SMSG (School Mathematics Study Group) Geometry: Teacher’s Commentary, Part I

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GEOMETRY

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School Mathematics Study Group

Geometry

Unit 15

Geometry Teacher's Commentary, Part I

Prepared under the supervision of the Panel on Sample Textbooks

of the School Mathematics Study Group: Frank B.Allen

Edwin C.Douglas Donald E. Richmond Charles E.Rckart Henry Swain Robert J. Walker

Lyons Township High School Taft School Williams College Yale University New Trier Township High School Cornell University

New Haven and London, Yale University Press

Copyright Q 1960, 1961 by Yale University. Printed in the United States of America. All rights reserved. This book may not be reproduced, in whole or in part, in any form, without written permission from the publishers; Financial support for the School Mathematics Study Group has been provided by the National Science Foundation.

Below are llated the names of a l l those who participated i n any of the w r i t i n g sessions

at which the following SMSO texts were prepared:

F i r ~ tCourse I n AlRebra, Geometry,

I n t e m e d i a t e Mathernatica, Eltmentarg Rurctiona, and Introduction to M a t r i x Algebra.

Alexander, Earlham College F.B. Allen, Lyons Township High School, La Grange, I l l i n o l a Alexander Beck, Olney tfigh School, Phllad t l p h l a , Penneylvania E . F . Benkenbach, University of C a l i i o r n l a at Loe h g e l e a 9 .O. Begle, School Mathematios Study Oroup, Y d e University Paul Berg, Stanford U n i v e r s i t y &I1 Berger, Monroe High School, S t . Paul, Minnesota Arthur Bernhart, University oi Oklahoma R.H. Bing, University of Wisconsin A.L. makers, U n i v e r s i t y of W e ~ t e r n H.W.

Australia Blank, New York University

A.A.

Shirley Boselly, Franklin H i g h School, Seattle, Washington K.E. Brown, Department of Health, Educat i o n , and Welfare, Washington, D.C. J.M. Calloway, Carleton College Hope Chipman, University High School, Ann Arbor, Wchigan R . R . C h r i s t i a n , University of Britiah Columbia R . J , Clark, St. Paulta School, Concord, New Hampshire P . H . Daus, Univeinity of California at La8 Angeles R.B. Davis, Syracuse U n i v e r s i t y Charlea IRPrima, C a l i f o r n i a 1nstitut;e of Technology Mary Dolcianl, Hunter College Edwin C . b u g l a a , The T a f t School, Watert o m , Connecticut Floyd D o m e , East High School, Denver, Cdlorado E.A. Dudley, North Haven Hlgh School, North Haven, Corneaticut

Lincoln h r s t , The Rice Institute Florence Elder, West Hempstead High School, West Hempatead, New York W.E. Ferguson, Newton High School, Newtonv l l l e , Maasachusetta N.J. Fine, University of Pennsylvania Joyce D. Fontalne, North Haven H i g h School, North Haven, C O M ~ C ~ ~ C U ~ P.L, Frledman, Haa6achusetts Institute of Technology Esther Q a a a e t t , Claramore High School, Claremore, Oklahoma R.K. Getoor, Univeraity or Washington V . H . Haag, Franklin and Marahall College R.R. Hartman, Edina-Morningaide Senior High School, Edina, Minnesota M.H. Heina, Unlveralty of I l l i n o i s Edwln Hewitt, University of Washlngton Martha H i l d e b ~ a n d , t Proviso Tomahip High School, Maywood, I l l i n o l a

.

Jurgenaen, Culver M i l i t a r y Academy, Culver, Indiana Joseph Lehner, Michigan S t a t e University Marguerite Lehr, Bryn Maw College Kenneth Ltlsenring, University o f Michigan Howard L e v l , Columbia Univeraity Eunice Lewis, Laboratorg High School, Unlveralty or Oklahoma M . A . L i n t o n , W i l l i a m Perm Charter School,

R.C

P h i l a d e l p h i a , Pennsylvania A.E. Livingston, University of Washington L.H. Looml3, Harvard University R .V. Lynch, Phillips Exeter Academy, F x e t e r , New Hampahire W.K. McNabb, Hockaday School, Dallas, Texas K.G. Michaels, North Haven High School, North Haven, Connecticut E.E. Moise, Univeraity of Michigan E.P. Northrop, Univeraity of Chicago O.J. Peteraon, Kansas S t a t e Teachera College, Bnporia, Kansas B.3. Pettis, University of North Carolina R . S . Pieters, P h i l l i p 8 Academy, Andover, Maasachuaett a H.O. Pollak, Bell Telephone Laboratorlee Walter Prenowitz, Brooklyn College O.B. P r i c e , University oP Kansas A.L. Putnam, Univeraity of Chicago P e r s i a 0. Redgrave, Norwich Free Academy, Norwlch, Connectlcut Mina Reea, Huntep College D.E. Richmond, Williams College C.E. Rickart, Yale Univeraity Harry Rudeman, Hunter College High School, New York C i t y J.T. Schwartz, New York U n i v e r e i t y 0 .E. Stanaitls, St. Olaf College Robert Starkey, Cubberley High Schovls, Palo Alto, C a l i f o r n i a P h i l l i p Stucky, Roosevelt High School, Seattle, Washington Henry Swain, New T r i e r Townahip High School, Winnetka, I l l l n o i a Henry Syer, Kent School, Kent, Connecticut Q.B. Thomas, Massachusetts Inatitute of Technology A.W. Tucker, Princeton U n i v e r s i t y H.E. Vaughan, University o f Illinola John Wagner, University of Texas R . 3 . Walker, Cornell U n i v e r s i t y A.D. Wallace, W a n e Univereity E.L. Waltera, Willlam Penn Senior H l g h Sohool, York, Penneylvania Warren White, North Hlgh School, Sheboygan, Wiaconsln D.V. Widder, Harvard Unlversity William Wooton, Pierce Junior College, Woodland H i l l a , Calffornla J .H. Zant Oklahoma State Univereity

,

GEOMETRY

Part I

Teachersf Commentary

Page

............... A WORD ABOUT THE: PROELEM SETS . . . . . . . . . . . . A GUIDE TO THE SELECTION OF PROBLEMS . . . . . . . . . INTRODUCTIOF!

......

.. , . . ...

lx XV

xvii 1

Chapter

1

Chapter

2

Chapter

3

Chapter

4

Chapter

5

Chapter

6

Chapter

7

. .. LINES, PLANES AND SEPARATION . . . ANGLES AND TRIANGLES , . . . . . . . . . CONGRUENCES . . . . . . . . . . . . . . . A CLOSER LOOK AT PROOF . . . . . . . . . GEOMETRIC INEQUALITIES . . . . . . . . .

Chapter

8

PERPEhmICULAR LINES AND PLANES IN SPACE

173

Chapter

9

PARALLEL LINES IN A PLANE

189

Chapter 10

COMMON SENSE ARD ORGANIZED KNOWLEDGE SETS, REAL NIIMBEFiS AND LINES

PLIALLELS IN SPACE

..

.

. ........ , . . . . . . .

11

37

55 75 127

147

235

INTRODUCTION The text t h a t you a r e about to teach from is t h e r e s u l t of

a c o l l a b o m t i o n between unlvers l t y rnathemticians and experienced high school teachers, The treatment of geometry in t h i ~Look I s very dlfferent, e s p e c i a l l y In the first few chapters, from t h e treatment t h a t nearly everybody I s used to. T h e r e is no question t h a t e v e r y change in teaching has I t s price: it c a l l s f o r a great deal of preparation when a treatment which has become second nature Is replaced by a treatment some of whose f e a t u r e s are new to the teacher as well a s t o t h e student. F o r t h i s reason, we have made changes only when we became convinced t h a t they were worth the p r i c e . It should be remembered a l s o t h a t w h l l e any change at a l l creates some trouble f o r t h e teacher, simply because it is a change, thls p r i n c i p l e does not apply to the student: for him 9 formal treatment of geometry would be new. T h i s mnual is intended to reduce your troubles to a minimum. It consists of t h r e e parts, The main part is a running commentary, r e f e r r i n g to particular s h o r t passages of the text. In this ~ r t , we try to explaln what we a r e driving at, and to warn of p o s s i b l e dlfflculties, ( A S of t h e F a l l of 1960, the t e x t has been revised a f t e r use In over one hundred classrooms, but it is n a t u r a l t o expect that there w i l l s t l l l be d i f f i c u l t i e s t h a t haven't been recognized and d l s cussed. ) In a v e r y large number of cases, we had trouble deciding what to put into t h e running commentary and what to leave out, We d e c i d e d a t length t h a t when in doubt we should put thlngs in. Thus we have no doubt i n c l u d e d many explanations which are unnecessary. These, however, should be easy to skip. Obviously, in a tenth-grade textbook many of the discussions have t o be l o g i c a l l y incomplete. We have cut some cornera, expecting t h e student's i n t u i t i o n to take over, and we b e l i e v e that t h i s l a as it s h o u l d be. A l l sorts of q u e s t i o n s can come up In c l a s s , however, and t h e chances are t h a t t h i s book w i l l provoke some questions t h a t student8 don't u s u a l l y ask in the

t r a d i t i o m l courses.

The running commentary is designed to h e l p

to be one up when this happens, We have a l s o indicated, at some points, t h e t h i n g s we think should be emphasized and the general s t y l e of preaentatlon that we had in m i n d . T h e r e a r e some topics that can't conveniently be d e a l t w i t h in connection w i t h a particular passage o f the t e x t . Some of these t o p i c s c u t across several chapters. We have therefore added a series of essays, under t h e general t i t l e , Talks t o Teachers. These include, in our opinion, some of the most important parta of the commentary. (These w i l l be referred t o , hereafter i n t h l s manual, simply as the Talks .) The first of t h e Talks, e n t i t l e d Facta and Theo~lea, we b e l i e v e you w i l l want to read right now dnd a t least once more after you have read w e l l into the t e x t , Finally, t o save you spade-work we have g i v e n anawers to a l l problems and solutiona to a l l but the simplest, These are interspersed in t h e running commentary a t the appropriate places, Answera have o f t e n been given in simplified radical form or as multiples of T r u t h e r t h a n in the form of decimal approximations. We b e l i e v e t h i s p o l i c y should be encouraged, but t h a t t h e student should be a b l e t o aupply a decimal approximation on demand, In addition to t h e Teacher's Comentary you should have a v a i l a b l e a copy o f S t u d i e s i n Mathemtlca, Volume 11, Euclidean Geometry Based on Ruler and P r o t r a c t o r Axioma, by C. K. Curtis, P. H. Daus, and R . J. Walker. This contains, especially in t h e f i r s t chapters, much material that could have been put in t h e T a l k s t o Teachers. It a l s o contains d e t a i l e d proofs of basic theorems t h a t are not mentioned in t h e t e x t . The propertlea s t a t e d i n t h e s e theorems are Intuitively obvious and are generdlly accepted by students without comment. A completely l o g i c a l development of geometry must, nevertheless, contaln proofs o f t h e s e theorems, and s o they are included here for whatever use you wish to make of them. This book w i l l be referred to frequently in this mnuttl. When we do so w e will speak of It as "Studies 11." you

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Some teachers may enjoy referring t o a lighter presentation of some geometric ideas. To t h e m we suggest S t u d i e s I n Mathematics, Volume V, Concepts of Informal Geometq.

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Although we felt it unwise to nuke our text logically complete In its proofs we d i d attempt t o g i v e a complete f o u n d a t i o n of postulates and d e f i n i t i o n s . On such a foundation a student can build as elabomte and complete a structure as his c a p a b i l i t i e s permit, with t h e h e l p of h i s teacher and of supplementary reading. The only difficulty apt to be met In l a y i n g this foundation is an apparent alownesa of t h e text in coming to grips with r e a l l y i n t e r e s t i n g geometric problems, However, you w i l l f l n d that the postulates, definitions and simple theorems In Chapters 2 , 3 and

4, although not p a r t i c u l a r l y interesting when you f i r e t s t u d y them, w i l l be of great value i n the later c h a p t e r s . Moreover, seen from the perspective of the later chapters the b a a l c material of the early chapter8 takes on a more interesting appearance as its importance to geometry becomes appreciated. If a student i s t o understand a complicated geometric s i t u a t i o n he must f i r s t have a clear picture of the fundamentals. Obviously you are going to llke some features of t h i s t e x t better than o t h e r s . In any case, we a s k t h a t you teach each chapter of this book as If you had f a i t h In t h e preaentation. If some features of It don't work, we want to know it, but we can't f l n d out, one way o r t h e other, unless they are given a fair try. A half-hearted experiment In t h e classroom has some of t h e disadvantages of a half-hearted back f l i p in a gymnaalum.

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USING THE: TIME AVAILABLE T h i s text was w r i t t e n so that very good classes wlll have

enough m t e r i a l t o challenge them for a year. It follows, then, t h a t some classes wlll not be a b l e to cover a l l t h e material. You may p r e f e r not to r u a h t h r o u g h important topics j u s t to cover pages, so this note w i l l suggest t h e kind of choices t h a t you can make. The choices mentioned are only samples, however, and you will find variations that f i t the needs of your own class. A f u l l course includes a l l exposition, and a substantial number of problems from each s e t . Few, if any, students will solve a l l t h e problems. An approximation to time allotment f o r classes which study every topic I s given in t h i s t a b l e , The names of chapters are topical and are not necessarily the actual c h a p t e r titles. Days 3 10,

I n t roduc t Ion.

Chapter Parallels fn Space

Pythag. Theorem

Days 6

11.

Area,

6

13.

Circles, Spheres

13

.

20

14.

Closer Look a t Proof

6

C h a r a c t e r i z a t i o n s of Sets. Constructions.

10

S e t s , Numbers, Lines.

10

10

/

6 Angles, Triangles

Congruences A

Inequalities

8

Perpendiculars in Space

9

Parallels in Plane Total

17

15-

Area of Circles

5

16.

volumes

8

17.

Coordinate Geometry

20

85

The list of days m u s t include t i m e u s e d f o r chapter reviews and t e a t s , Though such work Is important, a p r a c t i c a l observation l a in order: A class t h a t uses t w o days per chapter for reviewing and testing uses more t h a n one-sixth of the year in t h a t way, and xii

muat plan accordingly. We b e l i e v e that every course should include careful treatment of t h e f i r s t volume, regardless of the preceding t a b l e , Thla does not mean that proofs of theorems should be memorized or that all problems should be done, however. Selection of mterial, if necessary, can begin w i t h Volume 2, The table above ahowa that if you are not I n t o Chapter 10 by the end of the first aemeater, and rmny classes w i l l not be, you will want to plan ahead s o that you can study the chapters and t o p i c a most important for your students.

For example, you may decide to omit some mterlal in order to devote sufficient time to the chapter on coordinate geometry, A way to do this is to o m i t Chapter 10 and cover t h e ideas of Chapters 14, 15 and 16 i n t u i t i v e l y while doing selected problems. You m y a l s o d e c l d e t o take up Chapter 17 immediately a f t e r Chapter 12. Or you may decide to teach Chapters 8 and 10 largely on an i n t u i t i v e basis, using problema t o develop major concepts. Similarly f o r Chapters 14 and 15. Then omit Chapter 16 and treat most of Chapter 17. Certainly numerous such plans are p o s s i b l e , Ideally, the one basic plan Is to cover a11 mterial. R e a l i s t i c a l l y , due to f a c t o r s of t i m e and of individual and group d i f f e r e n c e s , several a l t e r n a t i v e plana must be considered, evaluated, and reviewed constantly We list here what can be omitted, In t h e o r d e r , very roughly, of preference in omission, the l a s t item being the one you should least consider omitting. Chapter 17 is not included in the list, p a r t l y because I t s place i n such a list is h i g h l y controversial and p a r t l y because a reason for omitting other topics is t o aasure adequate coverage of coordlnate geometry.

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Proofs in S e c t i o n 6-5 and in Chapters 16, 14, 10, 8, 15, 7 , 13 ( a f t e r Theorem 13-5), 12. A l l t e x t m t e r i a l (except for formulas) i n C h a p t e ~ s 16, 14, 15, 1 3 ( a f t e r d e f f n i t i o n of intercepted a r c ) , 10.

are not pmpoafng that anyone omit anythfng unnecessarily, f o r a l l the material i a worthwhile. We are merely proposing that, if pressed f o r time, you not r u s h through t o o much material w i t h your students but instead select the m t e r i a l b e s t suited to W2

t h e i r needa.

A WORD ABOUT THE PROBLEM SE3S

The problem sets In t h i s book are an extremely Important

part of the course. Many concepta are developed and expanded there. Careful assignment of the problema is e s s e n t i a l so a8 not t o exclude some of the important topics i n the development. Each problem a e t begins with some airnple exerciaea. Some of the more d i f f i c u l t problems, not necessarily to be found at the end of the set, dre starred. It is hoped t h a t the teacher w i l l read a l l of the problems in a s e t before making an asalgnment. In some caaes a sequence o f problema b u i l d s an important concept, and an assignment should contain a l l the problems t h a t develop the concept. In some instances a s p e c i a l comment about a problem occurs w i t h the answer

to the problem. We hope t h a t teachers w i l l use their own judgment about the number of problems t o assign. It is likely that no student will work a l l the problems, Certainly most students can be expected t o do only some of t h e large number provided. You have a good chance to allow for i n d i v i d u a l dlfferencea I n your assignments. Proofs, and reasons w i t h i n proofs, are given in varied form to suggest to the teacher that genelaal understanding of the problema is more important t h a n a rigid f o m of presentation. his applies e a p e c l a l l y to Chapter 5 and the following chapters, in which mny of the problems call for proofs of theorems. ) The s o l u t i o n s given are not always. the only possible s o l u t i o n s , and good original reasoning by atudents should be encouraged and commended. The f a c t t h a t we g i v e a proof, in o u r solutlona, In paragraph form for convenience and b r e v i t y does not mean t h a t we b e l l e v e that every student should give I t in t h i s form. The teacher can decide which form has the most educational value for his students a t t h e given time. On occaeion, students should be asked t o suggest and solve problems not In the t e x t .

A GUIDE TO THE SELECTION OF PROBUMS

Following is a t a b u l a t i o n of t h e problerna i n this t e x t . I t will be noted t h a t t h e problems are arranged Into t h r e e s e t s , I, 11, and 111. A t first glance, one might think t h a t these are i n order of difficulty THIS I S NOT THE MANNER IN WHICH THE PROBUMS ARE GROUPED! !!! Before explaining t h e grouping, it should be mentioned t h a t it is understood t h a t a t e a c h e r will s e l e c t from a l l of t h e problems those which he or s h e f e e l s are best f o r a p a r t i c u l a r c l a s s . However, careful attention should be given to the comments on t h e problems i n A Word About t h e Problem S e t s . Group I contains problems t h a t relate directly t o t h e materidl presented In the text. Group 11 contains two types of problems: ( 1) some t h a t are similar to t h o s e of Group I, and ( 2 ) some that are J u s t a l i t t l e more d i f f i c u l t than those In Group I. A t e a c h e r m y use t h i s group f o r t w o purposes: (1) f o r additional d r i l l m a t e r i a l , if needed, and ( 2 ) f o r problems a b i t more challenging t h a n t h o s e in Group I, t h a t could be used by a better c l a s s . Group I11 contains problems t h a t develop a n i d e a , using t h e Lnfomnation g i v e n in t h e text as a s t a r t i n g point. Many o f these problems are easy, interesting and challenging. The student m y find them more stimulating than t h e problems in Groups 1 or 11. However, if time is a f a c t o r , a student can very well not do any c , i them and s t i l l completely u h d e r s t a n d t h e material in t h e t e x t . These a r e enrichment problems. It is assumed t h a t a teacher will not feel t h d t he o r she must assign a l l of the problems in any s e t , o r all p a r t s of any one problem. It i s hoped t h a t t h l s llsting w i l l be h e l p f u l t o you in assigning problems f o r your students.

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We have included i n the problem s e t s r e s u l t s of theorems of the text whlch a r e important principles in t h e i r own r i g h t . I n t h i s respect we follow the precedent of most g e o m e t r y t e x t a . However, a l l essential and fundamental theorems are in the t e x t proper. T h e f a c t t h a t many importdnt and delightful theorems a r e to be found in the problem sets is v e r y desirable as enrichment. While no theorem s t a t e d in a problem set is used t o p r o v e any theorem in t h e t e x t proper, they are used In s o l v i n g numerical problems and o t h e r theorems in the problem s e t s . T h i s seems to be a p e r f e c t l y n o m l procedure. The d i f f i c u l t y (or d a n g e r ) , as moat teachers define it, is in allowing the r e s u l t of an i n t u i t i v e t y p e problem, o r a problem whose hypothesis assumes too much, to be used as a convincing argument f o r a theorem. The e a s i e s t and s u r e s t way to handle the s i t u a t i o n 1s t o make a blanket r u l e f o r b i d d i n g the use of any problem r e s u l t to prove another, Such a rule, however, tends t o overlook the economy of time and, often, the chance to foster t h e creative s p i r i t of the student. In t h i s t e x t we have t r i e d to eatablfsh a f l e x i b l e w t t e r n whlch w i l l allow a teacher and class t o s e t t h e i r own p o l i c y .

xvfii

GUIDE TO SELECTION OF PROBLEMS

Chapter 1

Set 1-1 1-2

2,5,6,7,9,101,3,4,

Chapter 2 2-1

1,3,7,12.

2-2

1,2,3,5.

2-3

1,3,4.

2-11

1,2,3.

2-6

2,3,4,5,7.

2-7a

1,2,3,6.

2-7b

1,2,3,7.

Chapter 3

.

3-la

1,2,3, b

3-lb

112,3,4,

3-lc

1,2,3,4,6.

3-2

1,2,3-

3-3

1,2s314,7,11,13.

Chapter 4 - 1

33495,69738,

9, 10, 11.

I Chapter 4 4-3

1,2,3,4,5,6,7,8-

4-4

1,2,3,4,5?

10,12.

Chapter 5

5-1

1,6,7,9.

Chapter 6 6-2a

1,2,3,8.

6-2b

1,?,4,5.

6-3

1Y2,3,5,7.

6-4

1,~.

6-5

1.

6,9,

I Chapter 7 Set

7-1

1,2,4,5.

7-3a

2,2,3.

7-3b

1,293#5*

7-3c

1,2,3,4,7*9,11*

7-3d

1,2,3.

7-Se

1,2,3,8-

7-4

1,2,3*

Chapter 8

8-1

1,2,&,5,6,899.

8-2a

1,2.

8-2k

1,

3-2~

1,2,3,4.

Chapter 9

9-1

1,2,3,4,5,7,9

9-3

192,39495.

9-4

1,4,5,6*

9-6

1,4,6,10. 2

Chapter 10 10-1

1,2,4.

10-2

1,293

10-3

1,2,4.

-

Chapter 1 COMMON SENSE AND ORGANIZED KNOWLEDGE Thls c h a p t e r should be treated as an introduction. It is not a review o f algebra or of the Pythagorean relation. The a l g e b r a i c problems and t h e Pythagorean r e l a t i o n are introduced t o illustrate mathematical method, not t o provide items for f o r g e t f u l students t o relearn during the first week of a new course. In this chapter I t i s d e s i r e d f i r s t t h a t the students see the d i s t i n c t i o n between a problem w i t h a n obvious aolution and one that requires thought and skill in i t s s o l u t i o n . Later the need f o r exact reasoning on the basis of p r e v i o u s l y defined o r accepted information is i l l u s t r a t e d . What should be impressed upon the student is the fact that once we e s t a b l i s h our basic information we I n t e n d to remain within the framework of our system to do the remainder of o u r work. We have o u r p o s t u l a t e s (which contain undefined terms), .and o u r definitions. On the basis of t h e s e (and these alone), we will b u i l d up a body of' geometrical information by the application of l o g i c a l reasoning. As polnted out in the t e x t , it I s impossible t o d e f f n e a l l terms, so we have to begin w i t h some undefined terms, Definitions are j u s t agreements that we make to allow us t o substitute a word, phrase o r symbol f o r other phrases that are, in general, longer and more tedious t o w r i t e out. A d e f i n i t i o n may be thought of as an abbrevlation f o r a longer phrase o r group of phrases. If P and Q r e p r e s e n t phrases such t h a t Q is taken as an a b b r e v i a t i o n f o r P, then the abbreviated form Q may be substituted f o r P in any discussion and t h e sense of the discussion remains the same. T h i s also works i n t h e reverse order. The expanded form P may replace t h e abbreviated form Q. For example, conalder the d e f i n i t i o n : A parallelogram I s a q u a d r i l a t e r a l whose opposite sides are parallel. If we h o w that the quadrilateral ABCD has and ADIIBC, then we can abbreviate this by saying t h a t ABCD is a parallelogram.

m1Im

--

On the other hand, if we know that ABCD is a parallelogram, then we can assert what this phrase s t a n d s f o r , namely: ABCD is a quadrilateral such t h a t AB( ICD and AD) ,

is a parallelogram" and the phrase, " ABCD is a quadrilateral and and AD\I E " can be used interchangeably. Since d e f i n i t i o n s are agreements t h a t a simple phrase means t h e same a s a more complicated phrase, there is no question about e v e r t r y i n g t o prove a d e f i n i t i o n . Only a very remarkable s t u d e n t w i l l fully understand the paragraphs about theorems, postulates, proofs and undefined terms, when he flrst s t u d i e s this c h a p t e r , These ideas will come i n t o sharp focus in t h e student's mind only when he h a s had some experience with them. Chapter 1 i s designed merely to give t h e student a s u f f i c i e n t l y good idea of what is going on so that he will be b e t t e r prepared f o r what follows. For t h i s purpose, s h o r t and simple statements to t h e class are probably b e s t . For example, if a s t u d e n t asks what a p r o o f is, a good a n s w e r is that a proof is a complete explanation of why a s t a t e ment is true. (Later t h e s t u d e n t will learn, by experience, the way all of us did, what s o r t of p r o o f is a c c e p t a b l e in mathematics.) In the same s p i r i t , a d e f i n i t i o n is simply an exact explanation o f what a word or phrase means. The explanation of the meaning o f p o s t u l a t e s has d e l i b e r ately been made a l i t t l e ambiguous. There are two p o s s i b l e viewpoints : 1. Until about 1800, everybody b e l i e v e d t h a t t h e p o s t u l a t e s of geometry were "self-evident t r u t h s " , and t h a t the theorems proved from them were statements of f a c t about the o u t s i d e world, learned by pure reason. So we see t h a t t h e phrase, "ABCD

m1

2. Since t h e discovery of non-Euclidean geometry, it has been p l a i n t h a t t h e postulates o f o r d l n a r y geometry are not

" s e l f - e v i d e n t t r u t h s " . There are many klnds of geometry; a l l o f them are equally v a l i d mathematically; some of the very 11 peculiar" ones are u s e f u l in physics; and ----each of them is desc r i b e d & i t s own s e t of postulates. P o s t u l a t e s , therefore, are simply d e s c r i p t i o n s of the kind of geometrical t h e o r y t h a t we propose to investigate a t a g i v e n time. And when we p r o v e a theorem, we are not showing that the theorem is "true" in the sense that it fits t h e f a c t s of t h e o u t s i d e world. When we prove a theorem, we are merely showing t h a t the theorem holds t r u e in t h e mathematical system described by our p o s t u l a t e s . ( s e e t h e remarks on non-Euclidian geometry In the c h a p t e r on parallels, and t h e Talks on Miniature Geometries and Non-Euclidean Geometry. ) It does not seem to us that this second viewpoint is s u i t a b l e f o r p r e s e n t a t i o n i n t h e second week of the tenth grade. The s t u d e n t would probably be completely bewlldered, and he might get t h e idea that Euclidean geometry is j u s t words, words, words. In Chapter 1 we have therefore been treading a r a t h e r f i n e line, explaining t o t h e s t u d e n t approximately as much as we think that he can understand, and being careful in the process not to make any statements t h a t w i l l have to be corrected l a t e r . What needs t o be emphasized, at the s t a r t , I s t h a t p o s t u l a t e s are not j u s t p u l l e d out of t h e a i r t o satisfy somebodyTs whlm. The space of Euclidean geometry is an extremely good approximation t o physical space. This is why it got i n v e n t e d , and t h i s 1s t h e most e f f e c t i v e way t o t h l n k about it. We can and we should use o u r intuition o f physical space t o h e l p us guess what can be proved and how we car? prove it, The proof itself, when we get it, should be l o g i c a l l y based on the postul a t e s . A mathematical system, like the geometry we are developi n g , t h a t c o n s i s t s of p o s t u l a t e s and theorems involving undefined and defined terms is called a deductive theory. T h i s t h e o r y

----

itself is given meaning and content by exhibiting an Pnterpret a t i a n of t h e undefined terms. When we give the usual i n t e r p r e t a t i o n of point, l i n e , and plane from physical space we g e t our physlcal geometry, which is an approximate model of our deductive theory. Other interpretations of t h e undefined terns lead to different models. A f u r t h e r discussion of mathematical models and how they work is given in the Talks. It might be well to return t o t h e Latter p a r t of this chapter a f t e r t h e student has had a f a l r amount of experience w i t h t h e concepts which we have been trying to explaln. After the class has f i n i s h e d Chapter 5 , the ideas o f p o a t u l a t e , theorem, proof and undefined term should have become entirely comprehensible. Chapter 6 will c l a r i f y some of the more troublesome problems Involved In some types of p r o o f s . The numbers in the left-hand margin refer t o the pages in t h e t e x t t h a t are being commented upon.

-

1

2

Some students may n o t remember how to s o l v e simultaneous equations. The thing t o do here, as far as t h e class aa a whale is concerned, is t o provide enough reminders so that the class understands the solution o f f e r e d by the book. Notice t h e manner in which t h e l e n g t h s of the s i d e s of the rectangles are discussed. The s i d e s of the rectangles are merely l i n e s e p e n t s , and each segment has a length t h a t fs a number of i n c h e s . Note that we write x = 8 and not x = 8 i n c h e s . There are times when w e want t o talk about x2 and we square numbers, for example, (812, b u t we do not square 8 inches. The problem here i s simply t o keep the u n i t s of measure o u t of the mathematical operations and use them in the interpretatLon of the res u l t s of these operations. The lower case l e t t e r s , x, y, are used to stand only f o r numbers which are lengths of the sides in some unit, for example: I f a r e c t a n g l e Is 8 Inches long and x stands f o r the length, t h e n x = 8.

[pages 1-21

Admittedly, this is a fine point, b u t we have been very c a r e f u l a b o u t it in the text, and it will be easier on the students if you back us up by being equally c a r e f u l about it i n the classroom. The usage t h a t we are f o l l o w i n g Its d i f f e r e n t f r o m that of physics and chemistry courses. Physicists have developed, t o a fine p o i n t , t h e art of handling unit signs as if t h e y were algebraic symbols. A simple example of thfs is 6 ft. x 6 ft. = 36 s q . ft. = 36 ft. 2 Fmrn here they move on t o measure accelerations in ft./sec.2 and p e r f o m cancellations between such expressions according to the ordinary laws governing f r a c t i o n s . We are not claiming f o r a moment t h a t there I s anything wrong w i t h t h i s . It Is n o t o n l y 2 very r i g h t , but very u s e f u l . It i s n o t , however, part of t h e natural subject matter of t h i s couyse, and so we are taking the more elementary viewpoint t h a t the things we h o w how t o add and multiply are numbers. This will be q u i t e adequate for o u r purposes, and the art of handling u n i t s algebraically can b e s t be learned in courses w h e r e it is needed.

2

You may have a student who will enjoy making apparatus t o illustrate t h e Egyptian method f o r c o n s t r u c t i n g a r i g h t angle. F i r s t he needs t o tie e l e v e n h o t s in a pfece of cord so tnat twelve equal lengths r e s u l t . Then he needs a b a r d and t w o t a c k s . Students can manipulate t h i ~simple apparatus to get a feeling f o r the o p e r a t i o n the Egyptians went through. Other s t u d e n t s may enjoy supplementery reading, for example, an encyclopedia account of the Egyptlan pyramids.

Yocr s t u d e n t s may insist that they do not have t o try " a l l " 2 2 2 c a s e s to be s u r e of getting a right angle when a + b = c

.

3

You will find it hard to argue against the p r i n c i p l e of reasoning they are u s i n g as long as you r e s t r i c t d l s c u s s ~ o nt o t h i s one case where the mathematical f a c t is correct in s p i t e of t h e reasoning, &t t r y such a thing a s the "formula" for primes

n-1,

when

p=41

n = 2 , p = 4 3 n = 3 , p = 47

n=4, p-53 n = 5, p = 61 n=6, p=71 The first s l x values f o r n, and many more, yield prime numbers f o r p . Your students may b e l i e v e t h a t t h i s is true f o r all values of n. If your c l a s s does n o t have anyone who h i t s upon the revealing number, 41, f o r whlch p i a n o t a prime, you can propose t h i s v a l u e yourself. Or, on a different level, mention the r i c h c h i l d who bel i e v e s -- because of several observations - - that every family has a C a d i l l a c . Problem Set 1-1

b.

d

Let

b e the number of miles between t h e cities.

d = i d + 7 .

5 "4.

a. b.

+ 21.

3d

=

d

d

=

1 10 2'

The distance is

1 mi. 10 3

4 in., 1 In. kt n be the number of Inches in the shorter piece and 5 - n the number of i n c h e s in the longer p i e c e . Then $ 2s the number af inches in the sides of the smaller square, and 5 - n is the number of i n c h e s in the slde of th large square. The problem t h e n r e q u i r e s 5 n 22. that ) = 4 (g)

(y

25 - 10n 16

+

n

4n = i E m

3n = 5 , or 2 n = 1 5'

n

=

-5 (which is meaningless h e r e ) .

7

J

1 In. long. This is a right triangle because ( 5 )2 + (12)2 = (13)2 Reason (d) I s l i k e l y . Reason ( b ) would account f o r large e r r o r s . Reason (a) is unlikely. Since l 2 - 2.1 + 2 = 1 the equation is true if n = 1. Y e s . No. No. a. The remainder is one. b. All of them. Comment: Each odd i n t e g e r can be represented by 2n -k 1 for some i n t e g e r n. If re expand (2n + 112 and divide by 4, the integral part af the quotient is n2 + n and t h e remainder is I. Hence, if 4 is divided I n t o the aquare of any odd integer, t h e remainder is 1. There a r e 31 (or in special cases, 30) regions formed, never 32. T h i s problem illustrates the danger of jumping t o haaty

The pieces are

2 and 1 3

3

.

conclusions. Y e s . b. Y e s . c , The areas are equal. d . The lengths a r e equal. The area of t h e rectangle is 63 w h i l e the sum of t h e areas of t h e pieces is 64. The f a l l a c y is that if the o t h e r measurements are c o r r e c t , t h e small triangles should have heights of r a t h e r than 4 . This can be shown by using similar triangles. The total time for the trip is the distance, 60, divided by a.

v

t h e average speed, 60, and is therefore 1 hour. Since thls hour is used up travelling the f i r s t 30 miles a t 3 0 miles per hour, our answer must be that the average speed of 60 m.p.h. is then Impossible to achieve.

8-10 This is a description o f w h a t Is i n v o l v e d in setting up a mathematical theory. It tock the human rzce a long time t o p e r f e c t this Idea. You cannot expect your students to grasp it f r o m an abstract d e s c r i p t i o n . The understanding of what is i n v o l v e d in l o g i c a l reasonfng will grow throughout the course as s t u d e n t s a c t i v e l y engage in loglcal reasoning. Nobody can learn loglcal reasoning in a vacuun. The i d e a t h e s t L d e n t needs tu get here is t h a t p o f n t , l i n e

and plane are basic terms In our system and t h a t we define more 10 complex terns l i k e t r i a n g l e , parallelogram, e t c . , in terms of p o i n t , line and p l a n e , You can draw d o t s of different size& on the blackboard t o h e l p g e t at t h e idea of p o i n t . Or you can mention a star, thousands of times a s large a s t h e earth, that is barely v i s i k , l e . Seen up c l o s e I t is tremendous. Seen from f a r t h e r and f a r t h e r away I t approximates more and more c l o s e l y the idea of a p o i n t . 10 It may be necessary to p o i n t o u t r e p e a t e d l y t h a t a line 11 does not stop1'.

The plane is t h e most d i f f i c u l t of the t h r e e terms f o r some s t u d e n t s to understand, This is revealed by such incorrect languagc as ''~ectangalarplane'' or " c i r c u l a r p l z n e " .

k plane is

suggested by such convenient o b j e c t s as the classroom f l o o r , the l o p of the teacher's desk, and a sheet of paper. Emphasize, whenever you use these o b j e c t s f o r illustrative purpoaes, that a mathematical plane 11 keeps on going", and move your hand in

appropriate d i r e c t i o n s . It may help the s t u d e n t if you occasionally, during the f i r s t months, s u g g e s t that they reread t h e t h i r d paragraph of page 31.

[pages 8-11]

--

Problem Set 1-2 12 1.

a.

measurement, size, dimension. b dimension, measurement, extent, size. plan, houaes, churches, schools. plane, bounded by, parallelogram, rectangle, space. a. Defining a term usually involves placing it i n a c l a s s and distinguishing it from o t h e r members of the class. The term "something" is an unnecessarily large class into which to place squares. The phrase "that is not round" does n o t distinguish I t from other "somethings". (one satisfactory definition at t h i s p o i n t : A square is a rectan~lewhose sides have t h e same length. ) b. Only one of the angles has a measure of 9 0 ' In a r i g h t triangle. ( A rlght triangle i a a triangle with one right angle. ) c. "When" refers to time, not t o geometric figures. A triangle is n o t a period o f tlrne. (An equilateral triangle is a triangle whose three sides are equal in length. ) d. llWhere'ldenotes a place. "~erimeter" is not a p l a c e . h he perimeter of a rectangle is a number equal t o the sum of t h e lengths of its s i d e s . ) e . This is a true statement, but it s t a t e s a process t o r computing circumference r a t h e r t h a n stating what circumference is. h he circumference of a c i r c l e is a number which i n d i c a t e s its length. ) A. False B. True C. False D. True

.

13

4.

5. 14 6.

*7.

[pages 12-14]

Chapter 2 SETS, REAL NUMBERS AND LINES

Some of t h e ways in which t h e material of this chapter d i f f e r s from that of a traditional text are: (1) sets a r e introduced and ( 2 ) the real numbers, and thereby arithmetic and algebra, are brought I n t o the course in a fundamental way, The reason for including s e t s becomes e v i d e n t when you r e a l i z e that every geometric f i g u r e is most simply s t u d i e d as a set of points T h i s book does n o t treat the t h e o r y of s e t s as an end i n i t s e l f b u t i n t r o d u c e s its ideas and terminology to the extent t h a t they contribtlte to t h e geometry course, The real numbers are needed in geometry for the measurement of segments, angles, areas and volumes. We introduce them exp l i c i t l y , rather than use them without any e x p l a n a t i o n . The immediate reason f o r i n t r o d u c i n g t h e r e a l numbers- in t h l s chapter is that they a r e needed Tor the statement of P o s t u l a t e s 2, 3, and 4. These p o s t u l a t e s guarantee i n e f f e c t t h a t l e n g t h s of segments are expressible as real numbers, and nave the familiar p r o p e r t i e s t h a t we expect. One important advantage of I n t r o d u c i n g real numbers so early is that we can u s e them to define betweenness f o r points on a llne. Then we can define segment, one o f t h e most important geometric figures, i n terms of between. Seeing numbers so strongly emphaslzed in a geometry course w i l l seem strange at first. A t the time when Euclid w r o t e , algebra hardly existed, except i n s o f a r as it was impllcit in geometry. In the following t w o thousand years or so algebra developed t o a high degree, but t h e teaching of elementary geometry has made r a t h e r l i g h t use o f it, In this book, algebra is used in two important ways. In the f i r s t p l a c e , it 1s used in the p o s t u l a t e s t o make them easier t o apply. If we take for granted that the real numbers are known, t l ~ e nit is possible to give a lokically complete s e t

of postulates, adequate for proving the theorems, avoiding some of t h e complications and d i f f i c u l t i e s involved i n , say, Hllbertfs Foundations Geometry, We will see also, as we go along, that a great deal of the traditional material of geanelzy was r e a l l y algebraic a l l along, and I s much easier t o handle when it is described algebraically. ( ~ h l sis especially true in t h e chapter on p r o p o r t i o n . ) We believe that f o r your students theae simplifications are-genuine simplifications, and will make geometry easier f o r them to understand in the long run. k t the algebraic apparatus used in t h i s c h a p t e r and later may very well call for more caref u l preparation t h a n you have aver given before to an early chapter of a textbook. 15 In t h e form in which we have presented it, the discussion of sets I s not really a mathematical theory but simply an explanation of the language i n which we propose to t a l k . As the "homely examples" in this s e c t i o n show, a l l of the basic ideas about s e t s -- with the sole exception of the empty set -- are already familiar. Only eome of the words in which we t a l k about them are new. The standard n o t a t i o n o f a set theory i s described i n Appendix I, e n t i t l e d A Convenient Shorthand. This is intended to be strictly optional and the title of the appendix I s meant to suggest the s p i r i t in which the notation was t o be regarded. There i s a serious danger in t a l k i n g t o o much, and t o o fancily, about s e t s , at the high school level: the impression may be conveyed that w r i t i n g things l i k e A ~ B ' C Cis a l o f t i e r occup a t i o n than proving meaty,theorems and solving hard problems In geometry and algebra. T h i s would be sad. We therefore b e l i e v e that t h e language o f s e t s should be introduced matterof-factly w l t h o u t fanfare, and that the notation of s e t t h e o r y should be taught to a given student only if and when the student is prepared to think of it as a matter of convenience.

or

-

As a matter of convenlence, however, t h e language of sets is going to be used continually. F o r example, an angle will be defined as t h e union of two non-collinear rays. Two lines in the same plane are parallel if t h e y do n o t intersect, and this means t h a t t h e l i n e s , considered as s e t s of points, have no member in common. 16 Notice t h a t we are referring t o t h e rectangles as the n o t t h e line segments plus union of t h e f o u r l i n e segments, t h e r e g i o n e n c l o s e d by them. Later we s h a l l be concerned with t h e i n t e r i o r of geometric f lgures

.

17

Such a statement as "...each o f the two lines is a set o f p o i n t s " seems to say something s p e c i f i c about "line", which is to be one of o u r undefined terms. This should not be cause f o r trouble, however, f o r the material here is informal and explanatory. It i s n o t part of our formal system of geometry.

.

Problem Set 2-1 19

1. 2.

The intersection i s [5, 9, 111. The union is E3, 4, 5, 6, 7, 9 , 10, 11, 121. S2 and S if you are a. S1 and S2; S1 and S3; S1 and S 5; 5 a boy, but S3 and S if you are a girl.

5

S1 ' '

The s e t consisting of all members of f a c u l t y and

students of your sbhool e.

20

3.

sl,

s2,

S3'

.

s5.

The set {A). The s e t {B,C]. T h e empty s e t .

4.

a. b.

Three committees: [A, 31, { A , C), EB,C 1. [A,B) and [ A , c ) have A in common. [ A , B ) and {B,c) have 3 in common. [A,c] and [B,c] have C in common. l1 Intersect l1 means ''have a member in common"

.

[pages 15-201

The s e t of a l l p o s i t i v e integers. The empty set. Or, the sets have no comrr,on member. The i n t e r s e c t i o n is t h e segment E. The union is t h e

triangle ABC. The s e t consisting of the one p a i r (2,1). The set consisting of the one palr (4,3). The empty set. Or, t h e r e are no common elements. a. The s e t of all p o s f t i v e integers diviaible by 6 ( l . e , , b.

c.

a.

21

23

by both 2 and 3 ) -- [ 6 , 12, 18, 24,.,.). 6n, where n is a positive integer.

The set of a l l p o s i t i v e integers divisible by e i t h e r 2 o r 3 , [2, 3, 4, 6 , 8, 9 , 10, 12, . . . I . 1.b. 3. c. 6,lo. a.1 p(n-1).

The material in t h i s s e c t i o n , too, i s informal. This i n t u i t i v e development is intended to canvlnce the student that to each p o i n t on a l i n e there corresponds a real number, and to each real number there corresponds a p o i n t on the l i n e . The f e e l i n g f o r the arrangement o f these real number^ on a line is Important to the student at this time, Pages 23 to 28 p o i n t out the properties of r e a l numbera concerning inequalLties and absolute values, and show their geometric interpretation on a l i n e . Proof of the f a c t that between any two rational numbers there 1s a t h i r d one is simple, and intereating to some. Int u i t i v e l y , the "average" seems to be such a number, The following argument justifies this intuitive notlon. 1. L e t a be the l a r g e r and b be the smaller of any a + b t w o rational. numbers. He show that is between a and b.

[pages 20-231

5. 6.

Hence a 2 is between a and b. Furthermore, 7 a is rational. +

+

For a more d e t a i l e d discussion of irrational numbers see Appendix 111, and a l s o Chapter 4 of Studies 11.

We Introduce here symbols that m i g h t be new to some students, namely , meaning greater than, 2 , meaning less than or equal to, and 2 , meaning greater than or equal to. To say that an inequality can be wrltten in reverse meana, for example, t h a t if 7 < 9 , then 9 > 7 . This is a statement in the form if x < y, then y > x. We a l s o have inequalities of the form x 5 y, or y 2 x . These could be i l l u s t r a t e d in the following manner: To say that x 5 8, means that x can be e i t h e r less than 8 or equal to 8, for example x can be -12, -a, 0, 3, 7.999 or 8 itself. For a more d e t a i l e d treatment of inequalities see Chapter 4, of Studies 11. There will also be some discussion of inequalities in Chapter 7 of the t e x t . 24 While the baslc algebraic postulates are put in Appendix I1 f o r completeness, the postulates (laws) for inequalitlea are included in the text proper, f o r many students are n o t acquainted with them. 25 Some students may be so used to saying "The square root of 9 is plus or minus 3" meaning that 9 has two square roots, 3 and - 3 , t h a t I t will be hard to convince them that t h e written statement I ' D = 3" is i n c o r r e c t . We h o w of no patent medicine to prescribe. Simply explain, move ahead, and remind l a t e r as necessary.

23

+

--

Problem S e t 2-2

26 1. 2,

All four are true. a. AB is less than CD, b. x is greater than y. [pages 23-26]

XY is greater than or equal to n Is l e s s than o r equal to 3.

YZ.

is l e s s t h a n 1 and 1 is leas than 2. 5 is greater than or equal to x and x is greater than or equal t o -5, or x I l e a between 5 and -5 Inclusive. x I s p o s i t i v e or x is g r e a t e r than 0 . k > 0. e. 2 < g < 3 . r < 0. f- 2 2 ~ 2 3 . 0

5

0.

g.

20. c, d, f, h.

or

t S

3.009,

3.05,

a < w < b . b < w < a .

3.1.

P a r t s ( d } and ( e ) are true f o r r > s but are not always t r u e f o r certain negative values. ) * 7 . a . S . b . T . c . 3 . d.T. e . T . ( ~ o t et o teacher.

27

>0

Most s t u d e n t s learn what "absolute valuei1 means by looking a t several examples. The method of " d e f i n i n g by p a i n t i n g t i helps t h e student t o grasp the concept, but it c e r t a i n l y is not a mathematical method. Assure your s t u d e n t s t h a t t h e i r notion of absolute value w i l l serve them satisfactorily in geometry. 28 P o i n t out t h a t this particular d e f i n i t i o n i s not intended to be explanatory i n the ordinary sense of the word. Awlmard though the d e f i n i t i o n may appear to be, it does p i n t h e idea down and 1s technically c o r r e c t , whereas superficially stated " d e f l n i t l o n s " that sound good o f t e n f a i l t o hold up under c l o s e inspection.

[pages 26-28]

--

Problem S e t 2-3

29 1. a, c, d. "2. b, c, d . 3 . a. r. b, -r. c. 0. 4. Drawings are o m i t t e d . a . The s e t of points t o the left of the zero mark. b. One point, a unit to t h e r i g h t of 0. c . The s e t o f p o l n t s t o the r i g h t of 1. d . The part of the l i n e t o t h e l e f t of and including e. f. g,

h.

30

5.

a. b,

30

1.

Two p o i n t s . The part of the l i n e between 1 and -1 inclusive. The union of the part of t h e line to the l e f t of -1

and the part to the r i g h t of I. The entire line. The first s e t includes 0; the second does n o t . The first set includes O and 1; the second does n o t .

Throughout this book, when we speak o f "two pointa", we r e a l l y mean two. That is, if A and B are two points, then A and B are d i f f e r e n t . The phrases "three polnts" , " t w o lines", and so on, are used in the aame way. O n the other hand, if we say merely that A and B are points of the line L, t h l s allows the possibility that A and B are the aame; If we mean t h a t they are d i f f e r e n t , we either say explicitly t h a t they a r e different or we say e x p l i c i t l y that there are two of

them. Some usages are matters of convention, and there is n o t unanimous agreement on them in the mathematical l i t e r a t u r e . example, most algebra textbooks say t h a t every quadratic equation has two roots; and thus t h e equation x2 - 2r + 1 = 0

ma or

has " t w o roots", which happen t o be the "two numbers" 1 and 1. ) We have therefore attempted t o write t h l s t e x t in such a way

t h a t the reader will understand what we mean w i t h o u t having to pay undue attention to the conventions t h a t we arc following. Sometimes we shall use t h e phrase "two d l f f e r e n t pointst' f o r emphasis -- even when the word " d i f f e r e n t " is not necessary logically. Postulate 1, f o r example, uses "different" In t h l s way. If you want to acquaint yourself in advance w i t h the notat i o n s t h a t are adopted in the t e x t , see the index of symbols at t h e end of t h e volume. Problem -Set 2-4

The numerical value of the l e n g 11 d l v i d e d - t h would be 1 by a g = I 1 7 or approximately 1.3; that of the width would be 1. b. The numerical value o f the w i d t h would be 8 72 d i v i d e d 1 7 or approximately 0.77; t h a t of the length by 11 = 22 would be 1. 4. 36 2 + 48 2 = 602 = 3600. 16 5. a. P = 4 - 4 8 = 192. b. P = 4. 4

3.

a.

A 33 +6.

1.

=

a*

3. ):(

48

2

5=-Tv

=

+

b2 = c2

2

+

b

1.

4 2 --16 9 . Given.

3.

Another form of S t e p 2.

2304.

=

A =

.

c 2 (fi) .

[pages 30-331

"7.

If t h e length o f any s i d e of t h e square is

s

u n i t s , It

5s given that

s2 = 4s a2 - 4 s

0 or s ( a - 4 ) = 0. The only meaningful s o l u t i o n to this equation is s = 4. Area and perimeter w i l l be numerically equal only if a s i d e 1s 4 units long, whatever t h e u n i t may be. Since any change In u n l t will change t h e 4 t o something else, t h e area and p e r i meter w i l l no longer be numerically equal. ( ~ o t et o t e a c h e r : & ready to commend o t h e r correct proofs students may give. The concept of g e n e r a l i z a t i o n i n mathematics is an important one.) from which

33

34

=

S e c t l o n 2-5 beglns by appeallng t o t h e s t u d e n t t s howledge. It then t l d e s c r i b e s this situation1'formally, in Postulate 2. The postulate is n o t a c a s u a l l y chosen group of words to use in playing a game. It I s on the c o n t r a r y a carefully chosen s t a t e ment t h a t g i v e s one 01" the basic p r o p e r t i e s of p o l n t s ; it forma l i z e s something w i t h which t h e student is already familiar at an informal o r intuitive level. Later postulates wlll continue the process of c h a r a c t e r i z i n g p o i n t , l i n e , and plane by formalizing properties which are Intuitively familiar or which h a v e been suggested by physical experience. Notice how t h e f i r s t strictly geometric d e f i n i t i o n is s e t off. This p a r t i c u l a r d e f i ' n i t i o n does not lend itself to a discussion of the nature o f mathematical d e f l n l t i o n as w e l l as some l a t e r ones do, so the text postpones such a discussion until a more suitable example appears. P o s t u l a t e 2 and the definition of "distance" use some words such as "anytt, " d i f f e r e n t 1 ' , llunlqueN which have not been defined, and t h i s may bother very d u t i f u l students who a r e t r y Ing t o be p r e c i s e , You can simply say that we are using the

(pages 3 3 - 3 4 ]

35

Engllsh language in t h e course, assuming t h a t the meanings of all s h p l e non-geometric terms are known. Such terms are used with t h e i r usual rneanlngs. In o t h e r words, t h e language o f ordinary speech is assumed. Geometric terms, words with t e c h n i c a l meanings, are the ones t h a t are treated carefully within the system of geometry. In Section 2-6 on the infinite ruler, we a r e trying to prepare t h e student In an i n t u i t i v e manner f o r Postulate 3 h he Ruler P o s t u l a t e . ) When investigating the general rule that the distance between the p o i n t t h a t corresponds to x and the point that corresponds to y is ly - x 1 it might be well t o check the r u l e f o r some whole numbers f i r s t , There are only three cases we have to consider: (1) both points correspond to p o s i t i v e numbers, (2) one p o i n t corresponds to a positive n m ber and the o t h e r corresponds to a negative number, and ( 3 ) b o b points correspond to negative numbers. The case when one p o i n t corresponds to zero has already been considered when discussing absolute values. P Q

10

0

Case 1

I

Case 2

-5

0

21

I

6

It Is clear t h a t the distance f r o m

P t o Q, (which I s the same as t h e ~ i s t a n c efrom Q t o P , ) I s 11 in all three cases above. Now l e t us check and see if the absolute value of the d l f f e r e n c e of the corresponding numbers will give the distance between these p o i n t s regardless of the o r d e r i n which we take t h e numbers in the formula, PQ = ly - xi. Case 1. Casee.

Case3.

10) = 11, and 110 - 211 = 11, P Q = 1 6 - ( - 5 ) l = 1 1 , and 1 - 5 - (611 = l l . P Q = 1-10-(-21)) = 1 1 , and )(-21) - (-10)l = 1 1 .

P&=

121

-

[pages 34-35]

37

Now t h e R u l e r P o s t u l a t e seems reasonable, because we have seen t h a t i t will give u s t h e r e s u l t s t h a t we would expect from t h e previous discussion. We now have a c o o r d i n a t e system on a line; t h e number corresponding t o a p o i n t is the c o o r d i n a t e of t h a t point. 'I3ough the book mentioned p r e v i o u s l y t h a t

a l i n e i s a set

of points, t h e r e was no formal statement a b o u t how many p o i n t s

a line contains.

P o s t u l a t e 3 gives us infinitely many p o l n t s on every l i n e . This i s so because we have assumed t h e real number system and are now postulating a one-to-one correspondence between the s e t of points on a line and t h e s e t of r e a l numbers. (The t e x t w i l l use t h e phrase "one-to-one correspondence" formally In Chapter 5 . ) When we say t h a t t h e p o i n t s on a l i n e are i n a one-to-one correspondence w i t h the real numbers, we mean: (I) to each p o i n t o f the l i n e there corresponds e x a c t l y one r e a l number and ( 2 ) to each real number t h e r e corresponds e x a c t l y one p o i n t o f t h e l i n e . One-to-one correspondences are n o t unlque to mathematics. For instance, how many times have you taken attendance In your class by looklng to see if each asslgned seat in the classroom is f i l l e d ? What you have done is t o e s t a b l i s h a one-to-one correspondence between assigned seats in your c l a s s room and s t u d e n t s in y o u r class. If you can match up a seat with each s t u d e n t , you know t h a t a l l of the s t u d e n t s are present. P o s t u l a t e 3 is a v e r y powerrul t o o l . Part (3) guarantees t h a t d i s t a n c e s on a l l n e behave In a way t h a t we would n o r m a l l y expect them to behave in. It would not be sufficient to postulate j u s t the existence of a one-to-one correspondence. We cannot have a n y t h i n g like t h i s :

because such an "undesirable" r u l e r does not satisfy Part (3) of P o s t u l a t e 3 . [page 371

If you are familiar with the foundations of geometry you may find Postulates 3 and 4, with their strong emphasis on algebra, rather strange. We have introduced real numbers in P o s t u l a t e s 2, 3, 4 a s a pedagogical d e v i c e at the t e n t h grade level to avoid very subtle and difficult d 1 S C ~ s S l b n Son the theory of measure of segments. (see the Talk on the Concept of Congruence f o r an indication of this. ) One ahould not infer t h a t we consider t h i s the best treatment at h i g h e r levels. In an advanced course In the foundations of geometry we would p r e f e r a treatment of t h e type given in Ailbertls Foundations o f Gearnetry or Veblents Monograph on the Foundations of Geometry ( ~ o n o g r a p h1 in Plonoaraphs on Topics of Modern Mathematics, edited by J. W. A . young.) In such a treatment t h e postulates would be more geometric, making no reference to a l g e b r a i c ent i t i e s , and o u r Postulates 2, 3 , 4 would appear as theorems -indeed d i f f i c u l t ones to prove. Note the c o n t r a s t w i t h the conventional treatment [and with ~ u c l i d )where betweenness is not even mentioned and betweenness r e l a t i o n s are taken, when needed, i n t u i t i v e l y from pictures. The e a r l y introduction of real numbers permits us to define betweenness. The mathematical treatments o f B i l b e r t and Veblen take betweenness as undefined and characterize it by postulates.

--

Problem Set 2-6 37 1.

a.

3.

d.

2.

1 L 1

I

II 1

II

1

I

( ~ o t et o teacher:

In

exact, w h i l e 1.732 - 1,414 result. )

b.

c.

38

out that - 6 is 0.318 is an approximate

( g ) point =

Ir - 31 by Jim's s c a l e . l(r - 5 ) - ( - 2 ) j = Ir - 31 by Pete's s c a l e . Ir - n 1 by Jim's s c a l e , 1(r - 5) - (n - 5 ) ( = Ir - nl by Pete's s c a l e .

4.

Subtrac-t

5.

a.

Yes.

b, c.

p

f r o m the value at

Q.

2 and q + 2. T h e d i s t a n c e , by definition, is Ip numbering

+

1 ( +~

2) - (q + 211 =

IP

ql.

For t h e new

ql.

-

Yes. Consider two p o i n t s with c o o r d i n a t e s n and r . After renumbering t h e o r i g i n a l scale, t h e coordinates will be d.

6.

( -n) and ( -r)

.

The d i s t a n c e between them is

In

It is a l s o true t h a t I(-n) 7.

a.

b.

c.

- r1 =

. Ir

-

nl

=

In - rl.

No. Gamma. 9 miles or 41 miles

Beta

l6

Alpha

Alpha

l6

Beta

Gamma

-

-

25

Gamma

Alpha.

d . Alpha. ( ~ o t eto teacher: This problem is leading up to the concept of betweennees.) [ pages 37-39

I

39

8.

9.

40

There are 2 p o s s i b l e arrangements. B

can be between

A

and

C.

A

can be between

C

and

B.

3

isbetween

and

C.

A

4

lo

B

15

C

14,

The concept of betweenness, though intuitively n a t u r a l , is

in h i g h school treatments of geometry. from the discussion in t h e text it can be seen that this can be a v e r y t r i c e concept if we c o n s i d e r the problem on a c l o s e d curve. Fortunately, l a t e r discussions and treatments in the text consider betweenness on a l i n e o n l y . In connection w i t h the idea of betweenness, It might be worthwhile t o propose the following problem to the c l a s s : Ln how many ways can f o u r round beads, of different c o l o r s , be one that has rarely been formalized

41

arranged in a s t r i n g so as to make a four-bead necklace? The answer is that there are only t h r e e different ways. The p o i n t is t h a t there i s only one way f o r t h e f i r s t t h r e e beads, A, B, C t o be arranged in t h e necklace; t h e six orders ABC, ACB, BAC, BCA, CAB, CBA all form the same necklace. The o n l y real c h o i c e is in the p o s i t i o n of the f o u r t h bead, D, and f o r t h i s t h e r e are t h r e e possibilities: D c a n be immediately between A a n d B, or immediately between B and C , o r immediately between k and C . The d e f i n l t l o n o f "between" is followed by a discussion of d e f i n i t i o n s in mathematics. A mathematical definition must be distinguished f r o m a dictionary definition which o f t e n gives only a synonym or description of the term d e f i n e d . A mathemati c a l d e f i n i t l o n Is, a s this manual mentioned in Chapter 1, a formal agreement to u s e -- when d e s i r e d - - one phrase a s an abbreviation f o r another phrase.

[pages 39- 41 3

Notlce that a d e f i n i t i o n I s l o g i c a l l y very different from a theorem. A typical theorem is in the form, if A, then B, where A and B are statements. It says t h a t statement B i s deducible from statement A . For example, l e t A be t h e statement ''a triangle has two congruent s l d e s " and B t h e statement "a triangle has two congruent anglea.ll These s t a t e ments mean d i f f e r e n t things, and we have learned a geometric fact when we prove that the second statement Inevitably f o l lows from the first. On the o t h e r hand a typical d e f i n i t i o n is of the form: P stands for (or I s an abbreviation of) Q, where F and Q are phrases. For example (see Chapter 1, commentary) let P be "parallelogram" and Q be ''a quadrilateral whose opposite s i d e s are p a r a l l e l . " No i m p l i c a t i o n i s involved h e r e -- P and Q are not even statements. Rather we are making an agreement, motivated by convenience, t h a t t h e short phrase P s h a l l stand for t h e long phrase Q. Sometimes, t o avoid awkwardness of language, we s t a t e a definition in " i f - - t h e n " form, for example: if the opposite sides of a quadrilateral are parallel, then we c a l l the quadrllateral a parallelogram. Don't be misl e d by this. No implication is involved. We are not stating a geometric fact, but an agreement about how geometric termlnology shall be used,namely that the word "parallelogram" s h a l l stand for t h e phrase "a quadrllateral whose oppoaite sldes are p a r a l l e l . l' You can d i s c u s s definitions in such down-to-earth terms as these: A mathematical d e f i n i t i o n i s a convenient handle for dealing with a mathematical idea just as t h e s e t o f f i n g e r h o l e s in a bowling b a l l is a convenient handle t o use when rolling the ball. You may want to present the idea of d e f i n i t i o n to your class like t h i s : Consider the following definitlon of "honor s t u d e n t . " " ~ t u d e n t sof East High w i t h a deportment grade of A and no academic grade below B are called honor students." [pages 41-42]

42

Knowledge that Cindy Marshall Is an honor student at East Hlgh is a l s o laowledge t h a t she has a deportment mark of A and has no academic mark below B definition of "honor student1'. On the o t h e r hand, Imowledge t h a t E r i c Hughes, a s t u d e n t at -st H l g h , had A in deportment and no mark below B is lmowledge t h a t he 1 s an honor student - - again b~ definition. ti on or s t u d e n t " is a convenient label that spares people all the words " a s t u dent with a deportment mark of A and no academic mark below B". A f i g u r e for Theorem 2-1 m i g h t lead t h e students to feel that they can "see" t h a t point B is between A and C. What must be realized 1 s t h a t a figure is not s u f f i c i e n t justification of a proof. To prove this theorem formally we must prove it on t h e basis of t h e d e f i n i t i o n and not the configuration, for the only formal lmowledge we have of betweenness is that provided by t h e d e f i n i t i o n . You might wonder why we prove theorems l i k e 2-1 at a l l ; t h e y seem so o b v i o u s . N o t l c e that according to our l o g i c a l program, a s outlined in S e c t i o n 1-2, every statement of o u r geometry must be either a theorem or a postulate. We could, of course, take as postulates all statements a s obvious as Theorem 2-1, and some text-books do thls. We choose, rather, to u s e as few p o s t u l a t e s as we feel a r e pedagogically necessary, and prefer to g i v e proofa of even the l'obvious" theorems. This does not mean that e i t h e r you or your students need spend much time on the proofs. We merely b e l i e v e that it is good f o r the students to h o w t h a t some "obvious" things can be proved, and t h a t mature rnathematlcians do n o t regard i t a waste of tlme to d e v i s e such proofs (and some of them are unimaginably difficult,) You will probably want t o point out to your s t u d e n t s t h a t t h e y are n o t expected t o " l e a r n " t h e proofs o f the theorems in t h l s c h a p t e r . The theorems may not seem meaty to beginning geometry students, and t h e p r o o f s are not at a l l typlcal of the kind of geometric reasoning they will usually be doing. We do n o t expect them to h o w how to write proofs of their own until Chapter 5 . The book gives p r o o f s f o r the sake of completeness.

and then go on, A s s u r e t h e students t h a t t h e tlme for mastering simple geometrfc proofs w i l l come, and t h a t t h e book will t h e n help t h e m get a s t a r t . The statement t h a t if x < y, t h e n y - x I s p o s i t i v e , might r e q u i r e some amplification. We can illustrate t h i s w i t h a s p e c i f i c example, letting x and y represent 2 and 7 r e s p e c t i v e l y . I f x < y, and we subtract t h e smaller number from t h e larger,then it Is certain t h a t t h e difference w i l l be a p o s i t i v e number (y - x > 0). If, on the other hand, x < y, and we s u b t r a c t y from x, we would have x - y < 0. If we s u b t r a c t 7 from 2 we get a negative number, w h i c h Is, of c o u r s e , less t h a n 0. In the theorem it is g i v e n t h a t x < y. Then y - x i s p o s i t i v e and, by definition of a b s o l u t e value, = y - x. Iy Go through them once,

42

XI

Problem Set

10. f. I x l - X g l . 2. It i s o n l y necessary t o read a single poaltive number if one u s e s the Ruler Placement P o s t u l a t e . N e i t h e r s u b t r a c t i o n nor computing an absolute value is necessary. 3. R S + S T = R T . 4. The coordinate of A is -2; t h a t of B is 11~. 5. c . See t h e Ruler P o s t u l a t e and definition of between. 6. The p o i n t having coordinate x l . Theorem 2-1. c.

43

7.

a.

By the Ruler P o s t u l a t e : 1 AE = 10 - 5 rl = 1

7

AE = EF

=

FB.

Therefore, E is between A and F. *8.

45

The inequality x > y > z can a l s o be written z < y < x , in which case y z, x - z, and x - y are a l l p o s i t i v e . Therefore, CB = y - z , CA = x - Z, and AB = x - y. From these t h r e e equations we observe t h a t C B + BA = x - z = C A . Therefore, B is between A and C. ( ~ o t e : A brief proof relates z < y < x to Theorem 2-1, )

-

The term "ray" mlght be new t o students. The text makes clear the distinction between ray and sejqnent. What should be pointed out to the students is that In t h e n o t a t i o n f o r a ray, -ir for i n s t a n c e AB, the first letter is t h e end p o i n t and the second 18 one of t h e Infinitely many p o i n t s t h r o u g h which t h e ray passes. I t is not c o r r e c t , t h e r e f o r e , to refer to t h e ray whose end p o i n t is A and which passes through point F as + + FA. The correct n o t a t l o n 1s AF, Observe t h a t In t h e f i g u r e for Theorem 2-4 the point P need not, in a p i t e of the dlagram, lie to the r l g h t of p o i n t B, P may be t h e same point as B, or P may be between A and B. However, P cannot be at A,,and A cannot be between P and B, since x is a positive number, Remarks on The L i n e Separation Theorem. The following theorem 3s not stated in the t e x t , but is often used tacitly later. It describes the separation of a line by a p o l n t , and is c l o s e l y analogous t o the l a t e r p o s t u l a t e s in Chapter 3 deal1ng w i t h t h e separation of a p l a n e by a Ilne and the separation of space by a plane.

---

[pages 43-46]

The Line --

Separation Theorem. L e t P be a p o l n t of t h e line L. Then L is the union of f and t w o s e t s HI and H2 not c o n t a i n i n g P, such t h a t (1) No p o i n t of L lies in both H1 and Hp. ( 2 ) If two p o i n t s Q and R are b o t h in the same set, H1 or H2, then P is not between Q and R, and (3) If Q Is in H,I and R is in H2, then P is between Q and R . Proof: L e t us s e t up a coordinate s y s t e m on L such t h a t P corresponds to 0 . Let HI b e t h e s e t of a l l p o i n t s of L w i t h negative coordinates and let !I2 be t h e set of a l l p o i n t s of L w i t h p o s i t i v e coordinates. Then L is the union o f P, H1 and H2, because every real number is positive, negatlve or zero. P Is n o t In either HI or H2 because 0 13 neither p o s i t i v e nor n e g a t i v e . (1) h o l d s because no number is b o t h p o s i t i v e and negative. It remains to verify (2) and (3). Let Q and R be p o i n t s with c o o r d i n a t e s x and y . Suppose that y is t h e larger; this 1s merely a choice of n o t a t i o n . If Q arid R are In HI, t h e n x < y < 0 ; by Theorem 2-1, R is between Q and P; and so P is not between Q and R . If Q and R a r e in H2, then 0 < x < y; Q I s between P and R; and so P is not between Q and R. This v e r i f l e s ( 2 ) . Let Q, R , x and y be as before, w i t h x < y. If Q is in H1 and R 1 3 In Hz, then x < 0 and y > 0 . Therefore, x ( 0 < y; and t h e r e f o r e , P is between Q and R. This v e r l f i e s (3). This theorem has Seen deliberately kept; o u t of t h e t e x t . It is so obvlous t h a t students can be expected to use it t a c i t l y and its p r o o f is not very interesting mathematically. Of course, the half-lines R1 and H2 are analogous to t h e half-planes and half-spaces t o be discussed In the next chapter. Notice that a half-line is d i f f e r e n t from a ray; a ray contains its end-point, but a h a l f - l l n e does n o t .

Notice t h a t the Line Separation Theorem guarantees that every ray has exactly one opposite ray.

-

Problem S e t 2-7b

Two. a. Theorem 2-1. b, Theorem 2-3. c. D e f i n i t i o n of between. 4+ a . Points X and Y and all points of XY between X and Y . b. P o i n t s of and all points Z of such t h a t Y is between X and 2. Caae 1. If A I s between B and C, then AB + AC = BC. Slnce AB = E, t h i s leads to the impossibility AC = 0.

8

If C is between B and A, then BC This leads t o the irnpossibillty C A = 0. Case 2.

+

CA = BA.

Case 3 .

B is between A and C, by Theorem 2-2, is t h e only rernalning possibility and must be true.

(Note: A proof baeed on setting up a coordinate system and using Theorem 2-1 is also possible. ) Theorem 2-4. Proof. Statements: 1. A B + BC = AC, 2. AC - PB = BC. 3. BC > 0. 4. AC > AB. a.

%!

Reasons: I k f l n i t i o n of between. h b t r a c t i n g AB f r o m each side. Distance Postulate. If AC - AB > 0 , AC > AB. and R but contains R . R belongs to X& but

contains p o i n t s Y neither p o i n t s Y nor Y does n o t . YZ + ZR = YR.

[pages 46-48]

R e v i e w Problems S1: S4:

Sg; S5j the empty a e t .

1; 2; no.

c.

Positive, Between 0 Negative.

a.

AB

+ BC

b,

AB

=

a. b.

and

2.

= AC.

DC.

There a r e 12 p o s s i b l e o r d e r s . I s t o t h e r i g h t of A . D A


B

A B t I32 = AC. % contains points A and C, but 68 contains n e l t h e r point A nor p o i n t C. A belangs t o b u t C does n o t . x = 9, y 4.

a

-

Perhaps they l i v e in the same house. H o ~ e v e r , since people a r e not always p r e c i s e fn every day l a n ~ ~ a gus.sge, e it may be t h a t t h e y o n l y live near each o t h e r - - as on opposite s i d e s of t h e street. I? - 2. a . AF and E. b. E and P. c.

d.

Triangle AFE. The empty s e t .

Trlangle AEF. 5. ( A X D , ABCE, ABDE, ACDE, BCDE.) b. 10. (AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.) c . 10. No. AC c o u l d only be 13 o r 7. e. a.

F ( ~ h o u l dbe 6). e. T 0. T f. T C . T g. T d. T h. F (~houldbe 7 ) . Yes. Since y is l a r g e r than x, the value o f y - x will bc t h e same as the value of Ix - y ( ( b ) i s n o t a c o o r d i n a t e system because t h e numbers 4, 3 , 2, 1 and 0 each correspond to more than one 3 o i n t . This is rlot permlsslble according t o P o s t u l a t e 3 . ( e ) is not e c o o r d i n a t e system because the dlstance between a.

.

poinSs with c o o r d i n a t e s 2 and 1 in t h e original numbering 1s 2 - 1 o r 1 In t h e numbering of ( e ) t h e distance between the sane two points is 12 - (-111 or 3. By Postulate 2 t h e same two points can correspond t o o n l y one number indicating d i s t a n c e . d: b, e; h; f.

[ pages 50-51 1

---

Illuatratlve Test Items f o r Chapter 2 A suitable chapter test might be made by selecting prob-

lems from the following list. These have been grouped i n t o s e t s of problems t h a t are similar with t h e idea that the teacher may w:sh to make a test by choosing none or more from each s e t , In compiling t h i s list and later lists, we generally have omitted items specifically calling for statements of d e f i n i t i o n s , postulates, theorems, and so on, in t h e b e l i e f t h a t each teacher on his own will draw on this wealth of test material, as well a s on his own ingenuity in constructing his own tests. A 1. a . L e t A be the set of squares of the f i r s t eight non-negative integers, L i s t the members of this set,

Let 3 be the s e t of the first e i g h t even p o s i t i v e i n t e g e r s . L i s t t h e members of t h i s set. c. I n a t is t h e intersection of sets A and B? d . What is the union of s e t s A and B? 2. Consider t h e set of a l l p o s i t i v e integers d i v i s i b l e by 5 . Consider the s e t of all p o s i t i v e integers d i v i s i b l e by 3. List the f i r s t five integers in the intersection of these two s e t s . 3. The i n t e r s e c t ion of ray 8 and ray 3 1s . The + union of ray AB and ray is 1. Arrange the f i v e colllnear p o i n t s E, L. M, S, T fn proper order if, LM + - M E = LE; SE + ET = TS; LS + SM =ML. H A number scale is placed on line RS w i t h -5 falllng at R and 6 at S . If the Ruler Placement Poatulate is applied w l t h 0 placed on R and a positive number on S , what will be the coordinate of S? Copy the following sentences and supply the appropriate missing symbols over each l e t t e r pbir, a. AB has no end points. b. The end p o i n t s of ME3 are M and R. c . RQ has one endpoint, R. b.

3

B.

4.

Three towns Lander, Manton and A m i t y are collinear but

n e c e s s a r i l y in t h a t order. It I s 9 mlles from Lander to Manton and 25 miles from Manton to Amity. a . Is it possible t o tell which town is between t h e other two? b. Which town is n o t between the o t h e r two? c . What may be t h e distance from Lander to Amity? d . I l l u s t r a t e with sketches. Given A , B, and C are t h r e e collinear p o i n t s with AB = 8 and CB = 5. If, also, the c o o r d i n a t e of B is -2, and the coordinate of A is less than t h a t of C, what are the coordinates of A and C? Draw two sketches giving different s e t s o f answers. A B F Ii T

not

C.

1.

In the f 'igure: the length o f AB l a t h e l e n g t h of AH is t h e length of is the length of 3 is e . t h e length of HT is or If A c.orresponds t o 0 and 3 t o 1 on a number l i n e , what set of numbers correspond t o the points o f + the ray Of the ray BA?

a. b, c. d.

3.

s?

D.

1.

a.

b. C.

1-71 + 131 = 1-71 - 131 = 1-71 1-31 =

-

2.

a. b.

3.

4.

How many square r o o t s does the number 49 have? &=

Write a s an inequality: K is a greater t h a n -10. b. Restate t h e following in words: Make a t r u e statement o u t of each of pressions by r e p l a c i n g each question t h e following symbols, , =, 5, 2 2 ? - 4 - 2 ; Ix + yl ? 1x1 + a.

Answers

4.

a.

No.

b.

Amity.

negative number 20

>x >

10.

the following exmark by one of :

13-61

lyl.

?

16-31;

2.

a.

7

b,

d.

~ + 7 , y. y - 3.

e.

y - x or ( x - y l or ly

c.

-XI.

3.

The set of numbers, x, such that numbera, x, such that x 1.

2.

Two; 7 .

3.

a. b.

4.

=;


K > -10. x is a number between 10 and 20.

>; 2 .

0. The s e t of

Chapter 3

LINES, PLANES A N D SEPARATION

The material of t h i s chapter differs from t h a t of t h e traditional text in several ways, F i r s t , some elementary solid geometry is introduced, f o r t h e authors believe t h a t there s h o u l d be no undue separatlon of s o l i d geometry from plane geometry. Second, t h e Impartant idea of convexity I s introduced. Most of t h e familiar geometric figures, such as triangular and r e c t a n g u l a r r e g i o n s , or the interiors o r circles and spheres, or rectangular s o l i d s and c f r c u l a r cones, a r c convex s e t s . Finally, t h e separation of a plane by one of i t s lines and the separatlon of space by a plane are studied. These ideas are t r e a t e d purely intuitively In most geometry texts as is indicated by phrases such as "two points are on opposite s i d e s of a line."

54

The t e x t description of the f i g u r e on page 54 asserts t h a t p o i n t s A, B, C and E are coplanar. Actually, F is contained f n the same p l a n e as A, B, C and E, and we can say t h a t A, 8, C, E and F are coplanar.

54

Most students will not see r e a d i l y t h a t Postulate 5a really does f i l l a p l a n e w i t h points. We do n o t belfeve that you should press t h e m a t t e r , f o r most students will not be interested in something so "obvious." You c a n show inquiring students this by using Postulates 1, 6 and 3 along with Postulate 5a as follows: A plane has

three non-colllnear p o i n t s A, B, C by P o s t u l a t e 5a. Then by P o s t u l a t e 1 there is a line L determined by B and C. The plane c o n t a i n s line L by Postulate 6 . Line L has infinitely many p o i n t s by Postulate 3. P o i n t A , in

combination wfth these points Individually, deterrnlnes infinitely many lines by Postulate 1. A l l of these lines (and t h e i r infinitudes of points) lie in the plane by Postulate 6. 56 P o s t u l a t e 6 assures us, as the text points out, that a plane I s flat. Your students can use a g l o b s in the classroom t o see t h a t is is not possible to f i n d two p o i n t s on a sphere such that the l i n e determined by them lies on the sphere. A sphere as a surface, then, does n o t s a t i s f y Postulate 6 . Other surfaces, for example cylindrical ones, a r e t r i c k i e r . Your students can find points on a steam p i p e in your room auch that the l i n e determined by them lfes on the p i p e . Pupils should r e a d i l y see, however, that finding some such pairs of points i s not enough. The question remains: do a l l pairs of points on the pipe satisfy t h e requirement? Since the answer is no, the c y l i n d r i c a l surface of t h e p i p e does n o t satiafy Postulate 6. A triangular region does n o t satfsfy Postulate 6 . Although the region contains the segment joining its points A and B, it does not c o n t a i n the l i n e AB which is determined by the points. 55 Theorem 3-1 could be a t a t e d I n the if-then form: If two l i n e s intersect, then they i n t e r s e c t in only one p o i n t . The t w o statements are equivalent. The students should be reminded of t h e fact that t h e "if . . then . . " r e l a t i o n s h i p is n o t unique t o geometry or mathematics. It is a cause and effect relationship common to science and everyday l i f e , for example:, "1f I do n o t sleep f o r two consecutive nights, then I will be tired." Statements auch as this o f t e n occur in conversation. F u l l use of the exercise material I n recognizing the hypothesis and conclusion of statements should be made when you reach S e c t i o n 3-2. Teachers w i l l recognize the proof of Theorem 3 - 1 as being 55 indirect. The t e x t does not wish t o describe indirect proof at t h i s p o i n t , o r even to describe proof at a l l . The thing to do, we b e l i e v e , 1s to go through the proof once with emphasis on

-

.

.

understanding and then go on without asking s t u d e n t s t o learn the proof. Theorem 3-1 and the method of indfrect proof are discussed in Chapter 6. One of t h e problems in the teaching of geometry is t h a t of keeping emphasis on t h e ideas of p r o o f s r a t h e r t h a n on rote memory. Teachers have their own ways of doing this, such as changing the labels on figures, encouraging students t o come up tuith different p r o o f s , going from paragraph form to twocolumn form and vice-versa, In other words, discourage mere mernorlzation of p r o o f s . ( ~ careful e n o t to discourage mental effort , however. ) The discuss~onin the t e x t of a way in whfch Theorem 3-2 could be proved suggests that you avoid a proof now--or at least avoid emphasizing one. The proof goes: It is impossible f o r a line and a p l a n e not containing the l i n e t o I n t e r s e c t in two d i f f e r e n t pofnts because then the line, by Postulate 6 , would l i e in the plane. 58 The text proves Theorems 3-2, 3-3, and 3-4 in Chapter 6 . Some t h e spent on t h e drawing of planes and lines in three-space is recommended. Some very smple demonstrations with a p i e c e of cardboard (representpart of a plane) and a pencil (representing p a r t of a line) might be performed to illustrate and clarify those postulates and theorems t h a t make reference t o three-apace. You m i g h t ask questions designed to c l a r i f y some of t h e postulates of thts chapter: f o r example, f o r Postulate 7 , "Why does a s t o o l with three legs tend t o be more stable than a c h a i r with f o u r legs?"

--

Problem S e t 3-la 53

1.

One. Infinitely many lines can be drawn.

2.

No. Three

.

3. 4.

Three No e n d - p o i n t s .

Tkfo end-poin ts.

--

Problem S e t 3 - l b

55 1. F and & are t h e 2.

3. 4.

same p o i n t by Theorem 3-1,

Infinitely many. One. None, if t h e p o i n t s are non-collinear; one, if c o l l i n e a r . Postulate 1. a, One l i n e , by Postulate I.

b.

Three l i n e s if the p o i n t s a r e non-collinear, There are three p a i r s of points, and each p a l r determines a line, by Postulate 1.

One l i ~ e if t h e p o i n t s are c o l l i n e a r .

56

5.

a.

b.

6.

a.

Six: AB, AC, AD, BC, BD, CD. One if D is c o l l i n e a r w i t h A, B, C. Four othervrise: %,

s, B,

8.

p o i n t s is colltnear if t h e r e is a line such that each p o i n t of t h e set l i e s on the line. A s e t of

b.

7.

A set of polnta is coplanar if t h e r e is a plane such t h a t each p o i n t of the s e t lies in the plane. For each plane there a r e at l e a s t three n o n - c o l l h e a r

p o i n t s which l i e in this plane. "Contains " fo m . Given any two d i f f e r e n t points, they lie on exactly one llne

.

Problem -S e t 3-lc

Infinitely many. Infinitely many. One, if the p o i n t s are non-collinear; i n f - l n i t e l y many, if the pointa are collinear. The ends of t h e three legs are always co-planar. The ends of the f o u r legs may n o t be coplanar. Point.

Line. Yes. Yes, if n > 2. A s e t of three or more p o i n t s Is non-collinear if there is no line which contains them a l l . Yes, if A , B, C are non-collinear. No, I f A, B, C a r e collinear. No.

Yes.

a.

A.

b.

C.

c. d.

E. Non-collinear,

o r coplmar .

60

of the tetrahedron is the l i n e segment determined by t w o vertTces. b. S i x ; ~ , ~ , ~ , ~ , ~ , c. Y e s , f'or example, the edges and CD have no p o i n t in common. d . No. The faces can b e paired six ways; each pair has an edge In common. Seven: A B C , B E , BCE, CDE, ADE, ACE, BDE.

11. a.

2 .

~n edge

Problem S e t 3-2 Hyp: Concl:

Concl:

John is ill. He should see a doctor. A peraon has red halr. The person 1s n l c e to h o w . P o u r ~ointslie on one l i n e . They a r e c o l l i n e a r , 1 do my homework well. I w i l l get a g o ~ dgrade.

WP: Concl:

A s e t of points lies in one plane, The points arc coplanar.

m: Concl: Ilyp;

Concl: Hyp :

m:

Two lines intersect. Concl: They determine a p l a n e , If two l i n e s are d i f f e r e n t , then they have a t most on? p o i n t in common. I f a s t u d e n t is a geometry student, then he lmows how t o add i n t e g e r s , If it rains, then it p o u r s . [pages 59-61 I

~

If a p o i n t is n o t on a line, t h e n t h e p o i n t and t h e l i n e are contained in exactly one plane. e . If a practice is dishonest, then it is unethical. f. If t w o lines are parallel, then they determine a plane. Postulate 1: If points P and Q, are different, then there Is e x a c t l y one l i n e which c o n t a i n s them. Theorem 3-1: If lines LI and L2 are different, they i n t e r s e c t f n at most one p o i n t . a , No. The theorem places the intersection of two l i n e s a s a condition for the conclusion while n o t asserting that any t w o lines must intersect. The statement in this problem a s s e r t s that t w o lines must always intersect. b. If two lines intersect i n a p o i n t , then there is exactly one plane containing them. d.

62

3.

4.

Before introducing the p o s t u l a t e s on s e p a r a t i o n it may be we11 to l o o k back arLd re-examine the p o s t u l a t e s we already have. Postulates 1, 5, 6 , 7 , 8 are sbnilar in t h a t they are pure1.y geometric and d e s c r l b e how points, l i n e s and planes l i e on o r are " i n c i d e n t with" each o t h e r . They are c a l l e d incidence p o s t u l a t e s . On the o t h e r hand, P o s t u l a t e s 2 , 3 , 4 involve algebra; they a r e concerned with properties of measurement, and so are c a l l e d metrical p o s t u l a t e s . The incidence p o s t u l a t e s are simple ones that logically_ form a n a t u r a l unit f o r beginning t h e course. But peda-

pogically t h i s does n o t seem attractive, f o r two reasons. Firsg the incidence postulates would confront . t h e student w i t h solid geometry in h f s first approach to a new subject, Second, the proofs of t h e basic incidence theorems ( f o r example Theorems 3-1, . , 3-4) i n v o l v e the indirect method, which causes difficulty f o r many students. To avoid these difficulties we have s p l i t o f f Postulate 1 from the incidence postulates and J o i n e d it to Postulates 2 , 3, and 4 t o f o m the basis of a beginning unit on measurement

..

[pages 61-62]

i n Chapter 2. This makes use of the student's h o w l e d g e of algebra, and involves, geometrically, only s e t s of p o i n t s contained in a l l n e . Then Chapter 3 discusses the incidence p r o p e r t i e s of p o i n t s , l i n e s and planes and s e p a r a t i o n properties. These a r e non-metrical in character. The discussion of the preceding three paragraphs suggests a r a t h e r b a s i c theoretical p o i n t , namely, t h e e f f e c t on a mathematical theory of introducing new p o s t u l a t e s . The next few paragraphs use a minfature geometry to illustrate t h i s baslc point oT theory. We intend t h i s as interesting background material for i t s broad effect, rather than f o r any m e d i a t e application t o the t e x t . Examine P o s t u l a t e s 1, 5, 6, 7 , 8. You see that they inolude familiar determination and intersectlon properties of points, lines and planes In Euclidean s o l i d geometry, and also, in Poatzulate 5, a minimal indication of how numerous points are. You probably have in mind, in my case, that a l i n e and a plane contain infinitely nany p o i n t s . But t h i s can n o t be proved on the basis of Po~tulates1, 5, 6, 7, 8. We show t h i s by exhibitfng an appropriate "model" for Postulates 1, 5 , 6 , 7 , 8. The model is a c o n c r e t e system of o b j e c t s which satisfy these postulates. Expressed differently, we g e t a model of our mathematical theory by assigning specific meaning to the undefined terms "point," " l i n e " and "plane," in such a way that the postulates become true statements. To construct o u r model, consl,der a s e t of' f o u r distinct objects, a a, b, c, d , For example, we can take four dots on a piece of paper as indicated in the diagram. Ue c a n think of them I f we wish a a the v e r t i c e s of a triangular '-,LA-' c pyramid. Interpret "point 'I to mean any one of t h e

objects a, b, c, d; "line1'to m e a n any p a f r of these objects; "plane" to mean any triple of them. Then our p o s t u l a t e s are no longer statements involving undefined or uninterpreted terms, but become definite statements ( t m e or false) about the objects a, b, c, d. Thus Postulate 1 now says: any two of the objects a, b, c, d are contained in a unfque p a f r of them. This is trivially true. Similarly, Postulate 6 says that if a triple of the objects contains two of them, then it contains the pair composed of these two. Thls I s also a trivial truth. Similarly it c a n be shown t h a t each of the Postulates 1, 5, 6, 7 , 8 is s a t i s f i e d when p o i n t , line and plane are interpreted in the given way. In virtue of t h i s t h e system composed of the four "points1' a, b, c, d, the six "lines" ( a , b), (a, c ) , ( a , d ) , ( b , c), (b, d), (c, d ) and the four planes (a, b, c ) , ( a , b, d ) , (a, c, d ) , (b, c, d) is called a model for p o s t u l a t e s 1, 5, 6 ,

7 , 8. Since t h e model satisfies Postulates 1, 5, 6, 7, 8 it must satisfy the theorems which are deduced from t h e s e postulates (using no others), for example, Theorems 3-1, 3-2, 3-3, 3-4. Thf s is easily v e r i f i e d . Now you c a n see t h a t the p r i n c i p l e that a line contains infinitely many points c a n ' t be deduced as a theorem from P o s t u l a t e s 1, 5, 6 , 7, 8. For if this could be done, our model would have to satisfy this principle -- and it doesn't, since each of its lines contains exactly two points. Now you can see the e f f e c t of introducing the metrical postulates, in particular Postulate 4, the Ruler Postulate. This guarantees t h a t a line is rich in points, and that its infinitude of points are arranged on the line and determine distances in j u s t t h e way we want for the kind of geometric theory we are constructing. The introduction of t h e m e t r i c postulates excludes finite models,of the type we have discussed, which do satisfy the incidence postulates. This illustrates the basic theoretical point we mentioned earlier: i n g e n e r a l , as new p o s t u l a t e s are added In a mathematical theory, t h e scope of its a p p l i c a t i o n , t h a t is the family of models which satisfy the postulates, is reduced. See the Talks: The Concept of

-

62

Cong;ruence and Miniature Geometries. Notice that in s e t s D, E, F there are infinftely many pairs of points such that the segments Joining them are contained In the s e t . The existence of a single p a i r of points

P, Q such t h a t PQ does not lie I n t h e s e t is sufficient to eliminate the possibility of convexity. Thus the union of the s e t of p o i n t s in t h e interior of a clrcle and one p o i n t outside t h e clrcle is n o t a convex s e t . Separation properties are n o t explicitly mentioned or explained in Euclld or in conventional texts. They appear in geometry in statements such as, "Consider two t r i a n g l e s which have t h e same base and a pair of vertices on opposite sides of the base." They appear in everyday llfe when we say, f o r example, that t h e town h a l l and the school are on the same s i d e 63 of the main highway. Notice how the text uaes the basfc idea of segment t o g l v e a preclse statement of what is Involved in the separation of a plane by one of i t s lines. The intuitive idea of two p o i n t s being on the "same side" of l i n e L is expressed precisely by the condition t h a t the segment j o i n i n g them does n o t i n t e r s e c t L. Notice how the p r e c i s e formulation of the separation postulate agrees with our Intuitive I d e a s about separation. 66 Postulate 10, t h e Space Separation Postulate, is entirely similar to Postulate 9, the Plane Separation Postulate. The corresponding result f o r a l i n e c a n be proved from the Ruler Placement Postulate, and was given at the end of C h a p t e r 2 of the Commentary.

Problem Set 3 -3

66

1.

a. b. c.

d.

Yes. The Ilne segment joining any t w o p o i n t s o f the l i n e lies entirely in t h e line. No. There Is one segment Joining the two p o i n t s and it does not l i e in the two p o i n t a . Yes. No. Any segment containing the removed p o i n t would not

-

[pages 62-66]

e.

lie entirely within t h e B e t even if I t s end-pofnts were w i t h i n the set. No. For any t w o points, R and S, of the s e t the segment does n o t lie in the surface. (ordinary 3-space is considered here.)

Yes. No. No. Yes. No. Yes. No. I.No. Yes. Yes. No. J . Two. Half-spaces. No. It i a necessary t h a t f o r every two p o l n t s , the entire

f. g. h.

67

2.

3.

4.

5.

segment jofning them lies in the set. V only. V is t h e only s e t in which the segment between any two p o i n t s is contained in t h e s e t . Y e s . Take any two points P and Q in t h e p l a n e . By Postulate 6 , we h o w t h a t t h e l i n e containing t h e s e p o i n t s l i e s in the plane. Hence Is contained in the plane, making t h e s e t convex. a. Y e s . For m y p o i n t s P and Q in the intersection:

b.

6,

7.

No.

Points P and Q may be selected as f o l l o w s :

No. Any segment containing the removed p o i n t would n o t lie entirely in the s e t even If its end-points were in the set. Yes. [pages 66-67]

8.

Any figures

of the following nature:

NOT CONVEX

CONVEX

9.

Yes. 10. No. 11.

a.

b.

12.

b.

-

68

a.

13.

No. The l i n e separating t h e half-planes is n o t contained in the unfon. No. A large region of the plane is still n o t covered, as I n the diagram.

Two. Half l i n e s . The Line Separation Statement: Given a p o i n t and a l i n e containing it. The p o i n t s of the l i n e d i f f e r e n t

from the given p o i n t form two sets suck that (1) each of the s e t s is convex and (2) if P is in one s e t and $ I s in the o t h e r , then t h e segment contains t h e given point. A r a y has an end-point, but a h a l f - l i n e has no end-point.

[pages 67-68]

14.

No. Yes. No.* Yes. Yes. ( + ~ h r e elines can separate a plane i n t o five regions if we a l l o w two parallels through a p o i n t t o a line. This would give :

However, if we should assume only one p a r a l l e l through a p o i n t to a l i n e , we could n o t g e t five regions. Note that within o u r postulational system so f a r developed we do not h o w which choice, if either, we dl1 accept, or which will be excluded.) 15. Four. Three. 16. E i g h t . Pour. "17. Consider the segment joining any two points P and Q of the Intersection. is contained in the first s e t , s i n c e it is convex. Is contained in the second s e t , since it is convex. By the definltlon of intersection, the intersection contains all points common to the two s e t s . Therefore, the intersection contalns and the intersection is a convex set. E *18. Two posslble figures:

m,

Review Problems

They may i n t e r s e c t In a p o i n t ( a s t h e corner of a room where t w o walls and the f l o o r meet) . Also, there may be no point common t o all three if there are three lines each of which is the lnteraection of t w o of t h e Yes.

No.

planes.

One p l a n e . a. If a zebra has polka dots, then it is dangerous. If a peetangle has s i d e s of equal l e n g t h s , then it is

.

c. d.

a square. If Oklahoma wins, then there will be a celebration. If t w o s t r a l g h t l i n e s intersect, then they determine a plane.

e . If a dog I s a cocker s p a n i e l , then it is sweet-tempered. E a c h half-plane is convex. Y e s .

From this statemen5 one g e t s the impression that a plane has boundaries. To h w e said, "The t o p of the t a b l e , if it were absolufely f l a t and smooth, would give a good idea of a small p a r t of a plane," would have been a b e t t e r s t a t e ment. Three n o n - c o l l i n e a r points. A l i c e and a p o i n t n o t on t h e l i n e . Two Intersecting l i n e s . In t h e set. Yes. No. Sfnce lles e n t l r e l y in plane E , if the two lines were to i n t e r s e c t , Ll would have to c o n t a i n some o t h e r point of plsne E. This is impossible by Theorem 3-2. a, One l i . n e c o n t a i n s a l l pointa of the s e t . b One plane c o n t a i n s a l l p o i n t s of t h e set. c. Yes. d. Yes.

Yes. f. No.

e.

g.

Yes.

[pages 6 8 - 6 9 ]

11.

Y e s , by P o s t u l a t e 7 .

12.

Yes.

---

Zllustrative Test Items f o r Chapter

3

I.

If two d i f f e r e n t lines i n t e r s e c t , t h e i r intersection is . If two d i f f e r e n t planes intersect, their intersection is . If a plane and a line n o t contained in the plane intersect, t h e i r intersection is

2,

Which of these regions, if

3,

Which, if any, of the f o l l o w i n g can separate a plane?

my, is not convex?

Segment b. P o i n t c . Line d . Ray F i l l in t h e blanks in the statements below on t h e basis of the figure shown. IMPORTANT: If none of the polnts given satisfies the condition, write NONE in the blank space. a.

4.

Points Points Points Points

P, and D, P, and P, D, B, and C, A, B, and A,

are collinear . a r e collinear. are coplanar. a r e coplanar.

5.

6.

7. I

8.

Write each o f the f o l l o w i n g statements In "if-then" form: a. Two d i f f e r e n t lines have at most one point in common. b. Any three non-collinear p o l n t s lie in exactly one plane. Complete: a . The s e t of a l l points in a plane which lie on one side of a given l i n e of t h e plane is a b. The two s e t s of points into which a separates space are each c a l l e d half spaces. How many planes can contain one given pofnt? Ttvo given points? Three non-collinear p o i n t s ? Indicate whether True or False: a. A line and a plane always have a t most one p o i n t in common. b. Two l t n e s alweys lie in the same plane. c. There a r e lines which do not intersect each other . d. If three polnts are c o l l f n e a r they are coplanar. e. A p o i n t and a line always lie in one and only one p l a n e . f. Given two d i f f e r e n t points A and 8 . There are a t least two d i f f e r e n t lines t h a t c o n t a i n both A and B , g Every two points are collinear. h. A line has t w o end-points. i There is a s e t of four p o i n t s whlch l i e in no plane. b

.

Given two points, t h e r e 1s mare than one plane containing them. S t a t e t h e Plane Separation P o s t u l a t e in your own words. j

9.

1. A p o i n t . A line. 2. d.

A point.

None. A.

P* j.

a. b.

6. a . 7.

8.

If L1 and L2 are two diff'erent l l n e g then they have at most one p o i n t in common. If A, B and C are three non-collinear p o i n t s , t h e n they lie i n exactly one plane. Half-plane.

b. Plane. Infinitely many. Infinitely many. a . F; b. F; c . ; d, T; e. h.

3;

f.

T; J .

T.

One.

F;

f.

F; g.

T;

Chapter 4 ANGLES AND TRIANGLES H i g h school geometries u s u a l l y take the n o t i o n of i n t e r i o r for granted. A person is supposed t o know from looking a t a figure when a point llea in the i n t e r i o r of an angle, for example. Most t h i n g s move along without undue difficulty unless somebody raises such a question as: k t what reason ca 7 you give t o support your claim that p o i n t B l i e s in the i n t e r i o r of a n g l e AOC? Such a q u e s t i o n can hardly be answered when there is no formal lolowledge from which t o reason. This book provides such formal howledge by t r e a t i n g n o t i o n s of betweenness, order and interior. Another way in which this book differs from almost every other text is in i t s careful treatment of angles: their d e f i nition, their ¶tion properties and their measure. This l a s t i s done i n a way t o suggest an analow with t h e measure of distance presented in Chapter 2. There i s a c l e a r - c u t d i s t i n c t i o n in t h l a t e x t between an angle and the measure of an angle. An a n g l e i s a s e t of p o i n t s ; i t s measure is a number. Such a d i s t i n c t i o n between t h e p o i n t s e t and t h e number i s u s u a l l y not made in t e x t books, t h e word 11 angle" being used f o r both. A t the end o f t h i s c h a p t e r you will see the beginning of something t h a t may strike you a& very peculiar. The use of the words equal and congruent in t h i s book i s different f r o m t h e common usage, and you should have early advance warning of t h i s , s o as t o be ready for i t . Near t h e end o f this chapter, i t i s explained that if m t A = mLB, then the a n g l e s are c a l l e d congruent, and we write L A 2,L 3. In Chapter 5 we will give a similar d e f i n i t i o n of congruence for sejgnents. That i s , i f AB = CD, t h e n t h e segments AB and are c a l l e d congruent, * and we w r i t e A 3 = CD.

71

12

73

74

ÿÿ any t e x t s also say t h a t two t r i a n g l e s are equal,, as an a b b r e v i a t i o n o f the statement t h a t the areas of t h e triangles a r e the same. In this book, t h i s a b b r e v i a t i o n w i l l be avoided; we shall simply say t h a t the triangles have t h e same area.) There is nothing complicated about our terminology, but you may f i n d it h a r d to get used to. To a v o i d t r o u b l e which might otherwise start soon, we recommend t h a t at the e a r l i e s t o p ~ o r t u n l t yyou read the t a l k on Equality, Congruence, Equivalence in which we explain what we have on o u r minds, and how and why we have departed from the traditional terminology, In thls c h a p t e r we have omitted -- r i g h t l y , we b e l i e v e - t h e proofs and even t h e statements of various simple and obvious theorems of a foundational character. Some of t h e s e w i l l be d i s c u s s e d in Chapter 6 , b u t f o r a thorough logical t r e a t m e n t of the material of this c h a p t e r , see Chapter 5 of Studies 11. No matter what t e x t is used, students must be cautioned that when using three l e t t e r s t o denote a n angle, t h e y must write t h e l e t t e r c o r r e s p o n d i n g to the vertex between the o t h e r two l e t t e r s . The three v e r t i c e s of a t r i a n g l e a r e the v e r t i c e s of t h e three angles of t h e triangle. To v e r i f y the statement that the angles o f A A S C a r e not contained in the t r l a n g l e , check t o see if t h e s e t of polnts I n L A B C is c o n t a i n e d in t h e s e t of p o i n t s of & A = . If we remember t h a t t h e set of polnts i n ~ A l X is the union of two rays, each of which extends infinitely f a r in one d i r e c t i o n , and t h e set of points of A ABC is the union of three segments, then we see that t h e triangle cannot possibly c o n t a i n its angles. We could deflne the i n t e r i o r ofLI3AC as the intersection of the s e t of p o i n t s on that side of containlng B with rH t h e set of p o i n t s on t h a t slde of AB c o n t a i n i n g C . This intersection is diagrammed on page 74. The interior of AABC may a l s o be defined as the intersection of t h r e e half-planes: (1) the s i d e of t h a t cone tains 3, (2) t h e s i d e o f BC that contains A, and ( 3 ) t h e

side of that c o n t a i n s C. A c r o s s hatchlng of t h e i n t e r section of these half-planes will graphically illustrate t h a t this r e g i o n is the same s e t of p o i n t s a s i n d i c a t e d in the t e x t .

--

Problem Set 4-1

75 1. union, rays, line. 2. 3.

4.

.

union, segments, non-collinear No. AC and AB are l i n e segments, but the s i d e s o f LA are rays. No. Although the union contains t h e t r i a n g l e , the union also c o n t a i n s therestofthesidesoftheangles. = / \ = A B

/ \

Union o f L A

76

andLB

5. 6. 7. 8.

L CEB, L,BED, L DEA. E i g h t . L A, L C, L A X , L ABD, L CBD,L ADC ,L ADB, L CDB.

9.

Two. LAMB, L B I C , L C M D , L DME, L EMF, L F W , L A W , L B M D ,

10.

LCME, L DMF, L EMA, L FMB. AABC, O A B F , A ECF, A ACD, A F C D , h AFD, A AGD,O GFD,

11.

Seven. L N P R , L NPT, L MPS,L MPT. LAEC,

AmD, A AEG, A EBD, A ABD, A E D , A GDC. a. D, F, M. b. E, G, H.

12. No. 13. 14. 15.

Yes. No. Yes.

16.

a.

17.

18.

I .

No.

It lies on the angle i t s e l f .

No. No. Yes. D

is such a p o i n t . b. Yes. E is such a point. P is in t h e interior of AABC. a. Y e s . b. Not necessarily. P and C c o u l d be on o p p o s i t e s i d e s f, of kB. A and C are in opposite haif-planes determined by l i n e m.

77

Section 4 - 2 Is n o t an i n t e g r a l p a r t of t h e course, and t h e information presented in it will not b e r e f e r r e d t o again in t h e t e x t . The material is made available to t h o s e classes whose t e a c h e r s consider it appropriate in t h e l o c a l curriculum.

79

You may wonder, a f t e r s e e i n g the m LA notation, why t h e t e x t u s e s AB instead of r n in Chapter 2. Actually r n E does have t h e advantage of consistency b u t we do not feel t h a t this is t o o f f s e t the advantages A 3 has: o f convenience and o f common usage. AB and AB are d i f f e r e n t symbols f o r different entities. So a r e L A and m L A .

79

treatment of measurement of angles, it is understood from the s t a r t t h a t t h e u n i t of measure is t h e degree. This is lmplicft in P o s t u l a t e 11, and in this respect the Angle Measurement P o s t u l a t e may seem more satisfyi n g t h a n Postulate 2 concerning distance, where a unit of measure was chosen but l e f t unspecified. There is nothing

80

81

It will be noted

t h a t In this

especially l o g i c a l , however, about the c h o i c e o f degree measure f o r angles: it merely happens to be customary and familiar. You may notice a similarity between t h e Angle Construction P o s t u l a t e and the Ruler Postulate. We again have a one-to-one correspondence, t h i s time between rays in a half-plane from a p o i n t on the edge of t h e h a l f - p l a n e and the numbers between 0 and 180. [pages 76-81 I

Some additional mention of the use of t h e degree s i g n may be necessary. When we label figures, as in t h e f i g u r e at t h e t o p o f page 80, the degree sign i s u s e d only to indicate t h a t t h e number a p p e a r i n g t o the left cf it is t h e degree measure of t h e angle, t o d i s t i n g u i s h from the use of a lower case letter t o i d e n t i f y the a n g l e . For example, we may have an 0 angle of a , and we must distinguish this from t h e angle that c o u l d be i d e n t i f i e d b y t h e l e t t e r " a 1 ' . We may speak o f L Q A B a s "a 110 degree angle" or we may say that L Q A B is an "angle w h o s e measure (now understood t o be degree measure) Is 40.'' One may a s k , " l ~ yeven mentian t h e degree once we h a v e establ i z h e d it as our unit of measure?" The reason is t h a t t h e degree i s n o t t h e o n l y unit by which we can measure angles. There is, o f course, t h e radian, whlch is f u n d a m e n t a l t o t r i g -

onometry, and we must be absolutely certain w i t h what unit we a r e working. 80 One difference in t h l s treatment of geometm is t h a t under o u r d e f i n i t i o n of a n aqgle there is no angle whose measure is 0 , n o r is t h e r e one whose measure is 18C. S i n c e t h e idea of a "180~angle" or "a straight angle" has been used in geometry f o r so long, it mlght be a little hard f o r us as teachers to become accustomed to t h i s usage. In t h i n k i n g of angles as point-sets it is apparent that an angle whose measure is 0 is indistinguishable f r o m a r a y , and an angle whose measure is 180 c a n n o t be distinguished from a l i n e . Hence, no such "angles" appear in this treatment. . A n o t h e r reason f o r not allowing t h e s e s p e c i a l angles is t h a t it is impossible t o determine the interior of an angle of zero measure or of o n e whose measure is 180. Incidental l.y, Euc l i d never used " s t r a i g h t angles. 'I --3 79,80 Note carefully how t h e r a y AC i n t h e f i g u r e on page 79 corresponds t o t h e number 180 and how t h i s be u s e d to determine the measures o f o t h e r angles as illustrated on page 80. -+ Note a l s o t h a t the r a y AB corresponds t o 0 . Although we do not allow the possibility of an angle of 180°, t h i s does not eliminate t h e possibility of two angles h a v i n g t h e sum of t h e i r

can

[pages 80-82 J

82

rncasures c q i ~ a lt o 180, and t:lus

we

&

nave suppiementary angles.

(See P s s t u l a t e i4.) The p n r a s e "linear p a i r " will p r o b a b l y be new to y o u . It Z s an e a s i l y rcmembercd name that simplifies t h e statement o f P o s t u l a t e 111 a n d some of Lne s ~ b s r - q u e n tdefinitions and proofs. On t h e o t h e r nand, we have not tcund it necessary to u s e the p h r a s e "adjacent angles". L i n e a r pair is ~ a s i l yd e f i n e d , f'or I t i n v c l i v e s only t h e r l o t i o n of o p p o s i t e r a y s . Tne i d e a of a d j a c c n t angles is more complicated, f o r it i n v ~ i v e st h e i d e a of s c p s r s t i m in a p l a n e . Two angles al>e a d j a c e n t if t h e y n a v e a cormon s i d e a n d t n e i r other t w o s i d e s a r e zontained in t h e opposite half-planes determln~dby tnz line c ~ n t a i n i n gtr.e cormon side.

Problem Set '4-3

a, p ; L.. m ; C . q; ( 3 , n; h , T h e r e m i n i n g a n g l e h s s a m r a s u r e 01' 50. 5. a . Bid or GHE. b . EFG or GFB. 6. 3 . XZY or YZX. . XZK or K Z X . c. Y or Y Z K . d. 180. . a 52, b = 128, c = 52, 8. 10"; T O @ ; ?!ikO; 1611.5~; (180 - n)', Sor 0 < i i < 180 n0 , ? o r o < 1-1 < 180 , ((40 I- n)", f o r 9 :: In1 < 90. 2.

8b

85

-

[ p a g e s 82-05]

,

12.

5

36

87

88

a. b.

One way by the Angle C o n s t r u c t i o n P o s t u l a t e . Two ways. There are t w o half-planes in E whose edges c o n t a i n Z.

Notice that the definition of r i g h t angle precedes any mention of perpendicularity. Various approaches would have been p o s s i b l e ; the one used seems t o be simplest logically, f o r it permits lines, rays and segments t o be included ' 7 one d e f i nitlon of p e r p e n d i c u l a r . The text p o i n t s out t h a t a ray o r a segment determines a unique l l n e which c o n t a i n s it. When two l i n e s intersect, f o u r rays a r e determined. These r a y s in t u r n dete,mIne f o u r angles. Sometimes we r e f e r to t h e angles a s angles formed by the l i n e s . ( A mathematical purist might want t o replace t h e phrase "if t h e t w o l i n e s containing the two s e t s determine a r i g h t angle" by "if t h e union of t h e two l i n e s containing the two s e t s has a r i g h t angle a s a subset'!) Theorem 4-4 c o ~ l d ,w i t h proper restatement, be taken a s t h e d e f i n i t i o n of r i g h t angles. I n t h a t case t h e definition of right a n g l e actually used in the text would be replaced by a theorem. Alternate proof for Theorem 11-7: + 4 4 Given that AC and AE are opposite rays, and AB and + AD are o p p o s i t e rays so that L1 andL 2 are vertical angles. Let m L 3 = r. Tnen by P o s t u l a t e 1 4 , m L 1 must b e 180-r, and rnL 2 must a l s o be 180-r. Therefore, m L 1 = mL 2, a n d L I 2 L 2 , whlch was to be proved.

[pages 85-88]

Problem Set 4-4 Only one. b , I n f i n i t e l y many. LRON andLSUN are supplementary and have equal measures. Therefore, each has a measure of 90, making a.

8

8ElE. a.

b. c.

d.

3.

and L RXB andLSXA. None occur. LRXB andLRXA. t SXA and L SXB

.

a. b.

80'.

d.

10'.

e.

c.

f.

a.

45.r0. 90. 45. Two pairs.

b.

7 0 , 110, 110.

c.

90.

r,

(180 - r),

a. b.

(90 - x)', xO, f o r O (x - go)',

far 0

CD. Some students may be conf'used by such a statement as: We suppose something I s f a l s e in order to prove it t r u e . It may h e l p to soft-pedal the word "false" and say that i f we don't know whether a statement I s true, it is reasonable t o t a k e its opposlte (or negative) and see what follows from it. Our approach 1s to explore possi billties, n o t to say categorically t h a t the given statement is false or equivalently that its o p p o s l t e is true. The very phrase "suppose so and ao" may be confuslng t o some students. The word "suppose" may suggest to them t h a t we are supposing it as a f a c t rather than c o n s i d e r i ~it as a

hypothesis. Remind them that in everyday life we often reason from premises wltfiout b o w i n g t h a t they are t r u e . For example, when n o t sure of today's date we might argue so: I know today is Saturday and I t h i n k the date is June l 5 t h , but I'm not sure. If today is June 15th, then June 1st also was a Saturday. W l t 1 remanber t h a t June 1st was a school day. Therefore, today can't be June 15th. Sometimes we actually reason from false premisea, as when we argue that if Lincoln had not been s h o t , the course of American h i s t o r y would have been such and such; or t h a t if the L u s i t a n i a had not been torpedoed, the United States would never have entered World War I. You may be able t o h e l p your s t u d e n t s by using, in informal classroom speech, such phrases a s : Assume f o r t h e sake of argument; Pretend, and see where you end up; Work on the theory t h a t . . . , and see t h e kind of jam you get i n t o .

--

Problem Set 6-2a

161

a. b, c.

2.

All. (1) T h i s set is not a stalnless steel product. (2) T h l s s e t is a s t a i n l e s s steel product. ( 3 ) This s e t will not m a t . ( 4 ) This s e t did rust. y is true, z is t r u e . w, u and x are not true. Yes, indirect reasoning is used i n reaching each conclusion.

3.

162

My Mother i a n o t c o l o r blind. My b r o t h e r is left-handed. Jane drank some hot chocolate,

1.

4.

5. 6.

7.

8.

Let Let k t

A

Let

D

B C

be l' someone is a member of the swimming c l u b t t . be "someone can play the plccolo". be "someone is a t u r t l e " . be "someone weare s t r i p e d trunks i n the c l u b pool".

Then the problem may be diagramed this way: (1) I f A Is true, then B I s true. (2) If 9 is t r u e , then C is not t r u e . (3) If D is true, then A is t r u e . (4) D is t r u e . The conclusion is t h a t "C is not true" is t r u e . Hence, in terns of t h e problem, the conclusion is " 3 am n o t a turtle". a . Red, white. b. Yes. A is n o t green. Given scalene A A B F . To prove F that t h e bisector of any angle, F, is not perpendicular to If we assume t h a t t h e bisector of L F is perpendicular t o

m.

m,

A

then AAF& 2 A BF& (A.S.A.) and AF = BF. The assumption that FQ I s perpendicular t o AB l e d to the contradiction that the scalene AABF is isosce3es. [pages 161-1621 Q

;

1

II

I

I

163

Notice f r o m the proofs In Section 6-2 that uniqueness is u s u a l l y established by indirect proof. Showing t h a t there is o n l y one of something can be accomplished by showing t h a t there cannot be t w o . Note t h a t I t is p o s s i b l e t o e s t a b l i s h uniqueness without, o r before, establishing existence. For example, the proof of uniqueness i n Theorem 3-3 can be made l o g i c a l l y independent of t h e q u e s t i o n of existence, as follows: Suppose t h a t there are two planes containing L and P, L e t Q and R be t w o points of L. Then both planes c o n t a i n P, Q, R which are non-collinear p o i n t s . This contradicts Postulate 7. Hence o u r supposition is f a l s e and there i s at most one plane cont a i n i n g L and P . I n ordinary life, t o o , knowledge of uniqueness c a n be Independent of' knowledge of exlstence. A person w i t h just one day of his v a c a t i o n left h o w s very w e l l t h a t he will not spend more t h a n one day sailing, But he does n o t m o w t h a t he will spend t h a t one day sailing. Existence means t k a t there I s at least one. Uniqueness means that there is at most one. Existence and uniqueness means that there is one and only one, o r exactly one.

Problem Set 6-2b

Postulates 6 and 7. 3 . WB and Q. %?and %?. fiff and H w e 6. =and N, A& and CQ, A61 a n d s , z a d n d , C* oe B& and DQ, CQ and D&. t* t* PQ and PT a r e t h e same line, Y e s . By Postulate 7 . Am. f, AB. B. If A, B, C, D are not coplanar, we list the planes ABC, ABD, ACD, E D . However, if A, 3, C, D are coplanaq there is only one plane determined.

1.66 1. Yes, 167

2. 3.

4.

5. 6.

s.

e

-

[pages 163-1671

168

Many students may feel that the formal proof of uniqueness in Theorem 6-1 la mere h a i r splitting. For them I t probably is b e s t not t o belabor t h e p o i n t . After they have had more contact with uniqueness principles they may better appreciate the point. Some students may object that the uniqueness proof is unnecessarily complicated, that t h e Angle Construction Postulate "guarantees" that there I s a unique l i n e M in plane B perpendicular to L a t P. This is not quite c o r r e c t . The Angle Conetruetion P o s t u l a t e asaerts that there is a unique + PY with Y in half-plane H such t h a t m L X P Y l a 90. Suppose then we apply the same process t o Then line the half-plane K opposite to H.

RLL.

A

, Y H P


L

-

--

Problem Set 6-3

174 1, EC I s t h e perpendicular bisector of

and so EB = ED by

Theorem 6-2. 2. x = 7, y = 5, z = 10. 3 . Since P and M are points which are each equidistant from A and , ia t h e perpendicular b i s e c t o r of by Theorem 6-2 and Postulate 1. Then QA = QB by Theoram 6-2. .

-

-

2.

PT = PR + RT. RT=RQ.

3.

PT

I.

175

7.

PR

=

+

1.

2. 3.

RQ.

No, Yes.

I

2. 3. 4.

5. 6.

5.

AF = BF. H

m.

HB b i s e c t a ABF

.

bisects

Base angles. Division, from Statement 2 From step 3 . If two angles of a triangle are congruent, t h e sides opposite them are congruent. Theorem 6-2 and Postulate 1.

I

+9.

Prove:

6

CF is perpendlcular

b i s e c t o r of Given:

D e f i n i t i o n of betweenness. Theorem 6-2. S u b s t i t u t i n g RQ f o r RT in Statement 1.

H

AH?? and

m. A

F

1.

x=y,

2. 3.

HB = HB.

2.

4.

HA

-

3. 4.

A ABH

P = S .

= AFBH.

HF, BA = BF.

-HB bisects

1.

RC

A.S.A.

5.

Corresponding parta. Theorem 6-2.

1.

Given.

SCA.

2.

AC = AC.

2.

Identity.

3.

A RCASASCA.

S.A.S.

4.

RA = 3 A .

3. 4.

5.

Q Is mid-point of Rs.

5.

Corresponding parts. Given.

AQ~E.

6

meorern 6-2.

6. 176

at

Definition of b i s e c t . Identity.

E.

= SC.

L RCA

1.

-

-

This dlscusaion of the introduction of auxiliary sets is a departure from the conventional treatment. It is important and deserves attention. Conaider how o f t e n students assume they can, by nconstrmction", justim referring to a line whoae existence hae not been proved and which, in f a c t , may not e x i s t (see Example 2 ) . Notice that we aay "introduce" line AB or se@nent PQ and avoid using such words as "draw" or "construct". As soon (or segment PQ) as we have shorn the existence of line we have the l o g i c a l r i g h t to reason about it and to derive properties of I t In our geometry. This I s independent of whether we choo8e t o dpaw or represent it In a diagram.

-

-

136 Having proved the existence and uniqueness of a certain geometric object in our theory, we sometimes ask how it could be constructed physically from given data using prescribed

operations or procedures. Thus the discussion of Theorem 6-4 g i v e s a precise description of the c o n s t r u c t i o n of the perpendicular t o a glven l i n e from a given external p o l n t using mler and protractor. In t h i s instance, the construction is glven before the proof t o help the student grasp it.

.

d i s t i n c t i o n between the common meanfng of draw" and the meaning of "introducett described above i a eatablished w i t h your students, it seems agreeable to use the term "draw" f o r convenience. An occasional reminder of the distinction should be made, however, so that the correct concept becomes the one suggested by whatever word is u s e d . ) (Once this important

tl

Notice in Section 6-4 t h a t we do not say that auxiliary segments always are shown as dotted segments. The d o t t e d segment seems necessary only when the figure becomes so complicated t h a t the method of proof becomes obscure. 178 If A, C , D and E are non-coplanar in Example 1, t h e 179 proof based on introducing does n o t h o l d . The proof in which is introduced does hold, however.

--

Problem S e t 6-4 18'

I.

n.

1. Conslder 2. AD = CD, 3. m LDAC = m L E A .

I

1. Two points determine a l i n e . 2. Given. 3. Base angles of an l s a s c e l e s triangle are congruent. 4. Given. 5 S u b t r a c t i o n using s t a t e ments 3 and 4 . 6. If two angles of a triangle are congruent, the sides opposite are congruent.

This proof does not work if points A, B, C and D are n o t coplanar. Step 5 would n o t be valid. [pages 178-1813

XA.

R

1. 2.

Two p o i n t s determine a line. Identical

XY = AB and AY = XB. A YXA 2 A BAX. rnL YXA = m L EAX. m L ECXA = m f YAX.

3. 4.

Given.

5

Corresponding p a r t s .

mLYXO = m L B A O .

6 7.

Subtraction in Statement 5 . Carresponding parts.

8.

A.S.A.

Draw

XA = XA.

LB. AXOY 2 A

LY

AOB.

.

S.S.S.

*A slmilar proof is possible if

m.

1.

maw iirS and

2.

YE = SA. L E

@LA.

3. Y S = S Y . 4. AYSA Z A SYE.

5,

LYSAZL

SYE.

is drawn.

1.

Two p o i n t s determine a l i n e .

2.

Given.

3. 4. 5.

Identity. S.A.S,

Corresponding parts.

1. 2.

3. 4,

5,

L e t M be t h e midpoint o f EA. Consider and MS. -"EM = AM. L E 2 LA and m2 A Y E M ~ PSAM.

m*

6. 7.

S T ZSM.

8, 9.

mLEm -- mLASW. mLEYS

1.

Consider AD.

2.

AC = AB.

mLMYS

=

=

rnLMSY.

1. 2.

3. 4.

5. 6 7,

8. 9.

rr.LASY.

A segment has exactly one midpoint. Two polnts determine a line, Definition of midpoint. Given. A.S.A.

Corresponding p a r t s

.

Base angles of an I s o s c e l e s t r i a n g l e are congment Corresponding parts. Addition of Statements 7 and 8.

.

5.

CD 3.

182

=

DB.

4.

AD. A ACD 2 & ABD,

5.

L A C D Z L ABD.

AD =

I

1.

Two p o i n t s determine a line.

2.

Given.

3. 4.

Identity. S.S.S. Corresponding parts.

5.

This is very unusual material for a tenth grade geometry t e x t , We introduce it t o indicate t h a t the assertions we make can be j u s t i f i e d f r o m o u r postulates (without recourse to diagrams), and to give some typical examplea of how we c a n logically j u s t i f y betweenness and separation properties which usually are read from flgures, There are t w o p i t f a l l s here. First, it is beat not to t r y to teach t h i a material to s t u d e n t s who are perfectly satisfied with the proofs as originally

given. They probably are n o t y e t ready f o r this k i n d of c r i t i c a l thinking m d t h e i r progress in geometrg will n o t be impeded by passing on to t h e next chapter. There is an o p p o s i t e danger for t h e very c r i t i c a l student. He may become d i s t r u s t f u l of diagrams and fall to develop a sound g e o m e t r l a intuition. He should be reminded that our theory o f geometry is suggested by physical space, is applicable to It, and t h a t many theorems can be discovered and most can be appreciated (see Chapter 7 , Section by t h e study of diagrams and models. 7-1, on making conjectures In geometry.) P o i n t out t h a t a geometric proof in which one step depend8 on the diagram, although not mathematically perfect, is still incomparably superior to what is considered l o g i c a l in most areas of human

discourse, Having clarified the basic p o l n t in this s e c t i o n we don't hesitate in later chapters to use the diagram to justify prope r t i e s of betweenness and separation. The complete j u s t i f i c a t i o n of a l l such properties used is still quite difficult and requires a deeper study of the foundations of geometry. (See Studies 11. ) As we mentioned earlier, you will have to decide how much time your c l a s s should devote to Section 6-5. If you do not chooae to have your class as a whole study t h e section, your b e t t e r students may f i n d t h a t the exposition and the problems provide excellent supplementary work.

Problem Set 6-5

1.

By Theorem 6-6. b. L C. Q Theorem 6-6. c L A, L B, and& C By the definition of t h e interior of a triangle.

a. L B.

.

.

-

185 *2. The argument (using the first drawing) depends on t h e assumption from the drawing t h a t E l a Inside L XBC'. In a c a r e f u l drawing ( s e e below) X will appear on the opposite side of B C f from E.

3.

The three p o s s f b i l i t i e s are: a. A is on L. In this case and

L

intersects both

m.

186

b,

HI. In this case A is on the same side o f B, and C is on t h e other side of L. In t h i s case L intersects This follows from the Plane S e p a r a t i o n P o s t u l a t e . A is In H2. In this case A is on the o t h e r slde of L from 8 ao L intersects S i n c e D is between A and C, D 1s In the Interior of L A X , by Theorem 6-6, and t h e definltion of the int e r i o r o f an angle implies that A and D are on w the same slde of BC. Theorem 6 2 implies t h a t D and F are on the same s i d e of BC. A is in

L as

.

c.

m.

4.

a.

[pages 185-1861

b.

187 5 .

a.

Since i n t e r s e c t s %?at D, it follows from the Plane Separation Postulate that F belongs t o Hz. w Since BE intersects AC at C, it f o l l o w s from the Plane Separation P o s t u l a t e that E belongs to Hz. C, A and D are on the same side of BC because it is given t h a t D is in the i n t e r i o r o f L A B C . Theorem

6-5. E is in H2 by t h e Plane Separation Postulate. Theorem 6-5. Each p o i n t of 3 with t h e exception of 3 l i e s in HI but no point of does. Also, B does n o t l i e on EC. Each p o i n t of EC other than E l l e s on the same fit side of AB as C and D, but each p o i n t on the ray 4 opposlte B w i t h t h e exception of B l i e s on the @ other side of AB. Note t h a t C and D ape on the same s i d e o f AB since D is in the interior of

-

b.

-

-

LAX. C,

It follows from Problem 3 that E3D intersects e i t h e r or . It f o l l o w s from parts a and b above that %? does not intersect E. d . Each p o i n t of AC other than A lies on the same side of with C and D by Theorem 6-5 and t h e P l a n e Separatfon P o s t u l a t e , b u t each point of the ray opposite w i t h the exception of B, lies on the o t h e r side of Since D is in the i n t e r i o r of LABC, it follows from t h e Angle Addition Postulate t h a t mLABD + m L D E = m L A B C . Since all of these measures are p o s i t i v e it is impossible that e i t h e r (1) rnLABD + rnLABC = m L D B C o r ( 2 ) mLAX + m L D E = rnLABD. S i n c e (1) is impossible, A is n o t in the i n t e r i o r ofLDB2. Since (2) is Tmpossible, C is not i n t h e i n t e r i o r of LAm. c.

a

3, E.

*6.

188~7. a,. D lies in the plane determined by A, B, C aince it is on the line E lies i n t h i s plane since it is C, on the line AB. u b. A and B are on the same side of ED and C is on M

s.

the opposite s i d e from A and 3 . Hence, ED sect8 at a point X between A and C.

c.

8.

e

Rc.

a.

True.

b.

c.

Tme. False.

d.

M e .

lnter-

Illustrative Test Items f o r Chapter 6

In A ABF, every p o i n t of except - and is In the i n t e r i o r of L of t h e p o i n t s of A 8 is In the in-

1.

-. -

t e r i o r o f L ABF. A

There is snow on Snow me1 t s at temperatures above 32'. Wrlte a the ground and the temperature o u t a i d e is 40'. logical conclusion. A

Given that PA = PB, H QA = QB and FQ meets M as shown in the figure. Without uslng corqruent triangles, prove A is the rnldpoint of AB. at

A

H

In this plane figure there are two isosceles triangles with f3. the same baae, HF AB and 18 to Every p o l n t of E? IS from A and B.

b~ m.

-

z.

B

If, f o r t h e sake of argument, you accept the following hypothesis, which of the f ollowlng are l o g i c a l conclusfons ? Every piece of Alpha candy 1s dlelicious. ConcLusions: a. Since t h l s piece of candy is delicious, I t must have been made by Alpha Company. b. ?%la Alpha caramel i s d e l i c i o u s . c . Since t h i s p i e c e of candy I s n o t delickous, it could not have been made by Alpha Company.

Hypothesis:

AF

=

AH

=

BF.

The p o i n t s given in t h e picture are coplanar.

BN. AK = BK. C, Does l i n e FH pass through K? If AT = 3 , t h e n BT = S t a t e a theorem which s u p p o r t s your conclusions.

A

FB

=

FQ WF

= A&. =

AB. WA.

W, Q and B necessarily collinear if these three p o i n t s are coplanar? Are

8.

Glven t h a t

9.

Prove that t h e perpendicular bisector o f one s i d e of a scalene t r i a n g l e cannot include the o p p o s i t e vertex o f t h e triangle.

B, C, F are f o u r non-coplanar p o i n t s , list all the planes determined by subsets of A, B, C, F. A,

--

In t h i s figure,

QB 1AF,

Prove:

AX XH HF

=

QA

=

= =

FX, XB. BF and QF.

Answera

and F; ABF; None. 2. The snow is melting. 3. Since PA = PB and QA = QB, P and Q l i e In the perpend i c u l a r bisector of AB. Therefore, %$ I s t h e perpendicular b i s e c t o r of a. 4. B i s e c t s . Perpendicular. Equidistant. 1.

A

-

5. 6. 7. 8. 9.

b, c . Yes. 3.

Theorem 6-2. No. Not unless the e n t i r e figure is a plane f i g u r e . A X , ABF, ACF, BCF.

Given that h ABF is scalene, Assume t h a t could b i s e c t and be perpendicular t o D. Then A A F H 2 h BFH by S.A.S. and FA 2 FB, so A AFB is i s o s c e l e s . The assumptions

F

A

H

w

E

lead to the contradiction t h a t a scalene t r i a n g l e is isosceles. Hence t h e assumptlona were false,

is t h e perpendicular b i s e c t o r of . Therefore, €@ = QF. Since XH = XB is given, X is the midpoint of C* HB and FX is its perpendicular bisector. Hence, HI? = m,

30. QB

Chapter 7 GEOMETRIC 1NEQUALITIES

The material covered in t N s chapter is quite similar to that found In corresponding chapters of o t h e r geornet~~j texts. The main difference i s that we compare t w o segments or t w o angles merely by comparing their lengths or measures. Thus, although our inequalitlee describe geometrlc r e l a t i o n s , they involve only real numbers. Thfs is another advantage of our early introduction of real numbers. Because atudents do not always mow p r i n c i p l e s of inequalities well, we restate t h e order p o s t u l a t e s f i r s t given in Section 2-2, giving examples to show how they are applied. The idea that a conjecture must not be considered true 189 until (unless) it has been proved, bears emphasis. To put it bluntly, a conjecture i s a guess. The kind of conjectures we pay attention t o are the ahrewd, reasonable ones that are based upon inductive t h i n k i n g or insight. WIt conjectures, no matter how reasonable they seem, remain guesses until they are proved. It may be good f o r your students to be reminded that conjecturing is an important p a r t , even if only the first stage, of mathematical work. After all, a m a n who develops new mathematics often muat t r y to decide what the t r u t h is before he can present a l o g i c a l proof of it. There i s no reason to look down on t h e a r t of making conjectures. There is, however, no excuse f o r confusing guessing with proving. Goldbachrs conjecture that every even number l a the sum of two primes is a simple non-geometric exmple that you can mention. After many generations the conjecture is s t i l l not a theorem. The example of Section 7-1 should suggest two t h i n g s to the student. F i r s t , he should try to make reasonable conject u r e ~ . Second, he should express his conjectures in good mathematical language. The aecond goal may be the more d i f f i cult to achieve.

--

Problem S e t 7-1 190 1.

2.

3.

4.

5.

7.

191 8.

The opposite s i d e a are unequal in length with the side opposite the largest angle having t h e greater length. AB + BC > AC. BC + AC > AB. AB + AC > B C . The a m of

the lengths of two sides of a triangle is greater than t h e length of t h e third side. RS ST + TQ > RQ. The sum of the lengths of three s i d e s of a quadrilateral i a greater than the length of the fourth

+

side. It increases. DF > XZ.

From B drop a perpendicular to E at a point D of E. Then + + D will l i e on aome AC and for this AC, rn L RAC = m L BAD 4 is a minimum. If AF is the opposite ray to 3, m L BAF is a maximum. The procedure doea not work since rn L DAE is larger than either m L BAD or m 1, EAC, T h i s shows up clearly If m L BAC is close to 180.

It may be helpf'ul to s t a t e the order principles In English a s w e l l as in algebraic syrnboliam.

For example, 0-2 may be stated: If the f i r s t of three numbers is l e s a than the second and the second is less than the third, then the first I s less than the t h i r d . Similarly, 0-3 asserts: If t h e same numbep is added to each of two unequal numbers, the sums are unequal in the same order. Or 0-3 may be stated: If the same number is added t o each side of an inequality, the Inequality remaLns [pages 190-1913

true.

You recognize that these o r d e r p r i n c i p l e s a r e essen-

t i a l l y t h e same as the ÿÿ xi oms o f ~ n e q u a l i t y " which appear in most geometry t e x t s . The order principles r e f e r t o real numbers rather than geometric quantities. 192 Example 6 , slmple as it seems, is quite important and

o f t e n used. I n many geometric problems it is necessary to prove a relation such as a < c or c > a . In the conventional treatment we r e f e r t o a diagram and conclude c > a by t h e p r i n c i p l e , h he whole fs greater t h a n any o f i t s parts" . C r d i n a r i l y , we prove a relatlon l i k e c > a by a p p l y i n g Examp l e 6 , t h a t i s , we show c = a + b , where b is p o s i t i v e . ( A c t u a l l y in our a p p l i c a t i o n s a, b and c will all be p o s i t i v e . ) We might reword t h l s as, a c b > a when b > 0 , since c = a + b. Even more s h p l y we can say, h he sum of two p c s l t i v e numbers is g r e a t e r t h a n e f t h e r number." Thus, t h e f l n a l j u s t i f i c a t i o n I s a p r o p e r t y of r e a l numbers. An important appllcatlon or Example 6 o c c u r s later i n S t e p 8 of t h e p r o o f o f Theorem 7-1. Example 6. If a f b = c a n d h Is posdtive t h e n a < c . Reasons o n l y : 1. Given. 4. Postulate 0-3. 2. & f i n i t i o n o f p o s i t i v e . 5. Substituting c f o r a + b. 3 . Relation between < and >.

Example.

If a + b


.

We have j u s t proved: If unequal numbers are multiplied by the same negative number, then the products are unequal in the oppoalte order. Actually all t h e famillar "Axioms of inequalityn can be derived f r o m t h e f o u r order postulates. Step 6 of the proof of Theorem 7-1 t a c i t l y assumes t h a t F 194 is In the i n t e r i o r of L B C D . This 1s justified i n Problem 4 of Problem Set 6-5. It is probably trme t h a t no kind of mathematics can be effectively presented in a completely rigorous form to a tenth-grade c l a s s . We should not feel guilty about teaching tenth-grade students merely aa much as they can learn. The betweenness problem here will probably go unnoticed by most students. It should be called to the attention only of very capable and c r i t i c a l students. (Such students will probably be rare.)

[pages 192-1941

The formal justification of Step 8 involves an application of Example 6 of Section 7-2: If a + b = c and b I s posltlve, ( s e e the Commentary above, ) We have Step 7, then a < c m L E C D = m L B + m L FCD, and m L FCD is positive ( a l l angle measure8 are p o s l t i v e by the Angle Measurement ~ o a t u l a t e ) Thus, by Hxample 6 m L B < m L B C D o r m L B C D > m L 13. Hereafter we usually apply Example 6 in such situations , d t h out explicit reference. The following lemma usually 1s applicable In proving an angle larger than another. Lemma. If D is in the interior of L ABC, then m L A X > m L ABD. Proof: The argument above applies. By the Angle Meaaurement Postulate m L A P Y : = m L A B D + m L CBD and m L ABC > m L ABD by Example 6. Similarly we can prove an analog f o r lengths of segments. Lemma. If C is between A and B, then AB > AC.

.

.

Problem Set 7-3a L ACB and L CAB. L FCB. L DBC and L EBA. m L DBC > m L A, by Theorem 7-1. m L DTX > m L C, by Theorem 7-1. m L DBC -t- rn L CEA = 180, by Postulate 14. 40.

is greater than 73. is equal to 112. is less than 112. I s equal to 30. is equal t o 90. w f.* This is h p o s s i b l e , since, by Theorem 6-3, AC and BC are not both perpendicular to ,

f;B

196 4.

*5.

6

No. It is not true for the exterior angle at each of the o t h e r v e r t i c e s . Another exception is a rectangle. By the Supplement Postulate, a + w = 180. But b < w, by the Exterior Angle Theorem. Adding a t o each s i d e of this inequality, we get a + b < a + w which becomes a -t b < 180, which was to be proved. Similarly, b + c < 180 and a -I- c < 180. Glven: A Am w i t h a7f =" '1Wr. To prove: m L A < 90. mLB

A

< 90.

Proof: By t h e previous problem we have mBut L the A + mbase L Bangles < 180.or

A

B

an isosceles triangle are congruent, so 2 ( m L A ) < 180, and m L A < 90. Also, m L B < 90, since the measures of the base angles are equal.

197

The S.A.A. Theorem usually is proved a f t e r the Parallel P o s t u l a t e Is introduced, since I t follows readily f r o m the theorem that the sum of the angle measures of a triangle is 180. Since the S,A.A, Theorem does not depend on the P a r a l l e l Postulate (chapter 9 ) we Introduce it here and can apply it whenever needed. An S . S . A . tneorem a l s o holda when the angle I s an obtuse angle, but there I s l i t t l e value Is bringing this fact t o t h e a t t e n t i o n of a class. Outstarudlng students mlght enjoy proving the f a c t , however.

--

Problem Set 7-3b

199 1. Since A&

A l s o AB =

by S.A.A.

38&,L QRA Z L QAB. AB and L H 2 L F.

=

Therefore, A AH-I

rCI

A BAF

2. 1.

2.

3. 4. 5.

6. 7. 3.

-

AB HI?, BF = BF.

Given. Identity.

AF =

Addition, Steps 1 and 2. Definition of perpendicular

HB. & K and Q are right angles. AK = HQ. A A K F ~ AHQB. KF=QB.

A F A H Z A AFX by Hypotenuse-Leg Theorem, hence L B F A 2 L FAB. Therefore, FB = AB.

-

5.

6.

200

.

lines Given. Hypotenuse - Leg Theorem. Corresponding parts.

C * ,

C*

1

BF, HB = QA. Given: HB 1 AF, &A Prove: A FAB is isosceles. Since AB = AB, A Am 2 A BAQ by Hypotenuse-Leg Theorem and so L HAB Z L QFU. It followa t h a t FA = FB and A FAB is isosceles. LAKF Z L AEQ (~upplementsof congment angles), L A = L A, A& = AF. Hence, A AQB ZA AFK by S.A.A. Then QB = FK. Since L s 2 L C , AB = FB. A l s o Ln A H 3 and F&H BH = BH, and & BAH and BFH are rfght angles. Therefore, these triangLes are congruent by the Hypotenuse-Leg Theorem Hence, AH = FH.

A

In the proof of Theorem 7-4, Statement ( 3 ) involves Example 6 of Section 7-2. (see comment above on Theorem 7-1, s t e p 8.)

>

One frequently sees Theorem

method: Given :

A ABC with To p ~ o v e : rn L C >

A8

7-4 proved by the following

> AC .

'\

0

\

E

C

Take D, between A and B, such that AD = AC. Bisect L A, and w let E be the i n t e r s e c t i o n of t h e bisector with the l i n e BC. Show t h a t A ADE 2 A ACE, by the S.A.S. Postulate, It follows that m L ADE = m L ACE. By the Exterior Angle Theorem, m L ADE > QIL D m . Therefore, m L C > m L B, which was to be proved. T h i s proof tacltly assumes t h a t the bisector of L A r e a l l y does C* Intersect BC In a point between B and C . See Problem 5 of Problem Set 6-5 f o r consideration of t h i s matter.

--

Problem Set 7-3c

203 1. L G. L K, 2. Em E. 204 3 . a. 40. b. 80.

4.

c.

AB.

a. b.

EPL>Kf;.

c.

d.

< MK.

KL ) ML None.

> MK.

ML > KL and ML > KM. f. MLkKM and ML > KL. In d A X , I s the longest side, s b c e it is opposite the angle w l t h the greatest measure. In A ADC, is the e.

5.

ML

-

longest side, for the same reason. Therefore, AD and is the longest of the f i v e aements. [pages 203-2043

> AC

6.

E,XE, XC'. You may expect to g e t a reaction from the student, to 'the effect t h a t the figure is incorrect, since m f A + m L B + rnL c < 180. his is a fine opportunity to p o i n t o u t that we cannot prove, on the bas1 s of t h e postulatea given so f a r , that t h e sum of the measures of the angles in a triangle is 180. When we g e t to the Parallel Postulate in Chapter 9, we w i l l be in a p o s i t i o n to prove the angle sum theorem. In any case, given the hypothesis that such a t r i a n g l e exists, we can assert the conclusion that i t s s i d e s are ordered in the given manner, ) Given: is the shortest s i d e . is the longest side. To prove: m L CFA > m L CBA. ( ~ o t eto the teacher:

7.

1. 2. 3.

205 *8.

I n A ABF, AB

rnL B F A > r n L ABF. InA BCF, CB > CF.

4. 5.

m L C F B > m L CBF. rnL BFA + mL. CFB ) m L ABF + m L CBF.

6.

rnLCFA>rnLCBA.

Given:

Step 5 and the Angle-

FA = FB.

To prove:

FH

m L B.

m L FAB

=

3. 4.

m L FAB

>mL

5.

FH

B > m L H .

>

FB.

H and 3.

> FB.

F A = FB.

mL

Adding Steps 2, 4.

Addit ion P o s t u l a t e . F

A is between

1. 2.

Given. Theorem 7-4. Given. Theorem 7-4.

) AF.

H.

1. 2. 3.

4.

5.

Given. Base angles of an Isosceles triangle are congruent. Theorem 7-1. Steps 2 and 3 . Theorem 7-5.

9.

a.

b. c. d. e. f. g.

h.

If a team can win some games, it has some spirit. If two angles are congment, they are right angles. Any two supplementary angles are congruent. The intersection of two half-planes is the i n t e r i o r of an angle. If Joe is seriously ill, he has scarlet fever. If a man Ilves in Ohio, he lives in Cleveland, Ohio. If two trimgles are congruent, then the t h r e e angles of one are congruent to the corresponding angles of the other. If the sum of the measures of two angles is 90, the angles are complementary.

St at ement

10.

11.

Statement

Converse

No. The converse should be, "lf f wlll be burned, 1 hold a lighted match too long." The hypothesis does n o t contain "if", and the conclusion does not contain "then". a. No. gb, gd, 9e, 9f. b.

206

Converse

Yes.

9a, gg.

Note t h a t the d l s t a n c e between a l i n e and a p a i n t l a a number. Theorem 7-7 r e a l l y involves three i n e q u a l i t i e s : (1) AB + PC > AC, (2) BC + AC > AB, (3) A C + A B > B C . A

The t e x t proves (15, and t h i s is s u f f i c i e n t since a relabeling of the f i g u ~ ewill give ( 2 ) and (3).

{pages 205-206 1

Problem -S e t 7-3d 207

4.

and AF. AT and TF. The statement of Theorem 7-6. HB < HA < HF. The statement of Theorems 7-6 and 7-5. 3 , 13. k - j < x < k + J .

5.

1.

1.

208 2.

3.

AT

DB DB CA CA

< CD + CB. < AD + AB. < CD + AD. < CB f AB.

1.

2.

3.

6

1. 2.

209

*7.

The sum of t h e lengths of two s l d e s of a triangle is greater than the length of the third s i d e of the triangle. Addition.

3. Division, D B + CA ( C D + AD + CB + AB. If the points are noncollinear, the inequality follows from Theorem 7-7. If the golnts are collinear, then either (1) B is on the segment z, in which case AB + BC = AC, or (2) A I s between B and C, in which case BC > AC, so AB + BC ) AC, or ( 3 ) C 1s between A and B, in which case AB > AC, ao AB + BC > AC.

We h o w from the preceding problem (problem 6 ) that the result is true in t h i s case; that is, A1A2 + A2A3 2 A1A3. Case 1.

(n = 3 ) .

Case 2.

(n = 4).

1.

A1A2

+

ApA3

+

;:A1A2 + A 2A 4 because A2A3 + A3Aq 2 A2A4.

A A 3 4

from Case 1 t h a t 2.

A1A2

+

A2A4

2 A1A4

by Case 1.

3.

A1A2

+

A2A3

+

> AIAq

A A

3 4 -

[pages 201-2091

it follows

follows from Steps 1 and

158

General Case (n i a arbitrarily large). 1. We continue aa in Cases 1 and 2 to show t h a t A 1A 2

2* A1*n-l

+8.

+

......

AZA3f

2

An

> A1\

- AIAnml

%-1

A r+

4.

A A 1 2

+ A2R3 +

Ph-2

A1\

+

3*

+

%-I 2 A l % - l '

by Case 1. by Step 2.

.... .....

+

%-2 An-1

-I-

Al*,

from Steps 1 and 3 . > PA + PC except when X is on the sement AC, in which case the equality sign holda. Similarly, XB + w > PB + PD except when x is on BD, in which case equality holds. Therefore,XA + XB + XC + XD > PA + PB + PC + PD except when X is on AC and also on and t h i s can happen only if X = P, which is excluded by XA

+

XC

a,

hypothesis

.

The result also holda If X is not in the plane of

B, C and D. Consider the reflection Q f of Q with respect to m. Then m is the 1-bisector of and i n t e r s e c t s at a point which we c a l l M. The point R on m t o make PR + RQ a minimum 1s the p o i n t where Intersects m. A,

"9.

mr

m'

L e t S be any point of m other than R, If S # M, then S Q 1 = SQ, so PS + SQ = PS + SQg. A SMQi Z A SMQ by S.A.S.

If S = M, then agaln PS + SQ = FS + SQt. I s l A PSQt, PS + SQ > PQ' = PR + RQ' = PR .'. PS + SQ > PR + RQ,

210

+ RQ.

The proof of theorem 7-8 is among the harder ones; you may want to s k i p it and merely authorize the use of the theorem in s o l v i n g problems. We have assumed properties f r o m the diagram trithout f o r mal j u s t i f i c a t i o n . This will be done often hereafter as we (page 2101

at the end of Chapter 6. The proof in the t e x t t a c i t l y assumes that K, M, C are noncollinear. The proof applies to the case indicated by the left-hand figure below as well as to t h a t shown Fn t h e t e x t . If K, M, C are collinear (see right-hand figure below), then B, K, C are c o l l i n e a r and K lies between B and C . Thus BC > CK and since CK = EF we have PC > EF. indicated in the Commentary

Proof of --

211

Restatement:

Theorem 7-9.

7

Given A ABC m d A DEF. If AB = DE, AC = DF and BC > EF, then m L A > m L D.

Proof: Slnce m L A and rn L D a r e numbera, there are only three possibilftles: (1) m L A = m L D ( 2 ) m L A < m L D , and (3) m L A > m L D. (1) If m L A = m L D, t h e n 4 BAC ZA EDF and BC = EF. This contradicts the hypothesis, t h e r e f o r e it is impossible that rnL A = m L D. (2) If m L A < m L D, then BC < EF by Theorem 7-8. The l a s t I s false. Therefore, I t is impossible that (page 211 3

m L A < m L D. Only possibility ( 3 ) remains, and the theorem is proved.

Problem -S e t 7-3e

If two triangles have two sides of one congruent to two sides of the o t h e r , the third side of the first is longer than the t h i r d s i d e of the second if and only if the included angle In the f i r s t is larger than the included angle in the second. ACD and E D , AC = BC, DC = DC and 3D < AD, and so 2. In m L x > m L y by Theorem 7-9.

212 1.

A

3. 1. 2.

RA = RF.

2. 2.

RB = RB.

3. m L ARB < m f ERF. 4. AB < BF.

4'

5.

RA = RF.

1.

RB

2. 3.

1. 2. 3.

m L F R B > m L ARB.

4

FB

5.

mL

In CQ

3. 4.

A

=

RB.

> A

4.

AB.

5.

> m L F.

ACQ and BCQ, AQ =

.

Definition of median. Identf ty. Supplement P o s t u l a t e Theorem 7-8. Theorem 7-4.

.

B2,

CQ and BC > AC Then by Theorem 7-9 L CQB > L CQA . Since t h e t w o angles are supplementary, ,L CQB is

-

=

Dl

obtuse. 213

Given. Identity. Given. Theorem 7-8.

6,

In

A

AHF and FQA, FH = AQ, AF = AF, and AH

Therefore, by Theorem 7-9, m L AFH A Am, AB > FB by Theorem 7-5.

7.

Q

A

Given: QR = QT, SR = ST. Prove : m L RQT > m L RST. [pages 212-2133

>mL

FAQ.

8

> FQ. Then in

RS

> RQ,

ST

> TQ*

m L 1 > m L 3, m L 2 > m L h. m L RQT > m L RST.

8*

1. 2. 3.

1.

AB = FB.

2.

BH = BH. mi. AEH > m L HBF. (or, m L HBF > m L ABH. See below. ) 4. AH > FH.

3,

4.

D e f i n i t i o n of median. Identity. Given.

Theorem 7-8.

A l s o , if the medlan were drawn so that L ABH

then AH < Alternate by S . S . S . , the given

9.

I. 2.

3. 4, 5.

AB

Proof: Assume that HA = HI?. ThenA AHB ~ A F H B ThLs contradicts ao L ABH 2 L F3H and Informat ion, so that HA # HI?.

1 m.

1.

L ACB>LABC. In A E D a n d A FBC, FC = DB.

2.

CB FB

4. 5.

>

FBH,

FH.

> AC.

=

t,

> r,

>

symbols by the

>

>

s by Theorem 1-6, and a+b+c r+stt. If t h e triangle is oblique the proof still holda. If the triangle is a r l g h t triangle, simply replace t w o of the

5.

b

c

-> symbol.

Given:AABC with AC=AB=CB.

C

CE LAB, A D l E B ,

BF Prove:

1 AC.

CE = IF

= AD.

A A B D ~ A B C FZ by S.A.A. and so AD = BF = CE.

A CAE

Review Problema

216

1.

Yes, if the trunk is perpendicular to the ground, There are really three congruent triangles by Hypotenuse-Leg Theorem.

2.

m.

3.

Given:

In A Amy

i a t h e shortest a i d e since I t is opposite the smallest angle. In A ACE, ( Ef o r the sane reason. Therefore, i a the shortest sepent in the figure. F

F8 1 E .

XE >E. Prove: 737 > E. h c a t e X on 2 so that Bx = E. LFXB > fore, AF

L A by Theorem

>

7-1.

L C = L FXB > L A .

CF.

4.

-

2.

AF = HB. BF = BF,

1. 2.

3.

AB =

HF.

3.

1.

4.

5. Yes.

There-

-

Given. Identity. Subtraction in Statements 1 and 2. Hypotenuse-kg Theorem. Corresponding parts.

There will be two triangles which are congruent by

S.A.A.

217

5.

SinceAC>AB,mLB>mLC. L ADC is an exterior angle of A ABD and so m L A E > m LB. Therefore, m L APC > m L C . Hence, AC > AD.

A

0

6. Let a, b and

c be the lengths of the sides as shown.

b

>r

7.

since is the shortest segment. y > s aince is t h e l o n g e s t segment. x -+ y > r + a, by addition. Therefore, m L F > m L A.

8.

Let a be the length of t h e l o n g e s t s i d e of the triangle and b and c the lengths of the o t h e r s i d e s .

x

0

I. Theorem 7-7.

I. a < b + c . 2.

a = a.

2.

3.

2 a < a + b + c . a + b + c a
AB). c < w ( L AFB l a an exterior angle of A ~ 1 3 ~And ~ )so, c ( a. Also, a ( a + x, which gives us that m L A < m L ABH. We now have rn L H < m L A < m L ABH and as a result we b o w that t h e three sides of A ABH are unequal.

+lo.

Since m L CAB

90. Then x < 90 and a f b + c < ( a + x) + x < 180 -k 90 = 270.

1a.

L CBA. m L X B Y = r n f A, L XYA Z L XYB, XY = XY, therefore A AXY 2 A BXY (s.A,A.) and AY = BY.

A

XBC 2 A XBY.

so BC = BY.

13.

A

XI3 bisects

and Therefore, AB = 2EC.

We prove that r 1.

(s.A.A.)

+

a

>x +

r + t > x + w .

x

B

y.

1.

Theorem 7-7.

2.

Addition.

3.

Subtraction.

4.

Statement 3 and the f a c t that t + v = s.

w + v > y . 2.

r + t + w + v > x + w + y .

r + t + v > x + y . 4. r + s > x + y .

3.

*14.

If L ABE I s a right angle, P = Q = B. Hence, we suppose, with no loss of generality, that L ABE is acute. Its vertical angle is also acute, so m L ABE: < 90, m L CBD < 90. We show that P is on the same side of B as E by showing t h a t it cannot be on the side with D. If P were on the s i d e with D, L ABE would be an exterior angle of P ABP. This leads to the contradiction that m L APB < m F ABE. However, this is impassible since m L APB = 90 and m L ABE: < 90. Hence, P is on the same s i d e of B as E. Similarly, I t may be shown that Q is on t h e same s i d e of B as D by considering A BCQ and showing that the assumption that it lies on the s i d e with E leads to the contradiction t h a t the acute exterior L C B D has measure less than the r i g h t L Cm.

---

Illustrative Test Items for Chapter 7 1.

A

Conslder t h i s figure and liat correct responses to rill blanks below.

K

.

a . x = b.

is the longest a i d e of A KBR. is the s h o r t e a t s i d e of A KRR.

c.

2.

In A m , if XY = 18, YZ = 10 and XZ triangle has the largest measure?

3.

A triangle has sides of

4.

Given:

=

15, whlch angle of the

lengths x and x + y. Can the third s i d e of the triangle be of length y? S t a t e a theorem t o support your conclusions,

A

AX.

E is a point between B and C . D 1s a p o i n t between A and E. Prove:

LADC>LB.

C

A

5

6

and m L ARB 0 As shown in t h i s figure, ABCD i a a squape and E is a p o i n t on AB such t h a t B 1s between A and E. Prove: ED > A c . Given A ABC with median

-

=

73.

Prove m L A

N\

A

B

E

> mLC.

7.

--*

If, in t h i s f i g u r e , BH b i s e c t s

* L A 3 F and L A = L F, prove t h e

ray o p p o s i t e

8.

bbisects L AHF.

Prove t h a t t h e perimeter of the pentagon (shown in this figure) is g r e a t e r than t h e perimeter of A ACE.

E

9.

For the given figure prove t h a t the sum of t h e altitudes is l e s s than the perimeter of t h e tpiangle. (use a, b, c,

as lengths of t h e

sides of the triangle and r, s, t, as lengths of the altitudes, as i n d i c a t e d . )

C

Indicate whether true o r falae. a . The b i s e c t o r of the vertex angle of an isosceles triangle b i s e c t s t h e base and Is perpendicular to it. b. The base angles of an Isosceles triangle are acute. c. Any e x t e r i o r angle of a t r i a n g l e is larger than any interior angle of the triangle. d . If an angle of one triangle is larger than an angle of a second triangle, then the s i d e o p p o s i t e the angle in the first triangle is l o n g e r than the s i d e o p p o s i t e t h e angle in the second. e. A t r i a n g l e can be formed w i t h sides of lengths 351, 513, 162.

f.

An a l t i t u d e of a triangle lies in the i n t e r l o r of t h e

triangle. g. h,

I. j.

k. 1.

11.

If AB > AC in A A X , then m L C > m L B. Two triangles are congruent if they have two angles and a s i d e of one congruent t o the corresponding parts or the other. If the three angles of a triangle have unequal measures, then no t w o s i d e s of t h e triangle are congruent. A median of a triangle is perpendicular to the side to which it is drawn. In A ABC both and AC can be perpendicular t o E. C, The shortest s e p e n t from P to AB i s the perpendicular from P t o E.

Prove: If D is a point between B and C, then i a shorter than one of E, Prove t h a t one of the congruent sides of an i s o s c e l e s t r i a n g l e I s longer than the segment which connects t h e v e r t e x w i t h any point in the base.

m.

12.

Answers

m.

KB.

1. 2.

x=45.

3. 4.

No. The sum of the lengths of two sides o f a triangle -is greater than the length of t h e t h i r d side. L A D Z is an e x t e r i o r angle of A DEC and so L ADC > L DEC. L DEC is an e x t e r i o r angle of A ABJ3 and so L DEC > L B. Therefore, 1 ADC > L B.

5.

BC

L 2.

>

AB by Theorem

7-8.

Since m L A = 90, L DBE is obtuse. BY corollary 7-1-1 L E is acute. Then in A DBE, DE > DB by Theorem 7-5. A A B D z A BCA by S.A,S. so AC = DB. Hence,

DE 7.

8.

>

AC.

Let G be a p o i n t on 3 beyond H so that 3 is the ray opposite HB. AABH Z A FflH by S.A.A. Theorem. T h e n L A H B z L F H B and hence L AHG 2 L GHF since supplements of congruent angles are congruent. 9

ED+CC>E AB+BC>A

EA = EA. ED + DC

9.

r

< c,

Theorem 7-7.

+ AB + BZ + t < b, s < a

EA

> EC +

AC

+ EA,

by addition.

+

by Theorem 7-6, then r

t

+

s


m L 3. Hence, L B is acute. Thus, AD < A 3 by Theorem 7-5.

If

AD

-

A

6

D

C

a

+

b

+

c

12.

Given: A AH??with AH = FH and B a point between A and F. To prove: AH > HB.

Proof: 1. m L H E A > m L F. I

3. 4.

m LHBA

AH

>

BH.

>m

L

A.

I 1. 2. 3. 4.

Theorem 7-1. Wse angles of an i s o s c e l e s t r i a n g l e are congruent. Substitution. Theorem 7-5.

Chapter 8

PERPENDICULAR LINES AND PLANES

219

IN SPACE

This is a good time to ask yourself whether it is l i k e l y that your class will cover all the topics in the text. You will want to plan ahead t o glve your class a suitable program. You could make, rather quickly i f necessary, an i n t u i t i v e p r e s e n t a t i o n o f the p r o p o s i t i o n s of Chapter 8 by using familiar physical objects. Having students draw some figures after looking at simple models will improve t h e i r a b i l i t y to handle three-dimensional problems. On the o t h e r hand, deductive work in three-space may seem more important to you than many alternatltves. f a r t of the time you plan to allot to deductive work can be spent on proofs in three-space, even i f this e n t a i l s omitting some deductive work in two-apace. It is worth spending time to make the basic definetion of the chapter meaningful. A sizeable model will make your demonstration more effective. Use the f l o o r as a plane, several pointers for concurrent l i n e s in t h e plane, and a window pole for the perpendicular. Have students concentrate on one particular pointer. Move the pole to show t h a t the pole can be in many positions, even In the plane, and be perpendicular to the p a r t i c u l a r p o i n t e r . But the pole - in a l l b u t one p o s i t i o n - is n o t perpendicular to the other pointers. When the pole is perpendicular to all of the pointers, it is perpendicular to the plane. I f some students d i s c o v e r the idea o f Theorem 8-3 at this time, that1s fine! Whlle such demonstrations can do much to a s s i s t students in understanding s p a t i a l relationships, a most e f f e c t i v e means is the assigning of smaller models to be constructed by each student. Coat hangers, thin wire, straws, string and cardboard can be used to make models of the next theorems to be studied. ( s e e Problem Set 8-la, Problem 10.)

One p a r t i c u l a r l y meaningml device whlch students can make at an early stage I s the following. Each student has a piece of cardboard on which he draws a segnent and marks a point on t h a t s e g e n t . Next he inserts several common pins such t h a t each p i n is perpendicular to the segment at the p o i n t . The teacher can check each model at a glance. The model helps to illustrate the basic definition of S e c t i o n 8-1 and Theorem 8-5. Some excellent materials , mainly s t i c k s and connectors, for c o n a t m c t l n g models in three-space are available f r o m suppliers of scientific and mathematics equipment. Many teachers find these to be advantageous over ready-made models

.

--

Problem S e t 8-1 220

1.

2

a.

Yea.

b.

No, there would be p o i n t s in apace which are not in plane B.

a.

b.

c.

Yes.

Each of the three l i n e s I s perpendicular to the original

line. {page 2203

4.

The statement ia true. ( ~ e f e rto the diacusaion of using t h e word if I n definitions, Chapter 2, page 41 of the t e x t . )

5.

LABR, LABS, LTBA.

6.

No. The definition requires that the line be perpendicul a r to every l i n e containing Q and lying in E

7.

a.

Yes.

8.

a.

Yes, three points are always coplanar.

b.

Not necesearily.

"9.

-

b

.

TSP

and

L TSR.

a.

b.

1.

Given.

&A.

2.

Given.

3.

BA = BA.

3.

Identity.

4.

A PAB Z A

5.

L

6.

AX = A X .

6

7.

h PAX r A QAX.

7. Statements 2, 5 , 6 and

8.

PX = QX.

8.

1.

PB = QB.

2.

PA

-

PAB

a. 4.

= L QAB.

S.S.S.

5 . Corresponding parts

.

Identity.

Corresponding parts

.

8-1 you can

use the t i p of a

l i g h t fixture and a spot on the f l o o r as pointa, and a window pole as a line. You can even t a g the pole with

B,

S .A.S.

No.

As a model f o r Theorem

222

.

and a movable

X.

[pages 220-2221

A,

Probleln S e t 8-2a

224 1. 2

Yes.

Statement of Theorem 8-1.

Yes.

Yes.

Yea.

6.

Statement of Theorem 8-1.

Some s t u d e n t s should enjoy making a model fcr Theorem 8 - 2 . We s u g g e s t a t h i n s t i c k punched through a s h e e t o f c a r d b o a r d , w i t h d i f f e r e n t c o l o r e d s t r i n g s leading from t h e ends of the s t i c k to A , B , and C , Then use thumb-tacks f o r p o i n t s X, Y a n d Z . You can d e v i s e a model f o r Theorem 8-3 by punching a 2 26 p o i r i t e r t h r o u g h a s h e e t of cardboard t o r e p r e s e n t L and E . Then lay p e n c i l s on the cardboard to r e p r e s e n t L1, L p , and L3. 225

Problem -S e t 8-2b 2

1.

This follows d i r e c t l y from Theorem 8-2.

2.

The l i n e of Intersection is p e r p e n d i c u l a r t o t h e f l o o r . Many, in f a c t , e v e r y line in t h e f l o o r g o i n g t h r o u g h t h e p o i n t at which L I n t e r s e c t s t h e floor w i l l be p e r p e a dicular to L. No. it is perpendicular o n l y to l i n e s of t h e f l o o r t h a t c o n t a i n t h e p o i n t of intersection o f L and t h e f l o o r .

223 3 .

a.

b.

Three. The - s i d e s o f t h e s q u a r e a l l l i e in a p1ar.e. AE and FE determine a n o t h e r plane, and and BH determine a third. - - - - - Weknow E H I H R , R R ~ R F , R F ~ F B , B F l z ( f r o m t h e square) and FB ( ~ i v e n) . . From t h e t , last two of these we note that one l i n e , F3, is perpendicular to two o t h e r lines at t h e i r p o i n t of C, i n t e r s e c t i o n so we know t h a t F'B p l a n e ABH. It IS a l s o t r u e that % plane ABH, but the s t u d e n t p r o b a b l y cannot prove t h i s now {page: ; 224- ~ P G !

1

1

1

4.

a. b

.

Three.

-

Planes

ABF,

1E.

RHB,

1

( ~ i v e n). % E. Postulate 2 . ) Therefore, HB f o ~ o w sfrom meorem 8-3.

HB

1. 2.

3. 4.

*6.

and

5. 6.

BR=BA. PB = m. A AaFr A BF.

7.

F A = 373.

8.

L FAR

4.

5.

FRA.

AHRF.

h he or ern 6-2

1

and plane AHRF. This

Given. Definition of a l l n e perpendkcular t o a plane. Definition of perpendicular lines. Given. Identity.

6. 7. 8.

S.A.S.

1.

Property of t h e edge of a cube. Same as Reason 1. Given. Subtraction, Steps 2 and 3.

Corresponding parts. Base angles of an isosceles t r i a n g l e .

Yes.

2. 3. 4.

5* 6

7.

8. 9.

BF.

2.

BR=BL.

3. 4.

AB

=

AR =

FL.

5. 6.

A ATR A FTL. TR=TL.

-

7.

K T l X

E -l m .

KT 1 plane ABFT. K T I and F -

KT

8.

9.

1x.

10. 11.

[pages 228-2291

S.A.S.

D e f i n i t i o n of congru-

ence . Property of a cube.

Theorem 8-3. D e f i n i t i o n of a line perpendicular to a plane. S.A.S.

Corresponding parts.

23 1

1.

Definition of a line perpendicular to a plane.

2. 3.

Given.

meorem8-3.

By the time you reach Theorem 8-4 it might be best to proceed without a complete or elaborate model. Students should be encouraged to perceive spatial relationships In a diagram rather than t o become completely dependent on spatial models You may use a spoked wheel and axle to make Theorem 8-5 intuitively familiar: any line perpendicular to the axle at the hub must be in the p l a n e of the wheel. Proof of Theorem 8-7 The perpendicular bisecting plane of a aegment I s the s e t of a l l points equidistant from the end-points of t h e

.

23 1

23 2

segment. Rea tatement : L e t E be the perpendicular bl secting plane of AB. L e t C be the mid-point o f Then

z.

(1)

If P

(2)

If

is in

PA =

PB,

E,

then P A ' = P B ,

then

P

is in

and

E.

Proof of (I): If P = C , then we already know t h a t PA = PB. If P # C, then % lies in E by Postulate 6 , by the definition of a l i n e perpendicular to a and plane. It follows t h a t L ACP z BCP, and, since CA = CB and CP = CP, we have A ACP Q A BCP by S . A . S . Therefore, PA = PB.

1

tA

[pages 230-2321

Proof of (2): If P = C , then c e r t a i n l y P is in E. If P C, then A ACP A BCP by S.S .S. Theorem. Thus C, L ACP EL BCP and E contains CP by Theorem 8-5, and P lies in E .

+

-1 E.

-

Alternate proof of (1): L e t C b e the mld-polct of AB. If P = C, then certainly PA = PB. If P # C, then % is the perpendicular bisector of AB ( in plane ABP) and t h e r e f o r e PA = PB by Theorem 6 - 2 . Alternate proof of (2) : If P = C , then certainly P is In E. If P C, then ;5if 1s the perpendicular bisector of (in plane ABP). Since E contains Fif by Theorem 8-5, P l i e s in E. The proof of part (2) of Theorem 8-7 as given above requires that Theorem 8-5 be proved p r e v i o u s l y . A simple indirect proof usea Theorem 8-1 and the Space Separation P o s t u l a t e in the following way:

Given P such t h a t PA = PB. Suppoee P doea n o t l i e in E. Then it l l e s in one of the two half-spaces Into which E separates space. A and 3 lie in opposite half-spaces, since AB intersects E at C, by hypothesis. Then P is in the half-space opposite to e i t h e r A or B, say B . Than % meeta E in a point Y. By (11, Y l a equidistant f r o m A and B, and by hypothesis, F is equidistant from A and B. Then by Theorem 8-1, B is equidistant f r o m A and B! This a b s u r d i t y implies that o u r supposition is false, and so P is in E.

-

--

Problem S e t 8-2c

233

1.

a.

InfinZtelymany.

b.

One.

2.

Yes.

3.

The conclusion follows d i r e c t l y from Theorem 8 - 5 .

4,

Points W, X, Y and Z are given equidistant from t h e ends of AB. By Theorem 8-7, they all belong t o the perpendicular bisecting plane of AB and are therefore coplanar.

Yes.

No.

-

5.

23 4

a. b.

-

BW.

-

BK.

-

BR.

Not n e c e s s a r i l y . in E.

90.

L BKF.

W, K

and

R

1.

Theorem 8 - b .

2.

Definition of a line perpendicular to a

could be any points

36. 1.

There e x i s t s a plane perpendicular t o L at M. If E . = E 1 , each l i n e in Et through M I s perpendicular

Et 2.

3.

If E # E f ,

the

intersection of E and Et is a l i n e

L' 4.

plane.

L.

to

.

L I L t .

1

3.

Postulate 8.

4.

D e f i n i t i o n of a line perpendicular to a plane.

23 4

The proof of Theorem 8-8 uses the word "let" in two

somewhat different senses. " ~ e Lines t L1 and L2 be perpendicular" means "call the two given perpendiculars L1 and L,". " ~ e t M be the mid-point of AB" means "~onsider C the mid-point of AB, and c a l l it M". h he mid-point exists by Theorem 2-51.

Review Problems

AR

>

RB.

~ L >B m L A

Theorem 8-8.

Yes.

No.

Yes,

Yes. No.

( ~ L =B 9 0 ) . No.

Yes.

Theorem 6-3. w

w

and WF are coplanar by Theorem 8-8, s o that M, Q , W and F are coplanar. If t w o p o i n t s a r e in a plane the line c o n t a i n i n g then is in the same C* w plane. Hence MW and QF a r e coplanar w i t h MQ and Only one.

MQ

-

WE". a.

b.

Three. Plane ABF, plane RHE a n d p l a n e RHF. Two i n t e r s e c t i n g lines determine a plane. - - A F L R H and A F ~ B Hso, A F L p 1 a n e R H . B by Theorem 8-3.

A XAP Y A XBP Hence Hence

by S.A.S.

Similarly we know XB = XC. is equidfstant from A , B, C.

XA = XB,

X L

1 planeABC. -

Ll&A,

&B,

QC.

PA = PB = PC,

A PA& 2 A PBQ 3 O PC&. &A = QB = QC. F o r any p o i n t X Q on L, A x A Q E h D Q Z A XCQ. XA = XB = XC.

{pages 236-2383

Given. Definition of a line perpendicular t o a

plane. Identity. Given. Hypotenuse-Leg Theorem. Corresponding p a r t s

.

S.A.S.

Corresponding parts.

10.

On the ray opposite to

that

QR = QB.

Then -

--*

&B l e t R be the p o i n t such APQR PQB by S . A . S .

:. PR = PB. AP LET, m a nd 1 I A P 1 plane PBC . Therefore, A APR Y A and AR = AB. . * . (EL*)

since APB

( S .A .S .)

by Theorem 6-2

and Postulate 1. 239 11.

Connect A with X, the point of such that BX = BH. Then A ABH E b ABX ( s . A . s . ) and AX = AH. Since LC, m L ABF > F, and since L AXF is an exterior angle of A AE!X, rnL AXF > 4 ABX > mL F. Then AF > AX and, substituting, we have AF > AH.

4

12.

+ AB were

perpendicular to each of the three rays AD, AE. Then by Theorem 8-3 and 8-5, the 4 + three rays would be coplanar. If AD and AE were + each perpendicular to AC and all were in a plane, + + then AD, AE would be o p p o s i t e rays and not perpend i c u l a r . Hence each ray cannot be perpendicular t o the other three.

Suppose

s,

13. 1. 2.

+ +

I

= In. YPLAB,

* H AB 1 YB.

or

I

I

1 1.

2.

3. 4.

5.

[ p a g e s 238-2391

Given. D e f i n i t i o n o f a line perpendicular to a plane. Given. Reason 2.

Statements 2, 4 and Theorem 8-3.

I

I

for Chapter 8 Z l l u s trative T e s t Items C a n the d i s t a n c e from a given point to a given plane

vary? Identify the s e t o f p o i n t s which are equidistant from two points A and B? Through a given p o i n t not In a plane, how many l 1 r . e ~can

be perpendicular to the plane?

At a point on a line how many lines can be perpendicular to the line? At a p o i n t on a l f n e how many planes can be perpendlcul a r to the line? Is it p o s s i b l e f o r a line which intersects a plane in only one p o i n t n o t t o be p e r p e n d i c u l a r t o any l i n e In the plane?

Can a l i n e be perpendicular to a line i n a plane and y e t n o t be perpendicular to t h e plane?

Three p o i n t s A , B, C are each equidistant from two points P and Q. F i l l in t h e blanks to make t r u e statements

.

a.

If A , B, C are collinear then equidistant from P and Q.

b.

If Is

is

are n o t collinear t h e n e q u i d i s t a n t from P and Q, A , B, C

Points A , B, C, and D are n o t coplanar, A C is i s o s c e l e s w i t h AB = A C . A DBC is isosceles with DB = DC 0 F Is t h e mid-point of E. In the figure at least one segment is perpendicular to a plane. What segment? What plane?

.

A

2.

Given

a.

jn

t h i s figure that

and a plane

determine

m ~ .IS %$

perpendicular to plane ABK? If your answer w a s "yes" s t a t e a theorem t h a t supports

,

your conc lus Ion. b.

Do

C-)f-*w

FB,

rta,

lie in plane Explain. c.

HB

all

KBQ? d i f f e r e n t planes determined

There will be by the given l i n e e .

3,

RQ

R

In this figure, plane E bisects

-

1

and E RQ. A l s o RT QT. Explaln why T l i e s in plane E.

Indicate whether true or false: I.

A l i n e perpendicular to a plane is perpendicular to

every line in the plane. 2.

If a line is perpendicular to two lines of a plane it must be perpendf c u l a r to the plane.

3,

Through a p o i n t on a plane only one plane can be passed.

4.

There are infinitely many lines perpendiculas to a given l i n e at a given p o i n t on the line.

5.

Two l i n e s perpendicular t o the same plane are coplanar.

Through a point on a line two planes can be passed perpendicular to the l i n e . Thlrteen points each equidistant from the end-points of a segment are coplanar.

If two lines L1 and L2 are each perpendicular to l i n e L, at a given point of L, there is a plane containing L1 and L2 t h a t is perpendicular t o L. A l l l i n e s perpendicular to a line at a given p o i n t of

the l i n e are coplanar. llne perpendicular to a l i n e in a plane is perpendicul a r t o the plane.

A

*

If AB and plane E are each perpendicular ta p o i n t P, then M lies in plane E.

In t h i s f i g u r e

E

is the perpendicular - bisecting plane of AB. 1f CF lies in B and CF = CB = FB, prove A ACF is

equilateral.

J+~/ 8

Given in this figure: * H K I E at 33.

Prove: A HBA,

d FAB,

A

and

are in one plane and are congruent to each o t h e r . KBA

E

w

F'H

at

3.

-

V is the mid-point of edge RW of the cube shown in this figure Prove

A.

.

VB

=

W.

1.

No, it is the length of the unique perpendicular sement From the p o i n t t o the plane.

2.

The perpendicular bisecting plane of

3.

One.

4.

Infinitely many.

5.

One.

7.

Yea.

8.

a.

If A , 3, C are coll-lnear then each point of the line containing A , B, C i a equidistant from P and Q.

b.

If A, 8, C are n o t collinear, then each p o i n t of the plane containing A, B, and C is equidistant from P and Q.

AB.

B.

1.

B C l p l a n e DFA.

2.

a.

No. BQ c m o t be proved perpendicular to the plane ABK on the basis of t h e information g i v e n .

b.

Yes, Theorem 8-5.

c.

Six;

3.

D.

-

ABK, ABQ, perpendicular to

ABH,

%8

ABR, at B.

and the plane

ABF,

This follows fromTheorem 8-7.

1.

1.

AC=C8.

AF 2. 3. 2.

%?

=

FB.

AC=CF=AF.

A

A C F Is equilateral. C*

2.

.

3.

HypothesLs and Step 1. Definition of equilateral triangle.

.

and F A are coplanar ( heo or en 8-8) 9 lnce a l l vertices of A HBA,d FAB and A K B A are p o i n t s of these l i n e s , the triangles are in one plane. HBA, L =A and L FAB a r e right angles (~efinitionof a l i n e perpendicular to a p l a n e ) . BA = RA (~dentity). d HBA s a FAB KBA ( ~ y p o t e n u s e - ~ e ghe or em).

U s e auxiliary segments

1. 2.

3. 4

A R A B S D WHF. RB = w . RV = VW.

L VRB

and

1 VWF

RB

I

and

T.

1.

S.A.S.

2.

3. 4.

Corresponding s i d e s . Definition of mid-point. planes of faces and WHIT,

5.

S.A.S.

6.

Corresponding sides.

are r i g h t angles.

Chapter 9 PARALLEL LINES IN A PLANE

In this chapter we introduce the Parallel Postulate and t h e familiar theorems on parallels and q u a d r i l a t e r a l s . The treatment is not significantly d i f f e r e n t from t h a t of most traditional texts, except in t h i s respect: The expllclt use of the postulates and theorems of our e a r l y chapters and the c a r e f u l formulation of d e f i n i t 9 o n s . By t h i s time the student should be quite adept at maklng proofs. Consequently, t h i s chapter s i m p l y s t a t e s t h e easier theorems and leave8 t h e i r proofs f o r the s t u d e n t to accomp l i s h . P r o o f s not supplied i n the text are provided in t h i s commentary. Please note, however, t h a t s t u d e n t s may o f t e n discover proofs d i f f e r e n t from t h e one given here, or in the t e x t , and, o f course, such proofs should recefve a p p r o p r i a t e recognition and acceptance. As we proceed to study more complicated material we shall relax t h e degree of precision with which we t r e a t it. We shall sometimes state definitions which are not wholly p r e c i s e and g i v e proofs that are not, logically complete, with the expectation that they will be understood w i t h t h e aid of diagrams. In succeeding c h a p t e r s t h i s is done more extens i v e l y . In the present chapter we point o u t s e v e r a l instances o f unprecise treatment and indicate appropriate c l a r i f i c a t i o n . The discussion of parallel lfnes in a plane, though by no means d i f f i c u l t , encompasses probably the rnos t significant property of Euclidean geometry, namely, the " ~ a r a l l e l ~ o s t u l a t e " , stated on page 262. By way o f introduction ask the students t o t e l l what they mean by parallel l i n e s . The answers will no doubt v a r y , and some will probably be inc o r r e c t . Most answers w i l l probably be descriptions, rather t h a n definitions. It is hoped t h a t from a d i s c u s s i o n of t h i s s o r t the class w i l l get the reeling that they a r e working with something that i s I n t u i t i v e l y very simple, b u t that a t

.

241

11

the same time the concept of parallelism is n o t one that can easily be 'pinned downtt by the s t u d e n t Point out to the students the definition of parallel lines gives t w o conditions t h a t m u s t be m e t by t h e lines, (1) they must l l e in the same plane and (2) they must n o t i n t e r s e c t . Ask the s t u d e n t f o r an example o f two lines that s a t i s f y condition ( 2 ) , but f a l l to s a t i s f y condition (1) and hence are n o t p a r a l l e l . Skew lines i s the example. Remind the students that parallel lines do n o t meet. You w i l l sometimes hear the expression: "Parallel l i n e s meet a t infinity". This does not mean t h a t t h e lines do meet, Mathematicians abhor exceptions, f o r example, two lines do n o t always meet in the Euclidean plane, and j u s t as it is convenfent to Introduce complex numbers i n t o algebra so t h a t every quadratic equation has a r o o t , so it is convenient to adjoin t o t h e p o i n t 8 of the plane, c e r t a i n "ideal" p o i n t s s o that we can say two lines always meet. Notlce, however, t h a t such lines are no longer Euclldean l f n e s . To each Euclldean line we adjoin an i d e a l polnt to form a new kind of line, called a projective l i n e , that is no longer a Euclidean line. T h i s is done in such a way that the same i d e a l p o i n t is adjoined t o each line of a f a m i l y of parallel l i n e s , If two Euclidean l i n e s are p a r a l l e l t h e n t h e i r associated p r o j e c t i v e l i n e s meet i n an ideal p o i n t . If two Euclidean lfnes are n o t p a r a l l e l they meet in a point P and their associated p r o j e c t i v e l i n e s meet in the same point P. T h i s avoids an exception, but all t h e propertlea of real p o i n t s do n o t carry over automatically t o i d e a l p o i n t s . When we say two projective lines meat at an ideal p o i n t , it f o l l o w s that t h e i r associated Euclidean lines do not meet a t a l l . If we adJoln these i d e a l p o i n t s to the s e t of r e a l p o i n t s in the Euclidean plane, we g e t a new "planet', whlch has d i f f e r e n t p r o p e r t i e s from the Euclldean plane, and which we may call a tt projective plane'' In the sense t h a t "point", " l i n e " , and II plane" would satisfy t h e s e t of Incidence postulates u s u a l l y

made f o r projective geometrg. But t h i s is not the geometry we are studying; In Euclidean Geometry we do have parallel lines, in P r o j e c t i v e Geometry there are no parallel l f n e e . Theorem 9-2 gave w one method f o r constructing a line 242 parallel to another line through an external point. The method was used in Theorem 9-3 to prove t h e existence of at leaat one llne parallel to a given line from a point not on the line. Some enterprising studenta will feel that Theorem 9 - 3 establishes uniqueness as well as existence of L2, especially in light of t h e paragraph following t h e proof. After a l l , Theorem 6 - 1 assures that Lp as a perpendicular to L1 at P is unique. Should t h i s arise you may counter with a statement of t h i s s o r t : "If t h i s seems astonishing to you, perhaps y o u are reading more meaning in Theorem 9-2 than is actually there. Notice that Theorem 9-2 does not say two lines in a plane are p a r a l l e l o n l y if they are both perpendicular to t h e same l i n e . Is it possible then that two lines could a l s o be parallel under some o t h e r conditions?" If more discussion seems necessary you may decide to present the following: L e t the figure be that of Theorem 9-3. From p o i n t R on L2 drop a perpendicular to L, u meeting L at S . Note that we do n o t know that -& Lg. H Fmm P make PZ RS. Now we have %?11 L and Lg 1 1 L by Theorem 9-2. We seem to have two l i n e s through P parallel to L.

1

1

L, >LQ

Y

Y

-1

The student w i l l probably claim t h a t PR RS and therefore %? and L2 colncide (Theorem 6 - 3 ) . While you nay agree with him t h a t t h i s aounds promislng, ask him to prove that the f a c t h i s argument is based on. Whatever he may PR r e f e r to as convincing evidence f r o m h i s general atore of knowledge you e a s i l y can maintain the essential polnt of t h e whole dlscusslon: t h a t n o t h i n g i n our p r e v i o u s p o s t u l a t e s o r theorems w i l l d i s p r o v e o u r argument, The s o r t of reasons which refute it - the sum of the measures of t h e angles of a quadrilateral is 360, or of a triangle is 180, alternate fnterior angles (Theorem 9 - 8 ) , corresponding angles h he or em 9-91, a n d so on - have not been proved y e t (and in fact, can ~ o be t u n t i l t h e P a r a l l e l P o s t u l a t e i s assumed). You would probably not want t o go f u r t h e r i n t o this w l t h your class, especially at t h i s t i m e - and probably not even t h i s far. But we should state the point to this diecussion, f o r t h e reader, a t l e a s t . The point is that t h e s t a t e m e n t s which would refute the above argument a r e all l o g i c a l l y e q u i v a l e n t t o Postulate 16. Neither P o s t u l a t e 16 nor any of t h e s e equivalent statements 1s deducible as a theorem from Postulates 1-15. It was t h e discovery of t h i s f a c t that finally l e d geometers to the r e a l i z a t i o n t h a t some p o s t u l a t e of p a r a l l e l i s m is necessary. ( s e e T a l k s on I n t r ~ a u c t i o nto Non-Euclidean Geometry and on Miniature Geometries.)

1z,

-

-

Notice t h a t we give a precise d e f i n i t i o n of alternate i n t e r i o r angles rather than a "definition" in terms of a picture. Obaerve that our d e f i n i t i o n depends on t h e separation concept as developed in Chapter 3 . Proof of Theorem 9 4 2 46 Given a t r a n s v e r a a l t o two l i n e s , if one p a i r of a l t e r n a t e i n t e r i o r angles are congment, then the o t h e r p a i r of alternate interior angles are also congruent.

2115

-

{pages 242-2463

Given: L ABC

Lines

= L BCD.

To Prove:

L1 and L2 c u t by tranaveral T such that

L x S L Y.

By the Supplement Postulate L ABC and 1 x are supplementary, as are A BCD and y Since

.

A

Am EL BCD, then x E L 7 , because

supplements of congruent angles are congruent.

< D

Problem Set 9-1

2481. 2

a.

No.

b.

No.

They do n o t intersect, they are both perpendicular t o a t h i r d line, they form alternate i n t e r i o r angles w i t h a transversal. ( ~ o t e : The t h L r d conditfon includes the second as a

special c a s e . ) 3.

No.

4.

Not necessarily.

5.

a.

No, the 80'

b,

Two sizes:

angles are not alternate i n t e r i o r angles, and the a l t e r n a t e I n t e r i o r angles are n o t equal. 80°and 1 0 0 ~ .

[pages 246-2483

249 7 .

S e l e c t any two points t , A , B on L. Draw PA. Draw L CPA 3 L 3AP ao t h a t C and B are on opposite s i d e s of Thep %$11 L by Theorem 9-5.

C

w.

8.

a.

9.

Yea.

10.

250 11.

E

Yes. b. No. c. y e s . d . Yea. e . Yes, s i n c e a line containing the center of the earth is perpendicular to c e r t a i n o t h e r lines containing the center. f. No. g. Yes. h. Yea.

-

(Such llnes are called skew lines. )

AABDEABAC byS.A.S. Then D B - C A . Then ADCB = h CDA b y 3 . S . S . and 4 B C D r d A D C . (1t i a not p o s s i b l e to prove that L BCD and L ADC met be right angles. Attempts t o do so suggest the need for some f u r t h e r postulate.) A A P R g d P B Q S h R Q C n A Q R P b y S . S . S . By corresponding parts 4 a = 4 A, b = 4 B and c = 4 C . Since the sum of the measures of a, b and L c is 180 by Postulates 13 and 14, the sum of the measures of L A, L 3 and C is 180.

Proof:

rnL

L

I t may seem surpr-laing t h a t we can prove that the sum of the measures of the angle8 of A A N I s 180 before we have introduced the Parallel Poatulate. In t h i s Problem the hypothesis assumes the existence of a trfangle in which the length of each segment j o i n i n g the mid-points of two sides 1s one-half the length of the t h i r d side. Thia cannot be proved before assuming the Parallel Postulate. We should note, however, t h a t if we do assume that such a triangle e x l s t s , and from t h i s show that the sum of t h e measures of the angles is 180, we can prove the Parallel Postulate. (see the commentary above on equivalence of atatementa t o the Parallel Postulate. See, a l s o , Talks on Introduction to Non-Euclidean Geometry, Corollary 7. ) {pages 249-2501

250 12.

Proof:

A P A R U A Q A R by S . A . S . Then By a nimilar proof using

Zl-.

and A ACD,

Wlw.

Then

%$ 11 %?

LARP~LARQ A

and by Theorem 9-2. ABD

( ~ o t e : A proof based on i s o s c e l e s triangles without

-

drawing

1. 2. 3. 4.

5.

AD

is a l s o p o a s i b l e .)

P D A T S A CBT. IYP=CT. mLDTA=dCTB. ADSTEACST. ~ M ' s = ~ c T s .

1.

S.A.S.

2.

Corresponding parts. Corresponding parts ,

3

.

4.

5. 6.

7. 8. 9. 10.

3.3.3. Cor~espondingparts Addition. Definition of perpend i c u l a r lines. Corresponding parts. Definition of perpendicular l i n e s . Theorem 9-2.

.

252

Proof of Theorem 9-6 Given two lines and a transversal, if one pair of corresponding angles a r e congruent, then the other three pairs of corresponding angles have the same property. Oiven: Lines L1 and L2 c u t by tranaveraal T such t h a t a p a i r of corresponding angles, L a and L a', are

congruent.

To Prove:

L e = L c t ,

L b GLbt,

Given t h a t /agL a t . By the Supplement Poa t u l a t e a is supplementary to < L, L b , and L a 1 is supplementary to L b t Sfnce supplements of congruent angles are 4 *& congruent, b PI bf Similarly we show L c a L c t and L d P L d t . T h e method of proof of Theorem 9-7 is merely to use the p r o p e r t y of vertical angles to establish a p a i r of alternate i n t e r i o r anglen congruent, and by Theorem 9-5, the lines a r e parallel. Because t h e converses of Theorems 9-5 and 9-7 a r e reasonable and are readily accepted by s t u d e n t s as intuitively true, you may find that the dependence on t h e P a r a l l e l P o s t u l a t e remains unrecognized, even a f t e r the converses have been proved. As preparation f o r the p r o o f of Theorem 9-8 and preliminary to the Paraf lel Postulate a c o n s i d e r a t i o n s imllar to the following could be discussed. It seems reasonable that t h e converse of Theo~ern9-5 is t r u e . Let's examine its reasonableness If we assume t h a t the parallel t o a line through a p o l n t n o t on the llne is n o t unique. Then we could suppose two such p a r a l l e l s e x i s t , as In the f i g u r e .

L

. .

25 2

Ld%Ldl,

Now how reasonable is the converse of Theorem 9-5? Accordi n g t o it, a = x and a = y, so that x = y. But by the Angle Construction Po8 tulate x # y. Thfs contradiction means that if we want t h e converse of Theorem 9-5, and many more such "reasonable" theorems, to hold, then we must accept the unfqueness of the parallel. Problems 7 and 8 of Problem S e t 9-3 present a more complete p i c t u r e of the situation by showing that the Parallel Postulate can be proved if Theorem 9-8 or Theorem 9-12 is assumed. From all of this the student should become convlnced some p o s t u l a t e of parallelism must be s t a t e d . The importance of the Parallel Postulate is best seen, perhaps after the sequence of theorems through Theorem 9-13 f a finlshed and the s t u d e n t c a n look a t the sequence, including the Postulate , in its entire development. 252 The Parallel Postulate seems reasonable on the basis of o u r experience In the world about us. There is no t h e o r e t i c a l reaaon why we could not assume the existence of two parallels to a given l i n e through an external p o i n t . From this point on, Parallel Postulates different from ours result in the development of d i f f e r e n t geometries, called Non-Euclidean Geometries. (See Chapter 1 of Studies I1 and the T a l k s on Miniature Geometries and Introduction to Non-Euclidean Geometry. )

253

254

Now that we have the uniqueness of a parallel through an external point it is p o s s i b l e t o prove the converse of Theorem 9-5. Note carefully in the proof in the t e x t how the fact t h a t t h i s parallel is unlque I s used to establish the validity of Theorem 9-8. Proofs of Theorems 9-9, 9-10, 9-11 and 9-12 Theorem B. If L1 11 Lp, we then b o w by Theorem 9-8 t h a t the altermate i n t e r i o r angles are congruent. By application of the property that vertical anglea are congruent, we can e s t a b l i s h the pairs of corresponding angles to be congruent.

The term " i n t e r i o r anglea on the same side of the transveraal'' can be defined formally as follows: L e t L be a transversal of L1 and L2, Lnteraacting them in P and Q. L e t A be a point of LI and B a point of L2 such t h a t A and B are on the same s i d e of L. Then L PQB and L WA are called interior angles on the same side of the transversal L. Compare this with the d e f i n i t i o n of alternate interior angles.

-----

-

-

25 4

Theorem 9-10. Given

L1 11 L2. Then it f o l l o w s from Theorem 9-8 that

La Ld

b. A l s o , L a and are supplementary. Hence, mL a + 4 d = 180 = 4 b + mL d . Therefore b and d are supplementary. In a l i k e manner e can be proved supplementary to L a .

L

L

L

Y

255

Theorem 9-11. Given: In a plane, L1 )( L3 and L~ 11 L ~ .TO Prove:

L1 1 1 L2. We use the ind i r e c t method of proof and assume that L1 is n o t parallel to L2. If t h i s is true, then these two lines will meet at some p o i n t P. This means t h a t there are now two lines through P ( L ~and L2) parallel to L3. This contradicts the Parallel Postulate, hence, L1 must be parallel t o L2.

-

Remark on proof of Theorem 9-11, proved directly as follows : Given: In a plane, L1 11 L3,

Lp

This theorem can be

I1

Lg. TO Prove: L1 It L2L e t T be a transveraal intersecting L1, L2 and L3. Such a transversal exists, since any line in the plane of L1, L2, L3 which meets L1 in only one point m u s t meet L2 and L3 by the Parallel Postulate. Consider the a l t e r nate I n t e r i o r angles formed as Indicated Ln the f i g u r e .

11 Lg, L2 1 1 L3, L1

hence

(1) L a g L b

byTheorem9-8.

hence

(2)

Lc Lb Laar&

by Theorem 9-8.

Therefore,

(3)

and

(4)

L1

11 L2

by Theorem 9-7.

255

-

Theorem 9-12. Lines L, L1 and L2 a r e ooplanar . Qiven: II II Lp and L Ll

1

at

P. L

To Prove:

1 L2.

L i n t e r s e c t s L2, othemise L and L1 would be parallel to L2 and contain P. This c o n t r a d i c t s the Parallel P o s t u l a t e . Therefore L is a transversal of L1 and Lg. By Theorem 9-8 it follows that I and L2 form a right angle. Thus L L2.

1

Problem -Set 9-3

255

1.

1.

Given.

2.

rnL AD I[ E.

2.

3.

mLc=mL~=go.

3.

Theorem 9 - 2 . Theorem 9-10.

1.

2.

mLA=mLB = C = 90.

3

Isosceles AABC w i t h AB = BC and t 11 AC and i n t e r setting AB and 8C at F and Q. Prove: A PBQ is i s o s c e l e s .

Given:

1. 2. 3.

4.

L A ~ L c . t i 1 E. L X LA and 1 y E L C so that L-X =-L y .

P B E B Q , or AFBQ is isoscefes.

3.

Theorem 5-2. Given. Theorem 9 - 9 ,

4.

Theorem 5-5.

1. 2.

255

3.

L PQT 1 S L RTS Z L S

by

Theorem 9 - 9 .

by Theorem 5-2.

From these two statements P Q E PT by Theorem 5-5.

256

4.

a.

b

5.

.

a.

b.

PQT

S

L RTS .

Then

Suppose M d o e s n o t i n t e r s e c t L2. Then, by deflnitlon, M 11 L2. But L1 is g i v e n 11 L2. Hence there are through P t w o parallels to L2 -- an impossibility by t h e Parallel P o s t u l a t e . The assumption that M does n o t i n t e r s e c t L2 is therefore f a l s e , so t h a t M does intersect L2.

L1 11 L2. R 11 Lg by the given i n f o r mation. Also by the given information both L1 and R c o n t a i n P. Since there cannot be two parallels to a line through a p o i n t , t h e assumption L1 )I Lp is false, and L1 i n t e r s e c t s L2.

Suppose

LY

-

BQY

Therefore,

257

L

and

LB

L B ELBQY

by Theorem 9-8.

L Y.

Consider YX forming L P Y Z with sides extending in the same d i r e c t i o n as those of L ABC. Then, f r o m part (a), m L PYZ = m L ABC. But mL PYZ + rnL XYZ = 180, and t h e r e f o r e

IIILABC + TILxrz

=

180.

It should be i n t u i t i v e l y c l e a r what is meant when we say t w o parallel rays extend in the same or opposite d i r e c t i o n s . A formal definition is e a s i l y given. If 3 11 3 and B and D are on the same s i d e (opposite sides) o f %? we say + + AB and CD extend in the same (opposite) directions. 6.

If the s i d e s of one angle a r e perpendicular ~ e s p e c t i v e l y to the s i d e s of another angle, then the angles are either congruent o r supplementary.

257 " 7 .

D r a w a transversal %? of L and M and a l s o of Le and M forming angles a, b and c as shown. If L p 1 1 M, then L b 3 L c; and since L1 11 M, L a 3t c by Theorem 9-8. Therefore, a at L b. But then L1 = L2 by the Angle Construction P o s t u l a t e , so there cannot be a second parallel to M through Q.

L

258 +8.

L

Consider a line

perpendicular to M from P. By Theorem 9-12, t L1. Assume L:, parallel to M. Then t L2. Since L1 and Lg cannot b a t h be perpendicular t o t a t P, L2 cannot be p a r a l l e l to M as was assumed.

1

t

1

258

Observe t h a t although the proof o t Theorem 9-13 is more precise than that given in most t e x t s , it s t i l l depends on the figure t o show that L x and L x 7 are alternate I n t e r i o r angles. Theorem 9-13 is the first major consequence of o u r Parallel P o s t u l a t e . The proof is directly related to the f a c t that there is but one line parallel to the base of the triangle through the opposite vertex. If there were more than one, or no parallels, the sum of t h e measures of the angles of a triangle would be less than 180 or greater than 180 as is the case in the Non-Euclidean Geometries. ( s e e Talk, I n t r o d u c t i o n to Non-Euclidean Geometry. ) It is interesting t h a t In Euclidean s p h e r i c a l geometry the sum of the measures o f the angles of a spherical triangle is greater than 180.

25 9

Proofs of the Corollaries Corollary 9-13-1. Given a correspondence between t w o triangles. If two angles of the first triangle are congruent t o t h e corresponding parts of the second, then the t h i r d angles are congruent.

Glven: A ACE and A BDF, such that A S L 3 and

LcP~LD. TO Prove: L E a L F. We now know, from Theorem 9-13 that the awn of the measures of the angles of a triangle is 180. Qiven that the sums of the measures of two angles in each triangle are equal, then the differences between t h i s sum and 180 in each case are equal. Thus

c

0

F

~ L E = ~andF L E S L F .

260

Corollary 3-13-2. his pmor follows d i r e c t l y from Theorem 9-13. If the sum of the measures of the angles of a t r i a n g l e is 180, and one angle has a measure of 9 0 , then the sum of t h e measures of the remaining two angles must be 90. By deflnitlon, then, these angles are complementary.

260

Corollary 9-13-3. Given: A ABC with exterior angle L BCR. To Prove: ~ L B C =R 4 A + mL B. By the Supplement P o s t u l a t e mA BCR = 180 BCA. From Theorem 9-13 it follows t h a t mL A + 4 B = 180 - m L BCA. Therefore 4 BCR = 4 A + mL J3.

[pages 259-2601

A

C

R

Problem S e t 9-4

261

3.

The Parallel Postulate assures us that L is the only parallel to AC through B. It ia a l s o used to prove that alternate i n t e r i o r angles are congruent when parallels are cut by a transversal, and this theorem in turn 1s used in the proof of the angle-sum theorem.

4

(~umbemin parentheses were given in the original problem.)

5.

a.

6.

By theorem on transversala of parallels LEBD Y L A and L DBC L C. B u t L EBD L DBC. Therefore A C. Hence AB = BC.

Yes.

b.

No.

1 7.

rnL

I. = 4 3 by hypothesis and 4 2 = mL 4 by Theorem 5-2. But m L I + r n L 3 = m L 2 + 4 4 by Corollary 9-13-3. Taking half of each swn we have mL 1 = 2, and t h e b i s e c t o r I s parallel to the base by Theorem 9-7. we have

rnL

[pages 260-261 1

/ \ / \ 2

4

For convenience we Indicate angles as shown in the figure

.

r=rnL2+44.

f.

1.

C o r o l l a r y 9-13-3.

r n l 2)

2.

Addition.

and

3.

Angle Addition Postulate

4.

Statements 2 and 3.

s =rnLl+rnL3.

r

+

+

(mL 3

2.

3.

s

=

(mL 1 -t

+

mL 4).

rnLl+mL2=t

rnL 3 + mL

4 = u. r + s = t + u .

4. *9.

Since QB = QA, L B E L 1 . Since L 2 and L 1 are complements, L 2 and L B are a l s o . But L B and C are complements, hence, 1 2 C because complements of the same angle are congruent. Now &A = QC, and, hence, &B = QC.

ZL

+lo.

In A ABC,

m A In

ATs,

A BTR,

4B mL

= 90

ATS =

mL BTR

=

mL STR

=

-

a.

180

-.

-

a =

180

= 180

-

9 0 ; a,

(mL ATS + mL BTR) 90 + a -( 180 - a + 7

proofs of Theorems 9-14 through 9-18 Theorem 9-14.

Given: Parallelogram ABCD To Prove: A ABC Y A CDA

with dlagonal

.

1. 2.

AD )I BC

and

AB lm.

I

LDCA~LCAB.

L DAC = L ACB. 3. 4.

AC

I

=a.

A A B C Z A CDA.

-

m.

1.

Definition of a

2.

parallelogram. Alternate i n t e r i o r angles.

3. 4.

Identity. A . S . A . Theorem.

The proof using diagonal BD is of course, similar to t h i s . Observe we are reading f r o m the f i g u r e t h a t D and B are on opposite sides o f %? 265

Theorem 9-15 1s an immediate consequence of Theorem 9-14: Since the triangles are congruent it follows that the corresponding sides are congruent.

Corollary 9-15-1. Given: L1 11 L p and P and Q on L1. To Prove: P and Q a r e equidistant f r o m L2. S

Epage 2651

T

From P and Q perpendiculam

Qr!

1.

Theorem 6-4 and d e f i n i t i o n of distance from a p o i n t to a line.

3.

Deflnltion of parallelo-

LQ.

to

PQTS

drop and

l a a parallelogram.

gram.

4.

266

Theoremg-15.

Theorem 9-16. Slnce the triangles i n t o which a diagonal divides a parallelogram are congruent, then t h e corresponding angles are congruent. In t h e figure of Theorem 9-14, Considering diagonal DB, we can show, in t h e same manner, A SL C.

LD=LB.

1

266

Theorem 9-17. Consider any two consecutive angles o f a parallelogram as the i n t e r i o r angles on the same side of a transversal cutting two parallel lines. Then Theorem 9-17 is immediate by Theorem 9-10 (given t w o parallel lines and a transversal, interior angles on the same side of the transversal are supplementary).

266

Theorem

9-18.

Given: Parallelogram ABCD with diagonals AC and E. (we assume from t h e figure that the diagonals i n t e r s e c t at P For a proof see answers to Problems' 19 and 20 of Problem Set 9-6.) To Prove: AC and b i s e c t each other.

.

[pages 265-266

3

1.

~ '

La %Lad.

AB

L b q b ' .

2. 3. 4.

5.

pt

m.

ABPZACDP.

AP ZCP. DP PL E. AC and

bisect

1.

Alternate interior

angles

.

2.

Theorem

3.

A.S.A.

9-15.

4.

Theorem. Corresponding parts.

5.

Definition of b i s e c t .

each other.

in the t e x t , there is a natural break, or summary p o i n t , a f t e r Theorem 9-18. Teachers should keep in mind that a c a r e f u l selection of problems can emphasize the common characteristic of Theorems 9-14 through 9-18, and similiarly f o r Theorems 9-19, 9-20, and 9-21. A t the same time, the fact that Theorems 9-14 through 9-25 all Involve quadrilaterals is strengthened by the arrangement of the t e x t . Thus Problem Set 9-6 supplies problems f o r both Section 9-5 and Section 9-6. P r o o f s of Theorems 9-19, 9-20, and 9-22 As is pointed out

266

Theorem 9-19.

Given: Quadrilateral ABCD with and r To Prove: ABCD is a parallelogram. 1. Draw diagonals AC and 2. By t h e S .S .S Theorem b ABC Q b CDA and A DAB 3. Therefore L a S L b and ELLd. 4. Then by Theorem 9-5, AB I( CD and 11 5, ABCD is a parallelogram by definition.

-

.

z.

LC

x.

m. d BCD .

266

-

Theorem 9 - 20. Given: Quadrilateral ABCD with m Cb and AB 11 m. To Prove: ABCD l a a parallelogram. Since 11 m, L a L b by alternate interior angles AC = m, and AABC E A CDA by the S . A . S . Postulate. Therefore BC and by Theorem 9-19 aBCD is a parallelo-

=

gram. 266

-

Theorem 9-21, Given: Quadrilateral ABCD with diagonals b i s e c t i n g each o t h e r a t P To Prove: ABCD is a parallelogram.

DB

and AC

.

1.

D P S m .

I

1.

Given.

2.

LCPBZLDPA.

1

2.

V e r t i c a l angles are

- AP P PC.

congruent.

L DPC a L BPA.

267

268

3.

A D P C ~ ~ B P A . ACPB r A APD.

4.

AB=CD.

5.

ABCD

n ~ m * is a parallelogram.

1

,

Postulate.

3.

S.A.S.

4.

Corresponding p a r t s .

5.

Theorem 9-19.

Theorem 9-22 a t a t e a a f a c t that surprises many students. Perhaps some students will enjoy making a model to d e m ~ n strate vi~ually, rather than just logically, that the length of the segment joining the mid-points of t w o sides is oneh a l f the length of the third a i d e . In some t e x t s a rectangle I s defined in the following way: If one angle of a parallelogram is a r i g h t angle then the figure is a rectangle. If this d e f t n i t i o n I s u s e d , you would want the Theorem. If one angle of a parallelogram l a a right angle then a l l four angles are right angles, which in effect is Theorem 9-23. Using t h i s theorem you see that the suggested d e f i n i t i o n is equivalent t o our definition of rectangle. {pages 266-268)

268

Proofs of Theorem 9-23, 9-24, and 9-25 Theorem 9 - 2 3 . By Theorem 9-17 the consecutive angles of a parallelogram are supplementary, and since one angle is a r i g h t angle I t s supplement must be a r i g h t angle. Two successive applications of Theorem 9-17 will establish that the o t h e r two angles are right angles. Or we could apply the theopem that opposite angles of a parallelogram are congru-

ent. Theorem 9-23 gives us an efficient way to prove that a quadrilateral is a rectangle. First prove that it is a parallelogram and then prove that one angle I s a r i g h t angle.

268

Theorem 9 -24. Given: Rhombua A3CD w i t h diagonals AC and -BB. TO Prove : AC BE.

-1

By the d e f i n i t i o n of rhombus AB = RD and CB = CD; A 6 that is, A and C are equidistant from B and D. Since A and C are coplanar with B and D, by Theorem 6-2 - B. 1s the perpendicular b i s e c t o r of , Hence, AC An alternate proof uaea the S .S .S. Theorem to get congruent any two of the triangles having a common s i d e . Then the angles of a linear pair are congruent, and the diagonals are perpendicular.

1

268

Theorem 9-25, Using the figure of Theorem 9-24 we have: Given: ABCD with and AC and BD bisecting each o t h e r . To Prove: ABCD is a rhombus. By hypothesis, %? is the perpendicular b i s e c t o r of so that AB = AD and CB = CD by Theorem 6-2. Similarly, AD = CD so that AB = AD = CD = CB. By definition, ABCD is a rhombus A n alternate proof uses the fact that A A P B Z A APD S A CPB A C P D by S . A . S .

BD,

.

After the class has become familiar w i t h the p r o p e r t i e s of quadrilaterals s t a t e d on the previous pages you might propose the following two problem for them to work. Neither of these can be solved since there f a a counter-(an example satisfying a l l of the glven conditions that J.oes not s a t i s f y the desired result) f o r each one. (1) Given quadrilateral ABCD auch t h a t 11 5 and AD E B C , prove t h i s quadrilateral is a parallelogram. Do

not inform the students that this cannot be proved. Let them search f o r themselves f o r a while and perhaps r e a l i z e that the counter-example is an isosceles trapezoid. This figure satisfies a l l of the given conditions, but c e r t a i n l y is n o t a parallelogram. Given a quadrilateral ABED auch that the diagonals are perpendicular to each o t h e r . Prove t h a t the quadrilateral I s a rhombus (or a aquare). T h i s problem, a l s o , cannot be solved. A counter-example is a k i t e , l i k e t h i s : It can be formed from two non-congment i s o s c e l e s triangles having the same base f i t t e d together as in the f i g u r e . A more general f i g u r e is also possible. ( 2

--

Problem Set 9-6 A l l four quadrilaterals. A 1 1 four.

Square, rhombus , A 1 1 four. Square, rhombus. All four. Square, rhombus All f o u r .

.

[pages 268-2691

D

269 2.

270

3.

Rectangle, square.

j.

Rectangle, aquare.

+ 30 + LA =

x

2x - 60 = 180 and x = 70. Therefore, F = 80; = = loo.

~ L B~ L H

3.

Since the opposite angles of a parallelogram are congruent, L H Z L A and a l s o L R g L A , so t h a t L R = L H. S k c e interior anglers on the aame a i d e of a transversal which c u t s parallel l i n e s are supplementary, L M is supplementary to L A . By s u b s t i t u t i o n we see t h a t 1 M is supplementary to L H.

4.

a. b. c.

d. e.

f. g. h.

I. j. k, 1.

5.

No. No. No. No. Yes. No. No. No. No. No. No. No. Yes. No. Yes. Yes. No. Yes. Yes, Yes. No. Yes. Y e s . Yes. Y e a . No. No. Yes. No. No,

Yes. Yes. Yes. Yes. No. No,

No, No. No, No. No.

No. No. No.

No. Yes. NO.

No.

and AB = DC s i n c e opposlte s i d e s o f a parallelogram are congruent, Then A APD % A CRB and AAPB E A CRD by S.A.S. Then by correspondlng parts R13 % PB and P D E. Having oppoaite sides congruent, DPBR is a parallelogram. AD = BC

=

l pages

269-270 ]

1. 2.

3.

5.

-

11 AID. FE 1 1 BC. AD BC. - I( FE

1.

Definition of a parallelogram. Theorem 9-ll*

FEEAD.

2. 3.

4. 5.

Statement 3. Statements 2 and 4

ABCD

i a a parallela-

gram.

1

and Theorem 9 - 2 0 .

1.

PXRY is a parallelogram.

1.

2.

PX=RY,

2.

RX=PY.

Theoremg-35.

Definition of a parallelogram. Theorem 9-15. Theorem 9-9. Theorem 5-2.

3. 4. 5.

LXPSSLT.

LSBLT.

3. 4.

L X P S q S .

5.

Angles congruent to the s m e angle Statements 3 and 4.

6.

PX

6-

Theorem 5-5.

7.

By steps similar t o steps 2-6. Statements 6 and 7 by

8.

SX.

PX+XR+RY+YP =

or

=

SX

+

XR

+

RY

+

XR

+

RY

+

RS

+

RT.

PX =

8.

.

addition,

+

YT, YP

1. 2.

Theorem 9- 18. Definition of a parallelogram.

EQ = FQ. EF is b i s e c t e d

by

3. 4.

Theorem 9-8. V e r t i c a l angles are congruent.

5. 6. 7.

A.S.A.

Q. [pages 270-2711

Corresponding p a r t s . D e f i n i t i o n of b i s e c t .

-

271 9.

Through D draw a parallel to CB meeting AB at X . Then DCBX is a parallelogram in which case CB = DX. Since it was glven that AD = CB, therefore DX = DA and L DXA Z L A . But, by corresponding angles L DXA 9~ L B. Therefore L A B.

21210.

a.

-

ADCQrAKBQ byA.S.A.orS.A.A. s o t h a t Q is mid-point of E. In A ADK, PQ ) I AK and 1 1 PQ = +K = $AB + BK) BK = CD since they are

.

corresponding parts of congruent triangles. PQ

$(AB

b.

8 inches.

1.

Draw

2.

RQ (1

3.

SP

-

4.

RQ

-

5. 6.

11

-

+

CD) . c.

E.

aement

.

2.

Theorem 9-22.

1 DE; S P u T D B .

3.

Theoremg-22.

4.

Statements 2 and 3 .

5.

Theorem 9-11.

6.

Theorem 9-20.

7,

Theorem 9 - 18.

;

SP.

be between B and C such that AD = BC1 so t h a t Then %? 1 1 ABC'D is a parallelogram and PLDCf = mL BCfD. Ct

Two p o i n t s determine a

R Q = V1D B .

gram. SQ and PR b i s e c t each o t h e r .

k t

G. 1.

R Q I ) S . SP&R is a parallelo-

7.

27313.

=

=

.

mL

Making this replacement in ADC'

4 BC'D

Hence,

< mL ADC, < m L ADC.

we have

A

By the

E x t e r i o r Angle Theorem C < 4 BC'D. Therefore

4

[pages 271-2733

C