piece of it, AC. And let CD be double AC. I say that if CD is cut in extreme and mean ratio then the greater piece is CB. For let the squares AF and CG have been described on each of AB and CD (respectively). And let the figure in AF have been drawn. And let BE have been drawn across. And since the (square) on BA is five times the (square) on AC, AF is five times AH. Thus, gnomon M N O (is) four times AH. And since DC is double CA, the (square) on DC is thus four times the (square) on CA—that is to say, CG (is four times) AH. And the gnomon M N O was also shown (to be) four times AH. Thus, gnomon M N O (is) equal to CG. And since DC is double CA, and DC (is) equal to CK, and AC to CH, [KC (is) thus also double CH], (and) KB (is) also double BH [Prop. 6.1]. And LH plus HB is also double HB [Prop. 1.43] Thus, KB (is) equal to LH plus HB. And the whole gnomon M N O was also shown (to be) equal to the whole of CG. Thus, the remainder HF is also equal to (the remainder) BG. And BG is the (rectangle contained) by CDB. For CD (is) equal to DG. And HF (is) the square on CB. Thus, the (rectangle contained) by CDB is equal to the (square) on CB. Thus, as DC is to CB, so CB (is) to BD [Prop. 6.17]. And DC (is) greater than CB (see lemma). Thus, CB (is) also greater than BD [Prop. 5.14]. Thus, if the straight-line CD is cut

507

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

ατος ¥κρον κሠµέσον λόγον τεµνοµένης τÕ µε‹ζον τµÁµα τÕ λοιπÕν µέρος ™στˆ τÁς ™ξ ¢ρχÁς εÙθείας· Óπερ œδει δε‹ξαι.

in extreme and mean ratio then the greater piece is CB. Thus, if the square on a straight-line is five times the (square) on a piece of itself, and double the aforementioned piece is cut in extreme and mean ratio, then the greater piece is (equal to) the remaining part of the original straight-line. (Which is) the very thing it was required to show.

ΛÁµµα.

Lemma

“Οτι δ ¹ διπλÁ τÁς ΑΓ µείζων ™στˆ τÁς ΒΓ, οÛτως δεικτέον. Ε„ γ¦ρ µή, œστω, ε„ δυνατόν, ¹ ΒΓ διπλÁ τÁς ΓΑ. τετραπλάσιον ¥ρα τÕ ¢πÕ τÁς ΒΓ τοà ¢πÕ τÁς ΓΑ· πενταπλάσια ¥ρα τ¦ ¢πÕ τîν ΒΓ, ΓΑ τοà ¢πÕ τÁς ΓΑ. Øπόκειται δ κሠτÕ ¢πÕ τÁς ΒΑ πενταπλάσιον τοà ¢πÕ τ¾ς ΓΑ· τÕ ¥ρα ¢πÕ τÁς ΒΑ ‡σον ™στˆ το‹ς ¢πÕ τîν ΒΓ, ΓΑ· Óπερ ¢δύνατον. οÙκ ¥ρα ¹ ΓΒ διπλασία ™στˆ τÁς ΑΓ. еοίως δ¾ δείξοµεν, Óτι οÙδ ¹ ™λάττων τÁς ΓΒ διπλασίων ™στˆ τÁς ΓΑ· πολλù γ¦ρ [µε‹ζον] τÕ ¥τοπον. `Η ¥ρα τÁς ΑΓ διπλÁ µείζων ™στˆ τÁς ΓΒ· Óπερ œδει δε‹ξαι.

And it can be shown that double AC (i.e., DC) is greater than BC, as follows. For if (double AC is) not (greater than BC), if possible, let BC be double CA. Thus, the (square) on BC (is) four times the (square) on CA. Thus, the (sum of) the (squares) on BC and CA (is) five times the (square) on CA. And the (square) on BA was assumed (to be) five times the (square) on CA. Thus, the (square) on BA is equal to the (sum of) the (squares) on BC and CA. The very thing (is) impossible [Prop. 2.4]. Thus, CB is not double AC. So, similarly, we can show that a (straightline) less than CB is not double AC either. For (in this case) the absurdity is much [greater]. Thus, double AC is greater than CB. (Which is) the very thing it was required to show.

γ΄.

Proposition 3

'Ε¦ν εÙθε‹α γραµµ¾ ¥κρον κሠµέσον λόγον τµηθÍ, τÕ œλασσον τµÁµα προσλαβÕν τ¾ν ¹µίσειαν τοà µείζονος τµήµατος πενταπλάσιον δύναται τοà ¢πÕ τ¾ς ¹µισείας τοà µείζονος τµήατος τετραγώνου.

If a straight-line is cut in extreme and mean ratio then the square on the lesser piece added to half of the greater piece is five times the square on half of the greater piece.

A

R J

D

G

B

O

H

X

K L

P

A

D

B

C P

M

R

G

M

O Q

Z S

N

H

E

K

L

F

S

N

E

ΕÙθε‹α γάρ τις ¹ ΑΒ ¥κρον κሠµέσον λόγον For let some straight-line AB have been cut in exτετµήσθω κατ¦ τÕ Γ σηµε‹ον, κሠœστω µε‹ζον τµÁµα treme and mean ratio at point C. And let AC be the τÕ ΑΓ, κሠτετµήσθω ¹ ΑΓ δίχα κατ¦ τÕ ∆· λέγω, Óτι greater piece. And let AC have been cut in half at D. I πενταπλάσιόν ™στι τÕ ¢πÕ τ¾ς Β∆ τοà ¢πÕ τ¾ς ∆Γ. say that the (square) on BD is five times the (square) on

508

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ ΑΕ, κሠκαταγεγράφθω διπλοàν τÕ σχÁµα. ™πεˆ διπλÁ ™στιν ¹ ΑΓ τÁς ∆Γ, τετραπλάσιον ¥ρα τÕ ¢πÕ τÁς ΑΓ τοà ¢πÕ τ¾ς ∆Γ, τουτέστι τÕ ΡΣ τοà ΖΗ. κሠ™πεˆ τÕ ØπÕ τîν ΑΒΓ ‡σον ™στˆ τù ¢πÕ τÁς ΑΓ, καί ™στι τÕ ØπÕ τîν ΑΒΓ τÕ ΓΕ, τÕ ¥ρα ΓΕ ‡σον ™στˆ τù ΡΣ. τετραπλάσιον δ τÕ ΡΣ τοà ΖΗ· τετραπλάσιον ¥ρα κሠτÕ ΓΕ τοà ΖΗ. πάλιν ™πεˆ ‡ση ™στˆν ¹ Α∆ τÍ ∆Γ, ‡ση ™στˆ κሠ¹ ΘΚ τÍ ΚΖ. éστε κሠτÕ ΗΖ τετράγωνον ‡σον ™στˆ τù ΘΛ τετραγώνJ. ‡ση ¥ρα ¹ ΗΚ τÍ ΚΛ, τουτέστιν ¹ ΜΝ τÍ ΝΕ· éστε κሠτÕ ΜΖ τù ΖΕ ™στιν ‡σον. ¢λλ¦ τÕ ΜΖ τù ΓΗ ™στιν ‡σον· κሠτÕ ΓΗ ¥ρα τù ΖΕ ™στιν ‡σον. κοινÕν προσκείσθω τÕ ΓΝ· Ð ¥ρα ΞΟΠ γνώµων ‡σος ™στˆ τù ΓΕ. ¢λλ¦ τÕ ΓΕ τετραπλάσιον ™δείχθV τοà ΗΖ· καˆ Ð ΞΟΠ ¥ρα γνώµων τετραπλάσιός ™στι τοà ΖΗ τετραγώνου. Ð ΞΟΠ ¥ρα γνώµων κሠτÕ ΖΗ τετράγωνον πενταπλάσιός ™στι τοà ΖΗ. ¢λλ¦ Ð ΞΟΠ γνώµων κሠτÕ ΖΗ τετράγωνόν ™στι τÕ ∆Ν. καί ™στι τÕ µν ∆Ν τÕ ¢πÕ τÁς ∆Β, τÕ δ ΗΖ τÕ ¢πÕ τÁς ∆Γ. τÕ ¥ρα ¢πÕ τÁς ∆Β πενταπλάσιόν ™στι τοà ¢πÕ τÁς ∆Γ· Óπερ œδει δε‹ξαι.

DC. For let the square AE have been described on AB. And let the figure have been drawn double. Since AC is double DC, the (square) on AC (is) thus four times the (square) on DC—that is to say, RS (is four times) F G. And since the (rectangle contained) by ABC is equal to the (square) on AC [Def. 6.3, Prop. 6.17], and CE is the (rectangle contained) by ABC, CE is thus equal to RS. And RS (is) four times F G. Thus, CE (is) also four times F G. Again, since AD is equal to DC, HK is also equal to KF . Hence, square GF is also equal to square HL. Thus, GK (is) equal to KL—that is to say, M N to N E. Hence, M F is also equal to F E. But, M F is equal to CG. Thus, CG is also equal to F E. Let CN have been added to both. Thus, gnomon OP Q is equal to CE. But, CE was shown (to be) equal to four times GF . Thus, gnomon OP Q is also four times square F G. Thus, gnomon OP Q plus square F G is five times F G. But, gnomon OP Q plus square F G is (square) DN . And DN is the (square) on DB, and GF the (square) on DC. Thus, the (square) on DB is five times the (square) on DC. (Which is) the very thing it was required to show.

δ΄.

Proposition 4

'Ε¦ν εÙθε‹α γραµµ¾ ¥κρον κሠµέσον λόγον τµηθÍ, If a straight-line is cut in extreme and mean ratio then τÕ ¢πÕ τÁς Óλης κሠτοà ™λάσσονος τµήµατος, τ¦ the sum of the squares on the whole and the lesser piece συναµφότερα τετράγωνα, τριπλάσιά ™στι τοà ¢πÕ τοà is three times the square on the greater piece. µείζονος τµήµατος τετραγώνου.

A

J

D

G

L

N

Z

H

B

M

A

B

C M

K

H

F

L

K

N

E

D

”Εστω εÙθε‹α ¹ ΑΒ, κሠτετµήσθω ¥κρον κሠµέσον λόγον κατ¦ τÕ Γ, κሠœστω µε‹ζον τµÁµα τÕ ΑΓ· λέγω, Óτι τ¦ ¢πÕ τîν ΑΒ, ΒΓ τριπλάσιά ™στι τοà ¢πÕ τÁς ΓΑ. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ Α∆ΕΒ, κሠκαταγεγράφθω τÕ σχÁµα. ™πεˆ οâν ¹ ΑΒ ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Γ, κሠτÕ µε‹ζον τµÁµά ™στιν ¹ ΑΓ, τÕ ¥ρα ØπÕ τîν ΑΒΓ ‡σον

G

E

Let AB be a straight-line, and let it have been cut in extreme and mean ratio at C, and let AC be the greater piece. I say that the (sum of the squares) on AB and BC is three times the (square) on CA. For let the square ADEB have been described on AB, and let the (remainder of the) figure have been drawn. Therefore, since AB has been cut in extreme and mean

509

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

™στˆ τù ¢πÕ τÁς ΑΓ. καί ™στι τÕ µν ØπÕ τîν ΑΒΓ τÕ ΑΚ, τÕ δ ¢πÕ τÁς ΑΓ τÕ ΘΗ· ‡σον ¥ρα ™στˆ τÕ ΑΚ τù ΘΗ. κሠ™πεˆ ‡σον ™στˆ τÕ ΑΖ τù ΖΕ, κοινÕν προσκείσθω τÕ ΓΚ· Óλον ¥ρα τÕ ΑΚ ÓλJ τù ΓΕ ™στιν ‡σον· τ¦ ¥ρα ΑΚ, ΓΕ τοà ΑΚ ™στι διπλάσια. ¢λλ¦ τ¦ ΑΚ, ΓΕ Ð ΛΜΝ γνώµων ™στˆ κሠτÕ ΓΚ τετράγωνον· Ð ¥ρα ΛΜΝ γνώµων κሠτÕ ΓΚ τετράγωνον διπλάσιά ™στι τοà ΑΚ. ¢λλ¦ µ¾ν κሠτÕ ΑΚ τù ΘΗ ™δείχθη ‡σον· Ð ¥ρα ΛΜΝ γνώµων κሠ[τÕ ΓΚ τετράγωνον διπλάσιά ™στι τοà ΘΗ· éστε Ð ΛΜΝ γνώµων καˆ] τ¦ ΓΚ, ΘΗ τετράγωνα τριπλάσιά ™στι τοà ΘΗ τετραγώνου. καί ™στιν Ð [µν] ΛΜΝ γνώµων κሠτ¦ ΓΚ, ΘΗ τετράγωνα Óλον τÕ ΑΕ κሠτÕ ΓΚ, ¤περ ™στˆ τ¦ ¢πÕ τîν ΑΒ, ΒΓ τετράγωνα, τÕ δ ΗΘ τÕ ¢πÕ τÁς ΑΓ τετράγωνον. τ¦ ¥ρα ¢πÕ τîν ΑΒ, ΒΓ τετράγωνα τριπλάσιά ™στι τοà ¢πÕ τÁς ΑΓ τετραγώνου· Óπερ œδει δε‹ξαι.

ratio at C, and AC is the greater piece, the (rectangle contained) by ABC is thus equal to the (square) on AC [Def. 6.3, Prop. 6.17]. And AK is the (rectangle contained) by ABC, and HG the (square) on AC. Thus, AK is equal to HG. And since AF is equal to F E [Prop. 1.43], let CK have been added to both. Thus, the whole of AK is equal to the whole of CE. Thus, AK plus CE is double AK. But, AK plus CE is the gnomon LM N plus the square CK. Thus, gnomon LM N plus square CK is double AK. But, indeed, AK was also shown (to be) equal to HG. Thus, gnomon LM N plus [square CK is double HG. Hence, gnomon LM N plus] the (sum of the) squares on CK and HG is three times the square HG. And gnomon LM N plus the (sum of the) squares on CK and HG is the whole of AE plus CK—which is the (sum of the) squares on AB and BC— and GH (is) the square on AC. Thus, the (sum of the) squares on AB and BC is three times the square on AC. (Which is) the very thing it was required to show.

ε΄.

Proposition 5

'Ε¦ν εÙθε‹α γραµµ¾ ¥κρον κሠµέσον λόγον τµηθÍ, κሠπροστεθÍ αÙτÍ ‡ση τù µείζονι τµήµατι, ¹ Óλη εÙθε‹α ¥κρον κሠµέσον λόγον τέτµηται, κሠτÕ µε‹ζον τµÁµά ™στιν ¹ ™ξ ¢ρχÁς εÙθε‹α.

If a straight-line is cut in extreme and mean ratio, and a (straight-line) equal to the greater piece is added to it, then the whole straight-line has been cut in extreme and mean ratio, and the greater piece is also the original straight-line.

D

L

A

J

G

B

D

K E

L

A

H

C

B

K

E

ΕÙθε‹α γ¦ρ γραµµ¾ ¹ ΑΒ ¥κρον κሠµέσον λόγον τετµήσθω κατ¦ τÕ Γ σηµε‹ον, κሠœστω µε‹ζον τµÁµα ¹ ΑΓ, κሠτÍ ΑΓ ‡ση [κείσθω] ¹ Α∆. λέγω, Óτι ¹ ∆Β εÙθε‹α ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Α, κሠτÕ µε‹ζον τµÁµά ™στιν ¹ ™ξ ¢ρχÁς εÙθε‹α ¹ ΑΒ. 'Αναγεγράφθω γ¦ρ ¢πÕ τÁς ΑΒ τετράγωνον τÕ ΑΕ, κሠκαταγεγράφθω τÕ σχÁµα. ™πεˆ ¹ ΑΒ ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Γ, τÕ ¥ρα ØπÕ ΑΒΓ ‡σον ™στˆ τù ¢πÕ ΑΓ. καί ™στι τÕ µν ØπÕ ΑΒΓ τÕ ΓΕ, τÕ δ ¢πÕ τÁς ΑΓ τÕ ΓΘ· ‡σον ¥ρα τÕ ΓΕ τù ΘΓ. ¢λλ¦ τù µν ΓΕ ‡σον ™στˆ τÕ ΘΕ, τù δ ΘΓ ‡σον τÕ ∆Θ· κሠτÕ ∆Θ ¥ρα ‡σον ™στˆ τù ΘΕ [κοινÕν προσκείσθω τÕ ΘΒ].

For let the straight-line AB have been cut in extreme and mean ratio at point C. And let AC be the greater piece. And let AD be [made] equal to AC. I say that the straight-line DB has been cut in extreme and mean ratio at A, and that the greater piece is the original straightline AB. For let the square AE have been described on AB, and let the (remainder of the) figure have been drawn. And since AB has been cut in extreme and mean ratio at C, the (rectangle contained) by ABC is thus equal to the (square) on AC [Def. 6.3, Prop. 6.17]. And CE is the (rectangle contained) by ABC, and CH the (square) on

510

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

Óλον ¥ρα τÕ ∆Κ ÓλJ τù ΑΕ ™στιν ‡σον. καί ™στι τÕ µν ∆Κ τÕ ØπÕ τîν Β∆, ∆Α· ‡ση γ¦ρ ¹ Α∆ τÍ ∆Λ· τÕ δ ΑΕ τÕ ¢πÕ τÁς ΑΒ· τÕ ¥ρα ØπÕ τîν Β∆Α ‡σον ™στˆ τù ¢πÕ τÁς ΑΒ. œστιν ¥ρα æς ¹ ∆Β πρÕς τ¾ν ΒΑ, οÛτως ¹ ΒΑ πρÕς τ¾ν Α∆. µείζων δ ¹ ∆Β τÁς ΒΑ· µείζων ¥ρα κሠ¹ ΒΑ τÁς Α∆. `Η ¥ρα ∆Β ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Α, κሠτÕ µε‹ζον τµÁµά ™στιν ¹ ΑΒ· Óπερ œδει δε‹ξαι.

AC. But, HE is equal to CE [Prop. 1.43], and DH equal to HC. Thus, DH is also equal to HE. [Let HB have been added to both.] Thus, the whole of DK is equal to the whole of AE. And DK is the (rectangle contained) by BD and DA. For AD (is) equal to DL. And AE (is) the (square) on AB. Thus, the (rectangle contained) by BDA is equal to the (square) on AB. Thus, as DB (is) to BA, so BA (is) to AD [Prop. 6.17]. And DB (is) greater than BA. Thus, BA (is) also greater than AD [Prop. 5.14]. Thus, DB has been cut in extreme and mean ratio at A, and the greater piece is AB. (Which is) the very thing it was required to show.

$΄.

Proposition 6

'Ε¦ν εÙθε‹α ·ητη ¥κρον κሠµέσον λόγον τµηθÍ, ˜κάτερον τîν τµηµάτων ¥λογός ™στιν ¹ καλουµένη ¢ποτοµή.

If a rational straight-line is cut in extreme and mean ratio then each of the pieces is that irrational (straightline) called an apotome.

D

A

G

B

D

”Εστω εÙθε‹α ·ητ¾ ¹ ΑΒ κሠτετµήσθω ¥κρον κሠµέσον λόγον κατ¦ τÕ Γ, κሠœστω µε‹ζον τµÁµα ¹ ΑΓ· λέγω, Óτι ˜κατέρα τîν ΑΓ, ΓΒ ¥λογός ™στιν ¹ καλουµένη ¢ποτοµή. 'Εκβεβλήσθω γ¦ρ ¹ ΒΑ, κሠκείσθω τÁς ΒΑ ¹µίσεια ¹ Α∆. ™πεˆ οâν εÙθε‹α ¹ ΑΒ τέτµηται ¥κρον κሠµέσον λόγον κατ¦ τÕ Γ, κሠτù µείζονι τµήµατι τù ΑΓ πρόσκειται ¹ Α∆ ¹µίσεια οâσα τÁς ΑΒ, τÕ ¥ρα ¢πÕ Γ∆ τοà ¢πÕ ∆Α πενταπλάσιόν ™στιν. τÕ ¥ρα ¢πÕ Γ∆ πρÕς τÕ ¢πÕ ∆Α λόγον œχει, Öν ¢ριθµÕς πρÕς ¢ριθµόν· σύµµετρον ¥ρα τÕ ¢πÕ Γ∆ τù ¢πÕ ∆Α. ·ητÕν δ τÕ ¢πÕ ∆Α· ·ητ¾ γάρ [™στιν] ¹ ∆Α ¹µίσεια οâσα τÁς ΑΒ ·ητÁς οÜσης· ·ητÕν ¥ρα κሠτÕ ¢πÕ Γ∆· ·ητ¾ ¥ρα ™στˆ κሠ¹ Γ∆. κሠ™πεˆ τÕ ¢πÕ Γ∆ πρÕς τÕ ¢πÕ ∆Α λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν, ¢σύµµετρος ¥ρα µήκει ¹ Γ∆ τÍ ∆Α· αƒ Γ∆, ∆Α ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΑΓ. πάλιν, ™πεˆ ¹ ΑΒ ¥κρον κሠµέσον λόγον τέτµηται, κሠτÕ µε‹ζον τµÁµά ™στιν ¹ ΑΓ, τÕ ¥ρα ØπÕ ΑΒ, ΒΓ τù ¢πÕ ΑΓ ‡σον ™στίν. τÕ ¥ρα ¢πÕ τÁς ΑΓ ¢ποτοµÁς παρ¦ τ¾ν ΑΒ ·ητ¾ν παραβληθν πλάτος ποιε‹ τ¾ν ΒΓ. τÕ δ ¢πÕ ¢ποτοµÁς παρ¦ ·ητ¾ν παραβαλλόµενον πλάτος ποιε‹ ¢ποτοµ¾ν πρώτην· ¢ποτοµ¾ ¥ρα πρώτη ™στˆν ¹ ΓΒ. ™δείχθη δ κሠ¹ ΓΑ ¢ποτοµή. 'Ε¦ν ¥ρα εÙθε‹α ·ητ¾ ¥κρον κሠµέσον λόγον τµηθÍ, ˜κάτερον τîν τµηµάτων ¥λογός ™στιν ¹ καλουµένη ¢ποτοµή· Óπερ œδει δε‹ξαι.

A

C

B

Let AB be a rational straight-line cut in extreme and mean ratio at C, and let the greater piece be AC. I say that AC and CB is each that irrational (straight-line) called an apotome. For let BA have been produced, and let AD be made (equal) to half of BA. Therefore, since the straightline AB has been cut in extreme and mean ratio at C, and AD, which is half of AB, has been added to the greater piece AC, the (square) on CD is thus five times the (square) on DA [Prop. 13.1]. Thus, the (square) on CD has to the (square) on DA the ratio which a number (has) to a number. The (square) on CD (is) thus commensurable with the (square) on DA [Prop. 10.6]. And the (square) on DA (is) rational. For DA [is] rational, being half of DA, which is rational. Thus, the (square) on CD (is) also rational [Def. 10.4]. Thus, CD is also rational. And since the (square) on CD does not have to the (square) on DA the ratio which a square number (has) to a square number, CD (is) thus incommensurable in length with DA [Prop. 10.9]. Thus, CD and DA are rational (straight-lines which are) commensurable in square only. Thus, AC is an apotome [Prop. 10.73]. Again, since AB has been cut in extreme and mean ratio, and AC is the greater piece, the (rectangle contained) by AB and BC is thus equal to the (square) on AC [Def. 6.3, Prop. 6.17]. Thus, the (square) on the apotome AC, applied to the rational (straight-line) AB, makes BC as width. And the (square) on an apotome, applied to a rational (straight-line), makes a first apotome, as width [Prop. 10.97]. Thus, CB is a first apotome. And CA was

511

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13 also shown (to be) an apotome. Thus, if a rational straight-line is cut in extreme and mean ratio then each of the pieces is that irrational (straight-line) called an apotome.

ζ΄.

Proposition 7

'Ε¦ν πενταγώνου „σοπλεύρου αƒ τρε‹ς γωνίαι ½τοι αƒ κατ¦ τÕ ˜ξÁς À αƒ µ¾ κατ¦ τÕ ˜ξÁς ‡σαι ðσιν, „σογώνιον œσται τÕ πεντάγωνον.

If three angles, either consecutive or not consecutive, of an equilateral pentagon are equal then the pentagon will be equiangular.

A

A

Z

B

G

E

D

F

B

C

Πενταγώνου γ¦ρ „σοπλεύρον τοà ΑΒΓ∆Ε αƒ τρε‹ς γωνίαι πρότερον αƒ κατ¦ τÕ ˜ξÁς αƒ πρÕς το‹ς Α, Β, Γ ‡σαι ¢λλήλαις œστωσαν· λέγω, Óτι „σογώνιόν ™στι τÕ ΑΒΓ∆Ε πεντάγωνον. 'Επεζεύχθωσαν γ¦ρ αƒ ΑΓ, ΒΕ, Ζ∆. κሠ™πεˆ δύο αƒ ΓΒ, ΒΑ δυσˆ τα‹ς ΒΑ, ΑΕ ‡σαι ™ισˆν ˜κατέρα ˜κατέρv, κሠγωνία ¹ ØπÕ ΓΒΑ γωνίv τÍ ØπÕ ΒΑΕ ™στιν ‡ση, βάσις ¥ρα ¹ ΑΓ βάσει τÍ ΒΕ ™στιν ‡ση, κሠτÕ ΑΒΓ τρίγωνον τù ΑΒΕ τριγώνJ ‡σον, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν, ¹ µν ØπÕ ΒΓΑ τÍ ØπÕ ΒΕΑ, ¹ δ ØπÕ ΑΒΕ τÍ ØπÕ ΓΑΒ· éστε κሠπλευρ¦ ¹ ΑΖ πλευρ´ τÍ ΒΖ ™στιν ‡ση. ™δείχθη δ κሠÓλη ¹ ΑΓ ÓλV τÍ ΒΕ ‡ση· κሠλοιπ¾ ¥ρα ¹ ΖΓ λοιπÍ τÍ ΖΕ ™στιν ‡ση. œστι δ κሠ¹ Γ∆ τÍ ∆Ε ‡ση. δύο δ¾ αƒ ΖΓ, Γ∆ δυσˆ τα‹ς ΖΕ, Ε∆ ‡σαι ε„σίν· κሠβάσις αÙτîν κοιν¾ ¹ Ζ∆· γωνία ¥ρα ¹ ØπÕ ΖΓ∆ γωνίv τÍ ØπÕ ΖΕ∆ ™στιν ‡ση. ™δείχθη δ κሠ¹ ØπÕ ΒΓΑ τÍ ØπÕ ΑΕΒ ‡ση· κሠÓλη ¥ρα ¹ ØπÕ ΒΓ∆ ÓλV τÍ ØπÕ ΑΕ∆ ‡ση. ¢λλ' ¹ ØπÕ ΒΓ∆ ‡ση Øπόκειται τα‹ς πρÕς το‹ς Α, Β γωνίαις· κሠ¹ ØπÕ ΑΕ∆ ¥ρα τα‹ς πρÕς το‹ς Α, Β γωνίαις ‡ση ™στίν. еοίως δ¾ δείξοµεν, Óτι κሠ¹ ØπÕ Γ∆Ε γωνία ‡ση ™στˆ τα‹ς πρÕς το‹ς Α, Β, Γ γωνίαις· „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ∆Ε πεντάγωνον. 'Αλλ¦ δ¾ µ¾ œστωσαν ‡σαι αƒ κατ¦ τÕ ˜ξÁς γωνίαι, ¢λλ' œστωσαν ‡σαι αƒ πρÕς το‹ς Α, Γ, ∆ σηµείοις· λέγω, Óτι κሠοÛτως „σογώνιόν ™στι τÕ ΑΒΓ∆Ε πεντάγωνον. 'Επεζεύχθω γ¦ρ ¹ Β∆. κሠ™πεˆ δύο αƒ ΒΑ, ΑΕ δυσˆ τα‹ς ΒΓ, Γ∆ ‡σαι ε„σˆ κሠγωνίας ‡σας περιέχουσιν, βάσις ¥ρα ¹ ΒΕ βάσει τÍ Β∆ ‡ση ™στίν, κሠτÕ ΑΒΕ τρίγωνον

E

D

For let three angles of the equilateral pentagon ABCDE—first of all, the consecutive (angles) at A, B, and C—-be equal to one another. I say that pentagon ABCDE is equiangular. For let AC, BE, and F D have been joined. And since the two (straight-lines) CB and BA are equal to the two (straight-lines) BA and AE, respectively, and angle CBA is equal to angle BAE, thus base AC is equal to base BE, and triangle ABC equal to triangle ABE, and the remaining angles will be equal to the remaining angles which the equal sides subtend [Prop. 1.4], (that is), BCA (equal) to BEA, and ABE to CAB. And hence side AF is also equal to side BF [Prop. 1.6]. And the whole of AC was also shown (to be) equal to the whole of BE. Thus, the remainder F C is also equal to the remainder F E. And CD is also equal to DE. So, the two (straight-lines) F C and CD are equal to the two (straight-lines) F E and ED (respectively). And F D is their common base. Thus, angle F CD is equal to angle F ED [Prop. 1.8]. And BCA was also shown (to be) equal to AEB. And thus the whole of BCD (is) equal to the whole of AED. But, (angle) BCD was assumed (to be) equal to the angles at A and B. Thus, (angle) AED is also equal to the angles at A and B. So, similarly, we can show that angle CDE is also equal to the angles at A, B, C. Thus, pentagon ABCDE is equiangular. And so let consecutive angles not be equal, but let the (angles) at points A, C, and D be equal. I say that pentagon ABCD is also equiangular in this case.

512

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

τù ΒΓ∆ τριγώνJ ‡σον ™στίν, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν· ‡ση ¥ρα ™στˆν ¹ ØπÕ ΑΕΒ γωνία τÍ ØπÕ Γ∆Β. œστι δ κሠ¹ ØπÕ ΒΕ∆ γωνία τÍ ØπÕ Β∆Ε ‡ση, ™πεˆ κሠπλευρ¦ ¹ ΒΕ πλευρ´ τÍ Β∆ ™στιν ‡ση. κሠÓλη ¥ρα ¹ ØπÕ ΑΕ∆ γωνία ÓλV τÍ ØπÕ Γ∆Ε ™στιν ‡ση. ¢λλ¦ ¹ ØπÕ Γ∆Ε τα‹ς πρÕς το‹ς Α, Γ γωνίαις Øπόκειται ‡ση· κሠ¹ ØπÕ ΑΕ∆ ¥ρα γωνία τα‹ς πρÕς το‹ς Α, Γ ‡ση ™στίν. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ØπÕ ΑΒΓ ‡ση ™στˆ τα‹ς πρÕς το‹ς Α, Γ, ∆ γωνίαις. „σογώνιον ¥ρα ™στˆ τÕ ΑΒΓ∆Ε πεντάγωνον· Óπερ œδει δε‹ξαι.

For let BD have been joined. And since the two (straight-lines) BA and AE are equal to the (straightlines) BC and CD, and they contain equal angles, base BE is thus equal to base BD, and triangle ABC is equal to triangle BCD, and the remaining angles will be equal to the remaining angles which the equal sides subtend [Prop. 1.4]. Thus, angle AEB is equal to (angle) CDB. And angle BED is also equal to (angle) BDE, since side BE is also equal to side BD [Prop. 1.5]. Thus, the whole angle AED is also equal to the whole (angle) CDE. But, (angle) CDE was assumed (to be) equal to the angles at A and C. Thus, angle AED is also equal to the (angles) at A and C. So, for the same (reasons), (angle) ABC is also equal to the angles at A, C, and D. Thus, pentagon ABCDE is equiangular. (Which is) the very thing it was required to show.

η΄.

Proposition 8

'Ε¦ν πενταγώνου „σοπλεύρου κሠ„σογωνίου τ¦ς κατ¦ τÕ ˜ξÁς δύο γωνίας Øποτείνωσιν εÙθε‹αι, ¥κρον κሠµέσον λόγον τέµνουσιν ¢λλήλας, κሠτ¦ µείζονα αÙτîν τµήµατα ‡σα ™στˆ τÍ τοà πεναγώνου πλευρ´.

If straight-lines subtend two consecutive angles of an equilateral and equiangular pentagon then they cut one another in extreme and mean ratio, and their greater pieces are equal to the sides of the pentagon.

A

A

J

E

D

B

H

E

G

D

Πενταγώνου γ¦ρ „σοπλεύρον κሠ„σογωνίου τοà ΑΒΓ∆Ε δύο γωνίας τ¦ς κατ¦ τÕ ˜ξÁς τ¦ς πρÕς το‹ς Α, Β Øποτεινέτωσαν εÙθε‹αι αƒ ΑΓ, ΒΕ τέµνουσαι ¢λλήλας κατ¦ τÕ Θ σηµεˆον· λέγω, Óτι ˜κατέρα αÙτîν ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Θ σηµε‹ον, κሠτ¦ µείζονα αÙτîν τµήµατα ‡σα ™στˆ τÍ τοà πενταγώνου πλευρ´. Περιγεγράφθω γ¦ρ περˆ τÕ ΑΒΓ∆Ε πεντάγωνον κύκλος Ð ΑΒΓ∆Ε. κሠ™πεˆ δύο εÙθεˆαι αƒ ΕΑ, ΑΒ δυσˆ τα‹ς ΑΒ, ΒΓ ‡σαι ε„σˆ κሠγωνίας ‡σας περιέχουσιν, βάσις ¥ρα ¹ ΒΕ βάσει τÍ ΑΓ ‡ση ™στίν, κሠτÕ ΑΒΕ τρίγωνον τù ΑΒΓ τριγώνJ ‡σον ™στίν, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται ˜κατέρα ˜κατέρv, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν. ‡ση ¥ρα ™στˆν ¹ ØπÕ ΒΑΓ γωνία τÍ ØπÕ ΑΒΕ· διπλÁ ¥ρα ¹ ØπÕ ΑΘΕ τÁς ØπÕ ΒΑΘ. œστι δ κሠ¹ ØπÕ ΕΑΓ τÁς ØπÕ ΒΑΓ διπλÁ,

B

C

For let the two straight-lines, BE and AC, cutting one another at point H, have subtended two consecutive angles, at A and B (respectively), of the equilateral and equiangular pentagon ABCDE. I say that each of them has been cut in extreme and mean ratio at point H, and that their greater pieces are equal to the sides of the pentagon. For let the circle ABCDE have been circumscribed about pentagon ABCDE [Prop. 4.14]. And since the two straight-lines EA and AB are equal to the two (straightlines) AB and BC (respectively), and they contain equal angles, the base BE is thus equal to the base AC, and triangle ABE is equal to triangle ABC, and the remaining angles will be equal to the remaining angles, respectively, which the equal sides subtend [Prop. 1.4]. Thus, angle BAC is equal to (angle) ABE. Thus, (angle) AHE (is)

513

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

™πειδήπερ κሠπεριφέρεια ¹ Ε∆Γ περιφερείας τÁς ΓΒ ™στι διπλÁ· ‡ση ¥ρα ¹ ØπÕ ΘΑΕ γωνία τÍ ØπÕ ΑΘΕ· éστε κሠ¹ ΘΕ εÙθε‹α τÍ ΕΑ, τουτέστι τÍ ΑΒ ™στιν ‡ση. κሠ™πεˆ ‡ση ™στˆν ¹ ΒΑ εÙθε‹α τÍ ΑΕ, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ΑΒΕ τÍ ØπÕ ΑΕΒ. ¢λλ¦ ¹ ØπÕ ΑΒΕ τÍ ØπÕ ΒΑΘ ™δείχθη ‡ση· κሠ¹ ØπÕ ΒΕΑ ¥ρα τÍ ØπÕ ΒΑΘ ™στιν ‡ση. κሠκοιν¾ τîν δύο τριγώνων τοà τε ΑΒΕ κሠτοà ΑΒΘ ™στιν ¹ ØπÕ ΑΒΕ· λοιπ¾ ¥ρα ¹ ØπÕ ΒΑΕ γωνία λοιπÍ τÍ ØπÕ ΑΘΒ ™στιν ‡ση· „σογώνιον ¥ρα ™στˆ τÕ ΑΒΕ τρίγωνον τù ΑΒΘ τριγώνJ· ¢νάλογον ¥ρα ™στˆν æς ¹ ΕΒ πρÕς τ¾ν ΒΑ, οÛτως ¹ ΑΒ πρÕς τ¾ν ΒΘ. ‡ση δ ¹ ΒΑ τÍ ΕΘ· æς ¥ρα ¹ ΒΕ πρÕς τ¾ν ΕΘ, οÛτως ¹ ΕΘ πρÕς τ¾ν ΘΒ. µείζων δ ¹ ΒΕ τÁς ΕΘ· µείζων ¥ρα κሠ¹ ΕΘ τÁς ΘΒ. ¹ ΒΕ ¤ρα ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Θ, κሠτÕ µε‹ζον τµÁµα τÕ ΘΕ ‡σον ™στˆ τÍ τοà πενταγώνου πλευρ´. еοίως δ¾ δείξοµεν, Óτι κሠ¹ ΑΓ ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Θ, κሠτÕ µε‹ζον αÙτÁς τµÁµα ¹ ΓΘ ‡σον ™στˆ τÍ τοà πενταγώνου πλευρ´· Óπερ ™δει δε‹ξαι.

double (angle) BAH [Prop. 1.32]. And EAC is also double BAC, inasmuch as circumference EDC is also double circumference CB [Props. 3.28, 6.33]. Thus, angle HAE (is) equal to (angle) AHE. Hence, straight-line HE is also equal to (straight-line) EA—that is to say, to (straight-line) AB [Prop. 1.6]. And since straight-line BA is equal to AE, angle ABE is also equal to AEB [Prop. 1.5]. But, ABE was shown (to be) equal to BAH. Thus, BEA is also equal to BAH. And (angle) ABE is common to the two triangles ABE and ABH. Thus, the remaining angle BAE is equal to the remaining (angle) AHB [Prop. 1.32]. Thus, triangle ABE is equiangular to triangle ABH. Thus, proportionally, as EB is to BA, so AB (is) to BH [Prop. 6.4]. And BA (is) equal to EH. Thus, as BE (is) to EH, so EH (is) to HB. And BE (is) greater than EH. EH (is) thus also greater than HB [Prop. 5.14]. Thus, BE has been cut in extreme and mean ratio at H, and the greater piece HE is equal to the side of the pentagon. So, similarly, we can show that AC has also been cut in extreme and mean ratio at H, and that its greater piece CH is equal to the side of the pentagon. (Which is) the very thing it was required to show.

θ΄.

Proposition 9

'Ε¦ν ¹ τοà ˜ξαγώνου πλευρ¦ κሠ¹ τοà δεκαγώνου If the side of a hexagon and of a decagon inscribed τîν ε„ς τÕν αÙτÕν κύκλον ™γγραφοµένων συντεθîσιν, ¹ in the same circle are added together then the whole Óλη εÙθε‹α ¥κρον κሠµέσον λόγον τέτµηται, κሠτÕ straight-line has been cut in extreme and mean ratio (at µε‹ξον αÙτÁς τµÁµά ™στιν ¹ τοà ˜ξαγώνου πλευρά. the junction point), and its greater piece is the side of the hexagon.†

E

B G

A

E

B

A

C

Z D

F D

”Εστω κύκλος Ð ΑΒΓ, κሠτîν ε„ς τÕν ΑΒΓ κύκλον ™γγραφοµένων σχηµάτων, δεκαγώνου µν œστω πλευρ¦ ¹ ΒΓ, ˜ξαγώνου δ ¹ Γ∆, κሠœστωσαν ™π' εÙθείας· λέγω, Óτι ¹ Óλη εÙθε‹α ¹ Β∆ ¥κρον κሠµέσον λόγον τέτµηται, κሠτÕ µε‹ζον αÙτÁς τµÁµά ™στιν ¹ Γ∆. Ε„λήφθω γ¦ρ τÕ κέντρον τοà κύκλου τÕ Ε σηµε‹ον,

Let ABC be a circle. And of the figures inscribed in circle ABC, let BC be the side of a decagon, and CD (the side) of a hexagon. And let them be (laid down) straighton (to one another). I say that the whole straight-line BD has been cut in extreme and mean ratio (at C), and that its greater piece is CD.

514

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

κሠ™πεζεύχθωσαν αƒ ΕΒ, ΕΓ, Ε∆, κሠδιήχθω ¹ ΒΕ ™πˆ τÕ Α. ™πεˆ δεκαγώνου „σοπλεύρον πλευρά ™στιν ¹ ΒΓ, πενταπλασίων ¥ρα ¹ ΑΓΒ περιφέρεια τÁς ΒΓ περιφερείας· τετραπλασίων ¥ρα ¹ ΑΓ περιφέρεια τÁς ΓΒ. æς δ ¹ ΑΓ περιφέρεια πρÕς τ¾ν ΓΒ, οÛτως ¹ ØπÕ ΑΕΓ γωνία πρÕς τ¾ν ØπÕ ΓΕΒ· τετραπλασίων ¥ρα ¹ ØπÕ ΑΕΓ τÁς ØπÕ ΓΕΒ. κሠ™πεˆ ‡ση ¹ ØπÕ ΕΒΓ γωνία τÍ ØπÕ ΕΓΒ, ¹ ¥ρα ØπÕ ΑΕΓ γωνία διπλασία ™στˆ τÁς ØπÕ ΕΓΒ. κሠ™πεˆ ‡ση ™στˆν ¹ ΕΓ εÙθε‹α τÍ Γ∆· ˜κατέρα γ¦ρ αÙτîν ‡ση ™στˆ τÍ τοà ˜ξαγώνου πλευρ´ τοà ε„ς τÕν ΑΒΓ κύκλον [™γγραφοµένου]· ‡ση ™στˆ κሠ¹ ØπÕ ΓΕ∆ γωνία τÍ ØπÕ Γ∆Ε γωνίv· διπλασία ¥ρα ¹ ØπÕ ΕΓΒ γωνία τÁς ØπÕ Ε∆Γ. ¢λλ¦ τÁς ØπÕ ΕΓΒ διπλασία ™δείχθη ¹ ØπÕ ΑΕΓ· τετραπλασία ¥ρα ¹ ØπÕ ΑΕΓ τÁς ØπÕ Ε∆Γ. ™δείχθη δ κሠτÁς ØπÕ ΒΕΓ τετραπλασία ¹ ØπÕ ΑΕΓ· ‡ση ¥ρα ¹ ØπÕ Ε∆Γ τÍ ØπÕ ΒΕΓ. κοιν¾ δ τîν δύο τριγώνων, τοà τε ΒΕΓ κሠτοà ΒΕ∆, ¹ ØπÕ ΕΒ∆ γωνία· κሠλοιπ¾ ¥ρα ¹ ØπÕ ΒΕ∆ τÍ ØπÕ ΕΓΒ ™στιν ‡ση· „σογώνιον ¥ρα ™στˆ τÕ ΕΒ∆ τρίγωνον τù ΕΒΓ τριγώνJ. ¢νάλογον ¥ρα ™στˆν æς ¹ ∆Β πρÕς τ¾ν ΒΕ, οÛτως ¹ ΕΒ πρÕς τ¾ν ΒΓ. ‡ση δ ¹ ΕΒ τÍ Γ∆. œστιν ¥ρα æς ¹ Β∆ πρÕς τ¾ν ∆Γ, οÛτως ¹ ∆Γ πρÕς τ¾ν ΓΒ. µείζων δ ¹ Β∆ τÁς ∆Γ· µείζων ¥ρα κሠ¹ ∆Γ τÁς ΓΒ. ¹ Β∆ ¥ρα εÙθε‹α ¥κρον κሠµέσον λόγον τέτµηται [κατ¦ τÕ Γ], κሠτÕ µε‹ζον τµÁµα αÙτÁς ™στιν ¹ ∆Γ· Óπερ œδει δε‹ξαι.

†

For let the center of the circle, point E, have been found [Prop. 3.1], and let EB, EC, and ED have been joined, and let BE have been drawn across to A. Since BC is a side on an equilateral decagon, circumference ACB (is) thus five times circumference BC. Thus, circumference AC (is) four times CB. And as circumference AC (is) to CB, so angle AEC (is) to CEB [Prop. 6.33]. Thus, (angle) AEC (is) four times CEB. And since angle EBC (is) equal to ECB [Prop. 1.5], angle AEC is thus double ECB [Prop. 1.32]. And since straight-line EC is equal to CD—for each of them is equal to the side of the hexagon [inscribed] in circle ABC [Prop. 4.15 corr.]— angle CED is also equal to angle CDE [Prop. 1.5]. Thus, angle ECB (is) double EDC [Prop. 1.32]. But, AEC was shown (to be) double ECB. Thus, AEC (is) four times EDC. And AEC was also shown (to be) four times BEC. Thus, EDC (is) equal to BEC. And angle EBD (is) common to the two triangles BEC and BED. Thus, the remaining (angle) BED is equal to the (remaining angle) ECB [Prop. 1.32]. Thus, triangle EBD is equiangular to triangle EBC. Thus, proportionally, as DB is to BE, so EB (is) to BC [Prop. 6.4]. And EB (is) equal to CD. Thus, as BD is to DC, so DC (is) to CB. And BD (is) greater than DC. Thus, DC (is) also greater than CB [Prop. 5.14]. Thus, the straight-line BD has been cut in extreme and mean ratio [at C], and its greater piece is DC. (Which is), the very thing it was required to show.

√ If the circle is of unit radius, then the side of the hexagon is 1, whereas the side of the decagon is (1/2) ( 5 − 1).

ι΄.

Proposition 10

'Ε¦ν ε„ς κύκλον πεντάγωνον „σόπλευρον ™γγραφÍ, ¹ If an equilateral pentagon is inscribed in a circle then τοà πενταγώνου πλευρ¦ δύναται τήν τε τοà ˜ξαγώνου κሠthe square on the side of the pentagon is (equal to) the τ¾ν τοà δεκαγώνου τîν ε„ς τÕν αÙτÕν κύκλον ™γγρα- (sum of the squares) on the (sides) of the hexagon and of φοµένων. the decagon inscribed in the same circle.†

K J

B

M A L N

M K H

E

H

L N

B

E

Z

G

A

F

D

D

C G 515

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

”Εστω κύκλος Ð ΑΒΓ∆Ε, κሠε„ς τÕ ΑΒΓ∆Ε κύκλον πεντάγωνον „σόπλευρον ™γγεγράφθω τÕ ΑΒΓ∆Ε. λέγω, Óτι ¹ τοà ΑΒΓ∆Ε πενταγώνου πλευρ¦ δύναται τήν τε τοà ˜ξαγώνου κሠτ¾ν τοà δεκαγώνου πλευρ¦ν τîν ε„ς τÕν ΑΒΓ∆Ε κύκλον ™γγραφοµένων. Ε„λήφθω γ¦ρ τÕ κέντρον τοà κύκλου τÕ Ζ σηµεˆον, κሠ™πιζευχθε‹σα ¹ ΑΖ διήχθω ™πˆ τÕ Η σηµε‹ον, κሠ™πεζεύχθω ¹ ΖΒ, κሠ¢πÕ τοà Ζ ™πˆ τ¾ν ΑΒ κάθετος ½χθω ¹ ΖΘ, κሠδιήχθω ™πˆ τÕ Κ, κሠ™πεζεύχθωσαν αƒ ΑΚ, ΚΒ, κሠπάλιν ¢πÕ τοà Ζ ™πˆ τ¾ν ΑΚ κάθετος ½χθω ¹ ΖΛ, κሠδιήχθω ™πˆ τÕ Μ, κሠ™πεζεύχθω ¹ ΚΝ. 'Επεˆ ‡ση ™στˆν ¹ ΑΒΓΗ περιφέρεια τÍ ΑΕ∆Η περιφερείv, ïν ¹ ΑΒΓ τÍ ΑΕ∆ ™στιν ‡ση, λοιπ¾ ¥ρα ¹ ΓΗ περιφέρεια λοιπÍ τÍ Η∆ ™στιν ‡ση. πενταγώνου δ ¹ Γ∆· δεκαγώνου ¥ρα ¹ ΓΗ. κሠ™πεˆ ‡ση ™στˆν ¹ ΖΑ τÍ ΖΒ, κሠκάθετος ¹ ΖΘ, ‡ση ¥ρα κሠ¹ ØπÕ ΑΖΚ γωνία τÍ ØπÕ ΚΖΒ. éστε κሠπεριφέρεια ¹ ΑΚ τÍ ΚΒ ™στιν ‡ση· διπλÁ ¥ρα ¹ ΑΒ περιφέρεια τÁς ΒΚ περιφερείας· δεκαγώνου ¥ρα πλευρά ™στιν ¹ ΑΚ εÙθε‹α. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΑΚ τÁς ΚΜ ™στι διπλÁ. κሠ™πεˆ διπλÁ ™στιν ¹ ΑΒ περιφέρεια τÁς ΒΚ περιφερείας, ‡ση δ ¹ Γ∆ περιφέρεια τÍ ΑΒ περιφερείv, διπλÁ ¥ρα κሠ¹ Γ∆ περιφέρεια τÁς ΒΚ περιφερείας. œστι δ ¹ Γ∆ περιφέρεια κሠτÁς ΓΗ διπλÁ· ‡ση ¥ρα ¹ ΓΗ περιφέρεια τÍ ΒΚ περιφερείv. ¢λλ¦ ¹ ΒΚ τÁς ΚΜ ™στι διπλÁ, ™πεˆ κሠ¹ ΚΑ· κሠ¹ ΓΗ ¥ρα τÁς ΚΜ ™στι διπλÁ. ¢λλ¦ µ¾ν κሠ¹ ΓΒ περιφέρεια τÁς ΒΚ περιφερείας ™στˆ διπλÁ· ‡ση γ¦ρ ¹ ΓΒ περιφέρεια τÍ ΒΑ. κሠÓλη ¥ρα ¹ ΗΒ περιφέρεια τÁς ΒΜ ™στι διπλÁ· éστε κሠγωνία ¹ ØπÕ ΗΖΒ γωνίας τÁς ØπÕ ΒΖΜ [™στι] διπλÁ. œστι δ ¹ ØπÕ ΗΖΒ κሠτÁς ØπÕ ΖΑΒ διπλÁ· ‡ση γ¦ρ ¹ ØπÕ ΖΑΒ τÍ ØπÕ ΑΒΖ. κሠ¹ ØπÕ ΒΖΝ ¥ρα τÍ ØπÕ ΖΑΒ ™στιν ‡ση. κοιν¾ δ τîν δύο τριγώνων, τοà τε ΑΒΖ κሠτοà ΒΖΝ, ¹ ØπÕ ΑΒΖ γωνία· λοιπ¾ ¥ρα ¹ ØπÕ ΑΖΒ λοιπÍ τÍ ØπÕ ΒΝΖ ™στιν ‡ση· ‡σογώνιον ¥ρα ™στˆ τÕ ΑΒΖ τρίγωνον τù ΒΖΝ τριγώνJ. ¢νάλογον ¥ρα ™στˆν æς ¹ ΑΒ εÙθε‹α πρÕς τ¾ν ΒΖ, οÛτως ¹ ΖΒ πρÕς τ¾ν ΒΝ· τÕ ¥ρα ØπÕ τîν ΑΒΝ ‡σον ™στˆ τù ¢πÕ ΒΖ. πάλιν ™πεˆ ‡ση ™στˆν ¹ ΑΛ τÍ ΛΚ, κοιν¾ δ κሠπρÕς Ñρθ¦ς ¹ ΛΝ, βάσις ¥ρα ¹ ΚΝ βάσει τÍ ΑΝ ™στιν ‡ση· κሠγωνία ¥ρα ¹ ØπÕ ΛΚΝ γωνίv τÍ ØπÕ ΛΑΝ ™στιν †ση. ¢λλ¦ ¹ ØπÕ ΛΑΝ τÍ ØπÕ ΚΒΝ ™στιν ‡ση· κሠ¹ ØπÕ ΛΚΝ ¥ρα τÍ ØπÕ ΚΒΝ ™στιν ‡ση. κሠκοιν¾ τîν δύο τριγώνων τοà τε ΑΚΒ κሠτοà ΑΚΝ ¹ πρÕς τù Α. λοιπ¾ ¥ρα ¹ ØπÕ ΑΚΒ λοιπÍ τÍ ØπÕ ΚΝΑ ™στιν ‡ση· „σογώνιον ¥ρα ™στˆ τÕ ΚΒΑ τρίγωνον τù ΚΝΑ τριγώµJ. ¢νάλογον ¥ρα ™στˆν æς ¹ ΒΑ εÙθε‹α πρÕς τ¾ν ΑΚ, οÛτως ¹ ΚΑ πρÕς τ¾ν ΑΝ· τÕ ¥ρα ØπÕ τîν ΒΑΝ ‡σον ™στˆ τù ¢πÕ τÁς ΑΚ. ™δείχθη δ κሠτÕ ØπÕ τîν ΑΒΝ †σον τù ¢πÕ τÁς ΒΖ· τÕ ¥ρα ØπÕ τîν ΑΒΝ µετ¦ τοà ØπÕ ΒΑΝ, Óπερ ™στˆ τÕ ¢πÕ τÁς ΒΑ, ‡σον ™στˆ τù ¢πÕ τÁς ΒΖ µετ¦ τοà

Let ABCDE be a circle. And let the equilateral pentagon ABCDE have been inscribed in circle ABCDE. I say that the square on the side of pentagon ABCDE is the (sum of the squares) on the sides of the hexagon and of the decagon inscribed in circle ABCDE. For let the center of the circle, point F , have been found [Prop. 3.1]. And, AF being joined, let it have been drawn across to point G. And let F B have been joined. And let F H have been drawn from F perpendicular to AB. And let it have been drawn across to K. And let AK and KB have been joined. And, again, let F L have been drawn from F perpendicular to AK. And let it have been drawn across to M . And let KN have been joined. Since circumference ABCG is equal to circumference AEDG, of which ABC is equal to AED, the remaining circumference CG is thus equal to the remaining (circumference) GD. And CD (is the side) of the pentagon. CG (is) thus (the side) of the decagon. And since F A is equal to F B, and F H is perpendicular (to AB), angle AF K (is) thus also equal to KF B [Props. 1.5, 1.26]. Hence, circumference AK is also equal to KB [Prop. 3.26]. Thus, circumference AB (is) double circumference BK. Thus, straight-line AK is the side of the decagon. So, for the same (reasons, circumference) AK is also double KM . And since circumference AB is double circumference BK, and circumference CD (is) equal to circumference AB, circumference CD (is) thus also double circumference BK. And circumference CD is also double CG. Thus, circumference CG (is) equal to circumference BK. But, BK is double KM , since KA (is) also (double KM ). Thus, (circumference) CG is also double KM . But, indeed, circumference CB is also double circumference BK. For circumference CB (is) equal to BA. Thus, the whole circumference GB is also double BM . Hence, angle GF B [is] also double angle BF M [Prop. 6.33]. And GF B (is) also double F AB. For F AB (is) equal to ABF . Thus, BF N is also equal to F AB. And angle ABF (is) common to the two triangles ABF and BF N . Thus, the remaining (angle) AF B is equal to the remaining (angle) BN F [Prop. 1.32]. Thus, triangle ABF is equiangular to triangle BF N . Thus, proportionally, as straight-line AB (is) to BF , so F B (is) to BN [Prop. 6.4]. Thus, the (rectangle contained) by ABN is equal to the (square) on BF [Prop. 6.17]. Again, since AL is equal to LK, and LN is common and at right-angles (to HA), base KN is thus equal to base AN [Prop. 1.4]. And, thus, angle LKN is equal to angle LAN . But, LAN is equal to KBN [Props. 3.29, 1.5]. Thus, LKN is also equal to KBN . And the (angle) at A (is) common to the two triangles AKB and AKN . Thus, the remaining (angle) AKB is

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ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

¢πÕ τÁς ΑΚ. καί ™στιν ¹ µν ΒΑ πενταγώνου πλευρά, ¹ δ ΒΖ ˜ξαγώνου, ¹ δ ΑΚ δεκαγώνου. `Η ¥ρα τοà πενταγώνου πλευρ¦ δύναται τήν τε τοà ˜ξαγώνου κሠτ¾ν τοà δεκαγώνου τîν ε„ς τÕν αÙτÕν κύκλον ™γγραφοµένων· Óπερ œδει δε‹ξαι.

†

If the circle is of unit radius, then the side of the pentagon is (1/2)

p

equal to the remaining (angle) KN A [Prop. 1.32]. Thus, triangle KBA is equiangular to triangle KN A. Thus, proportionally, as straight-line BA is to AK, so KA (is) to AN [Prop. 6.4]. Thus, the (rectangle contained) by BAN is equal to the (square) on AK [Prop. 6.17]. And the (rectangle contained) by ABN was also shown (to be) equal to the (square) on BF . Thus, the (rectangle contained) by ABN plus the (rectangle contained) by BAN , which is the (square) on BA [Prop. 2.2], is equal to the (square) on BF plus the (square) on AK. And BA is the side of the pentagon, and BF (the side) of the hexagon [Prop. 4.15 corr.], and AK (the side) of the decagon. Thus, the square on the side of the pentagon (inscribed in a circle) is (equal to) the (sum of the squares) on the (sides) of the hexagon and of the decagon inscribed in the same circle. 10 − 2

√

5.

ια΄.

Proposition 11

'Ε¦ν ε„ς κύκλον ·ητ¾ν œχοντα τ¾ν διάµετρον πεντάγωIf an equilateral pentagon is inscribed in a circle which νον „σόπλευρον ™γγραφÍ, ¹ τοà πενταγώνου πλευρ¦ has a rational diameter then the side of the pentagon is ¥λογός ™στιν ¹ καλουµένη ™λάσσων. that irrational (straight-line) called minor.

A

B

M

A

E

Z K

B F K

J

N G

L H

E

M

H N L

D

C

D G

Ε„ς γ¦ρ κύκλον τÕν ΑΒΓ∆Ε ·ητ¾ν œχοντα τ¾ν δίαµετρον πεντάγωνον „σόπλευρον ™γγεγράφθω τÕ ΑΒΓ∆Ε· λέγω, Óτι ¹ τοà [ΑΒΓ∆Ε] πενταγώνου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ™λάσσων. Ε„λήφθω γ¦ρ τÕ κέντρον τοà κύκλου τÕ Ζ σηµε‹ον, κሠ™πεζεύχθωσαν αƒ ΑΖ, ΖΒ κሠδιήχθωσαν ™πˆ τ¦ Η, Θ σηµε‹α, κሠ™πεζεύχθω ¹ ΑΓ, κሠκείσθω τÁς ΑΖ τέταρτον µέρος ¹ ΖΚ. ·ητ¾ δ ¹ ΑΖ· ·ητ¾ ¥ρα κሠ¹ ΖΚ. œστι δ κሠ¹ ΒΖ ·ητή· Óλη ¥ρα ¹ ΒΚ ·ητή ™στιν. κሠ™πεˆ ‡ση ™στˆν ¹ ΑΓΗ περιφέρεια τÍ Α∆Η περιφερείv, ïν ¹ ΑΒΓ τÍ ΑΕ∆ ™στιν ‡ση, λοιπ¾ ¥ρα ¹ ΓΗ λοιπÍ τÍ Η∆ ™στιν ‡ση. κሠ™¦ν ™πιζεύξωµεν τ¾ν

For let the equilateral pentagon ABCDE have been inscribed in the circle ABCDE which has a rational diameter. I say that the side of pentagon [ABCDE] is that irrational (straight-line) called minor. For let the center of the circle, point F , have been found [Prop. 3.1]. And let AF and F B have been joined. And let them have been drawn across to points G and H (respectively). And let AC have been joined. And let F K made (equal) to the fourth part of AF . And AF (is) rational. F K (is) thus also rational. And BF is also rational. Thus, the whole of BK is rational. And since circumference ACG is equal to circumference ADG, of which

517

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

Α∆, συνάγονται Ñρθαˆ αƒ πρÕς τù Λ γωνίαι, κሠδιπλÁ ¹ Γ∆ τÁς ΓΛ. δι¦ τ¦ αÙτ¦ δ¾ καˆ αƒ πρÕς τù Μ Ñρθαί ε„σιν, κሠδιπλÁ ¹ ΑΓ τÁς ΓΜ. ™πεˆ οâν ‡ση ™στˆν ¹ ØπÕ ΑΛΓ γωνία τÍ ØπÕ ΑΜΖ, κοιν¾ δ τîν δύο τριγώνων τοà τε ΑΓΛ κሠτοà ΑΜΖ ¹ ØπÕ ΛΑΓ, λοιπ¾ ¥ρα ¹ ØπÕ ΑΓΛ λοιπÍ τÍ ØπÕ ΜΖΑ ™στιν ‡ση· „σογώνιον ¥ρα ™στˆ τÕ ΑΓΛ τρίγωνον τù ΑΜΖ τριγώνJ· ¢νάλογον ¥ρα ™στˆν æς ¹ ΛΓ πρÕς ΓΑ, οÛτως ¹ τÁς ΜΖ πρÕς ΖΑ· κሠτîν ¹γουµένων τ¦ διπλάσια· æς ¥ρα ¹ τÁς ΛΓ διπλÁ πρÕς τ¾ν ΓΑ, οÛτως ¹ τÁς ΜΖ διπλÁ πρÕς τ¾ν ΖΑ. æς δ ¹ τÁς ΜΖ διπλÁ πρÕς τ¾ν ΖΑ, οÛτως ¹ ΜΖ πρÕς τ¾ν ¹µίσειαν τÁς ΖΑ· κሠæς ¥ρα ¹ τÁς ΛΓ διπλÁ πρÕς τ¾ν ΓΑ, οÛτως ¹ ΜΖ πρÕς τ¾ν ¹µίσειαν τÁς ΖΑ· κሠτîν ˜ποµένων τ¦ ¹µίσεα· æς ¥ρα ¹ τÁς ΛΓ διπλÁ πρÕς τ¾ν ¹µίσειαν τÁς ΓΑ, οÛτως ¹ ΜΖ πρÕς τÕ τέτατρον τÁς ΖΑ. καί ™στι τÁς µν ΛΓ διπλÁ ¹ ∆Γ, τÁς δ ΓΑ ¹µίσεια ¹ ΓΜ, τÁς δ ΖΑ τέτατρον µέρος ¹ ΖΚ· œστιν ¥ρα æς ¹ ∆Γ πρÕς τ¾ν ΓΜ, οÛτως ¹ ΜΖ πρÕς τ¾ν ΖΚ. συνθέντι κሠæς συναµφότερος ¹ ∆ΓΜ πρÕς τ¾ν ΓΜ, οÛτως ¹ ΜΚ πρÕς ΚΖ· κሠæς ¥ρα τÕ ¢πÕ συναµφοτέρου τÁς ∆ΓΜ πρÕς τÕ ¢πÕ ΓΜ, οÛτως τÕ ¢πÕ ΜΚ πρÕς τÕ ¢πÕ ΚΖ. κሠ™πεˆ τÁς ØπÕ δύο πλευρ¦ς τοà πενταγώνου Øποτεινούσης, οŒον τÁς ΑΓ, ¥κρον κሠµέσον λόγου τεµνοµένης τÕ µε‹ζον τµÁµα ‡σον ™στˆ τÍ τοà πενταγώνου πλευρ´, τουτέστι τÍ ∆Γ, τÕ δ µε‹ζον τµÁµα προσλαβÕν τ¾ν ¹µίσειαν τÁς ÓλÁς πενταπλάσιον δύναται τοà ¢πÕ τÁς ¹µισείας τÁς Óλης, καί ™στιν Óλης τ¾ς ΑΓ ¹µίσεια ¹ ΓΜ, τÕ ¥ρα ¢πÕ τÁς ∆ΓΜ æς µι©ς πενταπλάσιόν ™στι τοà ¢πÕ τÁς ΓΜ. æς δ τÕ ¢πÕ τÁς ∆ΓΜ æς µι©ς πρÕς τÕ ¢πÕ τÁς ΓΜ, οÛτως ™δείχθη τÕ ¢πÕ τÁς ΜΚ πρÕς τÕ ¢πÕ τÁς ΚΖ· πενταπλάσιον ¥ρα τÕ ¢πÕ τÁς ΜΚ τοà ¢πÕ τÁς ΚΖ. ·ητÕν δ τÕ ¢πÕ τÁς ΚΖ· ·ητ¾ γ¦ρ ¹ διάµετρος· ·ητÕν ¥ρα κሠτÕ ¢πÕ τÁς ΜΚ· ·ητ¾ ¥ρα ™στˆν ¹ ΜΚ [δυνάµει µόνον]. κሠ™πεˆ τετραπλασία ™στˆν ¹ ΒΖ τÁς ΖΚ, πενταπλασία ¥ρα ™στˆν ¹ ΒΚ τÁς ΚΖ· ε„κοσιπενταπλάσιον ¥ρα τÕ ¢πÕ τÁς ΒΚ τοà ¢πÕ τÁς ΚΖ. πενταπλάσιον δ τÕ ¢πÕ τÁς ΜΚ τοà ¢πÕ τÁς ΚΖ· πενταπλάσιον ¥ρα τÕ ¢πÕ τÁς ΒΚ τοà ¢πÕ τÁς ΚΜ· τÕ ¥ρα ¢πÕ τÁς ΒΚ πρÕς τÕ ¢πÕ ΚΜ λόγον οÙκ œχει, Öν τετράγωνος ¢ριθµÕς πρÕς τετράγωνον ¢ριθµόν· ¢σύµµετρος ¥ρα ™στˆν ¹ ΒΚ τÍ ΚΜ µήκει. καί ™στι ·ητ¾ ˜κατέρα αÙτîν. αƒ ΒΚ, ΚΜ ¥ρα ·ηταί ε„σι δυνάµει µόνον σύµµετροι. ™¦ν δ ¢πÕ ·ητÁς ·ητ¾ ¢φαιρεθÍ δυνάµει µόνον σύµµετρος οâσα τÍ ÓλV, ¹ λοιπ¾ ¥λογός ™στιν ¢ποτοµή· ¢ποτοµ¾ ¥ρα ™στˆν ¹ ΜΒ, προσαρµόζουσα δ αÙτÍ ¹ ΜΚ. λέγω δή, Óτι κሠτετάρτη. ú δ¾ µε‹ζόν ™στι τÕ ¢πÕ τÁς ΒΚ τοà ¢πÕ τÁς ΚΜ, ™κείνJ ‡σον œστω τÕ ¢πÕ τÁς Ν· ¹ ΒΚ ¥ρα τÁς ΚΜ µε‹ζον δύναται τÍ Ν. κሠ™πεˆ σύµµετρός ™στιν ¹ ΚΖ τÍ ΖΒ, κሠσυνθέντι σύµµετρός ™στι ¹ ΚΒ τÍ ΖΒ. ¢λλ¦ ¹ ΒΖ τÍ ΒΘ σύµµετρός ™στιν· κሠ¹ ΒΚ

ABC is equal to AED, the remainder CG is thus equal to the remainder GD. And if we join AD then the angles at L are inferred (to be) right-angles, and CD (is inferred to be) double CL [Prop. 1.4]. So, for the same (reasons), the (angles) at M are also right-angles, and AC (is) double CM . Therefore, since angle ALC (is) equal to AM F , and (angle) LAC (is) common to the two triangles ACL and AM F , the remaining (angle) ACL is thus equal to the remaining (angle) M F A [Prop. 1.32]. Thus, triangle ACL is equiangular to triangle AM F . Thus, proportionally, as LC (is) to CA, so M F (is) to F A [Prop. 6.4]. And (we can take) the doubles of the leading (magnitudes). Thus, as double LC (is) to CA, so double M F (is) to F A. And as double M F (is) to F A, so M F (is) to half of F A. And, thus, as double LC (is) to CA, so M F (is) to half of F A. And (we can take) the halves of the following (magnitudes). Thus, as double LC (is) to half of CA, so M F (is) to the fourth of F A. And DC is double LC, and CM half of CA, and F K the fourth part of F A. Thus, as DC is to CM , so M F (is) to F K. Via composition, as the sum of DCM (i.e., DC and CM ) (is) to CM , so M K (is) to KF [Prop. 5.18]. And, thus, as the (square) on the sum of DCM (is) to the (square) on CM , so the (square) on M K (is) to the (square) on KF . And since the greater piece of a (straight-line) subtending two sides of a pentagon, such as AC, (which is) cut in extreme and mean ratio is equal to the side of the pentagon [Prop. 13.8]— that is to say, to DC—-and the square on the greater piece added to half of the whole is five times the (square) on half of the whole [Prop. 13.1], and CM (is) half of the whole, AC, thus the (square) on DCM , (taken) as one, is five times the (square) on CM . And the (square) on DCM , (taken) as one, (is) to the (square) on CM , so the (square) on M K was shown (to be) to the (square) on KF . Thus, the (square) on M K (is) five times the (square) on KF . And the square on KF (is) rational. For the diameter (is) rational. Thus, the (square) on M K (is) also rational. Thus, M K is rational [in square only]. And since BF is four times F K, BK is thus five times KF . Thus, the (square) on BK (is) twenty-five times the (square) on KF . And the (square) on M K (is) five times the square on KF . Thus, the (square) on BK (is) five times the (square) on KM . Thus, the (square) on BK does not have to the (square) on KM the ratio which a square number (has) to a square number. Thus, BK is incommensurable in length with KM [Prop. 10.9]. And each of them is a rational (straightline). Thus, BK and KM are rational (straight-lines which are) commensurable in square only. And if from a rational (straight-line) a rational (straight-line) is subtracted, which is commensurable in square only with the

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ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

¥ρα τÍ ΒΘ σύµµετρός ™στιν. κሠ™πεˆ πενταπλάσιόν ™στι τÕ ¢πÕ τÁς ΒΚ τοà ¢πÕ τÁς ΚΜ, τÕ ¥ρα ¢πÕ τÁς ΒΚ πρÕς τÕ ¢πÕ τ¾ς ΚΜ λόγον œχει, Öν ε πρÕς ›ν. ¢ναστρέψαντι ¥ρα τÕ ¢πÕ τÁς ΒΚ πρÕς τÕ ¢πÕ τÁς Ν λόγον œχει, Öν ε πρÕς δ, οÙχ Öν τετράγωνος πρÕς τετράγωνον· ¢σύµµετρος ¥ρα ™στˆν ¹ ΒΚ τÍ Ν· ¹ ΒΚ ¥ρα τÁς ΚΜ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ. ™πεˆ οâν Óλη ¹ ΒΚ τÁς προσαρµοζούσης τÁς ΚΜ µε‹ζον δύναται τù ¢πÕ ¢συµµέτρου ˜αυτÍ, κሠÓλη ¹ ΒΚ σύµµετρός ™στι τÍ ™κκειµένV ·ητÍ τÍ ΒΘ, ¢ποτοµ¾ ¥ρα τετάρτη ™στˆν ¹ ΜΒ. τÕ δ ØπÕ ·ητÁς κሠ¢ποτοµÁς τετάρτης περιεχόµενον Ñρθογώνιον ¥λογόν ™στιν, κሠ¹ δυναµένη αÙτÕ ¥λογός ™στιν, καλε‹ται δ ™λάττων. δύναται δ τÕ ØπÕ τîν ΘΒΜ ¹ ΑΒ δι¦ τÕ ™πιζευγνυµένης τÁς ΑΘ „σογώνιον γίνεσθαι τÕ ΑΒΘ τρίγωνον τù ΑΒΜ τριγώνJ κሠεναι æς τ¾ν ΘΒ πρÕς τ¾ν ΒΑ, οÛτως τ¾ν ΑΒ πρÕς τ¾ν ΒΜ. `Η ¥ρα ΑΒ τοà πενταγώνου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ™λάττων· Óπερ œδει δε‹ξαι.

whole, then the remainder is that irrational (straight-line called) an apotome [Prop. 10.73]. Thus, M B is an apotome, and M K its attachment. So, I say that (it is) also a fourth (apotome). So, let the (square) on N be (made) equal to that (magnitude) by which the (square) on BK is greater than the (square) on KM . Thus, the square on BK is greater than the (square) on KM by the (square) on N . And since KF is commensurable (in length) with F B then, via composition, KB is also commensurable (in length) with F B [Prop. 10.15]. But, BF is commensurable (in length) with BH. Thus, BK is also commensurable (in length) with BH [Prop. 10.12]. And since the (square) on BK is five times the (square) on KM , the (square) on BK thus has to the (square) on KM the ratio which 5 (has) to one. Thus, via conversion, the (square) on BK has to the (square) on N the ratio which 5 (has) to 4 [Prop. 5.19 corr.], which is not (that) of a square (number) to a square (number). BK is thus incommensurable (in length) with N [Prop. 10.9]. Thus, the square on BK is greater than the (square) on KM by the (square) on (some straight-line which is) incommensurable (in length) with (BA). Therefore, since the square on the whole, BK, is greater than the (square) on the attachment, KM , by the (square) on (some straightline which is) incommensurable (in length) with (BA), and the whole, BK, is commensurable (in length) with the (previously) laid down rational (straight-line) BH, M B is thus a fourth apotome [Def. 10.14]. And the rectangle contained by a rational (straight-line) and a fourth apotome is irrational, and its square-root is that irrational (straight-line) called minor [Prop. 10.94]. And the square on AB is the rectangle contained by HBM , on account of joining AH, (so that) triangle ABH becomes equiangular with triangle ABM [Prop. 6.8], and (proportionally) as HB is to BA, so AB (is) to BM . Thus, the side AB of the pentagon is that irrational (straight-line) called minor.† (Which is) the very thing it was required to show.

p √ If the circle has unitqradius, then the side of the q pentagon is (1/2) 10 − 2 5. However, this length can be written in the “minor” form (see p √ √ √ √ Prop. 10.94) (ρ/ 2) 1 + k/ 1 + k 2 − (ρ/ 2) 1 − k/ 1 + k 2 , with ρ = 5/2 and k = 2. †

ιβ΄.

Proposition 12

'Ε¦ν ε„ς κύκλον τρίγωνον „σόπλευρον ™γγραφÍ, ¹ If an equilateral triangle is inscribed in a circle then τοà τριγώνου πλευρ¦ δυνάµει τριπλασίων ™στˆ τÁς ™κ the square on the side of triangle ABC is three times the τοà κέντρου τοà κύκλου. (square) on the radius of the circle.

519

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

A

A

D B

D

G E

C

B

E

'Εστω κύκλος Ð ΑΒΓ, κሠε„ς αÙτÕν τρίγωνον „σόπλευρον ™γγεγράφθω τÕ ΑΒΓ· λέγω, Óτι τοà ΑΒΓ τριγώνου µία πλευρ¦ δυνάµει τριπλασίων ™στˆ τÁς ™κ τοà κέντρου τοà ΑΒΓ κύκλου. Ε„λήφθω γ¦ρ τÕ κέντρον τοà ΑΒΓ κύκλου τÕ ∆, κሠ™πιζευχθε‹σα ¹ Α∆ διήχθω ™πˆ τÕ Ε, κሠ™πεζεύχθω ¹ ΒΕ. Κሠ™πεˆ „σόπλευρόν ™στι τÕ ΑΒΓ τρίγωνον, ¹ ΒΕΓ ¥ρα περιφέρεια τρίτον µέρος ™στˆ τÁς τοà ΑΒΓ κύκλου περιφερείας. ¹ ¥ρα ΒΕ περιφέρεια ›κτον ™στˆ µέρος τÁς τοà κύκλου περιφερείας· ˜ξαγώνου ¥ρα ™στˆν ¹ ΒΕ εÙθε‹α· ‡ση ¥ρα ™στˆ τÍ ™κ τοà κέντρου τÍ ∆Ε. κሠ™πεˆ διπλÁ ™στιν ¹ ΑΕ τÁς ∆Ε, τετραπλάσιον ™στι τÕ ¢πÕ τÁς ΑΕ τοà ¢πÕ τÁς Ε∆, τουτέστι τοà ¢πÕ τÁς ΒΕ. ‡σον δ τÕ ¢πÕ τÁς ΑΕ το‹ς ¢πÕ τîν ΑΒ, ΒΕ· τ¦ ¥ρα ¢πÕ τîν ΑΒ, ΒΕ τετραπλάσιά ™στι τοà ¢πÕ τÁς ΒΕ. διελόντι ¥ρα τÕ ¢πÕ τÁς ΑΒ τριπλάσιόν ™στι τοà ¢πÕ ΒΕ. ‡ση δ ¹ ΒΕ τÍ ∆Ε· τÕ ¥ρα ¢πÕ τÁς ΑΒ τριπλάσιόν ™στι τοà ¢πÕ τÁς ∆Ε. `Η ¥ρα τοà τριγώνου πλευρ¦ δυνάµει τριπλασία ™στˆ τÁς ™κ τοà κέντρου [τοà κύκλου]· Óπερ œδει δε‹ξαι.

Let there be a circle ABC, and let the equilateral triangle ABC have been inscribed in it [Prop. 4.2]. I say that the square on one side of triangle ABC is three times the (square) on the radius of circle ABC. For let the center, D, of circle ABC have been found [Prop. 3.1]. And AD (being) joined, let it have been drawn across to E. And let BE have been joined. And since triangle ABC is equilateral, circumference BEC is thus the third part of the circumference of circle ABC. Thus, circumference BE is the sixth part of the circumference of the circle. Thus, straight-line BE is (the side) of a hexagon. Thus, it is equal to the radius DE [Prop. 4.15 corr.]. And since AE is double DE, the (square) on AE is four times the (square) on ED—that is to say, of the (square) on BE. And the (square) on AE (is) equal to the (sum of the squares) on AB and BE [Props. 3.31, 1.47]. Thus, the (sum of the squares) on AB and BE is four times the (square) on BE. Thus, via separation, the (square) on AB is three times the (square) on BE. And BE (is) equal to DE. Thus, the (square) on AB is three times the (square) on DE. Thus, the square on the side of the triangle is three times the (square) on the radius [of the circle]. (Which is) the very thing it was required to show.

ιγ΄.

Proposition 13

Πυραµίδα συστήσασθαι κሠσφαίρv περιλαβε‹ν τÍ To construct a (regular) pyramid (i.e., a tetrahedron), δοθείσV κሠδε‹ξαι, Óτι ¹ τÁς σφαίρας διάµετρος δυνάµει and to enclose (it) in a given sphere, and to show that ¹µιολία ™στˆ τÁς πλευρ©ς τÁς πυραµίδος. the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid.

520

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

D

A K

E

G

B

C

A

B

E K

J Z

D

L

H L

H

F

'Εκκείσθω ¹ τÁς δοθείσης σφαίρας δίαµετρος ¹ ΑΒ, κሠτετµήσθω κατ¦ τÕ Γ σηµε‹ον, éστε διπλασίαν εναι τ¾ν ΑΓ τÁς ΓΒ· κሠγεγράφθω ™πˆ τÁς ΑΒ ¹µικύκλιον τÕ Α∆Β, κሠ½χθω ¢πÕ τοà Γ σηµείου τÍ ΑΒ πρÕς Ñρθ¦ς ¹ Γ∆, κሠ™πεζεύχθω ¹ ∆Α· κሠ™κκείσθω κύκλος Ð ΕΖΗ ‡σην œχων τ¾ν ™κ τοà κέντρου τÍ ∆Γ, κሠ™γγεγράφθω ε„ς τÕν ΕΖΗ κύκλον τρίγωνον „σόπλευρον τÕ ΕΖΗ· κሠε„λήφθω τÕ κέντρον τοà κύκλου τÕ Θ σηµε‹ον, κሠ™πεζεύχθωσαν αƒ ΕΘ, ΘΖ, ΘΗ· κሠ¢νεστάτω ¢πÕ τοà Θ σηµείου τù τοà ΕΖΗ κύκλου ™πιπέδJ πρÕς Ñρθ¦ς ¹ ΘΚ, κሠ¢φVρήσθω ¢πÕ τÁς ΘΚ τÍ ΑΓ εÙθείv ‡ση ¹ ΘΚ, κሠ™πεζεύχθωσαν αƒ ΚΕ, ΚΖ, ΚΗ. κሠ™πεˆ ¹ ΚΘ Ñρθή ™στι πρÕς τÕ τοà ΕΖΗ κύκλου ™πίπεδον, κሠπρÕς πάσας ¥ρα τ¦ς ¡πτοµένας αÙτÁς εÙθείας κሠοÜσας ™ν τù τοà ΕΖΗ κύκλου ™πιπέδJ Ñρθ¦ς ποιήσει γωνίας. ¤πτεται δ αÙτÁς ˜κάστη τîν ΘΕ, ΘΖ, ΘΗ· ¹ ΘΚ ¥ρα πρÕς ˜κάστη τîν ΘΕ, ΘΖ, ΘΗ Ñρθή ™στιν. κሠ™πεˆ ‡ση ™στˆν ¹ µν ΑΓ τÍ ΘΚ, ¹ δ Γ∆ τÍ ΘΕ, κሠÑρθ¦ς γωνίας περιέχουσιν, βάσις ¥ρα ¹ ∆Α βάσει τÍ ΚΕ ™στιν ‡ση. δι¦ τ¦ αÙτ¦ δ¾ κሠ˜κατέρα τîν ΚΖ, ΚΗ τÍ ∆Α ™στιν ‡ση· αƒ τρε‹ς ¥ρα αƒ ΚΕ, ΚΖ, ΚΗ ‡σαι ¢λλήλαις ε„σίν. κሠ™πεˆ διπλÁ ™στιν ¹ ΑΓ τÁς ΓΒ, τριπλÁ ¥ρα ¹ ΑΒ τÁς ΒΓ. æς δ ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ¢πÕ τÁς Α∆ πρÕς τÕ ¢πÕ τÁς ∆Γ, æς ˜ξÁς δειχθήσεται. τριπλάσιον ¥ρα τÕ ¢πÕ τÁς Α∆ τοà ¢πÕ τÁς ∆Γ. œστι δ κሠτÕ ¢πÕ τÁς ΖΕ τοà ¢πÕ τÁς ΕΘ τριπλάσιον, καί ™στιν ‡ση ¹ ∆Γ τÍ ΕΘ· ‡ση ¥ρα κሠ¹ ∆Α τÍ ΕΖ. ¢λλ¦ ¹ ∆Α ˜κάστV τîν ΚΕ, ΚΖ, ΚΗ ™δείχθη ‡ση· κሠ˜κάστη ¥ρα τîν ΕΖ, ΖΗ, ΗΕ ˜κάστV τîν ΚΕ, ΚΖ, ΚΗ ™στιν ‡ση· „σόπλευρα ¥ρα ™στˆ τ¦ τέσσαρα τρίγωνα

G

Let the diameter AB of the given sphere be laid out, and let it have been cut at point C such that AC is double CB [Prop. 6.10]. And let the semi-circle ADB have been drawn on AB. And let CD have been drawn from point C at right-angles to AB. And let DA have been joined. And let the circle EF G be laid down having a radius equal to DC, and let the equilateral triangle EF G have been inscribed in circle EF G [Prop. 4.2]. And let the center of the circle, point H, have been found [Prop. 3.1]. And let EH, HF , and HG have been joined. And let HK have been set up, at point H, at right-angles to the plane of circle EF G [Prop. 11.12]. And let HK, equal to the straight-line AC, have been cut off from HK. And let KE, KF , and KG have been joined. And since KH is at right-angles to the plane of circle EF G, it will thus also make right-angles with all of the straight-lines joining it (which are) also in the plane of circle EF G [Def. 11.3]. And HE, HF , and HG each join it. Thus, HK is at right-angles to each of HE, HF , and HG. And since AC is equal to HK, and CD to HE, and they contain right-angles, the base DA is thus equal to the base KE [Prop. 1.4]. So, for the same (reasons), KF and KG is each equal to DA. Thus, the three (straight-lines) KE, KF , and KG are equal to one another. And since AC is double CB, AB (is) thus triple BC. And as AB (is) to BC, so the (square) on AD (is) to the (square) on DC, as will be shown later [see lemma]. Thus, the (square) on AD (is) three times the (square) on DC. And the (square) on F E is also three times the (square) on EH [Prop. 13.12], and DC is equal to EH. Thus, DA (is)

521

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

τ¦ ΕΖΗ, ΚΕΖ, ΚΖΗ, ΚΕΗ. πυρᵈς ¥ρα συνέσταται ™κ τεσσάρων τριγώνων „σοπλέυρων, Âς βάσις µέν ™στι τÕ ΕΖΗ τρίγωνον, κορυφ¾ δ τÕ Κ σηµε‹ον. ∆ε‹ δ¾ αÙτ¾ν κሠσφαίρv περιλαβε‹ν τÍ δοθείσV κሠδε‹ξαι, Óτι ¹ τÁς σφαίρας διάµετρος ¹µιολία ™στˆ δυνάµει τÁς πλευρ©ς τÁς πυραµίδος. 'Εκβεβλήσθω γ¦ρ ™π' εÙθείας τÍ ΚΘ εÙθε‹α ¹ ΘΛ, κሠκείσθω τÍ ΓΒ ‡ση ¹ ΘΛ. κሠ™πεί ™στιν æς ¹ ΑΓ πρÕς τ¾ν Γ∆, οÛτως ¹ Γ∆ πρÕς τ¾ν ΓΒ, ‡ση δ ¹ µν ΑΓ τÍ ΚΘ, ¹ δ Γ∆ τÍ ΘΕ, ¹ δ ΓΒ τÍ ΘΛ, œστιν ¥ρα æς ¹ ΚΘ πρÕς τ¾ν ΘΕ, οÛτως ¹ ΕΘ πρÕς τ¾ν ΘΛ· τÕ ¥ρα ØπÕ τîν ΚΘ, ΘΛ ‡σον ™στˆ τù ¢πÕ τÁς ΕΘ. καί ™στιν Ñρθ¾ ˜κατέρα τîν ØπÕ ΚΘΕ, ΕΘΛ γωνιîν· τÕ ¥ρα ™πˆ τÁς ΚΛ γραφόµενον ¹µικύκλιον ¼ξει κሠδι¦ τοà Ε [™πειδήπερ ™¦ν ™πιζεύξωµεν τ¾ν ΕΛ, Ñρθ¾ γίνεται ¹ ØπÕ ΛΕΚ γωνία δι¦ τÕ „σογώνιον γίνεσθαι τÕ ΕΛΚ τρίγωνον ˜κατέρJ τîν ΕΛΘ, ΕΘΚ τριγώνων]. ™¦ν δ¾ µενούσης τÁς ΚΛ περιενεχθν τÕ ¹µικύκλιον ε„ς τÕ αÙτÕ πάλιν ¢ποκατασταθÍ, Óθεν ½ρξατο φέρεσθαι, ¼ξει κሠδι¦ τîν Ζ, Η σηµείων ™πιζευγνυµένων τîν ΖΛ, ΛΗ κሠÑρθîν еοίως γινοµένων τîν πρÕς το‹ς Ζ, Η γωνιîν· κሠœσται ¹ πυρᵈς σφαίρv περιειληµµένη τÍ δοθείσθÍ. ¹ γ¦ρ ΚΛ τÁς σφαίρας διάµετρος ‡ση ™στˆ τÍ τÁς δοθείσης σφαίρας διαµετρJ τÍ ΑΒ, ™πειδήπερ τÍ µν ΑΓ ‡ση κε‹ται ¹ ΚΘ, τÍ δ ΓΒ ¹ ΘΛ. Λέγω δή, Óτι ¹ τÁς σφαίρας διάµετρος ¹µιολία ™στˆ δυνάµει τÁς πλευρ©ς τÁς πυραµίδος. 'Επεˆ γ¦ρ διπλÁ ™στιν ¹ ΑΓ τÁς ΓΒ, τριπλÁ ¥ρα ™στˆν ¹ ΑΒ τÁς ΒΓ· ¢ναστρέψαντι ¹µιολία ¥ρα ™στˆν ¹ ΒΑ τÁς ΑΓ. æς δ ¹ ΒΑ πρÕς τ¾ν ΑΓ, οÛτως τÕ ¢πÕ τÁς ΒΑ πρÕς τÕ ¢πÕ τÁς Α∆ [™πειδήπερ ™πιζευγνµένης τÁς ∆Β ™στιν æς ¹ ΒΑ πρÕς τ¾ν Α∆, οÛτως ¹ ∆Α πρÕς τ¾ν ΑΓ δι¦ τ¾ν еοιότητα τîν ∆ΑΒ, ∆ΑΓ τριγώνων, κሠεναι æς τ¾ν πρώτην πρÕς τ¾ν τρίτην, οÛτως τÕ ¢πÕ τÁς τρώτης πρÕς τÕ ¢πÕ τÁς δευτέρας]. ¹µιόλιον ¥ρα κሠτÕ ¢πÕ τÁς ΒΑ τοà ¢πÕ τÁς Α∆. καί ™στιν ¹ µν ΒΑ ¹ τÁς δοθείσης σφαίρας διάµετρος, ¹ δ Α∆ ‡ση τÍ πλευρ´ τÁς πυραµίδος. `Η ¥ρα τÁς σφαίρας διάµετρος ¹µιολία ™στˆ τÁς πλευρ©ς τÁς πυραµίδος· Óπερ œδει δε‹ξαι.

†

also equal to EF . But, DA was shown (to be) equal to each of KE, KF , and KG. Thus, EF , F G, and GE are equal to KE, KF , and KG, respectively. Thus, the four triangles EF G, KEF , KF G, and KEG are equilateral. Thus, a pyramid, whose base is triangle EF G, and apex the point K, has been constructed from four equilateral triangles. So, it is also necessary to enclose it in the given sphere, and to show that the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid. For let the straight-line HL have been produced in a straight-line with KH, and let HL be made equal to CB. And since as AC (is) to CD, so CD (is) to CB [Prop. 6.8 corr.], and AC (is) equal to KH, and CD to HE, and CB to HL, thus as KH is to HE, so EH (is) to HL. Thus, the (rectangle contained) by KH and HL is equal to the (square) on EH [Prop. 6.17]. And each of the angles KHE and EHL is a right-angle. Thus, the semi-circle drawn on KL will also pass through E [inasmuch as if we join EL then the angle LEK becomes a right-angle, on account of triangle ELK becoming equiangular to each of the triangles ELH and EHK [Props. 6.8, 3.31] ]. So, if KL remains (fixed), and the semi-circle is carried around, and again established at the same (position) from which it began to be moved, it will also pass through points F and G, (because) if F L and LG are joined, the angles at F and G will similarly become right-angles. And the pyramid will have been enclosed by the given sphere. For the diameter, KL, of the sphere is equal to the diameter, AB, of the given sphere— inasmuch as KH was made equal to AC, and HL to CB. So, I say that the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid. For since AC is double CB, AB is thus triple BC. Thus, via conversion, BA is one and a half times AC. And as BA (is) to AC, so the (square) on BA (is) to the (square) on AD [inasmuch as if DB is joined then as BA is to AD, so DA (is) to AC, on account of the similarity of triangles DAB and DAC. And as the first is to the third (of four proportional magnitudes), so the (square) on the first (is) to the (square) on the second.] Thus, the (square) on BA (is) also one and a half times the (square) on AD. And BA is the diameter of the given sphere, and AD (is) equal to the side of the pyramid. Thus, the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid.† (Which is) the very thing it was required to show.

If the radius of the sphere is unity, then the side of the pyramid (i.e., tetrahedron) is

522

p 8/3.

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

D

G

A

E

D

B

C A

Z

E

B

F

ΛÁµµα.

Lemma

∆εικτέον, Óτι ™στˆν æς ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ¢πÕ τÁς Α∆ πρÕς τÕ ¢πÕ τÁς ∆Γ. 'Εκκείσθω γ¦ρ ¹ τοà ¹µικυκλίου καταγραφή, κሠ™πεζεύχθω ¹ ∆Β, κሠ¢ναγεγράφθω ¢πÕ τÁς ΑΓ τετράγωνον τÕ ΕΓ, κሠσυµπεπληρώσθω τÕ ΖΒ παραλληλόγραµµον. ™πεˆ οâν δι¦ τÕ „σογώνιον εναι τÕ ∆ΑΒ τρίγωνον τù ∆ΑΓ τριγώνJ ™στˆν æς ¹ ΒΑ πρÕς τ¾ν Α∆, οÛτως ¹ ∆Α πρÕς τ¾ν ΑΓ, τÕ ¥ρα ØπÕ τîν ΒΑ, ΑΓ ‡σον ™στˆ τù ¢πÕ τÁς Α∆. κሠ™πεί ™στιν æς ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ΕΒ πρÕς τÕ ΒΖ, καί ™στι τÕ µν ΕΒ τÕ ØπÕ τîν ΒΑ, ΑΓ· ‡ση γ¦ρ ¹ ΕΑ τÍ ΑΓ· τÕ δ ΒΖ τÕ ØπÕ τîν ΑΓ, ΓΒ, æς ¥ρα ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ØπÕ τîν ΒΑ, ΑΓ πρÕ τÕ ØπÕ τîν ΑΓ, ΓΒ. καί ™στι τÕ µν ØπÕ τîν ΒΑ, ΑΓ ‡σον τù ¢πÕ τÁς Α∆, τÕ δ ØπÕ τîν ΑΓΒ ‡σον τù ¢πÕ τÁς ∆Γ· ¹ γ¦ρ ∆Γ κάθετος τîν τÁς βάσεως τµηµάτων τîν ΑΓ, ΓΒ µέση ¢νάλογόν ™στι δι¦ τÕ Ñρθ¾ν εναι τ¾ν ØπÕ Α∆Β. æς ¥ρα ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ¢πÕ τÁς Α∆ πρÕς τÕ ¢πÕ τÁς ∆Γ· Óπερ œδει δε‹ξαι.

It must be shown that as AB is to BC, so the (square) on AD (is) to the (square) on DC. For, let the figure of the semi-circle have been set out, and let DB have been joined. And let the square EC have been described on AC. And let the parallelogram F B have been completed. Therefore, since, on account of triangle DAB being equiangular to triangle DAC [Props. 6.8, 6.4], (proportionally) as BA is to AD, so DA (is) to AC, thus the (rectangle contained) by BA and AC is equal to the (square) on AD [Prop. 6.17]. And since as AB is to BC, so EB (is) to BF [Prop. 6.1]. And EB is the (rectangle contained) by BA and AC—for EA (is) equal to AC. And BF the (rectangle contained) by AC and CB. Thus, as AB (is) to BC, so the (rectangle contained) by BA and AC (is) to the (rectangle contained) by AC and CB. And the (rectangle contained) by BA and AC is equal to the (square) on AD, and the (rectangle contained) by ACB (is) equal to the (square) on DC. For the perpendicular DC is the mean proportional to the pieces of the base, AC and CB, on account of ADB being a right-angle [Prop. 6.8 corr.]. Thus, as AB (is) to BC, so the (square) on AD (is) to the (square) on DC. (Which is) the very thing it was required to show.

ιδ΄.

Proposition 14

'Οκτάεδρον συστήσασθαι κሠσφαίρv περιλαβε‹ν, Î κሠτ¦ πρότερα, κሠδε‹ξαι, Óτι ¹ τÁς σφαίρας διάµετρος δυνάµει διπλασία ™στˆ τÁς πλευρ©ς τοà Ñκταέδρου. 'Εκκείσθω ¹ τÁς δοθείσης σφαίρας διάµετρος ¹ ΑΒ, κሠτετµήσθω δίχα κατ¦ τÕ Γ, κሠγεγράφθω ™πˆ τÁς ΑΒ ¹µικύκλιον τÕ Α∆Β, κሠ½χθω ¢πÕ τοà Γ τÍ ΑΒ πρÕς Ñρθ¦ς ¹ Γ∆, κሠ™πεζεύχθω ¹ ∆Β, κሠ™κκείσθω τετράγωνον τÕ ΕΖΗΘ ‡σην œχον ˜κάστην τîν πλευρîν

To construct an octahedron, and to enclose (it) in a (given) sphere, like in the preceding (proposition), and to show that the square on the diameter of the sphere is double the (square) on the side of the octahedron. Let the diameter AB of the given sphere be laid out, and let it have been cut in half at C. And let the semicircle ADB have been drawn on AB. And let CD be drawn from C at right-angles to AB. And let DB have

523

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

τÍ ∆Β, κሠ™πεζεύχθωσαν αƒ ΘΖ, ΕΗ, κሠ¢νεστάτω ¢πÕ τοà Κ σηµείου τù τοà ΕΖΗΘ τετραγώνου ™πιπέδJ πρÕς Ñρθ¦ς εÙθε‹α ¹ ΚΛ κሠδιήχθω ™πˆ τ¦ ›τερα µέρη τοà ™πιπέδου æς ¹ ΚΜ, κሠ¢φVρήσθω ¢φ' ˜κατέρας τîν ΚΛ, ΚΜ µι´ τîν ΕΚ, ΖΚ, ΗΚ, ΘΚ ‡ση ˜κατέρα τîν ΚΛ, ΚΜ, κሠ™πεζεύχθωσαν αƒ ΛΕ, ΛΖ, ΛΗ, ΛΘ, ΜΕ, ΜΖ, ΜΗ, ΜΘ.

D

A E

L

been joined. And let the square EF GH, having each of its sides equal to DB, be laid out. And let HF and EG have been joined. And let the straight-line KL have been set up, at point K, at right-angles to the plane of square EF GH [Prop. 11.12]. And let it have been drawn across on the other side of the plane, like KM . And let KL and KM , equal to one of EK, F K, GK, and HK, have been cut off from KL and KM , respectively. And let LE, LF , LG, LH, M E, M F , M G, and M H have been joined.

D

C

B

A L

H

H

E

K

F

B

C

K

M

G

G

F M

Κሠ™πεˆ ‡ση ™στˆν ¹ ΚΕ τÍ ΚΘ, καί ™στιν Ñρθ¾ ¹ ØπÕ ΕΚΘ γωνία, τÕ ¥ρα ¢πÕ τÁς ΘΕ διπλάσιόν ™στι τοà ¢πÕ τÁς ΕΚ. πάλιν, ™πεˆ ‡ση ™στˆν ¹ ΛΚ τÍ ΚΕ, καί ™στιν Ñρθ¾ ¹ ØπÕ ΛΚΕ γωνία, τÕ ¥ρα ¢πÕ τÁς ΕΛ διπλάσιόν ™στι τοà ¢πÕ ΕΚ. ™δείχθη δ κሠτÕ ¢πÕ τÁς ΘΕ διπλάσιον τοà ¢πÕ τÁς ΕΚ· τÕ ¥ρα ¢πÕ τÁς ΛΕ ‡σον ™στˆ τù ¢πÕ τÁς ΕΘ· ‡ση ¥ρα ™στˆν ¹ ΛΕ τÍ ΕΘ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΛΘ τÍ ΘΕ ™στιν ‡ση· „σόπλευρον ¥ρα ™στˆ τÕ ΛΕΘ τρίγωνον. еοίως δ¾ δείξοµεν, Óτι κሠ›καστον τîν λοιπîν τριγώνων, ïν βάσεις µέν ε„σιν αƒ τοà ΕΖΗΘ τετραγώνου πλευραί, κορυφαˆ δ τ¦ Λ, Μ σηµε‹α, „σόπλευρόν ™στιν· Ñκτάεδρον ¥ρα συνέσταται ØπÕ Ñκτë τριγώνων „σοπλεύρων περιεχόµενον. ∆ε‹ δ¾ αÙτÕ κሠσφαίρv περιλαβε‹ν τÍ δοθείσV κሠδε‹ξαι, Óτι ¹ τÁς σφαίρας διάµετρος δυνάµει διπλασίων ™στˆ τÁς τοà Ñκταέδρου πλευρ©ς. 'Επεˆ γ¦ρ αƒ τρε‹ς αƒ ΛΚ, ΚΜ, ΚΕ ‡σαι ¢λλήλαις ε„σίν, τÕ ¥ρα ™πˆ τÁς ΛΜ γραφόµενον ¹µικύκλιον ¼ξει κሠδι¦ τοà Ε. κሠδι¦ τ¦ αÙτά, ™¦ν µενούσης τÁς ΛΜ περιενεχθν τÕ ¹µικύκλιον ε„ς τÕ αÙτÕ ¢ποκατασταθÍ, Óθεν ½ρξατο φέρεσθαι, ¼ξει κሠδι¦ τîν Ζ, Η, Θ σηµείων, κሠœσται σφαίρv περιειληµµένον τÕ

And since KE is equal to KH, and angle EKH is a right-angle, the (square) on the HE is thus double the (square) on EK [Prop. 1.47]. Again, since LK is equal to KE, and angle LKE is a right-angle, the (square) on EL is thus double the (square) on EK [Prop. 1.47]. And the (square) on HE was also shown (to be) double the (square) on EK. Thus, the (square) on LE is equal to the (square) on EH. Thus, LE is equal to EH. So, for the same (reasons), LH is also equal to HE. Triangle LEH is thus equilateral. So, similarly, we can show that each of the remaining triangles, whose bases are the sides of the square EF HG, and apexes the points L and M , are equilateral. Thus, an octahedron contained by eight equilateral triangles has been constructed. So, it is also necessary to enclose it by the given sphere, and to show that the square on the diameter of the sphere is double the (square) on the side of the octahedron. For since the three (straight-lines) LK, KM , and KE are equal to one another, the semi-circle drawn on LM will thus also pass through E. And, for the same (reasons), if LM remains (fixed), and the semi-circle is car-

524

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

Ñκτάεδρον. λέγω δή, Óτι κሠτÍ δοθείσV. ™πεˆ γ¦ρ ‡ση ™στˆν ¹ ΛΚ τÍ ΚΜ, κοιν¾ δ ¹ ΚΕ, κሠγωνίας Ñρθ¦ς περιέχουσιν, βάσις ¥ρα ¹ ΛΕ βάσει τÍ ΕΜ ™στιν ‡ση. κሠ™πεˆ Ñρθή ™στιν ¹ ØπÕ ΛΕΜ γωνία· ™ν ¹µικυκλίJ γάρ· τÕ ¥ρα ¢πÕ τÁς ΛΜ διπλάσιόν ™στι τοà ¢πÕ τÁς ΛΕ. πάλιν, ™πεˆ ‡ση ™στˆν ¹ ΑΓ τÍ ΓΒ, διπλασία ™στˆν ¹ ΑΒ τÁς ΒΓ. æς δ ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ¢πÕ τÁς Β∆· διπλάσιον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς Β∆. ™δείχθη δ κሠτÕ ¢πÕ τÁς ΛΜ διπλάσιον τοà ¢πÕ τÁς ΛΕ. καί ™στιν ‡σον τÕ ¢πÕ τÁς ∆Β τù ¢πÕ τÁς ΛΕ· ‡ση γ¦ρ κε‹ται ¹ ΕΘ τÍ ∆Β. ‡σον ¥ρα κሠτÕ ¢πÕ τÁς ΑΒ τù ¢πÕ τÁς ΛΜ· ‡ση ¥ρα ¹ ΑΒ τÍ ΛΜ. καί ™στιν ¹ ΑΒ ¹ τÁς δοθείσης σφαίρας διάµετρος· ¹ ΛΜ ¥ρα ‡ση ™στˆ τÍ τÁς δοθείσης σφαίρας διαµέτρJ. Περιείληπται ¥ρα τÕ Ñκτάεδρον τÍ δοθείσV σφαίρv. κሠσυναποδέδεικται, Óτι ¹ τÁς σφαίρας διάµετρος δυνάµει διπλασίων ™στˆ τÁς τοà Ñκταέδρου πλευρ©ς· Óπερ œδει δε‹ξαι.

†

If the radius of the sphere is unity, then the side of octahedron is

√

ried around, and again established at the same (position) from which it began to be moved, then it will also pass through points F , G, and H, and the octahedron will have been enclosed by a sphere. So, I say that (it is) also (enclosed) by the given (sphere). For since LK is equal to KM , and KE (is) common, and they contain right-angles, the base LE is thus equal to the base EM [Prop. 1.4]. And since angle LEM is a right-angle—for (it is) in a semi-circle [Prop. 3.31]—the (square) on LM is thus double the (square) on LE [Prop. 1.47]. Again, since AC is equal to CB, AB is double BC. And as AB (is) to BC, so the (square) on AB (is) to the (square) on BD [Prop. 6.8, Def. 5.9]. Thus, the (square) on AB is double the (square) on BD. And the (square) on LM was also shown (to be) double the (square) on LE. And the (square) on DB is equal to the (square) on LE. For EH was made equal to DB. Thus, the (square) on AB (is) also equal to the (square) on LM . Thus, AB (is) equal to LM . And AB is the diameter of the given sphere. Thus, LM is equal to the diameter of the given sphere. Thus, the octahedron has been enclosed by the given sphere, and it has been simultaneously proved that the square on the diameter of the sphere is double the (square) on the side of the octahedron.† (Which is) the very thing it was required to show.

2.

ιε΄.

Proposition 15

Κύβον συστήσασθαι κሠσφαίρv περιλαβε‹ν, Î κሠτ¾ν πυραµίδα, κሠδε‹ξαι, Óτι ¹ τÁς σφαίρας διάµετρος δυνάµει τριπλασίων ™στˆ τÁς τοà κύβου πλευρ©ς. 'Εκκείσθω ¹ τÁς δοθείσης σφαίρας διάµετρος ¹ ΑΒ κሠτετµήσθω κατ¦ τÕ Γ éστε διπλÁν εναι τ¾ν ΑΓ τÁς ΓΒ, κሠγεγράφθω ™πˆ τÁς ΑΒ ¹µικύκλιον τÕ Α∆Β, κሠ¢πÕ τοà Γ τÍ ΑΒ πρÕς Ñρθ¦ς ½χθω ¹ Γ∆, κሠ™πεζεύχθω ¹ ∆Β, κሠ™κκείσθω τετράγωνον τÕ ΕΖΗΘ ‡σην œχον τ¾ν πλευρ¦ν τÍ ∆Β, κሠ¢πÕ τîν Ε, Ζ, Η, Θ τù τοà ΕΖΗΘ τετραγώνου ™πιπέδJ πρÕς Ñρθ¦ς ½χθωσαν αƒ ΕΚ, ΖΛ, ΗΜ, ΘΝ, κሠ¢φVρήσθω ¢πÕ ˜κάστης τîν ΕΚ, ΖΛ, ΗΜ, ΘΝ µι´ τîν ΕΖ, ΖΗ, ΗΘ, ΘΕ ‡ση ˜κάστη τîν ΕΚ, ΖΛ, ΗΜ, ΘΝ, κሠ™πεζεύχθωσαν αƒ ΚΛ, ΛΜ, ΜΝ, ΝΚ· κύβος ¥ρα συνέσταται Ð ΖΝ ØπÕ žξ τετραγώνων ‡σων περιεχόµενος. ∆ε‹ δ¾ αÙτÕν κሠσφαίρv περιλαβε‹ν τÍ δοθείσV κሠδε‹ξαι, Óτι ¹ τÁς σφαίρας διάµετρος δυνάµει τριπλασία ™στˆ τÁς πλευρ©ς τοà κύβου.

To construct a cube, and to enclose (it) in a sphere, like in the (case of the) pyramid, and to show that the square on the diameter of the sphere is three times the (square) on the side of the cube. Let the diameter AB of the given sphere be laid out, and let it have been cut at C such that AC is double CB. And let the semi-circle ADB have been drawn on AB. And let CD be drawn from C at right-angles to AB. And let DB have been joined. And let the square EF GH, having (its) side equal to DB, be laid out. And let EK, F L, GM , and HN have been drawn from (points) E, F , G, and H, (respectively), at right-angles to the plane of square EF GH. And let EK, F L, GM , and HN , equal to one of EF , F G, GH, and HE, have been cut off from EK, F L, GM , and HN , respectively. And let KL, LM , M N , and N K have been joined. Thus, a cube contained by six equal squares has been constructed. So, it is also necessary to enclose it by the given sphere, and to show that the square on the diameter of the sphere is three times the (square) on the side of the cube.

525

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

E K

N

Z L

A

J

N

K

H

F

M D

G

L

B

H

E

A

'Επεζεύχθωσαν γ¦ρ αƒ ΚΗ, ΕΗ. κሠ™πεˆ Ñρθή ™στιν ¹ ØπÕ ΚΕΗ γωνία δι¦ τÕ κሠτ¾ν ΚΕ Ñρθ¾ν εναι πρÕς τÕ ΕΗ ™πίπεδον δηλαδ¾ κሠπρÕς τ¾ν ΕΗ εÙθε‹αν, τÕ ¥ρα ™πˆ τÁς ΚΗ γραφόµενον ¹µικύκλιον ¼ξει κሠδι¦ τοà Ε σηµείου. πάλιν, ™πεˆ ¹ ΗΖ Ñρθή ™στι πρÕς ˜κατέραν τîν ΖΛ, ΖΕ, κሠπρÕς τÕ ΖΚ ¥ρα ™πίπεδον Ñρθή ™στιν ¹ ΗΖ· éστε κሠ™¦ν ™πιζεύξωµεν τ¾ν ΖΚ, ¹ ΗΖ Ñρθ¾ œσται κሠπρÕς τ¾ν ΖΚ· κሠδˆα τοàτο πάλιν τÕ ™πˆ τÁς ΗΚ γραφόµενον ¹µικύκλιον ¼ξει κሠδι¦ τοà Ζ. еοίως κሠδˆα τîν λοιπîν τοà κύβου σηµείων ¼ξει. ™¦ν δ¾ µενούσης τÁς ΚΗ περιενεχθν τÕ ¹µικύκλιον ε„ς τÕ αÙτÕ ¢ποκατασταθÍ, Óθεν ½ρξατο φέρεσθαι, œσται σφαίρv περιειληµµένος Ð κύβος. λέγω δή, Óτι κሠτÍ δοθείσV. ™πεˆ γ¦ρ ‡ση ™στˆν ¹ ΗΖ τÍ ΖΕ, καί ™στιν Ñρθ¾ ¹ πρÕς τù Ζ γωνία, τÕ ¥ρα ¢πÕ τÁς ΕΗ διπλάσιόν ™στι τοà ¢πÕ τÁς ΕΖ. ‡ση δ ¹ ΕΖ τÍ ΕΚ· τÕ ¥ρα ¢πÕ τÁς ΕΗ διπλάσιόν ™στι τοà ¢πÕ τÁς ΕΚ· éστε τ¦ ¢πÕ τîν ΗΕ, ΕΚ, τουτέστι τÕ ¢πÕ τÁς ΗΚ, τριπλάσιόν ™στι τοà ¢πÕ τÁς ΕΚ. κሠ™πεˆ τριπλασίων ™στˆν ¹ ΑΒ τÁς ΒΓ, æς δ ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ¢πÕ τÁς Β∆, τριπλάσιον ¥ρα τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς Β∆. ™δείχθη δ κሠτÕ ¢πÕ τÁς ΗΚ τοà ¢πÕ τÁς ΚΕ τριπλάσιον. κሠκε‹ται ‡ση ¹ ΚΕ τÍ ∆Β· ‡ση ¥ρα κሠ¹ ΚΗ τÍ ΑΒ. καί ™στιν ¹ ΑΒ τÁς δοθείσης σφαίρας διάµετρος· κሠ¹ ΚΗ ¥ρα ‡ση ™στˆ τÍ τÁς δοθείσης σφαίρας διαµέτρJ. ΤÍ δοθείσV ¥ρα σφαίρα περιείληπται Ð κύβος· κሠσυναποδέδεικται, Ðτι ¹ τÁς σφαίρας διάµετρος δυνάµει τριπλασίων ™στˆ τÁς τοà κύβου πλευρ©ς· Óπερ œδει δε‹ξαι.

G M D

C

B

For let KG and EG have been joined. And since angle KEG is a right-angle—on account of KE also being at right-angles to the plane EG, and manifestly also to the straight-line EG [Def. 11.3]—thus, the semi-circle drawn on KG will also pass through point E. Again, since GF is at right-angles to each of F L and F E, GF is thus also at right-angles to the plane F K [Prop. 11.4]. Hence, if we also join F K then GF will also be at right-angles to F K. And, again, on account of this, the semi-circle drawn on GK will also pass through point F . Similarly, it will also pass through the remaining (angular) points of the cube. So, if KG remains (fixed), and the semi-circle is carried around, and again established at the same (position) from which it began to be moved, then the cube will have been enclosed by a sphere. So, I say that (it is) also (enclosed) by the given (sphere). For since GF is equal to F E, and the angle at F is a right-angle, the (square) on EG is thus double the (square) on EF [Prop. 1.47]. And EF (is) equal to EK. Thus, the (square) on EG is double the (square) on EK. Hence, the (sum of the squares) on GE and EK—that is to say, the (square) on GK [Prop. 1.47]—is three times the (square) on EK. And since AB is three times BC, and as AB (is) to BC, so the (square) on AB (is) to the (square) on BD [Prop. 6.8, Def. 5.9], the (square) on AB (is) thus three times the (square) on BD. And the (square) on GK was also shown (to be) three times the (square) on KE. And KE was made equal to DB. Thus, KG (is) also equal to AB. And AB is the radius of the given sphere. This, KG is equal to the diameter of the given sphere. Thus, the cube has been enclosed by the given sphere. And it has simultaneously been shown that the square on the diameter of the sphere is three times the (square) on

526

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13 the side of the cube.† (Which is) the very thing it was required to show.

†

If the radius of the sphere is unity, then the side of the cube is

p 4/3.

ι$΄.

Proposition 16

Ε„κοσάεδρον συστήσασθαι κሠσφαίρv περιλαβε‹ν, To construct an icosahedron, and to enclose (it) in a Î κሠτ¦ προειρηµένα σχήµατα, κሠδε‹ξαι, Óτι ¹ τοà sphere, like the aforementioned figures, and to show that ε„κοσαέδρου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ™λάττων. the side of the icosahedron is that irrational (straightline) called minor.

D

A

G

D

B

A

'Εκκείσθω ¹ τÁς δοθείσης σφαίρας διάµετρος ¹ ΑΒ κሠτετµήσθω κατ¦ τÕ Γ éστε τετραπλÁν εναι τ¾ν ΑΓ τÁς ΓΒ, κሠγεγράφθω ™πˆ τÁς ΑΒ ¹µικύκλιον τÕ Α∆Β, κሠ½χθω ¢πÕ τοà Γ τÍ ΑΒ πρÕς ορθ¦ς γωνίας εÙθε‹α γραµµ¾ ¹ Γ∆, καί ™πεζεύχθω ¹ ∆Β, κሠ™κκείσθω κύκλος Ð ΕΖΗΘΚ, οá ¹ ™ν τοà κέντρου ‡ση œστω τÍ ∆Β, κሠ™γγεγράφθω ε„ς τÕν ΕΖΗΘΚ κύκλον πεντάγωνον „σόπλευρόν τε κሠ„σογώνιον τÕ ΕΖΗΘΚ, κሠτετµήσθωσαν αƒ ΕΖ, ΖΗ, ΗΘ, ΘΚ, ΚΕ περιφέρειαι δίχα κατ¦ τÕ Λ, Μ, Ν, Ξ, Ο σηµε‹α, κሠ™πεζεύχθωσαν αƒ ΛΜ, ΜΝ, ΝΞ, ΞΟ, ΟΛ, ΕΟ. ‡σόπλευρον ¥ρα ™στˆ κሠτÕ ΛΜΝΞΟ πεντάγωνον, κሠδεκαγώνου ¹ ΕΟ εÙθε‹α. κሠ¢νεστάτωσαν ¥πÕ τîν Ε, Ζ, Η, Θ, Κ σηµείων τù τοà κύκλου ™πιπέδJ πρÕς Ñρθ¦ς γωνίας εÙθε‹αι αƒ ΕΠ, ΖΡ, ΗΣ, ΘΤ, ΚΥ ‡σαι οâσαι τÍ ™κ τοà κέντρου τοà ΕΖΗΘΚ κύκλου, κሠ™πεζεύχθωσαν αƒ ΠΡ, ΡΣ, ΣΤ, ΤΥ, ΥΠ, ΠΛ, ΛΡ, ΡΜ, ΜΣ, ΣΝ, ΝΤ, ΤΞ, ΞΥ, ΥΟ, ΟΠ. Κሠ™πεˆ ˜κατέρα τîν ΕΠ, ΚΥ τù αÙτù ™πιπέδJ πρÕς Ñρθάς ™στιν, παράλληλος ¥ρα ™στˆν ¹ ΕΠ τÍ ΚΥ. œστι δ αÙτÍ κሠ‡ση· αƒ δ τ¦ς ‡σας τε κሠπαραλλήλους ™πιζευγνύουσαι ™πˆ τ¦ αÙτ¦ µέρη εÙθε‹αι ‡σαι τε κሠπαράλληλοί ε„σιν. ¹ ΠΥ ¥ρα τÍ ΕΚ ‡ση τε κሠπαράλληλός ™στιν. πενταγώνου δ „σοπλεύρου ¹ ΕΚ· πενταγώνου ¥ρα „σοπλεύρου κሠ¹ ΠΥ τοà ε„ς τÕν ΕΖΗΘΚ κύκλον ™γγραφοµένου. δι¦ τ¦ αÙτ¦ δ¾ κሠ˜κάστη τîν ΠΡ, ΡΣ, ΣΤ, ΤΥ πενταγώνου ™στˆν „σοπλεύρου τοà ε„ς τÕν ΕΖΗΘΚ κύκλον ™γγραφοµένου· „σόπλευρον ¥ρα τÕ ΠΡΣΤΥ πεντάγωνον. κሠ™πεˆ ˜ξαγώνου µέν ™στιν ¹ ΠΕ, δεκαγώνου δ ¹ ΕΟ, καί ™στιν Ñρθ¾ ¹ ØπÕ ΠΕΟ, πενταγώνου ¥ρα ™στˆν ¹

C

B

Let the diameter AB of the given sphere be laid out, and let it have been cut at C such that AC is four times CB [Prop. 6.10]. And let the semi-circle ADB have been drawn on AB. And let the straight-line CD have been drawn from C at right-angles to AB. And let DB have been joined. And let the circle EF GHK be set down, and let its radius be equal to DB. And let the equilateral and equiangular pentagon EF GHK have been inscribed in circle EF GHK [Prop. 4.11]. And let the circumferences EF , F G, GH, HK, and KE have been cut in half at points L, M , N , O, and P (respectively). And let LM , M N , N O, OP , P L, and EP have been joined. Thus, pentagon LM N OP is also equilateral, and EP (is) the side of the decagon (inscribed in the circle). And let the straight-lines EQ, F R, GS, HT , and KU , which are equal to the radius of circle EF GHK, have been set up at right-angles to the plane of the circle, at points E, F , G, H, and K (respectively). And let QR, RS, ST , T U , U Q, QL, LR, RM , M S, SN , N T , T O, OU , U P , and P Q have been joined. And since EQ and KU are each at right-angles to the same plane, EQ is thus parallel to KU [Prop. 11.6]. And it is also equal to it. And straight-lines joining equal and parallel (straight-lines) on the same side are (themselves) equal and parallel [Prop. 1.33]. Thus, QU is equal and parallel to EK. And EK (is the side) of an equilateral pentagon (inscribed in circle EF GHK). Thus, QU (is) also the side of an equilateral pentagon inscribed in circle EF GHK. So, for the same (reasons), QR, RS, ST , and T U are also the sides of an equilateral pentagon inscribed in circle EF GHK. Pentagon QRST U (is) thus equilat-

527

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

ΠΟ· ¹ γ¦ρ τοà πενταγώνου πλευρ¦ δύναται τήν τε τοà ˜ξαγώνου κሠτ¾ν τοà δεκαγώνου τîν ε„ς τÕν αÙτÕν κύκλον ™γγραφοµένων. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΟΥ πενταγώνου ™στˆ πλευρά. œστι δ κሠ¹ ΠΥ πενταγώνου· „σόπλευρον ¥ρα ™στˆ τÕ ΠΟΥ τρίγωνον. δι¦ τ¦ αÙτ¦ δ¾ κሠ›καστον τîν ΠΛΡ, ΡΜΣ, ΣΝΤ, ΤΞΥ „σόπλευρόν ™στιν. κሠ™πεˆ πενταγώνου ™δείχθη ˜κατέρα τîν ΠΛ, ΠΟ, œστι δ κሠ¹ ΛΟ πενταγώνου, „σόπλευρον ¥ρα ™στˆ τÕ ΠΛΟ τρίγωνον. δι¦ τ¦ αÙτ¦ δ¾ κሠ›καστον τîν ΛΡΜ, ΜΣΝ, ΝΤΞ, ΞΥΟ τριγώνων „σόπλευρόν ™στιν.

P

R L

F

N

O

aQ

T

Z

M a

P

W

V

K

J

E

F

Y

H

Q L

W

M

R

E

Z

S

eral. And side QE is (the side) of a hexagon (inscribed in circle EF GHK), and EP (the side) of a decagon, and (angle) QEP is a right-angle, QP is thus (the side) of a pentagon (inscribed in the same circle). For the square on the side of a pentagon is (equal to the sum of) the (squares) on (the sides of) a hexagon and a decagon inscribed in the same circle [Prop. 13.10]. So, for the same (reasons), P U is also the side of a pentagon. And QU is also (the side) of a pentagon. Thus, triangle QP U is equilateral. So, for the same (reasons), (triangles) QLR, RM S, SN T , and T OU are each also equilateral. And since QL and QP were each shown (to be the sides) of a pentagon, and LP is also (the side) of a pentagon, triangle QLP is thus equilateral. So, for the same (reasons), triangles LRM , M SN , N T O, and OU P are each also equilateral.

U

S

X

X

G

K

U

O

N H

Ε„λήφθω τÕ κέντρου τοà ΕΖΗΘΚ κύκλου τÕ Φ σηµε‹ον· κሠ¢πÕ τοà Φ τù τοà κύκλου ™πιπέδJ πρÕς Ñρθ¦ς ¢νεστάτω ¹ ΦΩ, κሠ™κβεβλήσθω ™πˆ τ¦ ›τερα µέρη æς ¹ ΦΨ, κሠ¢φVρήσθω ˜ξαγώνου µν ¹ ΦΧ, δεκαγώνου δ ˜κατέρα τîν ΦΨ, ΧΩ, κሠ™πεζεύχθωσαν αƒ ΠΩ, ΠΧ, ΥΩ, ΕΦ, ΛΦ, ΛΨ, ΨΜ. Κሠ™πεˆ ˜κατέρα τîν ΦΧ, ΠΕ τù τοà κύκλου ™πιπέδJ πρÕς Ñρθάς ™στιν, παράλληλος ¥ρα ™στˆν ¹ ΦΧ τÍ ΠΕ. ε„σˆ δ κሠ‡σαι· καˆ αƒ ΕΦ, ΠΧ ¥ρα ‡σαι τε κሠπαράλληλοί ε„σιν. ˜ξαγώνου δ ¹ ΕΦ· ˜ξαγώνου ¥ρα κሠ¹ ΠΧ. κሠ™πεˆ ˜ξαγώνου µέν ™στιν ¹ ΠΧ, δεκαγώνου δ ¹ ΧΩ, κሠÑρθή ™στιν ¹ ØπÕ ΠΧΩ γωνία, πενταγώνου ¥ρα ™στˆν ¹ ΠΩ. δι¦ τ¦ αÙτ¦ δ¾ κሠ¹ ΥΩ

T Let the center, point V , of circle EF GHK have been found [Prop. 3.1]. And let V Z have been set up, at (point) V , at right-angles to the plane of the circle. And let it have been produced on the other side (of the circle), like V X. And let V W have been cut off (from XZ so as to be equal to the side) of a hexagon, and each of V X and W Z (so as to be equal to the side) of a decagon. And let QZ, QW , U Z, EV , LV , LX, and XM have been joined. And since V W and QE are each at right-angles to the plane of the circle, V W is thus parallel to QE [Prop. 11.6]. And they are also equal. EV and QW are thus equal and parallel (to one another) [Prop. 1.33].

528

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

πενταγώνου ™στίν, ™πειδήπερ, ™¦ν ™πιζεύξωµεν τ¦ς ΦΚ, ΧΥ, ‡σαι κሠ¢πεναντίον œσονται, καί ™στιν ¹ ΦΚ ™κ τοà κέντρου οâσα ˜ξαγώνου. ˜ξαγώνου ¥ρα κሠ¹ ΧΥ. δεκαγώνου δ ¹ ΧΩ, κሠÑρθ¾ ¹ ØπÕ ΥΧΩ· πενταγώνου ¥ρα ¹ ΥΩ. œστι δ κሠ¹ ΠΥ πενταγώνου· „σόπλευρον ¥ρα ™στˆ τÕ ΠΥΩ τρίγωνον. δι¦ τ¦ αÙτ¦ δ¾ κሠ›καστον τîν λοιπîν τριγώνων, ïν βάσεις µέν ε„σιν αƒ ΠΡ, ΡΣ, ΣΤ, ΤΥ εÙθε‹αι, κορυφ¾ δ τÕ Ω σηµε‹ον, „σόπλευρόν ™στιν. πάλιν, ™πεˆ ˜ξαγώνου µν ¹ ΦΛ, δεκαγώνου δ ¹ ΦΨ, κሠÑρθή ™στιν ¹ ØπÕ ΛΦΨ γωνία, πενταγώνου ¥ρα ™στˆν ¹ ΛΨ. δι¦ τ¦ αÙτ¦ δ¾ ™¦ν ™πιζεύξωµεν τ¾ν ΜΦ οâσαν ˜ξαγώνου, συνάγεται κሠ¹ ΜΨ πενταγώνου, œστι δ κሠ¹ ΛΜ πενταγώνου· „σόπλευρον ¥ρα ™στˆ τÕ ΛΜΨ τρίγωνον. еοίως δ¾ δειχθήσεται, Óτι κሠ›καστον τîν λοιπîν τριγώνων, ïν βάσεις µέν ε„σιν αƒ ΜΝ, ΝΞ, ΞΟ, ΟΛ, κορυφ¾ δ τÕ Ψ σηµεˆον, „σόπλευρόν ™στιν. συνέσταται ¥ρα ε„κοσάεδρον ØπÕ ε‡κοσι τριγώνων „σοπλεύρων περιεχόµενον. ∆ε‹ δ¾ αÙτÕ κሠσφαίρv περιλαβε‹ν τÍ δοθείσV κሠδε‹ξαι, Óτι ¹ τοà ε„κοσαέδρου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ™λάσσων. 'Επεˆ γ¦ρ ˜ξαγώνου ™στˆν ¹ ΦΧ, δεκαγώνου δ ¹ ΧΩ, ¹ ΦΩ ¥ρα ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Χ, κሠτÕ µε‹ζον αÙτÁς τµÁµά ™στιν ¹ ΦΧ· œστιν ¥ρα æς ¹ ΩΦ πρÕς τ¾ν ΦΧ, οÛτως ¹ ΦΧ πρÕς τ¾ν ΧΩ. ‡ση δ ¹ µν ΦΧ τÍ ΦΕ, ¹ δ ΧΩ τÍ ΦΨ· œστιν ¥ρα æς ¹ ΩΦ πρÕς τ¾ν ΦΕ, οÛτως ¹ ΕΦ πρÕς τ¾ν ΦΨ. καί ε„σιν Ñρθαˆ αƒ ØπÕ ΩΦΕ, ΕΦΨ γωνίαι· ™¦ν ¥ρα ™πιζεύξωµεν τ¾ν ΕΩ εÙθεˆαν, Ñρθ¾ œσται ¹ ØπÕ ΨΕΩ γωνία δι¦ τ¾ν еοιότητα τîν ΨΕΩ, ΦΕΩ τριγώνων. δι¦ τ¦ αÙτ¦ δ¾ ™πεί ™στιν æς ¹ ΩΦ πρÕς τ¾ν ΦΧ, οÛτως ¹ ΦΧ πρÕς τ¾ν ΧΩ, ‡ση δ ¹ µν ΩΦ τÍ ΨΧ, ¹ δ ΦΧ τÍ ΧΠ, œστιν ¥ρα æς ¹ ΨΧ πρÕς τ¾ν ΧΠ, οÛτως ¹ ΠΧ πρÕς τ¾ν ΧΩ. κሠδι¦ τοàτο πάλιν ™¦ν ™πιζεύξωµεν τ¾ν ΠΨ, Ñρθ¾ œσται ¹ πρÕς τù Π γωνία· τÕ ¥ρα ™πˆ τÁς ΨΩ γραφόµενον ¹µικύκλιον ¼ξει κሠδˆα τοà Π. κሠ™¦ν µενούσης τÁς ΨΩ περιενεχθν τÕ ¹µικύκλιον ε„ς τÕ αÙτÕ πάλιν ¢ποκατασταθÍ, Óθεν ½ρξατο φέρεσθαι, ¼ξει κሠδι¦ τοà Π κሠτîν λοιπîν σηµείων τοà ε„κοσάεδρου, κሠœσται σφαίρv περιειληµµένον τÕ ε„κοσαέδρον. λέγω δή, Óτι κሠτÍ δοθείσV. τετµήσθω γ¦ρ ¹ ΦΧ δίχα κατ¦ τÕ α. κሠ™πεˆ εÙθε‹α γραµµ¾ ¹ ΦΩ ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Χ, κሠτÕ œλασσον αÙτÁς τµÁµά ™στιν ¹ ΩΧ, ¹ ¥ρα ΩΧ προσλαβοàσα τ¾ν ¹µίσειαν τοà µείζονος τµήµατος τ¾ν Χα πενταπλάσιον δύναται τοà ¢πÕ τÁς ¹µισείας τοà µείζονος τµήµατος· πενταπλάσιον ¥ρα ™στˆ τÕ ¢πÕ τÁς Ωα τοà ¢πÕ τÁς αΧ. καί ™στι τÁς µν Ωα διπλÁ ¹ ΩΨ, τ¾ς δ αΧ διπλÁ ¹ ΦΧ· πενταπλάσιον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΩΨ τοà ¢πÕ τÁς ΧΦ. κሠ™πεˆ τετραπλÁ ™στιν ¹ ΑΓ τÁς ΓΒ, πενταπλÁ ¥ρα ™στˆν ¹ ΑΒ τÁς ΒΓ. æς δ ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ¢πÕ τÁς

And EV (is the side) of a hexagon. Thus, QW (is) also (the side) of a hexagon. And since QW is (the side) of a hexagon, and W Z (the side) of a decagon, and angle QW Z is a right-angle [Def. 11.3, Prop. 1.29], QZ is thus (the side) of a pentagon [Prop. 13.10]. So, for the same (reasons), U Z is also (the side) of a pentagon—inasmuch as, if we join V K and W U then they will be equal and opposite. And V K, being (equal) to the radius (of the circle), is (the side) of a hexagon [Prop. 4.15 corr.]. Thus, W U (is) also the side of a hexagon. And W Z (is the side) of a decagon, and (angle) U W Z (is) a right-angle. Thus, U Z (is the side) of a pentagon [Prop. 13.10]. And QU is also (the side) of a pentagon. Triangle QU Z is thus equilateral. So, for the same (reasons), each of the remaining triangles, whose bases are the straight-lines QR, RS, ST , and T U , and apexes the point Z, are also equilateral. Again, since V L (is the side) of a hexagon, and V X (the side) of a decagon, and angle LV X is a rightangle, LX is thus (the side) of a pentagon [Prop. 13.10]. So, for the same (reasons), if we join M V , which is (the side) of a hexagon, M X is also inferred (to be the side) of a pentagon. And LM is also (the side) of a pentagon. Thus, triangle LM X is equilateral. So, similarly, it can be shown that each of the remaining triangles, whose bases are the (straight-lines) M N , N O, OP , and P L, and apexes the point X, are also equilateral. Thus, an icosahedron contained by twenty equilateral triangles has been constructed. So, it is also necessary to enclose it in the given sphere, and to show that the side of the icosahedron is that irrational (straight-line) called minor. For, since V W is (the side) of a hexagon, and W Z (the side) of a decagon, V Z has thus been cut in extreme and mean ratio at W , and its greater piece is V W [Prop. 13.9]. Thus, as ZV is to V W , so V W (is) to W Z. And V W (is) equal to V E, and W Z to V X. Thus, as ZV is to V E, so EV (is) to V X. And angles ZV E and EV X are right-angles. Thus, if we join straight-line EZ then angle XEZ will be a right-angle, on account of the similarity of triangles XEZ and V EZ. [Prop. 6.8]. So, for the same (reasons), since as ZV is to V W , so V W (is) to W Z, and ZV (is) equal to XW , and V W to W Q, thus as XW is to W Q, so QW (is) to W Z. And, again, on account of this, if we join QX then the angle at Q will be a right-angle [Prop. 6.8]. Thus, the semi-circle drawn on XZ will also pass through Q [Prop. 3.31]. And if XZ remains fixed, and the semi-circle is carried around, and again established at the same (position) from which it began to be moved, then it will also pass through (point) Q, and (through) the remaining (angular) points of the icosahedron. And the icosahedron will have been en-

529

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

ΑΒ πρÕς τÕ ¢πÕ τÁς Β∆· πενταπλάσιον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς Β∆. ™δείχθη δ κሠτÕ ¢πÕ τÁς ΩΨ πενταπλάσιον τοà ¢πÕ τÁς ΦΧ. καί ™στιν ‡ση ¹ ∆Β τÍ ΦΧ· ˜κατέρα γ¦ρ αÙτîν ‡ση ™στˆ τÍ ™κ τοà κέντρου τοà ΕΖΗΘΚ κύκλου· ‡ση ¥ρα κሠ¹ ΑΒ τÍ ΨΩ. καί ™στιν ¹ ΑΒ ¹ τÁς δοθείσης σφαίρας διάµετρος· κሠ¹ ΨΩ ¥ρα ‡ση ™στˆ τÍ τÁς δοθείσης σφαίρας διαµέτρJ· τÍ ¥ρα δοθείσV σφαίρv περιείληπται τÕ ε„κοσάεδρον. Λέγω δή, Óτι ¹ τοà ε„κοσαέδρου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ™λάττων. ™πεˆ γ¦ρ ·ητή ™στιν ¹ τÁς σφαίρας διάµετρος, καί ™στι δυνάµει πενταπλασίων τÁς ™κ τοà κέντρου τοà ΕΖΗΘΚ κύκλου, ·ητ¾ ¥ρα ™στˆ κሠ¹ ˜κ τοà κέντρου τοà ΕΖΗΘΚ κύκλου· éστε κሠ¹ διάµετρος αÙτοà ·ητή ™στιν. ™¦ν δ ε„ς κύκλον ·ητ¾ν œχοντα τ¾ν διάµετρον πεντάγωνον „σόπλευρον ™γγραφV, ¹ τοà πενταγώνου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ™λάττων. ¹ δ τοà ΕΖΗΘΚ πενταγώνου πλευρ¦ ¹ τοà ε„κοσαέδρου ™στίν. ¹ ¥ρα τοà είκοσαέδρου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ™λάττων.

closed by a sphere. So, I say that (it is) also (enclosed) by the given (sphere). For let V W have been cut in half at a. And since the straight-line V Z has been cut in extreme and mean ratio at W , and its lesser piece is ZW , then the square on ZW added to half of the greater piece, W a, is five times the (square) on half of the greater piece [Prop. 13.3]. Thus, the (square) on Za is five times the (square) on aW . And ZX is double Za, and V W double aW . Thus, the (square) on ZX is five times the (square) on W V . And since AC is four times CB, AB is thus five times BC. And as AB (is) to BC, so the (square) on AB (is) to the (square) on BD [Prop. 6.8, Def. 5.9]. Thus, the (square) on AB is five times the (square) on BD. And the (square) on ZX was also shown (to be) five times the (square) on V W . And DB is equal to V W . For each of them is equal to the radius of circle EF GHK. Thus, AB (is) also equal to XZ. And AB is the diameter of the given sphere. Thus, XZ is equal to the diameter of the given sphere. Thus, the icosahedron has been enclosed by the given sphere. So, I say that the side of the icosahedron is that irrational (straight-line) called minor. For since the diameter of the sphere is rational, and the square on it is five times the (square) on the radius of circle EF GHK, the radius of circle EF GHK is thus also rational. Hence, its diameter is also rational. And if an equilateral pentagon is inscribed in a circle having a rational diameter then the side of the pentagon is that irrational (straight-line) called minor [Prop. 13.11]. And the side of pentagon EF GHK is (the side) of the icosahedron. Thus, the side of the icosahedron is that irrational (straight-line) called minor.

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι ¹ τÁς σφαίρας διάµετρος δυνάµει πενταπλασίων ™στˆ τÁς ™κ τοà κέντρου τοà κύκλου, ¢φ' οá τÕ ε„κοσάεδρον ¢ναγέγραπται, κሠÓτι ¹ τÁς σφαίρας διάµετρος σύγκειται œκ τε τÁς τοà ˜ξαγώνου κሠδύο τîν τοà δεκαγώνου τîν ε„ς τÕν αÙτÕν κύκλον ™γγραφοµένων. Óπερ œδει δε‹ξαι.

So, (it is) clear, from this, that the square on the diameter of the sphere is five times the radius of the circle from which the icosahedron has been described, and that the the diameter of the sphere is the sum of (the side) of the hexagon, and two of (the sides) of the decagon, inscribed in the same circle.†

√ If the radius of the sphere is p unity, then the radius of the circle is 2/ 5, and the sides of the hexagon, decagon, and pentagon/icosahedron are √ √ √ √ 2/ 5, 1 − 1/ 5, and (1/ 5) 10 − 2 5, respectively.

†

ιζ΄.

Proposition 17

∆ωδεκάεδρον συστήσασθαι κሠσφαίρv περιλαβε‹ν, Î To construct a dodecahedron, and to enclose (it) in a κሠτ¦ προειρηµένα σχήµατα, κሠδε‹ξαι, Óτι ¹ τοà δω- sphere, like the aforementioned figures, and to show that δεκαέδρου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ¢ποτοµή. the side of the dodecahedron is that irrational (straightline) called an apotome.

530

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

E

U

N B H A

Y

R

O

J Q

T

W

M

Z

F

E

X

S

M

U

N

G

B

X

F V

R P

H T

Z

O

S

C

W

P L

K

G

D

A

'Εκκείσθωσαν τοà προειρηµένου κύβου δύο ™πίπεδα πρÕς Ñρθ¦ς ¢λλήλοις τ¦ ΑΒΓ∆, ΓΒΕΖ, κሠτετµήσθω ˜κάστη τîν ΑΒ, ΒΓ, Γ∆, ∆Α, ΕΖ, ΕΒ, ΖΓ πλευρîν δίχα κατ¦ τ¦ Η, Θ, Κ, Λ, Μ, Ν, Ξ, κሠ™πεζεύχθωσαν αƒ ΗΚ, ΘΛ, ΜΘ, ΝΞ, κሠτετηήσθω ˜κάστη τîν ΝΟ, ΟΞ, ΘΠ ¥κρον κሠµέσον λόγον κατ¦ τ¦ Ρ, Σ, Τ σηµε‹α, κሠœστω αÙτîν µείζονα τµήµατα τ¦ ΡΟ, ΟΣ, ΤΠ, κሠ¢νεστάτωσαν ¢πÕ τîν Ρ, Σ, Τ σηµείων το‹ς τοà κύβου ™πιπέδοις πρÕς Ñρθ¦ς ™πˆ τ¦ ™κτÕς µέρη τοà κύβου αƒ ΡΥ, ΣΦ, ΤΧ, κሠκείσθωσαν ‡σαι τα‹ς ΡΟ, ΟΣ, ΤΠ, κሠ™πεζεύχθωσαν αƒ ΥΒ, ΒΧ, ΧΓ, ΓΦ, ΦΥ. Λέγω, Óτι τÕ ΥΒΧΓΦ πεντάγωνον „σόπλευρόν τε κሠ™ν ˜νˆ ™πιπέδJ κሠœτι „σογώνιόν ™στιν. ™πεζεύχθωσαν γ¦ρ αƒ ΡΒ, ΣΒ, ΦΒ. κሠ™πεˆ εÙθε‹α ¹ ΝΟ ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Ρ, κሠτÕ µε‹ζον τµÁµά ™στιν ¹ ΡΟ, τ¦ ¥ρα ¢πÕ τîν ΟΝ, ΝΡ τριπλάσιά ™στι τοà ¢πÕ τÁς ΡΟ. ‡ση δ ¹ µν ΟΝ τÍ ΝΒ, ¹ δ ΟΡ τÍ ΡΥ· τ¦ ¥ρα ¢πÕ τîν ΒΝ, ΝΡ τριπλάσιά ™στι τοà ¢πÕ τÁς ΡΥ. το‹ς δ ¢πÕ τîν ΒΝ, ΝΡ τÕ ¢πÕ τÁς ΒΡ ™στιν ‡σον· τÕ ¥ρα ¢πÕ τÁς ΒΡ τριπλάσιόν ™στι τοà ¢πÕ τÁς ΡΥ· éστε τ¦ ¢πÕ τîν ΒΡ, ΡΥ τετραπλάσιά ™στι τοà ¢πÕ τÁς ΡΥ. το‹ς δ ¢πÕ τîν ΒΡ, ΡΥ ‡σον ™στι τÕ ¢πÕ τÁς ΒΥ· τÕ ¥ρα ¥πÕ τÁς ΒΥ τετραπλάσιόν ™στι τοà ¢πÕ τÁς ΥΡ· διπλÁ ¥ρα ™στˆν ¹ ΒΥ τÁς ΡΥ. œστι δ κሠ¹ ΦΥ τÁς ΥΡ διπλÁ, ™πειδήπερ κሠ¹ ΣΡ τÁς ΟΡ, τουτέστι τÁς ΡΥ, ™στι διπλÁ· ‡ση ¥ρα ¹ ΒΥ τÍ ΥΦ. еοίως δ¾ δειχθήσεται, Óτι κሠ˜κάστη τîν ΒΧ, ΧΓ, ΓΦ ˜κατέρv τîν ΒΥ, ΥΦ ™στιν ‡ση. „σόπλευρον ¥ρα ™στˆ τÕ ΒΥΦΓΧ πεντάγωνον. λέγω δή, Óτι κሠ™ν ˜νί ™στιν ™πιπέδJ. ½χθω γ¦ρ ¢πÕ τοà Ο ˜κατέρv τîν ΡΥ, ΣΦ παράλληλος ™πˆ τ¦ ™κτÕς τοà κύβου µέρη ¹ ΟΨ, κሠ™πεζεύχθωσαν αƒ ΨΘ, ΘΧ· λέγω, Óτι ¹ ΨΘΧ εÙθε‹ά ™στιν. ™πεˆ γ¦ρ ¹ ΘΠ ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Τ, κሠτÕ

Q

L

K

D

Let two planes of the aforementioned cube [Prop. 13.15], ABCD and CBEF , (which are) at right-angles to one another, be laid out. And let the sides AB, BC, CD, DA, EF , EB, and F C have each been cut in half at points G, H, K, L, M , N , and O (respectively). And let GK, HL, M H, and N O have been joined. And let N P , P O, and HQ have each been cut in extreme and mean ratio at points R, S, and T (respectively). And let their greater pieces be RP , P S, and T Q (respectively). And let RU , SV , and T W have been set up on the exterior side of the cube, at points R, S, and T (respectively), at right-angles to the planes of the cube. And let them be made equal to RP , P S, and T Q. And let U B, BW , W C, CV , and V U have been joined. I say that the pentagon U BW CV is equilateral, and in one plane, and, further, equiangular. For let RB, SB, and V B have been joined. And since the straight-line N P has been cut in extreme and mean ratio at R, and RP is the greater piece, thus the (sum of the squares) on P N and N R is three times the (square) on RP [Prop. 13.4]. And P N (is) equal to N B, and P R to RU . Thus, the (sum of the squares) on BN and N R is three times the (square) on RU . And the (square) on BR is equal to the (sum of the squares) on BN and N R [Prop. 1.47]. Thus, the (square) on BR is three times the (square) on KM . Hence, the (sum of the squares) on BR and RU is four times the (square) on RU . And the (square) on BU is equal to the (sum of the squares) on BR and RU [Prop. 1.47]. Thus, the (square) on BU is four times the (square) on U R. Thus, BU is double RU . And V U is also double U R, inasmuch as SR is also double P R—that is to say, RU . Thus, BU (is) equal to U V . So, similarly, it can be shown that BW , W C, CV are each equal to each

531

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

µε‹ζον αÙτÁς τµÁµά ™στιν ¹ ΠΤ, œστιν ¥ρα æς ¹ ΘΠ πρÕς τ¾ν ΠΤ, οÛτως ¹ ΠΤ πρÕς τ¾ν ΤΘ. ‡ση δ ¹ µν ΘΠ τÍ ΘΟ, ¹ δ ΠΤ ˜κατέρv τîν ΤΧ, ΟΨ· œστιν ¥ρα æς ¹ ΘΟ πρÕς τ¾ν ΟΨ, οÛτως ¹ ΧΤ πρÕς τ¾ν ΤΘ. καί ™στι παράλληλος ¹ µν ΘΟ τÍ ΤΧ· ˜κατέρα γ¦ρ αÙτîν τù Β∆ ™πιπέδJ πρÕς Ñρθάς ™στιν· ¹ δ ΤΘ τÍ ΟΨ· ˜κατέρα γ¦ρ αÙτîν τù ΒΖ ™πιπέδJ πρÕς Ñρθάς ™στιν. ™¦ν δ δύο τρίγωνα συντεθÍ κατ¦ µίαν γωνίαν, æς τ¦ ΨΟΘ, ΘΤΧ, τ¦ς δύο πλευρ¦ς τα‹ς δυνˆν ¢νάλογον œχοντα, éστε τ¦ς еολόγους αÙτîν πλευρ¦ς κሠπαραλλήλους εναι, αƒ λοιπሠεÙθε‹αι ™π' εÙθείας œσονται· ™π' εÙθείας ¥ρα ™στˆν ¹ ΨΘ τÍ ΘΧ. π©σα δ εÙθε‹α ™ν ˜νί ™στιν ™πιπέδJ· ™ν ˜νˆ ¥ρα ™πιπέδJ ™στˆ τÕ ΥΒΧΓΦ πεντάγωνον. Λέγω δή, Óτι κሠ„σογώνιόν ™στιν. 'Επεˆ γ¦ρ εÙθε‹α γραµµ¾ ¹ ΝΟ ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Ρ, κሠτÕ µε‹ζον τµÁµά ™στιν ¹ ΟΡ [œστιν ¥ρα æς συναµφότερος ¹ ΝΟ, ΟΡ πρÕς τ¾ν ΟΝ, οÛτως ¹ ΝΟ πρÕς τ¾ν ΟΡ], ‡ση δ ¹ ΟΡ τÍ ΟΣ [œστιν ¥ρα æς ¹ ΣΝ πρÕς τ¾ν ΝΟ, οÛτως ¹ ΝΟ πρÕς τ¾ν ΟΣ], ¹ ΝΣ ¥ρα ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Ο, κሠτÕ µε‹ζον τµÁµά ™στιν ¹ ΝΟ· τ¦ ¥ρα ¢πÕ τîν ΝΣ, ΣΟ τριπλάσιά ™στι τοà ¢πÕ τÁς ΝΟ. ‡ση δ ¹ µν ΝΟ τÍ ΝΒ, ¹ δ ΟΣ τÍ ΣΦ· τ¦ ¥ρα ¢πÕ τîν ΝΣ, ΣΦ τετράγωνα τριπλάσιά ™στι τοà ¢πÕ τÁς ΝΒ· éστε τ¦ ¢πÕ τîν ΦΣ, ΣΝ, ΝΒ τετραπλάσιά ™στι τοà ¢πÕ τÁς ΝΒ. το‹ς δ ¢πÕ τîν ΣΝ, ΝΒ ‡σον ™στˆ τÕ ¢πÕ τÁς ΣΒ· τ¦ ¥ρα ¢πÕ τîν ΒΣ, ΣΦ, τουτέστι τÕ ¢πÕ τÁς ΒΦ [Ñρθ¾ γ¦ρ ¹ ØπÕ ΦΣΒ γωνία], τετραπλάσιόν ™στι τοà ¢πÕ τÁς ΝΒ· διπλÁ ¥ρα ™στˆν ¹ ΦΒ τÁς ΒΝ. œστι δ κሠ¹ ΒΓ τÁς ΒΝ διπλÁ· ‡ση ¥ρα ™στˆν ¹ ΒΦ τÍ ΒΓ. κሠ™πεˆ δύο αƒ ΒΥ, ΥΦ δυσˆ τα‹ς ΒΧ, ΧΓ ‡σαι ε„σίν, κሠβάσις ¹ ΒΦ βάσει τÍ ΒΓ ‡ση, γωνία ¥ρα ¹ ØπÕ ΒΥΦ γωνίv τÍ ØπÕ ΒΧΓ ™στιν ‡ση. еοίως δ¾ δείξοµεν, Óτι κሠ¹ ØπÕ ΥΦΓ γωνία ‡ση ™στˆ τÍ ØπÕ ΒΧΓ· αƒ ¥ρα ØπÕ ΒΧΓ, ΒΥΦ, ΥΦΓ τρε‹ς γωνίαι ‡σαι ¢λλήλαις ε„σίν. ™¦ν δ πενταγώνου „σοπλεύρου αƒ τρε‹ς γωνίαι ‡σαι ¢λλήλαις ðσιν, „σογώνιον œσται τÕ πεντάγωνον· „σογώνιον ¥ρα ™στˆ τÕ ΒΥΦΓΧ πεντάγωνον. ™δείχθη δ κሠ„σόπλευρον· τÕ ¥ρα ΒΥΦΓΧ πεντάγωνον „σόπλευρόν ™στι κሠ„σογώνιον, καί ™στιν ™πˆ µι©ς τοà κύβου πλευρ©ς τÁς ΒΓ. ™¦ν ¥ρα ™φ' ˜κάστης τîν τοà κύβου δώδεκα πλευρîν τ¦ αÙτ¦ κατασκευάσωµεν, συσταθήσεταί τι σχÁµα στερεÕν ØπÕ δώδεκα πενταγώνων „σοπλεύρων τε κሠ„σογωνίων περιεχόµενον, Ö καλε‹ται δωδεκάεδρον. ∆ε‹ δ¾ αÙτÕ κሠσφαίρv περιλαβεˆν τÍ δοθείσV κሠδε‹ξαι, Óτι ¹ τοà δωδεκαέδρου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ¢ποτοµή. 'Εκβεβλήσθω γ¦ρ ¹ ΨΟ, κሠœστω ¹ ΨΩ· συµβάλλει ¥ρα ¹ ΟΩ τÍ τοà κύβου διαµέτρJ, κሠδίχα τέµνουσιν ¢λλήλας· τοàτο γ¦ρ δέδεικται ™ν τù παρατελεύτJ

of BU and U V . Thus, pentagon BU V CW is equilateral. So, I say that it is also in one plane. For let P X have been drawn from P , parallel to each of RU and SV , on the exterior side of the cube. And let XH and HW have been joined. I say that XHW is a straight-line. For since HQ has been cut in extreme and mean ratio at T , and its greater piece is QT , thus as HQ is to QT , so QT (is) to T H. And HQ (is) equal to HP , and QT to each of T W and P X. Thus, as HP is to P X, so W T (is) to T H. And HP is parallel to T W . For of each of them is at right-angles to the plane BD [Prop. 11.6]. And T H (is parallel) to P X. For each of them is at right-angles to the plane BF [Prop. 11.6]. And if two triangles, like XP H and HT W , having two sides proportional to two sides, are placed together at a single angle such that their corresponding sides are also parallel, then the remaining sides will be straight-on (to one another) [Prop. 6.32]. Thus, XH is straight-on to HW . And every straight-line is in one plane [Prop. 11.1]. Thus, pentagon U BW CV is in one plane. So, I say that it is also equiangular. For since the straight-line N P has been cut in extreme and mean ratio at R, and P R is the greater piece [thus as the sum of N P and P R is to P N , so N P (is) to P R], and P R (is) equal to P S [thus as SN is to N P , so N P (is) to P S], N S has thus also been cut in extreme and mean ratio at P , and N P is the greater piece [Prop. 13.5]. Thus, the (sum of the squares) on N S and SP is three times the (square) on N P [Prop. 13.4]. And N P (is) equal to N B, and P S to SV . Thus, the (sum of the) squares on N S and SV is three times the (square) on N B. Hence, the (sum of the squares) on V S, SN , and N B is four times the (square) on N B. And the (square) on SB is equal to the (sum of the squares) on SN and N B [Prop. 1.47]. Thus, the (sum of the squares) on BS and SV —that is to say, the (square) on BV [for angle V SB (is) a right-angle]—is four times the (square) on N B [Def. 11.3, Prop. 1.47]. Thus, V B is double BN . And BC (is) also double BN . Thus, BV is equal to BC. And since the two (straight-lines) BU and U V are equal to the two (straight-lines) BW and W C (respectively), and the base BV (is) equal to the base BC, angle BU V is thus equal to angle BW C [Prop. 1.8]. So, similarly, we can show that angle U V C is equal to angle BW C. Thus, the three angles BW C, BU V , and U V C are equal to one another. And if three angles of an equilateral pentagon are equal to one another then the pentagon is equiangular [Prop. 13.7]. Thus, pentagon BU V CW is equiangular. And it was also shown (to be) equilateral. Thus, pentagon BU V CW is equilateral and equiangular, and it is on one of the sides, BC, of the cube. Thus, if we make

532

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

θεωρήµατι τοà ˜νδεκάτου βιβλίου. τεµνέτωσαν κατ¦ τÕ Ω· τÕ Ω ¥ρα κέντρον ™στˆ τÁς σφαίρας τÁς περιλαµβανούσης τÕν κύβον, κሠ¹ ΩΟ ¹µίσεια τÁς πλευρ©ς τοà κύβου. ™πεζεύχθω δ¾ ¹ ΥΩ. κሠ™πεˆ εÙθε‹α γραµµ¾ ¹ ΝΣ ¥κρον κሠµέσον λόγον τέτµηται κατ¦ τÕ Ο, κሠτÕ µε‹ζον αÙτÁς τµÁµά ™στιν ¹ ΝΟ, τ¦ ¥ρα ¢πÕ τîν ΝΣ, ΣΟ τριπλάσιά ™στι τοà ¢πÕ τÁς ΝΟ. ‡ση δ ¹ µν ΝΣ τÍ ΨΩ, ™πειδήπερ κሠ¹ µν ΝΟ τÍ ΟΩ ™στιν ‡ση, ¹ δ ΨΟ τÍ ΟΣ. ¢λλ¦ µ¾ν κሠ¹ ΟΣ τÍ ΨΥ, ™πεˆ κሠτÍ ΡΟ· τ¦ ¥ρα ¢πÕ τîν ΩΨ, ΨΥ τριπλάσιά ™στι τοà ¢πÕ τÁς ΝΟ. το‹ς δ ¢πÕ τîν ΩΨ, ΨΥ ‡σον ™στˆ τÕ ¢πÕ τÁς ΥΩ· τÕ ¥ρα ¢πÕ τÁς ΥΩ τριπλάσιόν ™στι τοà ¢πÕ τÁς ΝΟ. œστι δ κሠ¹ ™κ τοà κέντρου τÁς σφαίρας τÁς περιλαµβανούσης τÕν κύβον δυνάµει τριπλασίων τÁς ¹µισείας τÁς τοà κύβου πλευρ©ς· προδέδεικται γ¦ρ κύβον συστήσασθαι κሠσφαίρv περιλαβε‹ν κሠδε‹ξαι, Óτι ¹ τÁς σφαίρας διάµετρος δυνάµει τριπλασίων ™στˆ τÁς πλευρ©ς τοà κύβου. ε„ δ Óλη τÁς Óλης, κሠ[¹] ¹µίσεια τÁς ¹µισείας· καί ™στιν ¹ ΝΟ ¹µίσεια τÁς τοà κύβου πλευρ©ς· ¹ ¥ρα ΥΩ ‡ση ™στˆ τÍ ™κ τοà κέντρου τÁς σφαίρας τÁς περιλαµβανούσης τÕν κύβον. καί ™στι τÕ Ω κέντρον τÁς σφαίρας τÁς περιλαµβανούσης τÕν κύβον· τÕ Υ ¥ρα σηµε‹ον πρÕς τÍ ™πιφανείv ™στι τÁς σφαίρας. еοίως δ¾ δείξοµεν, Óτι κሠ˜κάστη τîν λοιπîν γωνιîν τοà δωδεκαέδρου πρÕς τÍ ™πιφανείv ™στˆ τÁς σφαίρας· περιείληπται ¥ρα τÕ δωδεκαέδρον τÍ δοθείσV σφαίρv. Λέγω δή, Óτι ¹ τοà δωδεκαέδρου πλευρ¦ ¥λογός ™στιν ¹ καλουµένη ¢ποτοµή. 'Επεˆ γ¦ρ τÁς ΝΟ ¥κρον κሠµέσον λόγον τετµηµένης τÕ µε‹ζον τµÁµά ™στιν Ð ΡΟ, τÁς δ ΟΞ ¥κρον κሠµέσον λόγον τετµηµένης τÕ µε‹ζον τµÁµά ™στιν ¹ ΟΣ, Óλης ¥ρα τÁς ΝΞ ¥κρον κሠµέσον λόγον τεµνοµένης τÕ µε‹ζον τµÁµά ™στιν ¹ ΡΣ. [οŒον ™πεί ™στιν æς ¹ ΝΟ πρÕς τ¾ν ΟΡ, ¹ ΟΡ πρÕς τ¾ν ΡΝ, κሠτ¦ διπλάσια· τ¦ γ¦ρ µέρη το‹ς „σάκις πολλαπλασίοις τÕν αÙτÕν œχει λόγον· æς ¥ρα ¹ ΝΞ πρÕς τ¾ν ΡΣ, οÛτως ¹ ΡΣ πρÕς συναµφότερον τ¾ν ΝΡ, ΣΞ. µείζων δ ¹ ΝΞ τÁς ΡΣ· µείζων ¥ρα κሠ¹ ΡΣ συναµφοτέρου τÁς ΝΡ, ΣΞ· ¹ ΝΞ ¥ρα ¥κρον κሠµέσον λόγον τέτµηται, κሠτÕ µε‹ζον αÙτÁς τµÁµά ™στιν ¹ ΡΣ.] ‡ση δ ¹ ΡΣ τÍ ΥΦ· τÁς ¥ρα ΝΞ ¥κρον κሠµέσον λόγον τεµνοµένης τÕ µε‹ζον τµÁµά ™στιν ¹ ΥΦ. κሠ™πεˆ ·ητή ™στιν τÁς σφαίρας διάµετρος καί ™στι δυνάµει τριπλασίων τÁς τοà κύβου πλευρ©ς, ·ητ¾ ¥ρα ™στˆν ¹ ΝΞ πλευρ¦ οâσα τοà κύβου. ™¦ν δ ·ητ¾ γραµµ¾ ¥κρον κሠµέσον λόγον τµηθÍ, ˜κάτερον τîν τµηµάτων ¥λογός ™στιν ¢ποτοµή. `Η ΥΦ ¥ρα πλευρ¦ οâσα τοà δωδεκαέδρου ¥λογός ™στιν ¢ποτοµή.

the same construction on each of the twelve sides of the cube, then some solid figure contained by twelve equilateral and equiangular pentagons will have been constructed, which is called a dodecahedron. So, it is necessary to enclose it in the given sphere, and to show that the side of the dodecahedron is that irrational (straight-line) called an apotome. For let XP have been produced, and let (the produced straight-line) be XZ. Thus, P Z meets the diameter of the cube, and they cut one another in half. For, this has been proved in the penultimate theorem of the eleventh book [Prop. 11.38]. Let them cut (one another) at Z. Thus, Z is the center of the sphere enclosing the cube, and ZP (is) half the side of the cube. So, let U Z have been joined. And since the straight-line N S has been cut in extreme and mean ratio at P , and its greater piece is N P , the (sum of the squares) on N S and SP is thus three times the (square) on N P [Prop. 13.4]. And N S (is) equal to XZ, inasmuch as N P is also equal to P Z, and XP to P S. But, indeed, P S (is) also (equal) to XU , since (it is) also (equal) to RP . Thus, the (sum of the squares) on ZX and XU is three times the (square) on N P . And the (square) on U Z is equal to the (sum of the squares) on ZX and XU [Prop. 1.47]. Thus, the (square) on U Z is three times the (square) on N P . And the square on the radius of the sphere enclosing the cube is also three times the (square) on half the side of the cube. For it has previously been demonstrated (how to) construct the cube, and to enclose (it) in a sphere, and to show that the square on the diameter of the sphere is three times the (square) on the side of the cube [Prop. 13.15]. And if the (square on the) whole (is three times) the (square on the) whole, then the (square on the) half (is) also (three times) the (square on the) half. And N P is half of the side of the cube. Thus, U Z is equal to the radius of the sphere enclosing the cube. And Z is the center of the sphere enclosing the cube. Thus, point U is on the surface of the sphere. So, similarly, we can show that each of the remaining angles of the dodecahedron is on the surface of the sphere. Thus, the dodecahedron has been enclosed by the given sphere. So, I say that the side of the dodecahedron is that irrational straight-line called an apotome. For since RP is the greater piece of N P , which has been cut in extreme and mean ratio, and P S is the greater piece of P O, which has been cut in extreme and mean ratio, RS is thus the greater piece of the whole of N O, which has been cut in extreme and mean ratio. [Thus, since as N P is to P R, (so) P R (is) to RN , and (the same is also true) of the doubles. For parts have the same ratio as similar multiples (taken in corresponding

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ELEMENTS BOOK 13 order) [Prop. 5.15]. Thus, as N O (is) to RS, so RS (is) to the sum of N R and SO. And N O (is) greater than RS. Thus, RS (is) also greater than the sum of N R and SO [Prop. 5.14]. Thus, N O has been cut in extreme and mean ratio, and its greater piece is RS.] And RS (is) equal to U V . Thus, U V is the greater piece of N O, which has been cut in extreme and mean ratio. And since the diameter of the sphere is rational, and the square on it is three times the (square) on the side of the cube, thus N O, which is the side of the cube, is rational. And if a rational (straight)-line is cut in extreme and mean ratio then each of the pieces is the irrational (straight-line called) an apotome. Thus, U V , which is the side of the dodecahedron, is the irrational (straight-line called) an apotome [Prop. 13.6].

Πόρισµα.

Corollary

'Εκ δ¾ τούτου φανερόν, Óτι τÁς τοà κύβου πλευρ©ς So, (it is) clear, from this, that the side of the dodeca¥κρον κሠµέσον λόγον τεµνοµένης τÕ µε‹ζον τµÁµά hedron is the greater piece of the side of the cube, when ™στιν ¹ τοà δωδεκαέδρου πλευρά. Óπερ œδει δε‹ξαι. it is cut in extreme and mean ratio.† (Which is) the very thing it was required to show. †

If the radius of the circumscribed sphere is unity, then the side of the cube is

p

√ √ 4/3, and the side of the dodecahedron is (1/3) ( 15 − 3).

ιη΄.

Proposition 18

Τ¦ς πλευρ¦ς τîν πέντε σχηµάτων ™κθέσθαι κሠσυγκρ‹ναι πρÕς ¢λλήλας.

To set out the sides of the five (aforementioned) figures, and to compare (them) with one another.†

H

G

J

E

Z M

E H

N

A

K

G

D L

F

M N

B

A

'Εκκείσθω ¹ τÁς δοθείσης σφαίρας διάµετρος ¹ ΑΒ, κሠτετµήσθω κατ¦ τÕ Γ éστε ‡σην εναι τ¾ν ΑΓ τÍ ΓΒ, κατ¦ δ τÕ ∆ éστε διπλασίονα εναι τ¾ν Α∆ τÁς ∆Β, κሠγεγράφθω ™πˆ τÁς ΑΒ ¹µικύκλιον τÕ ΑΕΒ, κሠ¢πÕ τîν Γ, ∆ τÍ ΑΒ πρÕς Ñρθ¦ς ½χθωσαν αƒ ΓΕ, ∆Ζ, κሠ™πεζεύχθωσαν αƒ ΑΖ, ΖΒ, ΕΒ. κሠ™πεˆ διπλÁ ™στιν ¹ Α∆ τÁς ∆Β, τριπλÁ ¥ρα ™στˆν ¹ ΑΒ τÁς Β∆. ¢ναστρέψαντι ¹µιολία ¥ρα ™στˆν ¹ ΒΑ τÁς Α∆. æς δ ¹ ΒΑ πρÕς

K

C

D L

B

Let the diameter, AB, of the given sphere be laid out. And let it have been cut at C, such that AC is equal to CB, and at D, such that AD is double DB. And let the semi-circle AEB have been drawn on AB. And let CE and DF have been drawn from C and D (respectively), at right-angles to AB. And let AF , F B, and EB have been joined. And since AD is double DB, AB is thus triple BD. Thus, via conversion, BA is one and a half

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ELEMENTS BOOK 13

τ¾ν Α∆, οÛτως τÕ ¢πÕ τÁς ΒΑ πρÕς τÕ ¢πÕ τÁς ΑΖ· „σογώνιον γάρ ™στι τÕ ΑΖΒ τρίγωνον τù ΑΖ∆ τριγώνJ· ¹µιόλιον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΒΑ τοà ¢πÕ τÁς ΑΖ. œστι δ κሠ¹ τÁς σφαίρας διάµετρος δυνάµει ¹µιολία τÁς πλευρ©ς τÁς πυραµίδος. καί ™στιν ¹ ΑΒ ¹ τÁς σφαίρας διάµετρος· ¹ ΑΖ ¥ρα ‡ση ™στˆ τÍ πλευρ´ τÁς πυραµίδος. Πάλιν, ™πεˆ διπλασίων ™στˆν ¹ Α∆ τÁς ∆Β, τριπλÁ ¥ρα ™στˆν ¹ ΑΒ τÁς Β∆. æς δ ¹ ΑΒ πρÕς τ¾ν Β∆, οÛτως τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ¢πÕ τÁς ΒΖ· τριπλάσιον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς ΒΖ. œστι δ κሠ¹ τÁς σφαίρας διάµετρος δυνάµει τριπλασίων τÁς τοà κύβου πλευρ©ς. καί ™στιν ¹ ΑΒ ¹ τÁς σφαίρας διάµετρος· ¹ ΒΖ ¥ρα τοà κύβου ™στˆ πλευρά. Κሠ™πεˆ ‡ση ™στˆν ¹ ΑΓ τÍ ΓΒ, διπλÁ ¥ρα ™στˆν ¹ ΑΒ τÁς ΒΓ. æς δ ¹ ΑΒ πρÕς τ¾ν ΒΓ, οÛτως τÕ ¢πÕ τÁς ΑΒ πρÕς τÕ ¢πÕ τÁς ΒΕ· διπλάσιον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς ΒΕ. œστι δ κሠ¹ τÁς σφαίρας διάµετρος δυνάµει διπλασίων τÁς τοà Ñκταέδρου πλευρ©ς. κሠ™στιν ¹ ΑΒ ¹ τÁς δοθείσης σφαίρας διάµετρος· ¹ ΒΕ ¥ρα τοà Ñκταέδρου ™στˆ πλευρά. ”Ηχθω δ¾ ¢πÕ τοà Α σηµείου τÍ ΑΒ εÙθείv πρÕς Ñρθ¦ς ¹ ΑΗ, κሠκείσθω ¹ ΑΗ ‡ση τÍ ΑΒ, κሠ™πεζεύχθω ¹ ΗΓ, κሠ¢πÕ τοà Θ ™πˆ τ¾ν ΑΒ κάθετος ½χθω ¹ ΘΚ. κሠ™πεˆ διπλÁ ™στιν ¹ ΗΑ τÁς ΑΓ· ‡ση γ¦ρ ¹ ΗΑ τÍ ΑΒ· æς δ ¹ ΗΑ πρÕς τ¾ν ΑΓ, οÛτως ¹ ΘΚ πρÕς τ¾ν ΚΓ, διπλÁ ¥ρα κሠ¹ ΘΚ τÁς ΚΓ. τετραπλάσιον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΘΚ τοà ¢πÕ τÁς ΚΓ· τ¦ ¥ρα ¢πÕ τîν ΘΚ, ΚΓ, Óπερ ™στˆ τÕ ¢πÕ τÁς ΘΓ, πενταπλάσιόν ™στι τοà ¢πÕ τÁς ΚΓ. ‡ση δ ¹ ΘΓ τÍ ΓΒ· πενταπλάσιον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΒΓ τοà ¢πÕ τÁς ΓΚ. κሠ™πεˆ διπλÁ ™στιν ¹ ΑΒ τÁς ΓΒ, ïν ¹ Α∆ τÁς ∆Β ™στι διπλÁ, λοιπ¾ ¥ρα ¹ Β∆ λοιπÁς τÁς ∆Γ ™στι διπλÁ. τριπλÁ ¥ρα ¹ ΒΓ τÁς Γ∆· ™νναπλάσιον ¥ρα τÕ ¢πÕ τÁς ΒΓ τοà ¢πÕ τÁς Γ∆. πενταπλάσιον δ τÕ ¢πÕ τÁς ΒΓ τοà ¢πÕ τÁς ΓΚ· µε‹ζον ¥ρα τÕ ¢πÕ τÁς ΓΚ τοà ¢πÕ τÁς Γ∆. µείζων ¥ρα ™στˆν ¹ ΓΚ τÁς Γ∆. κείσθω τÍ ΓΚ ‡ση ¹ ΓΛ, κሠ¢πÕ τοà Λ τÍ ΑΒ πρÕς Ñρθ¦ς ½χθω ¹ ΛΜ, κሠ™πεζεύχθω ¹ ΜΒ. κሠ™πεˆ πενταπλάσιόν ™στι τÕ ¢πÕ τÁς ΒΓ τοà ¢πÕ τÁς ΓΚ, καί ™στι τÁς µν ΒΓ διπλÁ ¹ ΑΒ, τÁς δ ΓΚ διπλÁ ¹ ΚΛ, πενταπλάσιον ¥ρα ™στˆ τÕ ¢πÕ τÁς ΑΒ τοà ¢πÕ τÁς ΚΛ. œστι δ κሠ¹ τÁς σφαίρας διάµετρος δυνάµει πενταπλασίων τÁς ™κ τοà κέντρου τοà κύκλου, ¢φ' οá τÕ ε„κοσάεδρον ¢ναγέγραπται. καί ™στιν ¹ ΑΒ ¹ τÁς σφαίρας διάµετρος· ¹ ΚΛ ¥ρα ™κ τοà κέντρου ™στˆ τοà κύκλου, ¢φ' οá τÕ ε„κοσάεδρον ¢ναγέγραπται· ¹ ΚΛ ¥ρα ˜ξαγώνου ™στˆ πλευρ¦ τοà ε„ρηµένου κύκλου. κሠ™πεˆ ¹ τÁς σφαίρας διάµετρος σύγκειται œκ τε τÁς τοà ˜ξαγώνου κሠδύο τîν τοà δεκαγώνου τîν ε„ς τÕν ε„ρηµένον κύκλον ™γγραφοµένων, καί ™στιν ¹ µν ΑΒ ¹ τÁς σφαίρας διάµετρος, ¹ δ ΚΛ

times AD. And as BA (is) to AD, so the (square) on BA (is) to the (square) on AF [Def. 5.9]. For triangle AF B is equiangular to triangle AF D [Prop. 6.8]. Thus, the (square) on BA is one and a half times the (square) on AF . And the square on the diameter of the sphere is also one and a half times the (square) on the side of the pyramid [Prop. 13.13]. And AB is the diameter of the sphere. Thus, AF is equal to the side of the pyramid. Again, since AD is double DB, AB is thus triple BD. And as AB (is) to BD, so the (square) on AB (is) to the (square) on BF [Prop. 6.8, Def. 5.9]. Thus, the (square) on AB is three times the (square) on BF . And the square on the diameter of the sphere is also three times the (square) on the side of the cube [Prop. 13.15]. And AB is the diameter of the sphere. Thus, BF is the side of the cube. And since AC is equal to CB, AB is thus double BC. And as AB (is) to BC, so the (square) on AB (is) to the (square) on BE [Prop. 6.8, Def. 5.9]. Thus, the (square) on AB is double the (square) on BE. And the square on the diameter of the sphere is also double the (square) on the side of the octagon [Prop. 13.14]. And AB is the diameter of the given sphere. Thus, BE is the side of the octagon. So let AG have been drawn from point A at rightangles to the straight-line AB. And let AG be made equal to AB. And let GC have been joined. And let HK have been drawn from H, perpendicular to AB. And since GA is double AC. For GA (is) equal to AB. And as GA (is) to AC, so HK (is) to KC [Prop. 6.4]. HK (is) thus also double KC. Thus, the (square) on HK is four times the (square) on KC. Thus, the (sum of the squares) on HK and KC, which is the (square) on HC [Prop. 1.47], is five times the (square) on KC. And HC (is) equal to CB. Thus, the (square) on BC (is) five times the (square) on CK. And since AB is double CB, of which AD is double DB, the remainder BD is thus double the remainder DC. BC (is) thus triple CD. The (square) on BC (is) thus nine times the (square) on CD. And the (square) on BC (is) five times the (square) on CK. Thus, the (square) on CK (is) greater than the (square) on CD. CK is thus greater than CD. Let CL be made equal to CK. And let LM have been drawn from L at right-angles to AB. And let M B have been joined. And since the (square) on BC is five times the (square) on CK, and AB is double BC, and KL double CK, the (square) on AB is thus five times the (square) on KL. And the square on the diameter of the sphere is also five times the (square) on the radius of the circle from which the icosahedron has been described [Prop. 13.16 corr.]. And AB is the diameter of the sphere. Thus, KL is the radius of the circle from

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ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

˜ξαγώνου πλευρά, κሠ‡ση ¹ ΑΚ τÍ ΛΒ, ˜κατέρα ¥ρα τîν ΑΚ, ΛΒ δεκαγώνου ™στˆ πλευρ¦ τοà ™γγραφοµένου ε„ς τÕν κύκλον, ¢φ' οá τÕ ε„κοσάεδρον ¢ναγέγραπται. κሠ™πεˆ δεκαγώνου µν ¹ ΛΒ, ˜ξαγώνου δ ¹ ΜΛ· ‡ση γάρ ™στι τÍ ΚΛ, ™πεˆ κሠτÍ ΘΚ· ‡σον γ¦ρ ¢πέχουσιν ¢πÕ τοà κέντρου· καί ™στιν ˜κατέρα τîν ΘΚ, ΚΛ διπλασίων τÁς ΚΓ· πενταγώνου ¥ρα ™στˆν ¹ ΜΒ. ¹ δ τοà πενταγώνου ™στˆν ¹ τοà ε„κοσαέδρου· ε„κοσαέδρου ¥ρα ™στˆν ¹ ΜΒ. Κሠ™πεˆ ¹ ΖΒ κύβου ™στˆ πλευρά, τετµήσθω ¥κρον κሠµέσον λόγον κατ¦ τÕ Ν, κሠœστω µε‹ζον τµÁµα τÕ ΝΒ· ¹ ΝΒ ¥ρα δωδεκαέδρου ™στˆ πλευρά. Κሠ™πεˆ ¹ τÁς σφαίρας διάµετρος ™δείχθη τÁς µν ΑΖ πλευρ©ς τÁς πυραµίδος δυνάµει ¹µιολία, τÁς δ τοà Ñκταέδρου τÁς ΒΕ δυνάµει διπλασίων, τÁς δ τοà κύβου τÁς ΖΒ δυνάµει τριπλασίων, ο†ων ¥ρα ¹ τÁς σφαίρας διάµετρος δυνάµει ›ξ, τοιούτων ¹ µν τÁς πυραµίδος τεσσάρων, ¹ δ τοà Ñκταέδρου τριîν, ¹ δ τοà κύβου δύο. ¹ µν ¥ρα τÁς πυραµίδος πλευρ¦ τÁς µν τοà Ñκταέδρου πλευρ©ς δυνάµει ™στˆν ™πίτριτος, τÁς δ τοà κύβου δυνάµει διπλÁ, ¹ δ τοà Ñκταέδρου τÁς τοà κύβου δυνάµει ¹µιολία. αƒ µν οâν ε„ρηµέναι τîν τριîν σχηµάτων πλευραί, λέγω δ¾ πυραµίδος κሠÑκταέδρου κሠκύβου, πρÕς ¢λλήλας ε„σˆν ™ν λόγοις ·ητο‹ς. αƒ δ λοιπሠδύο, λέγω δ¾ ¼ τε τοà ε„κοσαέδρου κሠ¹ τοà δωδεκαέδρου, οÜτε πρÕς ¢λλήλας οÜτε πρÕς τ¦ς προειρηµένας ε„σˆν ™ν λόγοις ·ητο‹ς· ¥λογοι γάρ ε„σιν, ¹ µν ™λάττων, ¹ δ ¢ποτοµή. “Οτι µείζων ™στˆν ¹ τοà ε„κοσαέδρου πλευρ¦ ¹ ΜΒ τÁς τοà δωδεκαέδρου τÁς ΝΒ, δείξοµεν οÛτως. 'Επεˆ γ¦ρ „σογώνιόν ™στι τÕ Ζ∆Β τρίγωνον τù ΖΑΒ τριγώνJ, ¢νάλογόν ™στιν æς ¹ ∆Β πρÕς τ¾ν ΒΖ, οÛτως ¹ ΒΖ πρÕς τ¾ν ΒΑ. κሠ™πεˆ τρε‹ς εÙθε‹αι ¢νάλογόν ε„σιν, œστιν æς ¹ πρώτη πρÕς τ¾ν τρίτην, οÛτως τÕ ¢πÕ τÁς πρώτης πρÕς τÕ ¢πÕ τÁς δευτέρας· œστιν ¥ρα æς ¹ ∆Β πρÕς τ¾ν ΒΑ, οÛτως τÕ ¢πÕ τÁς ∆Β πρÕς τÕ ¢πÕ τÁς ΒΖ· ¢νάπαλιν ¥ρα æς ¹ ΑΒ πρÕς τ¾ν Β∆, οÛτως τÕ ¢πÕ τÁς ΖΒ πρÕς τÕ ¢πÕ τÁς Β∆. τριπλÁ δ ¹ ΑΒ τÁς Β∆· τριπλάσιον ¥ρα τÕ ¢πÕ τÁς ΖΒ τοà ¢πÕ τÁς Β∆. œστι δ κሠτÕ ¢πÕ τÁς Α∆ τοà ¢πÕ τÁς ∆Β τετραπλάσιον· διπλÁ γ¦ρ ¹ Α∆ τÁς ∆Β· µε‹ζον ¥ρα τÕ ¢πÕ τÁς Α∆ τοà ¢πÕ τÁς ΖΒ· µείζων ¥ρα ¹ Α∆ τÁς ΖΒ· πολλù ¥ρα ¹ ΑΛ τÁς ΖΒ µείζων ™στίν. κሠτÁς µν ΑΛ ¥κρον κሠµέσον λόγον τεµνοµένης τÕ µε‹ζον τµÁµά ™στιν ¹ ΚΛ, ™πειδήπερ ¹ µν ΛΚ ˜ξαγώνου ™στίν, ¹ δ ΚΑ δεκαγώνου· τÁς δ ΖΒ ¥κρον κሠµέσον λόγον τεµνοµένης τÕ µε‹ζον τµÁµά ™στιν ¹ ΝΒ· µείζων ¥ρα ¹ ΚΛ τÁς ΝΒ. ‡ση δ ¹ ΚΛ τÍ ΛΜ· µείζων ¥ρα ¹ ΛΜ τÁς ΝΒ [τÁς δ ΛΜ µείζων ™στˆν ¹ ΜΒ]. πολλù ¥ρα ¹ ΜΒ πλευρ¦ οâσα τοà ε„κοσαέδρου µείζων ™στˆ τÁς ΝΒ πλευρ©ς οÜσης τοà δωδεκαέδρου· Óπερ œδει δε‹ξαι.

which the icosahedron has been described. Thus, KL is (the side) of the hexagon (inscribed) in the aforementioned circle [Prop. 4.15 corr.]. And since the diameter of the sphere is composed of (the side) of the hexagon, and two of (the sides) of the decagon, inscribed in the aforementioned circle, and AB is the diameter of the sphere, and KL the side of the hexagon, and AK (is) equal to LB, thus AK and LB are each sides of the decagon inscribed in the circle from which the icosahedron has been described. And since LB is (the side) of the decagon. And M L (is the side) of the hexagon—for (it is) equal to KL, since (it is) also (equal) to HK, for they are equally far from the center. And HK and KL are each double KC. M B is thus (the side) of the pentagon (inscribed in the circle) [Props. 13.10, 1.47]. And (the side) of the pentagon is (the side) of the icosahedron [Prop. 13.16]. Thus, M B is (the side) of the icosahedron. And since F B is the side of the cube, let it have been cut in extreme and mean ratio at N , and let N B be the greater piece. Thus, N B is the side of the dodecahedron [Prop. 13.17 corr.]. And since the (square) on the diameter of the sphere was shown (to be) one and a half times the square on the side, AF , of the pyramid, and twice the square on (the side), BE, of the octagon, and three times the square on (the side), F B, of the cube, thus, of whatever (parts) the (square) on the diameter of the sphere (makes) six, of such (parts) the (square) on (the side) of the pyramid (makes) four, and (the square) on (the side) of the octagon three, and (the square) on (the side) of the cube two. Thus, the (square) on the side of the pyramid is one and a third times the square on the side of the octagon, and double the square on (the side) of the cube. And the (square) on (the side) of the octahedron is one and a half times the square on (the side) of the cube. Therefore, the aforementioned sides of the three figures—I mean, of the pyramid, and of the octahedron, and of the cube— are in rational ratios to one another. And (the sides of) the remaining two (figures)—I mean, of the icosahedron, and of the dodecahedron—are neither in rational ratios to one another, nor to the (sides) of the aforementioned (three figures). For they are irrational (straightlines): (namely), a minor [Prop. 13.16], and an apotome [Prop. 13.17]. (And), we can show that the side, M B, of the icosahedron is greater that the (side), N B, or the dodecahedron, as follows. For, since triangle F DB is equiangular to triangle F AB [Prop. 6.8], proportionally, as DB is to BF , so BF (is) to BA [Prop. 6.4]. And since three straight-lines are (continually) proportional, as the first (is) to the third,

536

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13 so the (square) on the first (is) to the (square) on the second [Def. 5.9, Prop. 6.20 corr.]. Thus, as DB is to BA, so the (square) on DB (is) to the (square) on BF . Thus, inversely, as AB (is) to BD, so the (square) on F B (is) to the (square) on BD. And AB (is) triple BD. Thus, the (square) on F B (is) three times the (square) on BD. And the (square) on AD is also four times the (square) on DB. For AD (is) double DB. Thus, the (square) on AD (is) greater than the (square) on F B. Thus, AD (is) greater than F D. Thus, AL is much greater than F B. And KL is the greater piece of AL, which is cut in extreme and mean ratio—inasmuch as LK is (the side) of the hexagon, and KA (the side) of the decagon [Prop. 13.9]. And N B is the greater piece of F B, which is cut in extreme and mean ratio. Thus, KL (is) greater than N B. And KL (is) equal to LM . Thus, LM (is) greater than N B [and M B is greater than LM ]. Thus, M B, which is (the side) of the icosahedron, is much greater than N B, which is (the side) of the dodecahedron. (Which is) the very thing it was required to show.

† If the radius of the given sphere is unity, then the sides of the pyramid (i.e.,q tetrahedron), octahedron, cube, icosahedron, and dodecahedron, p p √ √ √ √ √ respectively, satisfy the following inequality: 8/3 > 2 > 4/3 > (1/ 5) 10 − 2 5) > (1/3) ( 15 − 3).

Λέγω δή, Óτι παρ¦ τ¦ ε„ρηµένα πέντε σχήµατα οÙ συσταθήσεται ›τερον σχÁµα περιεχόµενον ØπÕ „σοπλεύρων τε κሠ„σογωνίων ‡σων ¢λλήλοις. `ΥπÕ µν γ¦ρ δύο τριγώνων À Óλως ™πιπέδων στερε¦ γωνία οÙ συνίσταται. ØπÕ δ τριîν τριγώνων ¹ τÁς πυραµίδος, ØπÕ δ τεσσάρων ¹ τοà Ñκταέδρου, ØπÕ δ πέντε ¹ τοà ε„κοσαέδρου· ØπÕ δ žξ τριγώνων „σοπλεύρων τε κሠ„σογωνίων πρÕς ˜νˆ σηµείJ συνισταµένων οÙκ œσται στερε¦ γωνία· οÛσης γ¦ρ τÁς τοà „σοπλεύρου τριγώνου γωνίας διµοίρου ÑρθÁς œσονται αƒ žξ τέσσαρσιν Ñρθα‹ς ‡σαι· Óπερ ¢δύνατον· ¤πασα γ¦ρ στερε¦ γωνία ØπÕ ™λασσόνων À τεσσάρων Ñρθîν περέχεται. δι¦ τ¦ αÙτ¦ δ¾ οÙδ ØπÕ πλειόνων À žξ γωνιîν ™πιπέδων στερε¦ γωνία συνίσταται. ØπÕ δ τετραγώνων τριîν ¹ τοà κύβου γωνία περιέχεται· ØπÕ δ τεσσάρων ¢δύνατον· œσονται γ¦ρ πάλιν τέσσαρες Ñρθαί. ØπÕ δ πενταγώνων „σοπλεύρων κሠ„σογωνίων, ØπÕ µν τριîν ¹ τοà δωδεκαέδρου· ØπÕ δ τεσσάρων ¢δύνατον· οÜσης γ¦ρ τÁς τοà πενταγώνου „σοπλεύρου γωνίας ÑρθÁς κሠπέµπτου, œσονται αƒ τέσσαρες γωνίαι τεσσάρων Ñρθîν µείζους· Óπερ ¢δύνατον. οÙδ µ¾ν ØπÕ πολυγώνων ˜τέρων σχηµάτων περισχεθήσεται στερε¦ γωνία δι¦ τÕ αÙτÕ ¥τοπον. ΟÙκ ¥ρα παρ¦ τ¦ ε„ρηµένα πέντε σχήµατα ›τερον σχÁµα στερεÕν συσταθήσεται ØπÕ „σοπλεύρων τε καˆ

So, I say that, beside the five aforementioned figures, no other (solid) figure can be constructed (which is) contained by equilateral and equiangular (planes), equal to one another. For a solid angle cannot be constructed from two triangles, or indeed (two) planes (of any sort) [Def. 11.11]. And (the solid angle) of the pyramid (is constructed) from three (equiangular) triangles, and (that) of the octahedron from four (triangles), and (that) of the icosahedron from (five) triangles. And a solid angle cannot be (made) from six equilateral and equiangular triangles set up together at one point. For, since the angles of a equilateral triangle are (each) two-thirds of a right-angle, the (sum of the) six (plane) angles (containing the solid angle) will be four right-angles. The very thing (is) impossible. For every solid angle is contained by (plane angles whose sum is) less than four right-angles [Prop. 11.21]. So, for the same (reasons), a solid angle cannot be constructed from more than six plane angles (equal to twothirds of a right-angle) either. And the (solid) angle of a cube is contained by three squares. And (a solid angle contained) by four (squares is) impossible. For, again, the (sum of the plane angles containing the solid angle) will be four right-angles. And (the solid angle) of a dodecahedron (is contained) by three equilateral and equian-

537

ΣΤΟΙΧΕΙΩΝ ιγ΄.

ELEMENTS BOOK 13

„σογωνίων περιεχόµενον· Óπερ œδει δε‹ξαι.

gular pentagons. And (a solid angle contained) by four (equiangular pentagons is) impossible. For, the the angle of an equilateral pentagon being one and one-fifth of right-angle, four (such) angles will be greater (in sum) than four right-angles. The very thing (is) impossible. And, on account of the same absurdity, a solid angle cannot be constructed from any other (equiangular) polygonal figures either. Thus, beside the five aforementioned figures, no other solid figure can be constructed (which is) contained by equilateral and equiangular (planes). (Which is) the very thing it was required to show.

A E

A

B

Z D

B

E F

G

D

C

ΛÁµµα.

Lemma

“Οτι δ ¹ τοà „σοπλεύρου κሠ„σογωνίου πενταγώνου γωνία Ñρθή ™στι κሠπέµπτου, οÛτω δεικτέον. ”Εστω γ¦ρ πεντάγωνον „σόπλευρον κሠ„σογώνιον τÕ ΑΒΓ∆Ε, κሠπεριγεγράφθω περˆ αÙτÕ κύκλος Ð ΑΒΓ∆Ε, κሠε„λήφθω αÙτοà τÕ κέντρον τÕ Ζ, κሠ™πεζεύχθωσαν αƒ ΖΑ, ΖΒ, ΖΓ, Ζ∆, ΖΕ. δίχα ¥ρα τέµνουσι τ¦ς πρÕς το‹ς Α, Β, Γ, ∆, Ε τοà πενταγώνου γωνίας. κሠ™πεˆ αƒ πρÕς τù Ζ πέντε γωνίαι τέσσαρσιν Ñρθα‹ς ‡σαι ε„σˆ καί ε„σιν ‡σαι, µία ¥ρα αÙτîν, æς ¹ ØπÕ ΑΖΒ, µι©ς ÑρθÁς ™στι παρ¦ πέµπτον· λοιπሠ¥ρα αƒ ØπÕ ΖΑΒ, ΑΒΖ µι©ς ε„σιν ÑρθÁς κሠπέµπτου. ‡ση δ ¹ ØπÕ ΖΑΒ τÍ ØπÕ ΖΒΓ· κሠÓλη ¥ρα ¹ ØπÕ ΑΒΓ τοà πενταγώνου γωνία µι©ς ™στιν ÑρθÁς κሠπέµπτου· Óπερ œδει δε‹ξαι.

It can be shown that the angle of an equilateral and equiangular pentagon is one and one-fifth of a rightangle, as follows. For let ABCDE be an equilateral and equiangular pentagon, and let the circle ABCDE have been circumscribed about it [Prop. 4.14]. And let its center, F , have been found [Prop. 3.1]. And let F A, F B, F C, F D, and F E have been joined. Thus, they cut the angles of the pentagon in half at (points) A, B, C, D, and E [Prop. 1.4]. And since the five angles at F are equal (in sum) to four right-angles, and are also equal (to one another), (any) one of them, like AF B, is thus one less a fifth of a right-angle. Thus, the remaining (angles), F AB and ABF , (in triangle ABF ) are one plus a fifth of a right-angle [Prop. 1.32]. And F AB (is) equal to F BC. Thus, the whole angle, ABC, of the pentagon is also one and one-fifth of a right-angle. (Which is) the very thing it was required to show.

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GREEK-ENGLISH LEXICON

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ΣΤΟΙΧΕΙΩΝ

GREEK–ENGLISH LEXICON

ABBREVIATIONS: act - active; adj - adjective; adv - adverb; conj - conjunction; fut - future; gen - genitive; imperat - imperative; impf - imperfect; ind - indeclinable; indic - indicative; intr - intransitive; mid - middle; neut - neuter; no - noun; par - particle; part - participle; pass - passive; perf - perfect; pre - preposition; pres - present; pro - pronoun; sg - singular; tr - transitive; vb verb.

¢ποτοµή, ¹ : vb, piece cut off, apotome. ¤πτω, ¤ψω, Âψα, —, µµαι, — : vb, touch, join, meet. ¢πώτερος -α -ον : adj, further off. ¥ρα : par, thus, as it seems (inferential). ¢ριθµός, Ð : no, number. ¢ρτιάκις : adv, an even number of times.

¥γω, ¥ξω, ½γαγον, -Ãχα, Ãγµαι, ½χθην : vb, lead, draw (a line). ¢δύνατος -ον : adj, impossible. ¢εί : adv, always, for ever. αƒρέω, αƒρήσω, ε[Œ]λον, Èρηκα, Èρηµαι, Åρέθην : vb, grasp.

αρτιόπλευρος -ον : adj, having a even number of sides. ¥ρχω, ¥ρξω, Ãρξα, Ãρχα, Ãργµαι, Ãρχθην : vb, rule; mid., begin. ¢σύµµετρος -ον : adj, incommensurable. ¢σύµπτωτος -ον : adj, not touching, not meeting.

¢ιτέω, α„τήσω, ½τησα, Éτηκα, Éτηµαι, Æτήθη : vb, postulate.

¥ρτιος -α -ον : adj, even, perfect.

α‡τηµα -ατος, τό : no, postulate. ¢κόλουθος -ον : adj, analogous, consequent on, in conformity with. ¥κρος -α -ον : adj, outermost, end, extreme. ¢λλά : conj, but, otherwise. ¥λογος -ον : adj, irrational.

¥τµητος -ον : adj, uncut. ¢τόπος -ον : adj, absurd, paradoxical. αÙτόθεν : adv, immediately, obviously. ¢φαίρεω : vb, take from, subtract from, cut off from; see αƒρέω. ¡φή, ¹ : no, point of contact.

¤µα : adv, at once, at the same time, together.

βάθος -εος, τό : no, depth, height.

¢µβλυγώνιος -ον : adj, obtuse-angled; τÕ ¢µβλυγώνιον, no, obtuse angle. ¢µβλύς -ε‹α -ύ : adj, obtuse.

βαίνω, -βήσοµαι, -έβην, βέβηκα, —, — : vb, walk; perf, stand (of angle). βάλλω, βαλî, œβαλον, βέβληκα, βέβληµαι, ™βλήθην : vb, throw.

¢µφότερος -α -ον : pro, both (of two).

βάσις -εως, ¹ : no, base (of a triangle).

¢ναγράφω : vb, describe (a figure); see γράφω.

γάρ : conj, for (explanatory).

¢ναλογία, ¹ : no, proportion, (geometric) progression.

γί[γ]νοµαι, γενήσοµαι, ™γενόµην, γέγονα, γεγένηµαι, — : vb, happen, become. γνώµων -ονος, ¹ : no, gnomon.

¢νάλογος -ον : adj, proportional. ¢νάπαλιν : adv, inverse(ly). αναπληρόω : vb, fill up.

γραµµή, ¹ : no, line.

¢ναστρέφω : vb, turn upside down, convert (ratio); see στρέφω. ¢ναστροφή, ¹ : no, turning upside down, conversion (of ratio). ¢νθυφαιρέω : vb, take away in turn; see αƒρέω.

γράφω, γράψω, œγρα[ψ/φ]α, γέγραφα, γέγραµµαι, ™ραψάµην : vb, draw (a figure). γωνία, ¹ : no, angle. δε‹ : vb, be necessary; δε‹, it is necessary; œδει, it was necassary; δέον, being necessary. δείκνυµι, δείξω, œδειξα, δέδειχα, δέδειγµαι, ™δείχθην : vb, show, demonstrate. δεικτέον : ind, one must show. δε‹ξις -εως, ¹ : no, proof.

¢νίστηµι : vb, set up; see ‡στηµι. ¥νισος -ον : adj, unequal, uneven. ¢ντιπάσχω : vb, be reciprocally proportional; see πάσχω. ¥ξων -ονος, Ð : vb, axis. ¤παξ : adv, once. ¤πας, ¤πασα, ¤παν : adj, quite all, the whole. ¥πειρος -ον : adj, infinite.

δείχνàµι, δείξω, œδειξα, δέδειχα, δέδειγµαι, ™δείχθην : vb, show, demonstrate. δεκαγώνος -ον : adj, ten-sided; τÕ δεκαγώνον, no, decagon.

¢πεναντίον : ind, opposite. ¢πέχω : vb, be far from, be away from; see œχω. ¢πλατής -ές : adj, without breadth.

δέχοµαι, δέξοµαι, ™δεξάµην, —, δέδεγµαι, ™δέχθην : vb, receive, accept. δή : conj, so (explanatory).

¢πόδειξις -εως, ¹ : no, proof.

δηλαδή : ind, quite clear, manifest.

¢ποκαθίστηµι : vb, re-establish, restore; see ‡στηµι.

δÁλος -η -ον : adj, clear.

¢πολαµβάνω : vb, take from, subtract from, cut off from; see λαµβάνω.

διάγω : vb, carry over, draw through, draw across; see ¥γω.

¢πότµηµα -ατος, τÕ : no, piece cut off, segment.

διαγώνιος -ον : adj, diagonal.

δηλονότι : adv, manifestly.

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ΣΤΟΙΧΕΙΩΝ

GREEK–ENGLISH LEXICON

διαλείπω : vb, leave an interval between. διάµετρος -ον : adj, diametrical; ¹ διάµετρος, no, diameter, diagonal. διαίρεσις -εως, ¹ : no, division, separation. διαιρέω : vb, divide (in two); διαρεθέντος -η -ον, adj, separated (ratio); see αƒρέω. διάστηµα -ατος, τό : no, radius.

™ναρµόζω : vb, insert; perf indic pass 3rd sg, ™νήρµοσται. ™νδέχοµαι : vb, admit, allow. ›νεκεν : ind, on account of, for the sake of. ™νναπλάσιος -α -ον : adj, nine-fold, nine-times. œννοια, ¹ : no, notion. ενπεριέχω : vb, encompass. ™νπίπτω : see ™µπίπτω.

διαφέρω : vb, differ; see φέρω.

™ντός : pre + gen, inside, interior, within, internal.

δίδωµι, δώσω, œδωκα, δέδωκα, δέδοµαι, ™δόθην : vb, give.

˜ξάγωνος -ον : adj, hexagonal; τÕ ˜ξάγωνον, no, hexagon.

διµοίρος -ον : adj, two-thirds.

˜ξαπλάσιος -α -ον : adj, sixfold.

διπλασιάζω : vb, double. διπλάσιος -α -ον : adj, double, twofold. διπλασίων -ον : adj, double, twofold. διπλοàς -Á -οàν : adj, double.

˜ξÁς : adv, in order, successively, consecutively. œξωθεν : adv, outside, extrinsic. ™πάνω : adv, above. ™παφή, ¹ : no, point of contact.

δίς : adv, twice. δίχα : adv, in two, in half.

™πεί : conj, since (causal). ™πειδήπερ : ind, inasmuch as, seeing that.

διχοροµία, ¹ : no, point of bisection. δυάς -άδος, ¹ : no, the number two, dyad. δύναµαι : vb, be able, be capable, generate, square, be when squared; δυναµένη, ¹, no, square-root (of area)—i.e., straight-line whose square is equal to a given area. δύναµις -εως, ¹ : no, power (usually 2nd power when used in mathematical sence, hence), square. δυνατός -ή -όν : adj, possible. δωδεκάεδρος -ον : adj, twelve-sided.

™πιζεύγνàµι, ™πιζεύξω, ™πέζευξα, —, ™πέζευγµαι, ™πέζεύχθην : vb, join (by a line). ™πιλογίζοµαι : vb, conclude. ™πινοέω : vb, think of, contrive. ™πιπέδος -ον : adj, level, flat, plane; τÕ ™πιπέδον, no, plane. ™πισκέπτοµαι : vb, investigate. ™πίσκεψις -εως, ¹ : no, inspection, investigation.

˜αυτοà -Áς -οà : adj, of him/her/it/self, his/her/its/own.

™πιτάσσω : vb, put upon, enjoin; τÕ ™πιταχθέν, no, the (thing) prescribed; see τάσσω.

™γγίων -ον : adj, nearer, nearest.

™πίτριτος -ον : adj, one and a third times.

™γγράφω : vb, inscribe; see γράφω.

™πιφάνεια, ¹ : no, surface.

εδος -εος, τό : no, figure, form, shape. ε„κοσάεδρος -ον : adj, twenty-sided.

œποµαι : vb, follow.

ε‡ρω/λέγω, ™ρî/ερέω, επον, ε‡ρηκα, ε‡ρηµαι, ™ρρήθην : vb, say, speak; per pass part, ειρηµένος -η -ον, adj, said, aforementioned. ε‡τε . . . ε‡τε : ind, either . . . or. ›καστος -η -ον : pro, each, every one.

œσχατος -η -ον : adj, outermost, uttermost, last.

˜κατέρος -α -ον : pro, each (of two).

œρχοµαι, ™λεύσοµαι, Ãλθον, ™λήλυθα, —, — : vb, come, go. ™τερόµηκης -ες : adj, oblong; τÕ ™τερόµηκες, no, rectangle. ›τερος -α -ον : adj, other (of two). œτι : par, yet, still, besides. εÙθύγραµµος -ον : adj, rectilinear; τÕ εÙθύγραµµον, no, rectilinear figure.

™κβάλλω, ™κβαλî, ™κέβαλον, ™κβέβίωκα, ™κβέβληµαι, ™κβληθήν εÙθύς -ε‹α -ύ : adj, straight; ¹ εÙθε‹α, no, straight-line; ™π' : vb, produce (a line). εÙθε‹ας, in a straight-line, straight-on. ™κθέω : vb, set out. εØρίσκω, εØρήσκω, ηáρον, εÛρεκα, εÛρηµαι, εØρέθην : vb, find. œκκειµαι : vb, be set out, be taken; see κε‹µαι. ™φάπτω : vb, bind to; mid, touch; ¹ ˜φαπτοµένη, no, tangent; ™κτίθηµι : vb, set out; see τίθηµι. see ¤πτω. ™κτός : pre + gen, outside, external. ™φαρµόζω, ™φαρµόσω, ™φήρµοσα, ™φήµοκα, ™φήµοσµαι, ™φήµόσθην : vb, coincide; pass, be applied. ™λά[σσ/ττ]ων -ον : adj, less, lesser. ™φεξÁς : adv, in order, adjacent.

™λλείπω : vb, be less than, fall short of. ™µπίπτω : vb, meet (of lines), fall on; see πίπτω.

™φίστηµι : vb, set, stand, place upon; see ‡στηµι.

œµπροσθεν : adv, in front.

œχω, ›ξω, œσχον, œσχηκα, -έσχηµαι, — : vb, have.

™ναλλάξ : adv, alternate(ly).

¹γέοµαι, ¹γήσοµαι, ¹γησάµην, ¹γηµαι, —, ¹γήθην : vb, lead.

541

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GREEK–ENGLISH LEXICON

½δη : ind, already, now.

κορυφή, ¹ : no, top, summit, apex; κατ¦ κορυφήν, vertically opposite (of angles). κρίνω, κρινî, œκρ‹να, κέκρικα, κέκριµαι, ™κρίθην : vb, judge.

¼κω, ¼ξω, —, —, —, — : vb, have come, be present. ¹µικύκλιον, τό : no, semi-circle. ¹µιόλιος -α -ον : adj, containing one and a half, one and a half times. ¼µισυς -εια -υ : adj, half.

κύβος, Ð : no, cube. κύκλος, Ð : no, circle. κύλινδρος, Ð : no, cylinder.

½περ = ½ + περ : conj, than, than indeed.

κυρτός -ή -όν : adj, convex.

½τοι . . . ½ : par, surely, either . . . or; in fact, either . . . or.

κîνος, Ð : no, cone. λαµβάνω, λήψοµαι, œλαβον, ε‡ληφα ε‡ληµµαι, ™λήφθην : vb, take. λέγω : vb, say; pres pass part, λεγόµενος -η -ον, no, so-called; see œιρω.

θέσις -εως, ¹ : no, placing, setting, position. θεωρηµα -ατος, τό : no, theorem. ‡διος -α -ον : adj, one’s own. „σάκις : adv, the same number of times; „σάκις πολλαπλάσια, the same multiples, equal multiples. „σογώνιος -ον : adj, equiangular.

λείπω, λείψω, œλιπον, λέλοιπα, λέλειµµαι, ™λείφθην : vb, leave, leave behind. ληµµάτιον, τό : no, diminutive of λÁµµα.

„σόπλευρος -ον : adj, equilateral.

λÁµµα -ατος, τό : no, lemma.

„σοπληθής -ές : adj, equal in number.

λÁψις -εως, ¹ : no, taking, catching.

‡σος -η -ον : adj, equal; ™ξ ‡σου, equally, evenly.

λόγος, Ð : no, ratio, proportion, argument.

„σοσκελής -ές : adj, isosceles.

λοιπός -ή -όν : adj, remaining.

‡στηµι, στήσω, œστησα, —, —, ™σταθην : vb tr, stand (something).

µανθάνω, µαθήσοµαι, œµαθον, µεµάθηκα, —, — : vb, learn.

καθάπερ : ind, according as, just as. κάθετος -ον : adj, perpendicular.

µέσος -η -ον : adj, middle, mean, medial; ™κ δύο µέσων, bimedial. µεταλαµβάνω : vb, take up.

καθόλου : adv, on the whole, in general.

µεταξύ : adv, between.

καλέω : vb, call. κ¢κεινος = κሠ™κε‹νος . κ¥ν = κሠ¥ν : ind, even if, and if. καταγραφή, ¹ : no, diagram, figure.

µετέωρος -ον : adj, raised off the ground.

µέγεθος -εος, τό : no, magnitude, size. ‡στηµι, στήσω, œστην, ›στηκα, ›σταµαι, ™σταθην : vb intr, stand µείζων -ον : adj, greater. up (oneself); Note: perfect I have stood up can be taken µένω, µενî, œµεινα, µεµένηκα, —, — : vb, stay, remain. to mean present I am standing. µέρος -ους, τό : no, part, direction, side. „σοϋψής -ές : adj, of equal height.

µετρέω : vb, measure. µέτρον, τό : no, measure.

καταγράφω : vb, describe/draw, inscribe (a figure); see γράφω. κατακολουθέω : vb, follow after. καταλείπω : vb, leave behind; see λείπω; τ¦ καταλειπόµενα, no, remainder. κατάλληλος -ον : adj, in succession, in corresponding order.

µηδείς, µηδεµία, µηδέν : adj, not even one, (neut.) nothing. µηδέποτε : adv, never. µηδέτερος -α -ον : pro, neither (of two). µÁκος -εος, τό : no, length. µήν : par, truely, indeed.

καταµετρέω : vb, measure (exactly).

µονάς -άδος, ¹ : no, unit, unity.

καταντάω : vb, come to, arrive at. κατασκευάζω : vb, furnish, construct. κε‹µαι, κε‹σοµαι, —, —, —, — : vb, have been placed, lie, be made; see τίθηµι.

µοναχός -ή -όν : adj, unique.

κέντρον, τό : no, center. κλάω : vb, break off, inflect. κλίνω, κλίνω, œκλινα, κέκλικα, κέκλιµαι, ™κλίθην : vb, lean, incline. κλίσις -εως, ¹ : no, inclination, bending. κο‹λος -η -ον : adj, hollow, concave.

µοναχîς : adv, uniquely. µόνος -η -ον : adj, alone. νοέω, —, νόησα, νενόηκα, νενόηµαι, ™νοήθην : vb, apprehend, conceive. οŒος -α -ον : pre, such as, of what sort. Ñκτάεδρος -ον : adj, eight-sided. Óλος -η -ον : adj, whole. еογενής -ές : adj, of the same kind. Óµοιος -α -ον : adj, similar.

542

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GREEK–ENGLISH LEXICON

еοιοπληθής -ές : adj, similar in number.

παραλληλεπίπεδος, -ον : adj, with parallel surfaces; τÕ παραλληλεπίπεδον, no, parallelepiped.

еοιοταγής -ές : adj, similarly arranged.

παραλληλόγραµµος -ον : adj, bounded by parallel lines; τÕ παραλληλόγραµµον, no, parallelogram.

еοιότης -ητος, ¹ : no similarity. еοίως : adv, similarly. еοταγής -ές : adj, ranged in the same row or line.

παράλληλος -ον : adj, parallel; τÕ παράλληλον, no, parallel, parallel-line. παραπλήρωµα -ατος, τό : no, complement (of a parallelogram).

еόλογος -ον : adj, corresponding, homologous. еώνυµος -ον : adj, having the same name.

παρατέλυετος -ον : adj, penultimate.

Ôνοµα -ατος, τό : no, name; ™κ δύο Ñνοµάτων, binomial.

παρέκ : prep + gen, except.

Ñξυγώνιος -ον : adj, acute-angled; τÕ Ñξυγώνιον, no, acute angle.

παρεµπίπτω : vb, insert; see πίπτω.

Ñξύς -ε‹α -ύ : adj, acute. Ðποιοσοàν = Ðπο‹ος -α -ον + οâν : adj, of whatever kind, any kind whatsoever. Ðπόσος -η -ον : pro, as many, as many as.

πάσχω, πείσοµαι, œπαθον, πέπονθα, —, — : vb, suffer. πεντάγωνος -ον : adj, pentagonal; τÕ πεντάγωνον, no, pentagon. πενταπλάσιος -α -ον : adj, five-fold, five-times. πεντεκαιδεκάγωνον, τό : no, fifteen-sided figure.

Ðποσοσδηποτοàν = Ðπόσος -η -ον + δή + ποτέ + οâν : adj, of whatever number, any number whatsoever.

πεπερασµένος -η -ον : adj, finite, limited; see περαίνω.

Ðποσοσοàν = Ðπόσος -η -ον + οâν : adj, of whatever number, any number whatsoever.

περαίνω, περανî, ™πέρανα, —, πεπέρανµαι, ™περανάνθην : vb, bring to end, finish, complete; pass, be finite.

Ðπότερος -α -ον : pro, either (of two), which (of two).

πέρας -ατος, τό : no, end, extremity.

Ñρθογώνιον, τό : no, rectangle, right-angle.

περατόω, —, —, —, —, — : vb, bring to an end.

Ñρθός -ή -όν : adj, straight, right-angled, perpendicular; πρÕς Ñρθ¦ς γωνίας, at right-angles. Ôρος, Ð : no, boundary, definition, term (of a ratio).

περιγράφω : vb, circumscribe; see γράφω. περιέχω : vb, encompass, surround, contain, comprise; see œχω.

Ðσαδηποτοàν = Óσα + δή + ποτέ + οâν : ind, any number whatsoever. Ðσάκις : ind, as many times as, as often as. Ðσαπλάσιος -ον : pro, as many times as.

περιλαµβάνω : vb, enclose; see λαµβάνω.

Óσος -η -ον : pro, as many as.

περιφέρεια, ¹ : no, circumference.

περιλείποµαι : vb, remain over, be left over. περισσάκις : adv, an odd number of times. περισσός -ή -όν : adj, odd.

Óσπερ, ¼περ, Óπερ : pro, the very man who, the very thing which. Óστις, ¼τις, Ó τι : pro, anyone who, anything which.

περιφέρω : vb, carry round; see φέρω. πηλικότης -ητος, ¹ : no, magnitude, size.

Óταν : adv, when, whenever. Ðτιοàν : ind, whatsoever. οÙδείς, οÙδεµία, οÙδέν : pro, not one, nothing.

πίπτω, πεσοàµαι, œπεσον, πέπτωκα, —, — : vb, fall.

οØδέτερος -α -ον : pro, not either.

πλευρά, ¹ : no, side.

οØθέτερος : see οØδέτερος.

πλÁθος -εος, τÕ : no, great number, multitude, number.

οÙθέν : ind, nothing.

πλήν : adv & prep + gen, more than.

οâν : adv, therefore, in fact. οÛτως : adv, thusly, in this case.

ποιός -ά -όν : adj, of a certain nature, kind, quality, type.

πάντως : adv, in all ways.

πολλαπλασιασµός, Ð : no, multiplication.

παρ¦ : prep + acc, parallel to.

πολλαπλάσιον, τό : no, multiple.

παραβάλλω : vb, apply (a figure); see βάλλω.

πολύεδρος -ον : adj, polyhedral; τό πολύεδρον, no, polyhedron. πολύγωνος -ον : adj, polygonal; τό πολύγωνον, no, polygon.

παραβολή, ¹ : no, application. παράκειµαι : vb, lie beside, apply (a figure); see κε‹µαι.

πλάτος -εος, τό : no, breadth, width. πλείων -ον : adj, more, several.

πολλαπλασιάζω : vb, multiply.

πολύπλευρος -ον : adj, multilateral. παραλλάσσω, παραλλάξω, —, παρήλλαχα, —, — : vb, miss, fall πόρισµα -ατος, τό : no, corollary. awry. ποτέ : ind, at some time.

543

ΣΤΟΙΧΕΙΩΝ

GREEK–ENGLISH LEXICON

πρ‹σµα -ατος, τÕ : no, prism.

συναποδείκνυµι : no, demonstrate together; see δείκνυµι.

προβαίνω : vb, step forward, advance.

συναφή, ¹ : no, point of junction.

προδείκνυµι : vb, show previously; see δείκνυµι.

σύνδυο, οƒ, αƒ, τά : no, two together, in pairs.

προεκτίθηµι : vb, set forth beforehand; see τίθηµι.

συνεχής -ές : adj, continuous; κατ¦ τÕ συνεχές, continuously.

προερέω : vb, say beforehand; perf pass part, προειρηµένος -η -ον, adj, aforementioned; see ε‡ρω.

σύνθεσις -εως, ¹ : no, putting together, composition.

προσαναπληρόω : vb, fill up, complete.

σύνθετος -ον : adj, composite. συ[ν]ίστηµι : vb, construct (a figure), set up together; perf imperat pass 3rd sg, συνεστάτω; see ‡στηµι.

προσαναγράφω : vb, complete (tracing of); see γράφω.

συντίθηµι : vb, put together, add together, compound (ratio); see τίθηµι.

προσαρµόζω : vb, fit to, attach to. προσεκβάλλω : vb, produce (a line); see ™κβάλλω. προσευρίσκω : vb, find besides, find; see εØρίσκω.

σχέσις -εως, ¹ : no, state, condition.

προσλαµβάµω : vb, add.

σχÁµα -ατος, τό : no, figure.

προκειµαι : vb, set before, prescribe.

σφα‹ρα -ας, ¹ : no, sphere.

πρόσκειµαι : vb, be laid on, have been added to; see κε‹µαι.

τάξις -εως, ¹ : no, arrangement, order.

προσπίπτω : vb, fall on, fall toward, meet; see πίπτω.

ταράσσω, ταράξω, —, —, τετάραγµαι, ™ταράχθην : vb, stir, trouble, disturbe; τεταραγµένος -η -ον, adj, disturbed, perturbed. τάσσω, τάξω, œταξα, τέταχα, τέταγµαι, ™τάχθην : vb, arrange, draw up.

προτασις -εως, ¹ : no, proposition. προστάσσω : vb, prescribe, enjoin; τÕ τροσταχθέν, no, the thing prescribed; see τάσσω. προστίθηµι : vb, add; see τίθηµι. πρότερος -α -ον : adj, first (comparative), before, former.

τέλειος -α -ον : adj, perfect.

προτίθηµι : vb, assign; see τίθηµι.

τέµνω, τεµνî, œτεµον, -τέτµηκα, τέτµηµαι, ™τµήθην : vb, cut; pres/fut indic act 3rd sg, τέµει.

προχωρέω : vb, go/come forward, advance.

τεταρτηµοριον, τÕ : no, quadrant.

πρîτος -α -ον : adj, first, prime.

τετράγωνος -ον : adj, square; τÕ τετράγωνον, no, square.

πυραµίς -ίδος, ¹ : no, pyramid.

τετράκις : adv, four times.

·ητός -ή -όν : adj, expressible, rational.

τετραπλάσιος -α -ον : adj, quadruple.

·οµβοειδής -ές : adj, rhomboidal; τÕ ·οµβοειδές, no, romboid. ·όµβος, Ð no, rhombus.

τετράπλευρος -ον : adj, quadrilateral.

σηµε‹ον, τό : no, point.

τετραπλόος -η -ον : adj, fourfold. τίθηµι, θήσω, œθηκα, τέθηκα, κε‹µαι, ™τέθην : vb, place, put. τµÁµα -ατος, τό : no, part cut off, piece, segment.

σκαληνός -ή -όν : adj, scalene.

τοίνυν : par, accordingly.

στερεός -ά -όν : adj, solid; τÕ στερεόν, no, solid, solid body.

τοιοàτος -αύτη -οàτο : pro, such as this.

στοιχε‹ον, τό : no, element.

τοµεύς -έως, Ð : no, sector (of circle).

στρέφω, -στρέψω, œστρεψα, —, ™σταµµαι, ™στάφην : vb, turn. σύγκειµαι : vb, lie together, be the sum of, be composed; συγκείµενος -η -ον, adj, composed (ratio), compounded; see κε‹µαι.

τοµή, ¹ : no, cutting, stump, piece. τόπος, Ð : no, place, space. τοσαυτάκις : adv, so many times.

σύγκρίνω : vb, compare; see κρίνω.

τοσαυταπλάσιος -α -ον : pro, so many times.

συµβαίνω : vb, come to pass, happen, follow; see βαίνω.

τοσοàτος -αύτη -οàτο : pro, so many.

συµβάλλω : vb, throw together, meet; see βάλλω.

τουτέστι = τοàτ' œστι : par, that is to say.

σύµµετρος -ον : adj, commensurable.

τραπέζιον, τό : no, trapezium.

σύµπας -αντος, Ð : no, sum, whole.

τρίγωνος -ον : adj, triangular; τÕ τρίγωνον, no, triangle.

συµπίπτω : vb, meet together (of lines); see πίπτω.

τριπλάσιος -α -ον : adj, triple, threefold.

συµπληρόω : vb, complete (a figure), fill in.

τρίπλευρος -ον : adj, trilateral.

συνάγω : vb, conclude, infer; see ¥γω.

τριπλ-όος -η -ον : adj, triple.

συναµφότεροι -αι -α : adj, both together; Ð συναµφότερος, no, sum (of two things).

τρόπος, Ð : no, way.

544

ΣΤΟΙΧΕΙΩΝ

GREEK–ENGLISH LEXICON

τυγχάνω, τεύξοµαι, œτυχον, τετύχηκα, τέτευγµαι, ™τεύχθην : vb, hit, happen to be at (a place). Øπάρχω : vb, begin, be, exist; see ¥ρχω. Øπεξαίρεσις -εως, ¹ : no, removal. Øπερβάλλω : vb, overshoot, exceed; see βάλλω. Øπεροχή, ¹ : no, excess, difference. Øπερέχω : vb, exceed; see œχω. Øπόθεσις -εως, ¹ : no, hypothesis. Øπόκειµαι : vb, underlie, be assumed (as hypothesis); see κε‹µαι. Øπολείπω : vb, leave remaining. Øποτείνω, Øποτενî, Øπέτεινα, Øποτέτακα, Øποτέταµαι, Øπετάθην : vb, subtend. Ûψος -εος, τό : no, height. φανερός -ά -όν : adj, visible, manifest. φ絈, φήσω, œφην, —, —, — : vb, say; œφαµεν, we said. φέρω, ο‡σω, ½νεγκον, ™νήνοχα, ™νήνεγµαι, ºνέχθην : vb, carry. χώριον, τό : no, place, spot, area, figure. χωρίς : pre + gen, apart from. ψαύω : vb, touch. æς : par, as, like, for instance. æς ›τυχεν : par, at random. æσαύτως : adv, in the same manner, just so. éστε : conj, so that (causal), hence.

545