SMSG (School Mathematics Study Group) Geometry: Student’s Text, Part I

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-1,000,000; x means that x < y.

T h e express i o n X I 3'

means t h a t x Is leas than o r equal t o y . cause 3 (

For example, 3

l. x 1.

g.

Ix) 1.

x

b.

c. d.

a. b.

m L XYZ. Compare DF and X Z . Regarding A PIX and h J U N such that m LPIQ = m t JUN, PD > JU, and QD = NU, a hasty person might conclude t h a t PQ > JN. Draw a f i g u r e showing t h a t t h e conclusion is n o t

H e r e are some experiments

1.

2.

3. 4.

5. 6.

justified.

7.

is a p o i n t in p l a n e E, ,B is a ray n o t lying in E, and + AC is a ray lying in E. Con sidering different positions + o f AC, d e s c r i b e a s accurately + a s you can t h e position of AC which makes L BAC as small as possible; as l a r g e a s possible. N o p r o o f is expected b u t you a r e asked to guess the answer on t h e basis of your knowledge o f space. A

8.

O n t h e basis

of drawings decide whether o r n o t an angle can be

t r i s e c t e d by the following procedure: L e t A A X be an i s o s c e l e s triangle with congruent sides A 3 and E . T r i s e c t side BC with p o i n t s D, E so t h a t BD = DE = E C . Is L BAD L DAE Z L EAC?

-. 1-2.

-

A l g e b r a o f 1-nequalities.

B e f o r e considering geometric inequalities we r e v i e w some of t h e facts concerning i n e q u a l i t i e s between real numbers.

Note first

t h a t a < b and b > a a r e merely t w o ways of writing t h e same t h i n g ; we u s e whichever is more c o n v e n i e n t , e . g . 3 < 5 or 5 > 3 .

D e f i n i t i o n s . A real number is positive If it i s greater t h a n z e r o ; it is negative If it is less t h a n z e r o . We now r e s t a t e t h e o r d e r p o s t u l a t e s , giving examples of their use.

0-1. (uniqueness of order.) For every x and y, one and only one of t h e following r e l a t i o n s h o l d s : x < y, x = y, x > y. 0-2. (Transitivity o f o r d e r . ) If x < y and y < z, then

x


3 , we can conclude t h a t a < b. Proof: If a < 3 and 3 < b, t h e n a < b. Example 3 . Any p o s i t i v e number is greater than any negative number. Given: p is p o s i t i v e , n I s n e g a t i v e .

TO prove:

p

>

n.

Proof: 1.

p

2.

p

is p o s i t i v e .

>

0.

1.

Given.

2.

D e f i n i t i o n of positive.

3. 4.

0

n

5. 6

n p

< < < >

Relation between < and Definition of n e g a t i v e . Postulate 0-2. Relation between < and

3. 4.

p. 0.

5

p. n.

6.

>. >.

( ~ d d l t i o nf o r inequalities.) If x < y, then y + z, f o r every z . Example 4. Since 3 < 5 it f o l l o w s t h a t 3 + 2 < 5 + 2, or 5 < 7 ; t h a t 3 + (-3) < 5 + ( - 3 1 , or 0 < 2; t h a t 3 + ( - 8 ) < 5 + (-8), or -5 < - 3 . Example 5. If a < b then -b < -a. P r o o f : a + (-a-b)< b + ( - a - b ) , o r -b < -a. Example 6 . If a + b = c and b is positive, then a < c. Proof:

0-3.

x

+

z


0. < b.

Why? my?

Fxample 7. If a + b < c t h e n a < c - b. Proof left to t h e s t u d e n t . Example 8. If a < b, t h e n c - a > c - b f o r e v e r y c . Proof l e f t t o t h e s t u d e n t . 0-4. z

>

( r u l t i p l f c a t i o n for 1ne~ualities.1 If

xz < y z . Example 9 . From

x


AC, then rn L C > rn L B. Tilecrern 7 - 5 . Given A A E . If mL C > m L 13, t h e n AB > A C . We have seen lots of p a i r s of theorems t h a t a r e r e l a t e d this

way.

F o r example, we showed t h a t i f a t r i a n g l e i s i s o s c e l e s , t h e n

its base angles are c o n g r u e n t : and l a t e r we showed t h a t if t h e uase a n g l e s of a t r i a n g l e are congruent, t h e n t h e triangle is [ s e c . 7-31

i s o s c e l e s . Each of t h e s e theorems is the converse of t h e o t h e r . We showed t h a t ?very equilateral triangle is equiangular; and l a t e r we proved t h e c o n v e r s e , which s t a t e s t h a t every equiangular triangle i s equilateral. It is very important to remember t h a t t h e converse of a true theorem is not necessarily t r u e at all. F o r exaiple, the theorem It vertical angles a r e congruent'' is always t r u e , but t h e converse, "congruent angles a r e v e r t i c a l " is c e r t a i n l y not t r u e in a l l cases. If t w o t r i a n g l e s are congruent, then they have the same a r e a , but if t w o t r i a n g l e s have the sane area, it does n o t f o l l o w that they are congruent. If x = y, t h e n it follows t h a t r2 = y 2 ; b u t if 2 2 x = y , I t does n c t f o l l o w t h a t x = y . (The other possibility is t h a t x = -y.) It is true t h a t every physicist is a s c i e n t i s t , but it is not true t h a t every scientist is a physicist. If a theorem and its converse are both t r u e , they can be conv e n i e n t l y combined into a s i n g l e statement by u s i n g t h e phrase "if and only if". Thus, if we say: Two angles of a triangle a r e congruent if and only if t h e opposite s i d e s are congruent; we are i n c l u d i n g in one statement b o t h theorems on i s o s c e l e s t r i a n g l e s . The f i r s t half of this d o u b l e statement: Two angles of a t r i a n g l e are congruent if the opposite s i d e s are c o n g r u e n t ; is Theorem 5-2; and t h e second half: Two angles o f a triangle a r e congruent only i f the o p p o s i t e s i d e s are congruent;

is a restatement of Theorem 5-5.

Problem S e t 7 - 3 c

5 , HK

1 4 , KG = 11. Name t h e largest angle.

1.

In A G H K , GH

2.

Name t h e smallest angle. In A ABC, m L A = 36, mL B = 74, and m L C = 7 0 . longest side. Name t h e s h o r t e s t side.

=

=

[sec. 7-31

Name the

3.

Glven t h e f i g u r e w i t h HA = HB, m L H 3 K = l h O , and m L ABB = 100, fill in t h e blanks below: a. m L A = - - b. m L R H B = - - A C . - - - is the longest s i d e o f A ABH.

Y

AB

-

4. What c o n c l u s i o n can you reach a b o u t the length of ML A KLM if: a. m L K > m L M? b. c.

d.

e. f.

5.

K

in

rnLK m L K > m L L? mLM>mLL? m L K > m L 14 and

m L K > m L L? m L K 2 m . L L and m L M L m L L?

If the f i g u r e were c o r r e c t l y drawn whlch segment would be t h e longest?

6.

Name the sides o f t h e f i g u r e in order o f i n c r e a s i n g l e n g t h . A

7.

If in the f i g u r e

AF

is t h e

shortest s i d e and CB i s t h e l o n g e s t side, p r o v e t h a t m L F > rn L E. ( ~ i n t : u s e diagonal FB.)

4

C

*8.

If t h e base o f a n i s o s c e l e s triangle is e x t e n d e d , a segment tthich joins t n e v e r t e x of t h e t r i a n g l e with any p o i n t in t h i s

extension is g r e a t e r than one of t h e congruent s i d e s of t h e triangle. F

3.

Write the c o n v e r s e of each statement. Try to d e c i d e whether each s t a t e m e n t , and each c o n v e r s e , is t r u e or f a l s e . a . If a team has some spirit, it can win some games. b.

I f two angles a r e r i g h t angles, they are c o n g r u e n t .

Any t w o congruent angles a r e supplementary. The interior o f a n a n g l e i s t h e i n t e r s e c t i o n o f two half - p l a n e s , e. If Joe has scarlet f e v e r , he is s e r i o u s l y i l l . f. If a man l l v e s in Cleveland, Ohio, i ~ el i v e s in Ohio. g. If the three angles of one t r i a n g l e are congruent t o t h e corresponding angles o f a n o t h e r triangle, t h e t r i a n g l e s are congruent. h. If two angles a r e complementary, t h e sum of t h e i r measures i s go. 10. men asked t o g i v e the converse of t h l s statement, "If I h o l d a l l g h t e d m a t c h t o o long, 7 will be burned", J o h n s a i d , "I will be burned if' I h o l d a l i g h t e d match too l o n g . " Was John's s e n t e r ~ c et h e c o n v e r s e o f the o r i g i n a l statement? Disc;iss. 11. a . Is a converse of a t r u e s t a t e ~ ~ e nalways t t!=ue? Which p a r t s of Problem 9 illustrate y o u r answer? b. May a converse of a f a l s e statement be true? Wk-~ichparts c. d.

of Problem 9 illustrate your answer?

Isec. 7-31

The s h o r t e s t segment joining a point t o a line

Theorem 7-6.

is the perpendicular segment,

Restatement: Let Q be the f o o t of the p e r p e n d i c u l a r t o the line L through t h e point P, and l e t R be any o t h e r p o i n t on

L. Then PQ Proof: L e t


mL PRQ. B u t m L PQS = m L PQR = 90, and so m L PQ,R > m L PRQ. By Theorem 7 - 5 it follows that PQ < PR, whlch was t o be proved,

D e f i n i t i o n . The d i s t a n c e between --a line and a p o i n t not on it i s the length of the p e r p e n d i c u l a r segment from the p o i n t t o the l i n e . The distance between a line and a p o i n t on the l i n e is defined t o b e zero.

Theorem 7-7. ( ~ n eTriangle Inequality. ) The sum of t h e l e n g t h s of any two s i d e s o f a t r i a n g l e i s g r e a t e r than t h e length of the third s i d e . Restatement: In any triangle A A X , we have AB + BC > AC.

Let

Proof: that

DB

=

AB.

D be a p o i n t of the ray opposite to Since B is between C and D

I X

=

+

D B + B!2.

(1) X = A B + B C . ( 2 ) m L DAB < m L DAC,

because B is in the I n t e r i o r of L DAC. Since A DAB is i s o s c e l e s , w i t h AB = DB, it follows t h a t (3)

m L ADB

rn L DAB.

=

By ( 2 ) and ( 3 ) we ilave

rn L ADB

< mL

DAC.

Applying T,leorem 7-5 t o A ADC, we see t h a t

(4) Dc

>

AC.

Ey (1) 2nd ( 4 ) i t follows t h a t

AB

+ E

)

which was t o be p r o v e d .

Set Problem -

1.

Here AH


BH. Prove: AF

>

AH, --*

12.

f*

13.

+

4

+

Prove t h a t each of f o u r rays AE, AC, AD and AE cannot be perpendicular t o the o t h e r t h r e e . H

f*

Given: XB and YB are two lines in plane E; m is a plane 1 XB C* @ at B; n is a plane YE a t B; AB f s t h e intersectton of rn and

1

n. Prove:

e AB

1 E.

Chapter 9

PARALLEL LINES IN A PLANE

-

Conditions Which Guarantee Parallelism. Thus f a r in o u r geometry we have been mainly concerned w i t h what happens when lines and planes intersect i n c e r t a i n ways. We a r e now going to see what happens when they do n o t i n t e r s e c t . It will turn out that many more i n t e r e s t i n g things c a n be proved. We f i r s t c o n s i d e r the case of two lines. Theorem 3-3 g i v e s us some information right away, since it says t h a t if two lines intersect they l i e in a p l a n e . Hence, if two lines are n o t coplanar they cannot i n t e r s e c t . Definition: Two lines which a r e n o t coplanar a r e s a i d to be skew. You can e a s i l y f i n d examples of skew l i n e s in your classroom. This still leaves open t h e question as t o whether two coplanar lines must always i n t e r s e c t . In Theorem 9-2 we shall prove t h e existence of coplanar lines t h a t do n o t intersect, b u t are parallel, like t h i s : -

1

-

L e t us f i r s t make a precise definition. D e f i n i t i o n : Two lines a r e parallel if they are coplanar and do n o t Intersect. Note that f o r two l i n e s to be parallel two conditions must be s a t l s f i e d : they must not i n t e r s e c t ; they must both l i e i n the

same plane.

Theorems.

Two p a r a l l e l lines l i e in exactly one plane.

Proof': I f LL and L2 are p a r a l l e l lines it follows from t h e above definition that there is a plane E containing L1 and L ~ . If P is any p o i n t of L2 it follows from Theorem 3-3 t h a t there is only one plane containing L1 and P. Hence, E is the o n l y plane c o n t a i n i n g L1 and L2. We will use the a b b r e v l a t t o n

L1

11

Lp

t o mean that the

lines L1 and L2 a r e parallel. As a matter of convenience we will say t h a t two segments a r e p a r a l l e l if the lines t h a t contain them are parallel. We will speak slmllarly of a line and a segment, o r a line and ray, and so on. For example, suppose we have given that L1 11 L p , in the f i g u r e below:

+

--*

-+

z,

Then w e can also w r l t e I[=, AB 11 Lp, L1 1 1 CD, BA 11 and s o on. Each of t h e s e statements is equivalent to t h e s t a t e ment that L1 I I L2. It does n o t seem easy t o tell from the definition whether t w o lines which seem t o be p a r a l l e l r e a l l y are parallel. Every lfne stretches o u t i n f i n i t e l y far in two d i r e c t i o n s , and t o tell whether two lines do not intersect, we would have to look at all o f each o f the two lines. There is a simple condition, however, which is sufficient to guarantee that two lines are parallel. It goes like t h l s :

-

Theorem 9-2. Two lines i n a p l a n e are parallel if they a r e both p e r p e n d i c u l a r to t h e same line.

Proof: Suppose t h a t L1 and L2 are two lines in plane each perpendicular t o a line L, a t p o i n t s P and Q.

There a r e now two possibilities: (1) L1 and L2 i n t e r s e c t in a p o i n t (2) L1

and

L2

R.

do n o t i n t e r s e c t .

In Case (1) we would have two lines, L1 and L2, each perpendicular to L and each passing t h r o u g h R. T h i s is impossible by Theorem 6-1 i f R l i e s on L, and by Theorem 6-3 i f R is n o t on L. Hence, Case (2) is t h e only possible one, and so, by definition, L1 11 L2 Theorem 9-2 enables us t o prove the following important e x i s t e n c e theorem.

E,

on

Theorem 3-3. L e t L be a l i n e , and l e t P be a p o i n t n o t L. Then t h e r e is a t least one llne through P, parallel t o

Proof: L e t L1 be a line through P, perpendicular to L. (BY Theorem 6-1, there is such a llne.) L e t L:, be a line through P, perpendicular t o L1 in the plane of L and P. By Theorem 9-2, L2 11 L. It m i g h t seem n a t u r a l , at this p o i n t , to t r y t o prove t h a t the parallel given by Theorem 9-3 is unlque; that is, we m i g h t try t o show t h a t in a plane through a given point n o t on a given line there is only one p a r a l l e l to the given line. Astonishing as it may seem, this cannot t h e postulates proved the basis t h a t we have s t a t e d -so f a r ; it must be taken as a new p o s t u l a t e . --We will d i s c u s s this in more d e t a i l in S e c t i o n 9 - 3 , In t h e meantime, b e f o r e we g e t to work on the basis of this new p o s t u l a t e we shall prove some additional theorems which, like Theorem 9-2, t e l l us when t w o llnes are parallel, We f i r s t give some d e f i n i t i o n s .

-

-

Definition: A transversal of two coplanar Lines is a llne which i n t e r s e c t s them in two d i f f e r e n t p o i n t s . W e say t h e two lines a r e "cut" by the transversal.

D e f i n i t i o n : L e t L be a t r a n s v e r s a l of L1 and L2, intersecting them in P and Q, L e t A be a p o i n t of Ll and B a p o i n t of Lg such t h a t A and S a r e on opposite s l d e s of L. Then L PQB and L &PA are alternate i n t e r i o r angles formed by the t r a n s v e r s a l to the two lines.

Notice t h a t in the d e f i n i t i o n of a transversal, the two lines that we s t a r t w i t h may or may not be parallel. But if they I n t e r s e c t , then the t r a n s v e r s a l is n o t allowed to i n t e r s e c t them at their common p o i n t . The situation in the f i g u r e below is n o t allowed:

-

That is, in this figure L is not a transversal t o the lines L1 and L2. Notice a l s o t h a t a common perpendicular to two lines in a plane, as in Theorem 9-2, is always a t r a n s v e r s a l .

Theorem 9-4.

IF t w o lines are c u t by a t r a n s v e r s a l , and 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 congruent, t h e n t h e other p a i r of alternate i n t e r i o r angles a r e a l s o congruent.

1 a E L a t , then L b Z L b r . And if L b E 1 bt, L a ~1 a l . The p r o o f Is l e f t t o the student.

That is, if

then

The following theorem is a generalization of Theorem 9 - 2 , that i s , i t includes Theorem 9-2 as a s p e c i a l case:

,Theorem 9-5. If two lines are c u t by a transversal, and if a p a i r o f alternate i n t e r i o r angles are c o ~ g r u e n t , then t h e lines are para1 l e l

.

d

[ s e c . 9-11

Proof: L e t L be a transversal t o L1 and L2, i n t e r secting them in P and &. Suppose that a p a i r of alternate i n t e r i o r angles are congruent. There are now two p o s s i b i l i t i e s :

(1)

L1 and

(2)

Li

II

L:,

intersect in a polnt

R.

L2.

In Case (1) the f i g u r e looks l i k e t h i s :

be a p o i n t of L1 on t h e opposite side of L from R . Then SPQ is an exterior angle of A PQR, and L PQR I s one of the remote interior angles. By Theorem 7-1, t h i s means t h a t Let

S

L

-

B u t we know by hypothesis t h a t one pair of a l t e r n a t e i n t e r i o r

angles a r e congruent. By-t h e preceding theorem, b o t h pairs o f alternate i n t e r i o r angles are congruent. Therefore,

m LSPQ = m

PQR.

Since S t a t e m e n t (1) leads t o a contradiction of o u r hypothesis, Statement (1) is f a l s e . T h e r e f o r e Statement ( 2 ) is true.

[sec.

9-11

-9-I

Problem Set 1.

2.

a.

Does the definition o f parallel l i n e s s t a t e that the lines m u s t remain t h e same d i s t a n c e a p a r t ?

b,

I f two g i v e n lines do n o t l i e in one p l a n e , can t h e lines be parallel?

, or

!I'wo lines in a p l a n e are parallel if

if

9

o r if 3.

If two l i n e s in a plane are intersected by a t r a n s v e r s a l , are t h e a l t e r n a t e i n t e r i o r angles always congruent?

4.

In space, if two lines a r e perpendicular the two lines p a r a l l e l ?

5.

a.

b.

6.

t o a third

If the 80' angles were correctly drawn, would L1 be parallel to L2 according to Theorem 9 - 57 Explain.

.f

I

I

How many d i f f e r e n t measures of angles would occur in t h e drawing? What measures?

In t h e figure, if t h e angles were o f the size indicated, which lines would be p a r a l l e l ?

ALI

4

L2

+

90' 91

Ma' V

9-11

'L3 89.

00.

M24

[sec.

line, are

v

'Dl*

7.

Given a l i n e L and a p o i n t P not on L, show how p r o t r a c t o r and mlcr can be used t o draw a parallel to L

through

8.

P.

Suppose the following two definitions are agreed upon: A vertical l i n e is

one contafning t h e c e n t e r of t h e

earth. A horfzontal line is one which is perpendicular to some

v e r t i c a l line.

9. 10.

a.

Could two horizontal lines be parallel?

b.

Could two v e r t i c a l l i n e s be parallel?

c.

Could two h o r i z o n t a l lines be perpendicular?

d.

Could two v e r t i c a l lines be perpendicular?

e.

Would every v e r t i c a l l i n e a l s o b e horizontal?

E.

Would every horizontal line also be v e r t i c a l ?

g.

C o u l d a horizontal line be parallel to a v e r t i c a l line?

h.

Would every line be horizontal?

Is it possible t o f i n d two lines in space which are neither parallel nor intersecting? m L CBA = 90, AD = CB. Prove: m L ADC = m BCD. Can you also prove m L ADC = m L BCD = go? Given:

1 DAB

m and

0

=

C \

1

'\., , '

\

/ // /

0

A#

I'

'\

\

'.

\B

11.

Given the AR = RC = AP = PB = BQ = QC = Prove :

figure with PQ, RQ, PR.

rnL~+rnL~+mLC=180. B (~int: Prove m L a = m L A ,

12.

13.

Given:

AB = AC, AP = A&. e Prove: PQ 1 BC (~int: L e t the b i s e c t o r of L A i n t e r s e c t PQ at R and at D.) #

.

Given: The figure with

L A ~ B , AD =

BC,

SD = SC. Prove : - -

ST

-

1 AB. -

[ sec. 9-11

C

Correspondfng Angles. In the figure below, t h e angles marked called corresponding a n g l e s : 9-2.

Similarly, c,

c1

and

and

b

d,

d'

a

and

a1

are

a r e corresponding angles; and the p a i r s are also corresponding angles.

bt

D e f i n f t i o n : If two lines are c u t by a transversal, if L x ' and 1 y a r e a l t e r n a t e i n t e r l o r angles, and if L y and L z ; are v e r t i c a l angles, then L x and L z are corresponding ; mules.

You should prove the following theorem.

Theorem 9-6. If two lines are c u t by a transversal, and if one pair of corresponding angles are congruent, then the o t h e r t h r e e palrs of corresponding angles have the same p r o p e r t y . The proof is only a little longer than that of Theorem 9-4.

T h e o ~ e m9-7. If two lines are cut by a transversal, and I T

a p a i r of corresponding angles are congruent, then the lines a r e p a r a l l e l . The p r o o f is left t o t h e s t u d e n t .

It looks as though the converses of Theorem 9-5 and Theorem 9-7 o u g h t to be t r u e . The converse of Theorem 9-5 would say that if two parallel llnes are c u t by a t r a n s v e r s a l , then the a l t e r nate i n t e r i o r angles a r e congruent. T h e converse of Theorem 9-7 would say that if two parallel lines are c u t by a t r a n s v e r s a l , then corresponding angles ape congruent. These theorems, however, cannot be proved on t h e basis of the p o s t u l a t e s that we have s t a t e d so f a r . To prove them, we shall need t o u s e t h e Parallel P o s t u l a t e , which w i l l be stated in the next section. The Parallel P o s t u l a t e is essential t o t h e p r o o f s of many o t h e r theorems o f o u r geometry as well. Some of these you are already familiar with rrom your work in other grades. F o r example, you have known f o r some time that t h e sum o f t h e measures o f the angles of any t r i a n g l e is 180. Yet, without the Parallel P o s t u l a t e I t is impossible t o prove this very important theorem. L e t us go on, then, t o the P a r a l l e l P o s t u l a t e .

The Parallel

9-3.

-

Postulate.

-

P o s t u l a t e 16. he Parallel Postulate.) Through a given e x t e r n a l p o i n t t h e r e is a t most one l i n e p a r a l l e l t o a given line,

Notice t h a t w e don't need t o s a y , i n t h e p o s t u l a t e , t h a t t h e r e is at least one such parallel, because w e already know this by Theorem 9-3.

It might seem n a t u r a l to suppose that we already have enough

,

p o s t u l a t e s to be able to prove anythlng t h a t is "reasonabletT;and slnce t h e Parallel Postulate is reasonable, we might try to prove it i n s t e a d of calling it a p o s t u l a t e . A t any rate, some v e r y c l e v e r people felt t h i s way about the p o s t u l a t e , over a period of a good many centuries. None of them, however, was a b l e t o find a proof. Finally, in the last century, I t was discovered that no such proof 1s possible. The point is that there are some mathematical systems that are a l m o s t like t h e geometry t h a t we are studying, but n o t q u i t e . In these mathematical systems, nearly all of the postulates of ordinary geometry are satisfied, b u t t h e P a r a l l e l Postulate is n o t , These "Non-Euclidean Geometries" may seem strange, and in fact they are. (FOP example, in theae It geometries" ---t h e r e is no such thing as-a square.) N o t only do they lead to i n t e r e s t i n g nathematical theories, b u t they a l s o have important applications t o p h y s i c s . Now t h a t we have t h e P a r a l l e l P o s t u l a t e we can go on to prove numerous important theorems we could n o t prove without it. We s t a r t by p r o v i n g t h e converse of Theorem 9-5.

Theorem 9-8. If two parallel lines are cut by a t r a n s v e r s a l , then a l t e r n a t e i n t e r i o r angles are congruent. Proof:

transversal

We have g i v e n parallel lines L3, intersecting them in P

L1 and L2, and a and Q.

Suppose that, through

P

-

1a

and L b are not congruent. Let L be a line f o r which a l t e r n a t e i n t e r i o r angles are congruent.

-

(BY the Angle Construction P o s t u l a t e , t h e r e is such a l f n e . )

# I,,, -

Lb

L

are n o t congruent. NOW l e t us s e e what we have. By hypothesis, L1 11 L2. And by Theorem 9-5, we know t h a t L ) I L2. Therefore there a r e two lines through P, p a r a l l e l t o L2. This is impossible, because it contradicts the Parallel Postulate. Therefore L a L b, which was t o be proved. The p r o o f s of t h e following theorems are short, and you s h c u l d write them f o r yourself: Then

L

because

and

c

Theorem 9_9. If two parallel lines are cut by a transversal, each p a i r o f corresponding angles are congruent. Theorem 9-10. If two p a r a l l e l lines are c u t by a t r a n s v e r s a l , i n t e r i o r angles on the same side of the t r a n s v e r s a l a r e supplementary. Restatement: Lg. P m v e t h a t supplementary to

Given

L

L

11

and T is supplementary to

b e.

L1

Lp

intersects L1 and L d and L a is

-

meorem 9-11. In a plane, two lines parallel to the same line are parallel t o each o t h e r . Theorem 9-12. I n a plane, if a line is p e r p e n d i c u l a r to one of two parallel lines it is perpendicular to t h e other.

Problem Set 9-3 I.

Given: r n L A = m L B = m L C = 9 0 . Prove: m L D = 9 0 .

2.

Prove that a line parallel to the base of an isosceles triangle and intersectfng the o t h e r two sides o f the t r i a n g l e forms another Isosceles triangle. Given:

In t h e f i g u r e , RT = RS, )I

Prove:

PQ

=

PT.

z.

4,

Review indirect proof as Illustrated by t h e proof of Theorem 9 - 8 . Give an i n d i r e c t proof of each of the following statements, showing a contradiction of the Parallel P o s t u l a t e . a.

In a plane, i f a t h i r d line M i n t e r s e c t s one o f two parallel lines L1 a t P, it also i n t e r s e c t s the o t h e r

M L2

T

b.

In a plane, if a line R i n t e r s e c t s only one of two other lines L1 and L2, then t h e lines L1 and L2 intersect. Given:

Prove:

a.

at

R

intersects

R

does not i n t e r s e c t

L1

Intersects

L1

P.

L2.

L2

P r o v e : Two a n g l e s i n a plane which have t h e i r s i d e s respectively parallel and extending both in t h e same ( o r both in o p p o s i t e ) directions a r e congruent.

I( 5 , BE 1 1 YZ

Given:

BA

Prove :

LABC2

8

b.

P r o v e : Two angles in a plane which have t h e i r s i d e s r e s p e c t i v e l y p a r a l l e l b u t h a v e only one p a i r e x t e n d i n g in t h e same direction a r e supplementary.

-----

+

1] , 3 JIYZ.

P

Given : BA Prove :

LABC2 LXYZ . + ~&YZ = (b) ~LABC

In ( a )

In

180.

Y7 X

r7

z

C

( ~ o t e : Only c e r t a i n c a s e s are illustrated and proved

A l l o t h e r cases can a l s o be proved easily. The term " d i r e c t i o n " is undefined but should be understood. )

here.

6.

Make drawings of v a r i o u s p a i r s of angles ABC and DEF such 4 + t h a t BA ED and State a theorem that you think may be t r u e about the measures of such a n g l e s .

1

7.

3 1 3.

If Theorem 9-8 is assumed

as

a p o s t u l a t e , then t h e Parallel

h hat is, it must be P o s t u l a t e can be proved as a theorem. shown t h a t there cannot be a second parallel t o a line through a p o i n t not on it.) Given: L1 and L2 are two lines containing P, and L1 11 M .

Prove: to M.

L2 n o t parallel M

*8.

Show t h a t if Theorem 9-12 (1f a transversal is perpendicular t o one of two parallel l i n e s , I t is perpendicular to the other.) is assumed as a p o s t u l a t e , the Parallel P o s t u l a t e can be proved as a theorem.

Given: L1 11 M L2 contaln P . Prove: M.

and

L1 and

( L ~ L1.)

L2 not parallel t o

9-4. Triangles. Theorem 9-13. The sum of the measures of t h e angles o f a

triangle is 180. Proof: Given ABC, l e t L parallel to A C , L e t L x, / X I , in t h e figure

.

be the line through 8, L y, L y t and L z be as

H

be a point of L on the same s i d e of AB as C. H Since AC I( A is on the same side of BD as C. T h e r e f o r e C is in t h e i n t e r i o r of L ABD ( d e f i n i t i o n of interlor of an angle), and so, by the Angle Addition P o s t u l a t e , we have m L A B D = m L z +mLyt. By t h e Supplement P o s t u l a t e , x1 + m ABD = 180. Let D -

%8,

rnL

1

Therefore

m L x ' + m L z + m L y T = 180. But we know by Theorem 9-8 that m L x = m L xl and m y = rnL y l , because these are a l t e r n a t e i n t e r i o r angles. By substitutlon we g e t m L x + m L z +mLy=180, which was to be proved. From t h i s we g e t a number of important corollaries:

1

C o r o l l a r y 9-13-1. Given a correspondence between two triangles. If two pairs of corresponding angles are congruent, then the t h i r d p a i r of corresponding angles are also congruent.

The corollary says that if A GL A t and B EL B1, then C S L C 1 . AS the figure suggests, t h e c o r o l l a r y applies t o cases where the correspondence given is n o t a congruence, as well as to cases where ABC A AIB1C'

L

1

=

[aec.

.

9-41

Corollary 9-13-2. The a c u t e angles of a right t r i a n g l e a r e complementary.

Corollary 9-13-3. For any triangle, t h e measure o f an e x t e r i o r angle is the sum of the measures of t h e two remote I n t e r i o r angles.

Problem -Set 9-4

1.

2.

If the measures of t w o angles of a t r t a n g l e are as follows, what is the measure of the third angle?

a.

37and58.

d.

r and s.

b.

149 and 30.

e.

45+a

c.

n

f.

90

and

n.

and

45-a.

and $k.

To f i n d the distance from a p o i n t A to a d i s t a n t point P, a s u r v e y o r may A -. measure a small d i s t a n c e AB and also measure L A and L B . From this information he can c o m ~ u t e B the measure of L P and by a p p r o p r i a t e formulas then compute AP. If m L A = 87.5 and rn B = 88.3, compute m P.

Dp

1

3.

Why I s the Parallel P o s t u l a t e essential to the proof of Theorem 9-13?

On a drawing like t h e one on the right fill in the values of a l l of t h e angles.

Given:

A

Z

LX

and

B E LY, can you correctly conclude t h a t :

Given: and

4

BD

% 11

Prove:

AB

bisects e

L

EBC,

AC. =

BC.

The bisector of an e x t e r i o r angle at the vertex of an isosceles triangle is parallel t o t h e base. Prove this.

8.

Given:

The f i g u r e .

Prove:

s

+

r

=

t

+

u.

Draw

*9.

Given : In the figure, L BAC is a r i g h t angle and QB = QA

.

Prove:

*lo.

Q3 =

QC.

In ~ A B C , L C I s a r i g h t angle, AS = AT and BR = BT. Given:

Prove:

m

L STR

=

,-

45.

( ~ i n t : Suppoae m L A = a. Write formulas in turn f o r the measures of other angles in t h e figure in terms o f a . )

A

B

--

9-5.

Quadrilaterals in Plane. A quadrilateral is a plane figure with four sides, like one of the following:

The two figures on the bottom illustrate what we might c a l l the most general case, in which no t w o s i d e s are congruent, no two s i d e s are parallel, and no two angles

are congruent.

We can s t a t e the definition of a quadrilateral more precisely, in the following way. D e f i n i t i o n : L e t A, B, C and D be f o u r p o i n t s l y i n g in the same plane, such t h a t no three of them a r e collinear, and such t h a t t h e segments AB, BC, CD and DA i n t e r s e c t only in t h e i r end-points. Then t h e union of these f o u r s e g n e n t s is a quadrilateral.

F o r s h o r t , we will denote this figure by ABCD. Notice t h a t in each of t h e examples above, w i t h the exception of the l a s t one, t h e quadrilateral plus its i n t e r i o r f o m a convex s e t , in t h e senee which was defined in Chapter 3. This is n o t trme of t h e f i g u r e a t t h e lower right, but this figure is still a quadrllatera1 under o u r d e f i n i t i o n . N o t i c e , however, t h a t under o u r definition o f a quadrilateral, figures l i k e the following one a r e ruled o u t . [set*

9-51

-

Here the figure is n o t a quadrilateral, because t h e segnents BC and DA i n t e r s e c t in a point which is n o t an end-point of e i t h e r o f them. N o t i c e a l s o , however, that a quadrilateral can be formed, u s i n g t h e s e same f o u r p o i n t s as vertices, like this:

-

Here

ABDC

is a quadrilateral.

D e f i n i t i o n s : Opposite sides of a q u a d r i l a t e r a l a r e two s l d e s t h a t do n o t I n t e r s e c t . Two of its angles are opposlte if they do n o t c o n t a i n a common side. Two s l d e s a r e called consecutive if they have a common v e r t e x . S i m i l a r l y , two angles a r e c a l l e d c o n s e c u t i v e if t h e y contain a common s i d e . A diaeonal is a segment joining two non-consecutive v e r t i c e s .

In a quadrilateral AB and-3 are -opposite sides, - mCD, as a r e 3C and AD. AD and CD or AD and AB are consecutive s i d e s . and % a r e the diagonals o f ABCD. Which angles are opposite? Which consecutive?

=

a quadrilateral in which two, and only two, opposite sides are parallel, Definition:

A trapezoid is

Definition: A parallelo~ram is a quadrilateral in which both p a i r s of o p p o s i t e s i d e s are parallel.

You should n o t have much trouble in proving the basic theorems on trapezoids and parallelograms: Theorem 9-14. Either diagonal separates a parallelogram into two congruent t r i a n g l e s . That is, if ABCD is a parallel agram, t h e n ABC A CDA.

Theorem 9-35. a r e congruent.

In a parallelogram, any two opposite sfdes

Corollary 9-15-1. If L1 11 Lg and if P and Q are any t w o points on L1, then the distances of P and Q from L2 are equal.

Thls p r o p e r t y of parallel lines is sometimes abbreviated by saying that " p a r a l l e l lines a r e everywhere equidistant".

The d l s t a n c e between two parallel Lines is the d i s t a n c e from any p o i n t of one l i n e t o t h e other l l n e . Definition:

Theorem

9-16.

I n a parallelogram, any two opposite angles

are congruent. Theorem 9-17.# In a parallelogram, any two consecutive angles are supplementary.

Theorem 9-18. The diagonals of a parallelogram b i s e c t each other.

In Theorems 9-14 through 9-18 we a r e concerned w i t h several p r o p e r t i e s of a parallelogram; t h a t is, If we know t h a t a quadrilateral i s a parallelogram we can conclude certain f a c t s about it. In the following three t h e o r e m we provide f o r the converse r e l a t i o n s h i p ; that is, if we know c e r t a i n f a c t s about a quadril a t e r a l we can conclude t h a t i t is a parallelogram.

-

Theorem 9-19. Given a quadrilateral in which b o t h p a i r s of opposite sides are congruent. Then t h e q u a d r i l a t e r a l is a parallelogram.

Theorem 9-20, Lf two sides of a quadrilateral are parallel and congruent, then the quadrilateral is a parallelogram. Theorem 9-21. If the diagonals of a quadrilateral b i s e c t each other, then the quadrilateral is a parallelogram.

The following theorem s t a t e s t w o u s e f u l f a c t s , this theorem is given in full,

The proof of

-

Theorem 9-22. The segnent between t h e mid-points o f t w o sides of a triangle I s parallel to the third side and h a l f as long as the t h i r d s i d e .

.

Restatement: Given A R B C L e t D and E points o f A8 and E. Then DE 11 A C , and DE

be the mid1 AC. =

Using the P o i n t P l o t t i n g Theorem, l e t

F be t h e p o i n t of the ray opposite t o ED such t h a t EF = DE, We give the rest of t h e proof in the t w o - c a l m form. The notation f o r angles is t h a t of the f i g u r e . Proof:

1.

EF

=

Statements ED.

+

Reasons 1.

F

was chosen so as to make

true. E is the mid-point of E. Vertical angles are congruent, The S . A . S . Postulate. Corresponding p a r t a of congruent triangles, this

2. 3.

4 5.

7. 8. 9.

AD

=

ADFC -

DE

FC. is a parallelogram.

11 5.

6. 7. 8, 9.

m e o r e m 9-5. AD = DB, DB = FC,

by hypothesis

by statement

4.

and

Theorem 9-20. D e f i n i t i o n of a parallelogram. 10. DE = 21D F , by statement 1, and DF = A C , by Theorem 9-15.

-

9-6.

Rhombus, Rectangle and Square.

Definitions: A rhombus is a parallelogram a l l of whose sides a r e congruent.

A rectan~leis a parallelogram all of

n i g h t angles

.

B

~

whose angles are

c

Finally, a square is a rectangle a l l o f whose s i d e s are cohgruent.

As b e f o r e , we l e a v e the proofs of t h e following theorems f o r the s t u d e n t .

-

If a parallelogram has one right angle, then it has f o u r right angles, and t h e parallelogram is a rectangle.

Theorem 9 - 2 3 .

Theorem 9-24. t o one

In a ri.,ombu, the diagonals are perpendicular

another.

Theorem 9 - 2 5 . If t h e d i a g o n a l s of a quadrilateral b i s e c t each o t h e r and a r e perpendicular, then t h e quadrilateral is a rhombus .

--

Problem S e t 9-6 1.

1 I

--

For which of the q u a d r i l a t e r a l s rectangle, square, rhombus, parallelogram -- can each o f t h e following p r o p e r ties be proved? a.

B o t h pairs of opposite angles a r e congruent.

b.

Both pairs o f o p p o s i t e s i d e s a r e congruent.

c.

Each diagonal b i s e c t s two angles.

d.

The diagonals bisect each other.

e.

The diagonals are perpendicular.

f.

Each p a i r o f consecutive angles is supplementary.

g.

Each p a i r of consecutive sides is congruent.

h.

The f i g u r e is a parallelogram.

i.

Each p a i r of consecutive angles I s congruent.

j.

The diagonals are congruent.

2.

With t h e measures of t h e angles a s given In parallelogram ABFH, g i v e t h e degree measure of each angle.

3.

In t h i s figure ABHQ and are paralfelograms. What is the relationship of L M t o H? of L R to & H? Prove your answer. APRM

L

A

0

F

4.

Would t h e f o l l o w i n g information about a q u a d r i l a t e r a l b e s u f f i c i e n t t o prove it a parallelogram? a rectangle? a rhombus? a square? Consider each item of information separately, a.

Both pairs of its opposite sides are parallel.

b.

Both p a i r s of i t s opposite sides are congruent.

c

Three of its angles are r i g h t angles.

d.

Its diagonals b i s e c t each o t h e r .

e.

I t s diagonals are congruent.

f.

I t s diagonals are perpendicular and congruent.

g.

Its diagonals a r e perpendicular b i s e c t o r s o f each o t h e r .

h.

It is equilateral.

i.

It is equiangular.

j

5.

.

.

It is equilateral and equiangular.

k,

Both pairs of I t s o p p o s i t e angles are congruent.

1.

Each p a i r of its consecutive angles is supplementary.

Given: ABCD is a p a r a l l e l o gram with diagonal A C . A P = RC. Prove:

is a p a r a l l e l -

DPBR

ogram.

6.

c

0

A

B

Given: Parallelograms AFED and FBCE, as shown in t h i s plane f I gure

.

Prove: ogram.

ABCD

is a parallel-

C A

Ir lines a r e d r a m parallel to the legs of an isosceles trfangle through a p o i n t in t h e base of the t r i a n g l e , then a parallelogram is formed and its p e r i m e t e r i s equal t o the sum

of the lengths of t h e l e g s . Given: Ln the f i m r e - RS S RT, pX 11 $,

PY 11 RX. Prove:

a.

is a Parallelogram.

PXRY

= RS

+ RT.

In this figure, if ABCD is a parallelogram w i t h dfagona l s AC and BD i n t e r s e c t ing in Q and EF is drawn t h r o u m Q, prove that EP is bisected by &.

-

9.

Given t h e i s o s c e l e s trapezoid ABCD in which AD = GB and

/?y---i,

- CD

11 AB.

Prove

1 A L B.

1'

A

X

B

10.

The median of a t r a p e z o i d is t h e segment joining t h e m i d -

p o i n t s of its non-parallel s i d e s .

a.

Prove the foll~wlngt h e o r e m : The median o f a t r a p e z o i d is parallel t o the bases and e q u a l in length t o h a l f

the sum of the l e n g t h s o f t h e bases. Given:

Trapezoid -

ABCD P the and Q -

CD 1) %,

with

midpoint o f AD the mldpoint of BC. Prove: PQ I I PQ = (AB + CD)

AB

5

( ~ i n t : Draw e A% at K.)

.

H

DQ meeting

If A 3 = gin. and DC = T i n , t h e n c.

If

D C = %

AE=7,

FQ

1

A

B

and

then

=

11. A Tonvex quadrflateral with v e r t i c e s labeled consecutively AEGD is called a kite if A 8 = BC and CD = DA. S k e t c h some kites. S t a t e as many t h e o r e m s about a k i t e as you can and p r o v e a t least one of them,

Glven:

Quadrilateral ABCD w i t h P, Q, R , S t h e rnidpohts o f the sides, Prove: RSPQ is a parallelograrn, and PR and b i s e c t each o t h e r .

- - -

(Hint: D r a w RQ, DR and P&. )

RS,

SP,

13.

Given: Inthe figure AD < BC, DA E,

1

Prove:

m

C

< m L D.

*14. Prove that t h e sum of the l e n g t h s of the perpendiculars drawn from any p o i n t in t h e base o f an isosceles t r i a n g l e t o t h e legs is equal t o the l e n g t h o f the altitude upon e i t h e r of the l e g s . (~int: D r a w

1E.

Then t h e f i g u r e s u g g e s t s t h a t PX and are congruent, and that PY and are congruent.) 1

.

C

B

Prove t h a t t h e sum o f t h e lengths o f t h e perpendiculars drawn f r o m any p o i n t in t h e i n t e r i o r o f an e q u i l a t e r a l t r i a n g l e t o t h e t h r e e sides is equal to t h e length of an a l t i t u d e .

(~int: Draw a segment, per~endicular to the altitude used, f r o m t h e i n t e r i o r point.) Given a hexagon as fn t h e f i g u r e w i t h AB 11 E ,

Prove:

-

FA

1 1 E.

17. a .

-

II=,

as

- -

BC ~ [ B T C ~ in figure.

Prove: b,

-

Given E l , B B 1 , C C 1 are p a r a l l e l and

AC IIA'C'.

Is the f i g u r e necessarily a plane f i g u r e . Will your proof apply if it

18. Given ABCD is a s q u a r e and the p o i n t s K , L, M N d i v i d e the s i d e s as shown, a and b being l e n g t h s of the indicated segments.

Prove:

*lg.

*20.

KLMN

is a square.

Show t h a t if

ABCD

i n t e r i o r of

L ABC.

is a parallelogram then

D

is in the

Show that the diagonals of a parallelogram i n t e r s e c t each other.

9-7.

Transversals To Many Parallel Lines.

Definitions: In p o i n t s A and t h e segment

AB

If a transversal intersects t w o lines L1, L2 B, then we say t h a t L1 and L2 i n t e r c e p t on the t r a n s v e r s a l .

Suppose t h a t we have given three lines L1, L2, L3 and a transversal intersecting them in points A , B and C . If AB = BC, then we say t h a t the t h r e e l i n e s i n t e r c e p t congruent segments on

the t r a n s v e r s a l .

We shall prove the following:

Theorem 9-26.

If three p a r a l l e l lines i n t e r c e p t congruent segments on one t r a n s v e r s a l , then t h e y i n t e r c e p t congruent segments on any other t r a n s v e r s a l . P r o o f : L e t LI, L2 and L3 be parallel l i n e s , c u t by a transversal T1 in p o i n t s A , B and C. L e t T2 be another t r a n s v e r s a l , c u t t i n g these lines in D, E, and F . We have

given t h a t AB = BC;

and we need t o prove t h a t

DE and

=

El?.

We will f i r s t prove the theorem f o r the case in which T:, a r e not parallel, and A # D, as in-the f i g u r e :

T1

L e t T3 be the line through A , parallel to T2, intersec ting L2 and L3 in G and H; and l e t T4 be t h e line

through

L X,

B, y,

p a r a l l e l t o T2, intersecting L3 in I . L e t L w and L z be as i n d i c a t e d in the f i g u r e .

S tatements

a

1. 2.

Lx

3. 4.

T3 II T q . LwzLy.

5.

AABG = A B C I . AG

7.

AGED

8.

2.

AB=BC.

6.

Reasons

=

BI.

and BIFE are parallelograms. A G = D E and B I = E F .

1. Theorem 9-9. 2. Hypothesis 3 . Theorem 9-11. 4 . Theorem 9-9. 5. A. S . A . 6 . Definition of congruent triangles. 7 . Definition of parallelograms.

.

8. 9.

O p p o s i t e sides of a parallelogram a r e congruent, S t e p s 6 and 8.

This proves t h e theorem f o r t h e c a s e i n which t h e two t r a n s v e r s a l s a r e n o t parallel, and intersect L1 in two different p o i n t s . The other c a s e s are easy. ( 1 ) If the two transversals are parallel, like T2 and Tj in t h e f i g u r e , then t h e theorem holds, because opposite sides o f a parallelogram are congruent, (Thus, if AG = GH, i t follows t h a t DE = El?.)

If the t w o transversals i n t e r s e c t a t A, like T1 and T3 in t h e f i g u r e , then the theorem h o l d s ; i n f a c t , we have a l r e a d y proved t h a t if AB = BC, then AG = GH. (2)

The following c o r o l l a r y generalizes Theorem 9-26.

Corollary 9-26-1. If t h r e e o r more parallel l i n e s intercept congruent segments on one t r a n s v e r s a l , then t h e y i n t e r c e p t cong r u e n t segments on any o t h e r t r a n s v e r s a l .

Tbat is, given t h a t A A

1 2

= A A = A 3A 4 = 2 3

it follows t h a t BIB2 = B2B3 = B3B4

.."

- ...,

and so on. This follows by repeated applications o f the theorem t h a t we have j u s t p r o v e d .

D e f i n i t i o n : Two or more s e t s a r e concurrent if t h e r e is a p o i n t which b e l o n g s to a l l of the s e t s .

In particular, three or m o r e lines a r e concurrent i f t h e y all pass t h r o u g h one p o i n t .

The f o l l o w i n g theorem is an i n t e r e s t i n g application of

Corollary 9-26-1.

Theoren 9-27. The medians of a triangle a r e c o n c u r r e n t in a p o i n t t w o - t h i r d s the way from any vertex t o t h e mid-point of the cppos 1t e side

.

Given: I n AABC, D, E and F a r e the m i d - p o i n t s of BC, C A and A 8 r e s p e c tively

C

.

There is a p o i n t - which lies on AD, BE and 2 CF; and AP = 3 AD,

To P r o v e :

BP

=

2

g BE, CP

P

A

F

8

2

= T CF.

S k e t c h of p r o o f :

( 1)

Let L1, L2, L3, parallel lines d i v i d i n g (a)

L3,

Lq,

and so (b)

L5

E

L, and L5, w i t h L3 = AD be five CB i n t o f o u r congruent segments. Then i n t o t w o congruent segments,

dlvlde

iies on

L4.

L1, L2, Lg, L4 divide BE segments, and s o if P AD and BE, then BP

-

into t h r e e congruent is the point o f intersection of

=

BE.

if

In the same way, w i t h lines parallel to P i is t h e intersection of BE and E ,

%, then

we f i n d that BPI =

(3) From (1) and ( 2 ) and Theorem 2-4 it follows that and t h e r e f o r e the t h r e e medians a r e concurrent,

PI

2 7 aE. =

P,

(4) S i n c e we now know that get AF =

2

CP = 3 AD

passes through P we c a n easily CF f r o m the figure in ( I ) , and similarly g e t f r o m the rigure in ( 2 ) .

Definition: The centroid o f a t r i a n g l e is the point of concurrency of the medians.

Problem S e t 9-7 1.

Given: AR ]I R x 11 a.

b.

AB =

)(

su 1 1

Prove w AC,

BC.

E.

TZ. ZY

=

*

YX. t ,

Do TR and ZX have to be coplanar t o carry o u t t h e proof?

The procedure a t t h e right

rule a s h e e t o f paper, B, i n t o columns o f e q u a l width. If A is an o r d i n a r y sheet of r u l e d paper and B is a second s h e e t placed over it as shown, e x p l a i n why c a n be used t o

3.

Dlvide a given segment following method:

+ AR

(1) Draw ray 2

into f i v e congruent p a r t s by t h e

AB.)

(not collinear with

-

Use rulert o mark o f f congruent segments AN1, - your N1N2, 1112E13, N3N4 and NIIN5 of any convenient l e n g t h .

(3) Draw (4)

A3

N5B

L

Measure AMGB and use your protractor to draw c o r r e 2 sponding angles congruent t o L AN B with v e r t i c e s at

5

Nb,

N3,

N2

E x p l a i n why

4.

The medians of

and AB

N1. is divided i n t o ' congruent p a r t s .

AABCmeet C

a t Q, as shown in this figure

.

If BF = 18, AQ = 10, CM = 9, then BQ = QH = , CQ =

I

5.

+6.

In e q u i l a t e r a l AABC if one median is 15 Inches long, what is the -d i s t a n c e from-the c e n t r o i d to A? To the midp o i n t o f AB? To side AC?

Given: M.

BQ

-

b i s e cts b i s e c t s CM at

at P.

CM

B

Prove: Q is a trisection point oi AC; that is, A Q = 2QC. ( ~ i n t : On the ray o p p o s i t e + to CB take p o i n t E auch that CE = CB and show t h a t BQ i s contained in a median o f ~ A B E) .

"7.

A

What is t h e smallest number of congruent segments i n t o which AC can be divl.ded by some set of equally spaced parallels which will include t h e parallels and # CT if:

8,

a.

b. c. d. e.

AB = 2 and 1a n d AB = l T AB = 21 and

BC = l? BC = l? BC = 6?

and BC = l? and BC = l?

AB = 1.414 AB=

fi

[sec. 9-7

1

Q

C

*8,

Prove t h a t the l i n e s through o p p o s i t e v e r t i c e s of a parallelogram and t h e midpoints o f t h e opposite s i d e s t r i s e c t a diagonal.

(Hint: Through an extremity of t h e d i a g o n a l , consider parallel to one of the lines.) Given:

a

ABCD

X

is a parallelogram. and Y a r e midpoints.

Prove : AT = TQ

=

QC ,

Review Problems

1.

I n d i c a t e whether each of t h e following statements is t r u e in ALL c a s e s , true in SOME cases and f a l s e in o t h e r s , o r t r u e in NO case, using the l e t t e r A , S or N: a.

Llne segments In t h e same plane which have no p o t n t In common are parallel.

b.

If two sides of a quadrilateral then ABCD is a trapezoid.

c

.

d.

ABCD

are parallel,

Two angles in a plane which have thelr s i d e s r e s p e c t i v e ly perpendicular a r e congruent.

If two p a r a l l e l l i n e s are c u t by a transversal, then p a i r o f a l t e r n a t e exterior angles are congruent.

3

e.

If t w o lines are cut by a t r a n s v e r s a l , then t h e rays b i s e c t i n g a pair of alternate i n t e r i o r angles are parallel.

p l a n e , if a Line is parallel to one of t w o parallel l i n e s , it is p a r a l l e l t o t h e o t h e r .

f.

In

g.

In a plane two lines are e i t h e r p a r a l l e l or they

a

intersect.

In a parallelogram t h e opposite angles a r e supplementary. The diagonals of a rhombus b i s e c t each o t h e r .

All three e x t e r l o r angles of a triangle a r e acute. A quadrilzteral having t w o o p p o s i t e angles which a r e

right angles Is a rectangle.

The diagonals of a rhombus a r e congruent.

If a q u a d r i l a t e r a l is e q u i l a t e r a l , then a l l o f its angles a r e congruent.

If two o p p o s i t e s i d e s of a quadrilateral are congruent and the o t h e r two sides are parallel, the quadrilateral is a parallelogram. The diagonals o f a rhombus b i s e c t the angles of the rhombus .

If t h e diagonals of a p a r a l l e l o g r a m are perpendicular, t h e parallelogram is a s qtlare

.

If a median to one s i d e of a trlangle I s n o t an altitude, t h e o t h e r two s i d e s are unequal in length. E i t h e r diagonal of a parallelogram makes two congruent

triangles with the sides.

If a diagonal o f a quadrilateral divides I t I n t o two c o n p u e n t triangles, the quadrilateral is a parallelogram

If two lines a r e intersected by a t r a n s v e r s a l , the alternate i n t e r i o r angles are congruent.

u.

All f o u r s i d e s of a rectangle are congruent.

v.

A l l f o u r angles of a rhombus are congruent.

w

.

x. 2.

A square is a rhombus

.

A square is a rectangle.

Would the followtng information about a quadrilateral be sufficient to p r o v e it a parallelogram?

rhombus? A rectangle? separately.

I t s diagonals b f s e c t each o t h e r .

b.

Its diagonals are congruent,

3.

.

A

Consider each i t e m o f information

a.

c

A square?

It is e q u i l a t e r a l .

d.

It is equilateral and equlangular.

e.

A diagonal bisects two angles.

f.

Every two o p p o s i t e sides are congruent.

g.

Some two consecutive sides a r e congruent and perpendicular.

h.

The diagonals are perpendicular.

i.

Every two opposite angles a r e congruent.

J.

Each diagonal b i s e c t s two zngles.

k.

Every two consecutive angles are supplementary.

1.

Every two consecutive sides a r e c o n g r u e n t .

LA

and

a.

If only one pair of corresponding sldes extend in t h e same direction the angles a r e

b.

If corresponding sides extend in o p p o s i t e directfons, then the angles a r e

LB

have t h e i r sides r e s p e c t i v e l y parallel.

In Problems 4, 5 and 6 below s e l e c t the one word or phrase t h a t makes t h e statement true, 4.

The b i s e c t c r s of the opposite angles of a non-equilateral

parallelogram ( a ) coincide, ( b ) are perpendicular, ( c ) intersect b u t a r e n o t perpendicular, ( d ) a r e p a r a l l e l .

5.

*6.

f i g u r e formed by joining the consecutive mid-points of t h e sides of a rhombus is ( a ) a rhombus, (b) a rectangle, (c) a square, (d) none of these answers.

The

The figure f o m e d by joining t h e consecutive m i d - p a i n t s of the s i d e s o f quadrilateral ABCD is a square ( a ) if, and ofily if, the d i a g o n a l s of ABCD a r e congruent and perpendicular, (b) if, and o n l y if, t h e d i a g m a l s of ABCD a r e congruent, ( c ) i f , and only if, ABCD I s a square, (d) if, and o n l y if, t h e diagonals of ABCD are perpendicular,

7.

In the left-hand c o l ~ u n nbelow, c e r t a i n conditions a r e specif l e d . In t h e r i g h t - h a n d column, some deducible conclusions are l e f t f o r you t o compiete. Given: MW and KR are diagonals of HKWR.

Conditions : a.

b. c.

MKWR rnLa = n r w MKWR mLa = MKWR

mLa=

Conclusions:

is a parallelogram, 3 0 , and = 110. is a rectangle and 30.

rnLd

is a rhombus, 30 and M = 6 .

rnk

=

and rn@K

mLd =

-and mfi and

=

RK

=

=

=

A

8.

Given: In the figure AE = EB, GF = 8 , @ @ C F = FB, DE 11 CB. Find:

9.

10.

C

DG.

F

If t h e perimeter (sum of l e n g t h s of s i d e s ) of a triangle is 18 i n c h e s , what is the perimeter of the t r i a n g l e formed by J o i n i n g t h e mid-points of sides of the f i r s t t r i a n g l e ? a.

If

rnL

and m L C = 25, what is t h e measure of A = 30

L CBD? b.

If m L A = a and a mL C = F, what is mL CBD?

mL

ABC?

A 11.

Show t h a t the measure of E, formed by t h e b i s e c t o r of L ABC and the blsector o f e x t e r i o r L ACD of A ABC, is equal to 1 A.

L

rnL

E

12.

#

In the figure AB

+

)I

8

CD, EG b i s e c t s L BEF, rnL G = 90. If the measure of L GEF = 25, what is the measure of L GFD?

G


i?dl

(I .

11'L.?!L-

L l ~ q c s;lave

equal radii, t h e i r

. o i;k!ci_l tnelsur.es. ' 2 )

An

. 3 ~ t0:'

t;1e2~ -ly1e q and radius r

.

42

Theoren 15-5. (F. 52'() The a r e a of a sector p r o d u c t oi' i',s r a d i u s l-g 1fn~::rl of i t s a r c

Ls h a l f t h e

.

Theorem 13-6. ( F . 527) The area oi' a s e c t o r of r a d i u s 2 and a r c ineasyre g is r*

7

Theorem

C 3

l

l

0

1.6-1, (P. 535) -

1

-

r

.

A ~ Ic r o s s - s e c t i o n s of a triangular

( c , 536)

The uppei' and l o w e r bases o r

con~ruen: . (P. 536) ( ~ r i s xCross-Sectlon heo or em.)

triangular y ~ - ~ i s .,ire .l ,

I?

I

1 . -

All

r,r30zs-sqction;: oi' 3 prism have % h e s m e area.

o r 1 1 1 - 2 equ:il aress . Theoren

16-3.

(P. 537) The 5x0 bases of a prism have

(F . 537)

The l a t e r a l faces of a prisn

are

parallelogr~n1=egic!13, 3t1d t h e l a t e r a l f a c e s of a r i p h t prisrn are rec'canpular ~ e ~ i a n s .

Theorem 16-4. (P. 540) A cross-section of a triangular p y r a m i d , by a p l a n e between the vertex and t h e base, is a t r i a n g u l a r region similar to the base. If t h e distance from the vertex t o the cross-section plane is k and t h e a l t i t u d e is h, t h e n t h e r a t i o 0% the area of the cross-sectfon to the area of k the base is (i;)

.

Theorem 16-5. ( P . 542) In any pyramid, t h e ratlo f t h e 9 ales of a c~oss-sectlon and t h e area of the base is , where h is t h e a l t i t u d e of t h e pyramid and k 9 s the d i s t a n c e from t h e v e r t e x to t h e plane of t h e cross-section.

(g)

(P . 543) (The Pyramid Cross-Sect i o n heo or em. ) Given t w o pyramids w i t h the same altitude. If t h e bases have t h e same area, then cross-sec Lions equidistant from t h e bases a l s o have t h e same area. Theorem 16-6.

.

Theorem 16-7. ( F 548) The volume of any prism is the p r o d u c t of the a l t i t u d e and the area of t h e base. Theorem 16-8.

549) If t w o pyramids

(P.

have t h e same alti-

tude and t h e same base a r e a , t h e n they have t h e same volume. Theorem 16-9, (P, 550) The volume of a trfangular p y r a m i d i s o n e - t h i r d t h e p r o d u c t of i t s a l t f t u d e and i t s base a r e a .

.

Theorem 16-10. ( P 551) The volume of a pyramid is one-third the product o f i t s a l t i t u d e and i t s base a r e a .

.

Theorem 16-11. ( P 555) A cross-section o f a c l r c u l a r cylinder Ls a c i r c u l a r r e g i o n congruent t o t h e base. Theorea 1G-12.

(P.

555)

The area of a cross-section of a

c i r c u l a r c y l i n d e r is e q u a l to t h e area of the b a s e , Theorem 16-13, (P. 555) A cross-section of a cone of a l t i t u d e h, made by a plane at a d i s t a n c e k from t h e v e r t e x , is a c i r c u l a s r e g i o n whose area h a s a r a t i o t o the a r e a of t h e k base of (K)

.

.

Theorem 16-14. ( P 557) The volume of a c i r c u l a r c y l i n d e r is t h e p r o d u c t of t h e a l t i t u d e and t h e area of the base.

Theorem 16-15. (P. 553) The volume of a c i r c u l a r cone I s one-third the product of t h e altitude and t h e area of t h e base.

559) The volume of a sphere of r a d i u s

Theorem 16-16.

(P.

Theorem.

(P. 5 6 2 )

4 - 3 . is p

radius

r

is

S

=

4m

Theorem 17-1. ( P have t h e same slope. Theorem 17-2.

(P

r

The surface area of a sphere of

2

. 577)

On a non-vertlcal l i n e , a l l segments

. 584)

Two n o n - v e r t i c a l l i n e s a r e p a r a l l e l

if and only if t h e y have t h e same s l o p e .

.

Theorem 17-3. ( P 586) Two n o n - v e r t i c a l lines are perpendicular if a n d o n l y if their s l o p e s a r e t h e n e g a t i v e reciprocals of each o t h e r ,

Theorem.

.

( P 589) (Tne Distance orm mu la. ) The d i s t a n c e between t h e p o i n t s (xl,yl) and Is equal t o 2 2 - xl) + (y2 - yl)

.

Theorem 17-5. (7. 593) (The Mid-Point Formula .) L e t PI = ( x l , y l ) and let P p = (x2.y2). Then t h e mid-point x + x 2 yl+Y2) of PIP2 is t h e p o i n t P = ( 1 2 2 Theorvcm 17-6. (P 6 0 5 ) L e t L be a n o n - v e r t i c a l line w i t h w i t h s l o p e m, and let ? be a p o i n t of L, with c o o r d i n a t e s (xl,yl ) FOP every p o i n t Q = (x,y) of L, t h e e q u a t i o n y - yl = m(x - xl) is s a t i s f i e d .

-

.

.

y

Theorem, (P. 6 0 7 ) The graph of t h e equation - yl = m(x - xl) IS the l i n e t h a t passes through the

(xlT

)

Y ~

m.

and has s l o p e

Tneorem 17-8. (F is t h e l i n e with s l o p e

point

. 611) m

The graph of the e q u a t i o n and y - i n k e r c e p t b .

(P. 613) m.

of a l i n e a r equation in

x

y =

mx + b

Every l i n e i n t h e plane is the graph

and

y.

.

Theorem 17-10. ( P 613) The graph o f a l i n e a r e q u a t i o n in x and y i s always a l i n e .

Theorem 17-11. (x

- a)2 +

and r a d i u s

(y

-

b12 =

. 623)

r2

The graph of t h e equation

is the c i r c l e w i t h c e n t e r at

(a,~)

r.

Theorem 17-12. equation of the, f o m

x2

(P

(P. 624) x

2

+

y2

Every c i r c l e is t h e graph of an

+

PLX

+

BY

+c

Theorem, ( F . 625 ) Given t h e + y2 + Ax + By C C = 0 . The graph of

= 0.

e q u a t i on t h i s e q u a t i o n is

(1) a circle, ( 2 ) a p o i n t or ( 3 ) t h e empty s e t .

Index of --

Definitions

F o r precisely d e f i n e d geometric terms t h e reference is t o t h e formal definition. F o r o t h e r terms t h e r e f e r e n c e is to an informal deflnltion or to t h e m o s t prominent discussion.

a b s o l u t e v a l u e , 27 a c u t e a n g l e s , 86 a l t e r n a t e i n t e r i o r angles, 245 altitude o f p r i s m , 535 of pyramid,. 540 . of t r i a n g l e , 214, 215 angle(s), 71 acute,

86

alternate i n t e r i o r , 245 bisector o f , 129 c e n t r a l , 429

complementary, 86 congruent, 86, 109

consecutive , 264 corresponding, 251 d i h e d r a l , 299 exterior, 293 exterior of, 73 i n s c r i b e d , 432 i n t e r c e p t s an arc, 433 i n t e r i o r o f , 73 measure of, 79, 80 o b t u s e , 86 o f p o l y g o n , 506 o p p o s i t e , 264

r e f l e x , 78

remote i n t e r i o r , 193 right, 85 right dihedral,. .301 s i d e s o f , 71 s t r a i g h t , 78 supplementary, 82

vertex o f , 71 v e r t i c a l , 88 apothem, 512

a r c ( s ) , 429 c e n t e r or, 437 c o n g r u e n t , 441 degree measure of, 430 end-points o f , 429 l e n g t h of, 525 major, 429 m i n o r , 429 of s e c t o r , 527

area, 320 ,, c i r c l e , 521, 5 2 2

parallelogram, 330 polygonal r e g i o n , 3 20 rectangle, 322 right triangle, 328 s p h e r e , 562 trapezoid, 331 t r i a n g l e , 328 unit o f , 321 arithmetic mean, 364 auxiliary s e t s ,

176

base of pyramid, 540 between, 41, 182 b i s e c t o r of an angle, 129 bisector of a segment, 169 b i s e c t s , 47, 129 Cavalierils P r i n c i p l e , 548 c e n t e r of a r c , 437 c i r c l e , 409 sphere, 409 c e n t r a l a n g l e , 429 c e n t r o i d , 280, 621 c h o r d , 410

circle(s), 409 area o r , 521, 522 circumference of, 516 congruent , 417 e q u a t i o n o f , 623, 624, 625 e x t e r i o r o f , 412 g r e a t , 410 i n t e r i o r of, 412 segment or, 528 tangent,

417

circular cone, 554 c y l i n d e r , 553 reasoning, 119 region, 520 area o f , 521

circumference, 516 circumscribed

c i r c l e , 490 t r i a n g l e , 490 collinear, 54 complement, 86 complementary angles, 86 concentric c i r c l e s , 409 s p h e r e s , 409 conclusion, 60

1

I

concurrent sets, 278, 469 cone, circular, 554 right circular, 555 volume of, 557 congruence , 97 congruent angles, 86, 109 a r c s , 441 c i r c l e s , 417 segments, 109 t r i a n g l e s , 98, 111 c o n s e c u t i v e angles, 264 consecutive sides, 264 constructions, 477 converse, 202 convex polygon, 507 convex s e t s , 62 coordfnate system, 37, 571 coordinates of a point, 37, 569 c o - p l a n a r , 54 corollary, 128 correspondence, 97 corresponding angles, 251 cross-sec t l o n of a prism, 535 of a pyramid, 540 cube, 229 cylinder c i r c u l a r , 553

volume of', 557 d i a g o n a l , 264, 509 diameter, 410 dihedral a ngle, 299 edge of, 299 f a c e o f , 299 measure of, 301 plane angle o f , 300 d i s t a n c e , 34 d i s t a n c e between a point and a line, 206 a p o i n t and a plane, 235 t w o p a r a l l e l lines, 266 distance formula, 589 edge of half p l a n e , 64 end-point ( s ) of a r c , 429 of r a y , 46 o f segment, 45 empty s e t , 18

equation of c i r c l e , 623 of l i n e , 605, 611 e q u i a n g u l a r t r i a n g l e , 128 equilateral t r i a n g l e , 1 2 8 E u l e r , 327 e x i s t e n c e p r o o f s , 165 e x t e r i o r angle, 193 ejcse .-