CHAPTER 5 - Mr Jaffemrjaffe.net/GeomRef/unit5solutions.pdfAnswers may vary. Sample answer: At position B, the goalkeeper is closer to both sides of the imaginary triangle formed by

CHAPTER 5

80 GeometryChapter 5 Worked-out Solution Key

Copyright © McDougal Littell Inc. All rights reserved.

Think & Discuss (p. 261)

1. Answers may vary.

Sample answer:

At position B, the goalkeeper is closer to both sides ofthe imaginary triangle formed by the goal posts and theopponent. The goal keeper has an equal space to defendto the left and right and so has a greater chance of catch-ing or deflecting a ball shot to either side than she wouldhave at position A or position C.

2. about an opponent could move closer to the goal toincrease the shooting angle.

Skill Review (p. 262)

1. Sample answer:

3.

4.

5.

6. The slope of the line perpendicular to is because

Lesson 5.1

Developing Concepts Activity 5.1 (p. 263)

Exploring the Concept

1.–3. Sample answer

5. m�CMA � 90�

BMA

C

�12 � 2 � �1.

�12BC

� 2��4�2

�0 � 4

�2 � 0m �

y2 � y1

x2 � x1

� 5

� �25

� �9 � 16

� ��3�2 � 42

AB � ��0 � 3�2 � �4 � 0�2

� ��1, 2�� 22

, 42�M � �0 � ��2�

2,

4 � 02 �

A BM

40�;

Drawing Conclusions

1. CMvxxxy is perpendicular to ABxxx since and

CMvxxxy intersects ABxxx at its midpoint.

2. Sample answers:

3. Sample answer:

The distances from D to the endpoints A and B are equal,the distances from E to the endpoints A and B are equal,the distances from F to the endpoints A and B are equal,and the distances from G to the endpoints A and B areequal.

4. Any point on the perpendicular bisector is the same distance from either of the endpoints of the segment.

5.1 Guided Practice (p. 267)

1. If D is on the perpendicular bisector of then D isequidistant from A and B.

2. Point G must be on the bisector of by Theorem 5.4.

3.

4. and are both right angles and are congruent.

5. because C is on the perpendicular bisector of

6.

7. The distance from M to is equal to the distance fromM to

5.1 Practice and Applications (pp. 268–271)

8. No; C is not on the perpendicular bisector of becauseAC and BC are not equal.

AB

→PN.

→PL

m�LPM � m�NPM

AB.AC � BC

�BDC�ADC

AD � BD

�HJK

K

J

H

G

AB,

BMA

CD

EF

G

m�CAB � 90�

2. Sample answer:

P

4. Sample answer:

MA � MB � 32 mm

6. Sample answer:

CA � CB � 37 mm

Point D Point E Point F Point G

GB � 67 mmFB � 58 mmEB � 48 mmDB � 43 mm

GA � 67 mmFA � 58 mmEA � 48 mmDA � 43 mm

MCRBG-0501-SK.qxd 6-14-2001 11:37 AM Page 80

Geometry 81Chapter 5 Worked-out Solution Key


Chapter 5 continued

9. No; the diagram does not show that

10. No; along with the information given, we would alsoneed

11. No; since P is not equidistant from the sides of P isnot on the bisector of

12. No; since we do not know for sure that one of the dis-tances given is a perpendicular distance.

13. No; the diagram does not show that the segments withequal length are perpendicular segments.

14.

16. 17.

18. Point U must be on SVvxxy, the perpendicular bisector of RT___

.

19.

20. Point M must be on JNwy, the bisector of �HJK. 21. B

22.

A

23.

m�VWU � 40�

m�VWU � m�VWX � 40�

C

24. F 25. D 26. E

27. Given: P is on m.

Prove:

Statements Reasons

1. P is on line m. 1. Given

2. 2. By construction

3. 3. Reflexive Property of Congruence

4. 4. SSS Congruence Postulate

5. 5. Corresponding parts of congruenttriangles are congruent.

6. 6. Theorem 3.1↔CP � AB

�CPA � �CPB

�CPA � �CPB

CP � CP

CA � CB

PA � PB

↔CP�AB.

m�VWU � 50� � 90�

m�VWU � m�UVW � 90�

m�XTV � 60�

m�XTV � 30� � 90�

m�XTV � m�TVX � 90�

NQ � 2

SR � 17VT � 8

AD � BD � 5 cm

A B

D

4 cm 4 cm

3 cm

�A.�A,

AP � PB.

CA � CB. 28.

Statements Reasons

1. , 1. Given

.

2. 2. Definition of segment bisector

3. and 3. Definition of perpendicular are right angles. lines

4. 4. Reflexive Property ofCongruence

5. 5. SAS Congruence Postulate

6. 6. Corresponding parts of congru-ent triangles are congruent.

7. 7. Definition of congruent segments

8. C is equidistant 8. Definition of equidistant from A and B.

29.

Statements Reasons

1. Construct inter- 1. Perpendicular Postulatesecting at a point P.

2. and are 2. Definition of perpendicular right angles. lines

3. and are 3. Definition of right trianglesright triangles.

4. 4. Given; Definition of con-gruence


6. 6. HL Congruence Theorem

7. 7. Corresponding parts of con-gruent triangles are congru-ent.

8. is the perpendicular 8. Definition of perpendicularbisector of and C is bisectoron the perpendicular bisector of

30.

Statements Reasons

1. is the perpendicular 1. Givenbisector of

2. 2. Definition of perpendicular bisector of a segment

3. 3. Perpendicular Bisector Theorem

4. 4. Definition of congruent seg-ments


6. 6. SSS Congruence Postulate�GMH � �GMK

GM � GM

GH � GK, MH � MK

GH � GK, MH � MK

GJ � HK, HJ � JK

HK.GJ

AB.

AB

↔CP

AP � BP

�CPA � �CPB

CP � CP

CA � CB, or CA � CB

�CPB�CPA

�CPB�CPA

AB

↔CP � AB

CA � CB

CA � CB

�APC � �BPC

CP � CP

�CPB�CPA

AP � BP

↔CP bisects AB

↔CP � AB

15.

D is inches from eachside of �A.

112

A

D3 in.


Chapter 5 continued



31. The post is the perpendicular bisector of the segmentbetween the ends of the wires.

32.

Statements Reasons

1. D is in the interior of 1. Given

2. D is equidistant from 2. Given

and

3. 3. Definition of equidistant

4. 4. Definition of distance froma point to a line

5. and are 5. If two lines are then theyright angles. form 4 right angles.

6. are 6. Definition of right triangleright triangles.

7. 7. Reflexive Property of Congruence

8. 8. HL Congruence Theorem

9. 9. Corresponding parts of congruent triangles are congruent.

10. bisects and 10. Definition of angle bisectorpoint D is on the bisector of

33. Line l is the perpendicular bisector of

34. should bisect to give the goalie equal distances

to travel on each side of .

35. increases as the puck gets closer to the goal. Thischange makes it more difficult for the goalie because thegoalie has a greater area to defend since the distances fromgoalie to the sides of (the shooting angle) increase.

36. Answers may vary.

Sample answer:

This demonstrates the Perpendicular Bisector Theorembecause D is on the perpendicular bisector of and D isequidistant from A and B.

AB

D2A � D2B � 48 mm

D1A � D1B � 40 mm

BA C

D1

D2

�APB

m�APB

→PG

�APB→PG

AB.

�ABC.

�ABC→BD

�ABD � �CBD

�DAB � �DCB

BD � BD

�DAB and �DCB

�,�DCB�DAB

DA � →BA, DC �

→BC

DA � DC

→BC.

→BA

�ABC.

37. a–c.

The perpendicular bisectors meet in one point.

d. The fire station at A should respond because it is closest to the house at X.

38.

because the product of their slopes is

because the product of their slopes is

39.

40. bisects because the perpendicular distances

from W to and are equal. We know these are

perpendicular distances because in problem 38 it was

shown that and

5.1 Mixed Review (p. 271)

41.

r � 6 cm

2r � 12

d � 12 cm

WT�→YZ.WS�

→YX

→YT

→YS

�XYZ→YW

� �10

� �1 � 9

� ��1�2 � ��3�2

WT � ��5 � 6�2 � �1 � 4�2

� �10

� �9 � 1

� ��3�2 � 12

WS � ��3 � 6�2 � �5 � 4�2

�3 � ��13� � �1�.

�1.WT�→YZ

� �13

��26

slope of YZ � m �0 � 28 � 2

� 3��3�1

slope of WT � m �1 � 45 � 6

��13

� 3 � �1�.

�1.WS�→YX

� 3�62

slope of →YX � m �

8 � 24 � 2

�1

�3� �

13

slope of WS � m �5 � 43 � 6

A

C

X

Closest to C

Closest to B

Closest to A

B

42.

37.68 cm

2 � 3.14 � 6

C � 2�r




Chapter 5 continued

43.

45.

46. 47.

48.

49.

50.

52.

Lesson 5.2

Developing Concepts Activity (p. 272)

1.–4. Yes, all three bisectors intersect at the same point.

Conjecture: For any acute scalene triangle, the threeperpendicular bisectors of the three sides will intersectat the same point.

The three segments that are formed by connecting thevertices of the triangle to the point of intersection ofthe perpendicular bisectors of the sides have the samelength.


1. If three or more points intersect at the same point, thelines are concurrent.

2. The word incenter is made up of the words in and center.The incenter is the “center” of the circle that is “in” thetriangle. The word circumcenter is made up of the partscircum and center. Circum can be short for circumference,which is the distance around the circle, and can help us toremember that the circumcenter is the “center” of thecircle that is “around” the triangle.

3. 4. MK � MJ � 5GC � GA � 7

AP � BP � CP � 38 mm

x � 8

6x � 48

10x � 48 � 4x

�10x � 22�� 70� � 4x�

x � 59

x� � 31� � 90�

� 0�0

11m �

�8 � ��8�8 � ��3�

� �113

m �12 � 11�10 � 3

�87

��8�7

m �0 � 8

�7 � 0� 0�

05

m �5 � 59 � 4

� �45

�8

�10m �

5 � ��3��6 � 4

113.04 cm2

3.14 � 36

3.14�6�2

A � �r2 5.2 Practice and Applications (pp. 275–278)

5.

The perpendicular bisectors intersect outside the obtuse triangle.

7.

The perpendicular bisectors intersect at a point on the right triangle.

10. always 11. always 12. never 13. sometimes

14. 15.

16. by the Pythagorean Theorem.

17. Let the midpoint of be called point R. Then

by the Pythagorean Theorem.

18. The student’s conclusion is false because D is not thepoint of intersection of the angle bisectors. D is the pointof intersection of the perpendicular bisectors of the sidesof the triangle. So

19. The student’s conclusion is false because the angle bisec-tors of a triangle intersect in a point that is equidistantfrom the sides of the triangle, but MQ and MN are notnecessarily distances to the sides; M is equidistant from

and LJ.JK, LK,

DA � DC � DB.

QN � PQ � 25.

25 � PQ

625 � �PQ�2

576 � 49 � �PQ�2

�24�2 � �7�2 � �PQ�2

�PR�2 � �QR�2 � �PQ�2

PR � RM �PM2

�482

� 24.

PM

KB � CK � 3

CK � 3

�CK�2 � 9

16 � �CK�2 � 25

42 � �CK�2 � 52

�JC�2 � �CK�2 � �KJ�2

WB � WC � 20DR � SD � 9

51.

x � 34

2x � 34 � x

�2x � 6�� 40� � x�

6.

The perpendicular bisectorsintersect at a point inside anacute triangle.

8. and 9.

Sample answer:

The segments are congruent. This confirmsTheorem 5.6.

A B

C

D

44.

� �5

�5

�1

m �10 � 5

�2 � ��1�


Chapter 5 continued



20. To find the point that is equidistant from each location,draw the triangle and construct the perpendicular bisectorof two sides. The point of intersection is the point that isequidistant from each location.

21. Point H is the point of intersection of the perpendicular bisectors. So H is equidistant fromeach location. H would bethe best location for the new home.

22.

Statements Reasons

1. ¤ the bisectors of 1. Givenand

2. 2. bisects so D isequidistant from the sidesof

3. 3. bisects so D isequidistant from the sidesof

4. 4. Transitive property ofequality

5. D is on the bisector 5. Converse of the Angle of Bisector Theorem

6. D is equidistant from 6. Givens and Steps 2, 3,and and 4

23. She could construct the perpendicular bisectors to findthe point that is equidistant from the vertices of the trian-gle. By doing so, she would see that the perpendicularbisectors do intersect at a point on the hypotenuse. Sincethe point on the hypotenuse would be the point of inter-section of the perpendicular bisectors, then it would beequidistant from the vertices.

24.–25. The radius isapproximately

26. ft inches

30 in. in. per year years

The mycelium is approximately 3.75 years old.

� 3.75� 8

� 30212

212 ft.

1

1

y

x

A (2, 5)

B (6, 3)

C (4, 1)

CA.BC,AB,

�C.

DF � DG

�ABC.

�ABC,→BDDE � DF

�BAC.

�BAC,→ADDE � DG

DG � CA

DF�BC,DE�AB,�C,�B,�A,

ABC,

School

Office

FactoryH

27.

E

28.

(Pythagorean Theorem)

because T is the midpoint of

C

29. The midpoint of is

The midpoint of is

The midpoint of is

The slope of

The perpendicular bisector has slope because

is an equation of the perpendicular bisector of

The slope of

—CONTINUED—

BC �0 � 6

18 � 12�

�66

� �1.

AB.y � �2x � 15

y � 3 � �2x � 12

y � 3 � �2�x � 6�

�2 �12

� �1.

�2

AB �6 � 0

12 � 0�

612

�12

.

�0 � 182

, 0 � 0

2 � � �182

, 02� � �9, 0�.

AC

�12 � 182

, 6 � 0

2 � � �302

, 62� � �15, 3�.

BC

�0 � 122

, 0 � 6

2 � � �122

, 62� � �6, 3�.

AB

XY � 10

XY � 2 � 5

XY.

XY � 2�XT�

XT � 5

�XT�2 � 25

�XT�2 � 144 � 169

�XT�2 � 122 � 132

�XT�2 � �TW�2 � �XW�2

XW � WZ � 13

m�ADC � 140�

m�ADC � 40� � 180�

m�ADC � m�DCA � m�CAD � 180�

m�DCA � m�CAD � 40�

12 �m�BCA� �12 �m�CAB� � 40�

12 �m�BCA � m�CAB� �12 � 80�

m�BCA � m�CAB � 80�

100� � m�BCA � m�CAB � 180�

m�ABC � m�BCA � m�CAB � 180�




Chapter 5 continued

29. —CONTINUED—

The perpendicular bisector of has slope 1 because

is an equation of the perpendicular bisectorof

The slope of so is

horizontal.

So the perpendicular bisector is the vertical line

30. The lines and intersect at the pointbecause

To show is also on substitute x � 9and y � �3 in the equation.

Since is true, the point is on the line

31. Let P be the point

Since P is equidistant from A,B, and C.

AP � BP � CP � 3�10,

� 3�10

� �9 � �10

� �90

� �81 � 9

� ��9�2 � ��3�2

CP � ��9 � 18�2 � ��3 � 0�2

� 3�10

� �9 � �10

� �90

� �9 � 81

� ��3�2 � ��9�2

BP � ��9 � 12�2 � ��3 � 6�2

� 3�10

� �9 � �10

� �90

� �81 � 9

� �92 � ��3�2

AP � ��9 � 0�2 � ��3 � 0�2

�9, �3�.

y � x � 12.�9, �3��3 � �3

�3 � �3

�3 � 9 � 12

y � x � 12

y � x � 12,�9, �3�y � �2�9� � 15 � �18 � 15 � �3.�9, �3�

x � 9y � �2x � 15

x � 9.

ACAC �0 � 0

18 � 0�

018

� 0,

BC.y � x � 12

y � 3 � x � 15

y � 3 � 1 � �x � 15�

�1 � 1 � �1.BC


32.

34. j has slope because

An equation of j is


An equation of is


An equation of j is


An equation of j is

38. There is enough information to prove byusing the SAS Congruence Postulate.

39. There is not enough information given to proveOne pair of congruent sides, one side

congruent to itself, and one pair of congruent angles aregiven. But the angles must be the included angles and theyare not.

40. There is enough information given to proveOne pair of congruent legs and one

pair of congruent hypotenuses are given. The HLCongruence Theorem can be used to prove

Lesson 5.3


1. The centroid of a triangle is the point where the threemedians intersect.

2. The legs, and of right are also altitudesof because is the perpendicular segment fromK to and is the perpendicular segment from L toKM.

LMLMKM�KLM

�KLMLM,KM

�PMN � �KML.

�PMN � �KML.

�GJF � �GJH.

�ABC � �DEC

y � �1110 x �

565 .

y � �1110 x �

565

y � 9 � �1110 x �

115

y � 9 � �1110 �x � 2�

y � ��9� � �1110 �x � ��2��

�1110 � 10

11 � �1.�1110

y �32 x � 5.

y �32 x � 5

y � 8 �32 x � 3

y � 8 �32 �x � 2�

�23 � 3

2 � �1.32

y �12 x �

52.j

y �12 x �

52

y � 6 �12 x �

72

y � 6 �12 �x � 7�

�2 � 12 � �1.1

2

y � �13 x �

133 .

y � �13 x �

133

y � 4 � �13 x �

13

y � 4 � �13 �x � 1�

�13 � 3 � �1.�

13

� 22.5 square units

� 452

� 12 � 9 � 5

A �12 bh 33.

� 77 square units

� 1542

� 12 � 22 � 7

A �12 bh


Chapter 5 continued



3. indicates that G is the midpoint of Therefore, is a median of

4. indicates that is an altitude of

5. indicates the is bisected. So is an angle bisector of

6. and indicates G is the midpoint ofand is a perpendicular bisector of but it is

also a median, an altitude, and an angle bisector sinceby the SAS Congruence Postulate so

7. indicates and In this case is a

perpendicular bisector, an angle bisector, a median, andan altitude of


8.

9. 10.

11.

12.

13.–14. Sample answer:

is an acute triangle.

15. Yes, they all met at the same point. It is labeled as pointD.

�ABC

A

LD

N

M B

C

PHEP

�48

�12

PHEH

�4

12�

13

� 48 units

� 15 � 15 � 18

� 2 � 7.5 � 15 � 2 � 9

� 2�DG� � 15 � 2�DH�

Perimeter of �DEF � DE � EF � FD

15 � EF

225 � �EF�2

144 � 81 � �EF�2

122 � 92 � �EF�2

�EH�2 � �HF�2 � �EF�2

12 � EH

PH � 4 32 � 8 � EH

PH � 8 � 12 8 �23 EH

PH � PE � EH EP �23 EH

FH � DH � 9

�DEF.

EGDG � FG.�DGE � �FGE,�DEG � �FEG,�DGE � �FGE

�DEG � �FEG.�DGE � �FGE

�DEF,EGDFDG � FGEG�DF

�DEF.EG�DEF�DEG � �FEG

�DEF.EGEG�DF

�DEF.EGDF.DG � FG 16. Sample answer:

The distance from the centroid to a vertex is two thirds ofthe distance from that vertex to the midpoint of the oppo-site side.

17.

18.

19.

The coordinates of T are

20.

21. � �4, 4�� 82

, 82�M � �5 � 3

2,

2 � 62 �

NTNR

�69

�23

NR � 11 � 2 � 9

NT � 11 � 5 � 6

� �2, 2�� 42

, 42� R � ��1 � 5

2,

�2 � 62 �

�5, 6 � 4� � �5, 2�.

PT � 4

PT �23 � 6

PT �23 PQ

� 6

� �36

� �0 � 36

� �02 � ��6�2

PQ � ��5 � 5�2 � �0 � 6�2

� �5, 0�� 102

, 02�Q � ��1 � 11

2,

�2 � 22 �

40 � 40

40 �23 � 60

CD �23 �CM�

CM � 60 mm

CD � 40 mm

54 � 54

54 �23 � 81

BD �23 �BL�

BL � 81 mm

BD � 54 mm

64 � 64

64 �23 � 96

AD �23 �AN�

AN � 96

AD � 64 mm




Chapter 5 continued

22.

23.

24. Sample answer:

P is the orthocenter of

25. Sample answer:

The orthocenter of ispoint G.

26. Sample answer:

P is the orthocenter of

27.C

FA B

D

GH

E

�KLM.

P

K

ML

�EFGE

G F

�ABC.

A

P

B

C

2�5 � 2�5

2�5 �23 � 3�5

JP �23 JM

� 3�5

� �9 � �5

� �45

� �9 � 36

� ��3�2 � ��6�2

JM � ��4 � 7�2 � �4 � 10�

� 2�5

� �4 � �5

� �20

� �4 � 16

� ��2�2 � ��4�2

JP � ��5 � 7�2 � �6 � 10�2 28. G and H are the same point.

29. When is measured, it is found that Sincethen G and H must be the same point; therefore

the lines containing the three altitudes intersect at onepoint.

30.–32.

33. The measure of the angle between r and is approxi-mately

34. a.

b.

c. Check drawings.

d.

The length of the altitude is equal to twice the area divid-ed by the base.

12 � BE

90 � 7.5 � BE

90 �12

� 15 � BE

90 �12

� CA � BE

A �12

bh

� 90 square units�12

� 20 � 9A �12

bh

CD � 9

�CD�2 � 81

144 � �CD�2 � 225

122 � �CD�2 � 152

�AD�2 � �CD�2 � �AC�2

20�.MP

R

Sr

B

L

D

C

M

P

GH � 0,GH � 0.GH

e.

2Ab

� h

2A � bh

A �12

bh


Chapter 5 continued



35.

Statements Reasons

1. is isosceles; 1. Givenis a median to base

2. D is the midpoint of 2. Definition of median

3. 3. Definition of midpoint

4. 4. Definition of isosceles triangle


6. 6. SSS Congruence Postulate

7. 7. Corresponding parts ofcongruent triangles arecongruent.

8. and are a 8. Definition of linear pairlinear pair.

9. 9. If two lines intersect toform a linear pair of con-gruent angles, then thelines are perpendicular.

10. is also an altitude. 10. Definition of altitude

36. No, medians to the legs of an isosceles triangle are notperpendicular to the legs (unless the triangle is actuallyequilateral).

37. Yes, the medians of an equilateral triangle are also alti-tudes because the proof for the isosceles triangle could beused for the equilateral triangle.

Yes, the medians would be contained in the angle bisec-tors. By looking at the proof in Exercise 35, it can beseen that the median was also the angle bisector since thetwo triangles are congruent.

Yes the medians would be contained in the perpendicularbisectors because it was shown in Exercise 35 that themedian was perpendicular to the side at the midpoint.

38. The median of an equilateral triangle is also a perpendic-ular bisector of a side, an altitude, and an angle bisector.


39. The parallel line would also have slope

An equation of the line through P that is parallel tois y � �x � 8.y � �x � 3

y � �x � 8

y � 7 � �x � 1

y � 7 � �1�x � 1�

�1.

BD

BD�AC

�BDA�BDC

�BDC � �BDA

�BDC � �BDA

BD � BD

AB � CB

AD � CD

AC.

AC.BD�ABC


An equation of the line through P that is parallel tois

41. The parallel line would also have slope 3.




43. because you need the angles which do nothave or as a side.

44. because you need the angles which have oras a side.

45. Sample answer:

Quiz 1 (p. 285)

1.

x � 16

4x � 3x � 16

4x � 9 � 3x � 25

� 5�10

� �25 � �10

� �250

� �169 � 81

� �132 � 92

h � ��13 � 0�2 � �0 � 9�2

1

1

y

x

(0, 9)

(13, 0)

HJEF�F � �J

GJDF�E � �H

y � �12 x.y � �

12 x � 1

y � �12 x

y � 2 � �12 x � 2

y � ��2� � �12 �x � 4�

�12.

y � 3x � 21.y � 3x � 5

y � 3x � 21

y � 9 � 3x � 12

y � ��9� � 3�x � 4�

y � �2x � 14.y � �2x � 3

y � �2x � 14

y � 8 � �2x � 6

y � 8 � �2�x � 3�

y � ��8� � �2�x � ��3��

�2.

2.

y � 12

2y � 24

3y � y � 24




Chapter 5 continued

3.

4. because the perpendicular bisectors intersect at a point equidistant from the vertices of the triangle.

5. The balancing point would be at point G because that isthe centroid of the triangle.

5.3 Math and History (p. 285)

1. You need to go to the post office (P), then the market(M), then the library (L) or in reverse order.

2. The goalie’s position on the angle bisector optimizes thechance of blocking a scoring shot because the distancethe goalie would have to travel to protect either side ofthe goal would be the same.

Technology Activity 5.3 (p. 286)

Investigate

1. �

So is the angle bisector of

2. F was the point of intersection of the angle bisectors of angles and by construction. Since

AFxxxy is the angle bisector of Because angle bisector AFxxxy passes through the

intersection point, F, of BDxxxxy and CExxxy, the three anglebisectors are concurrent.

3. This makes a median also.

4. F was the point of intersection of two medians by con-struction. Since G is the midpoint of and

is the third median of the triangle. Since F is on by construction, F is on all three medians and the medi-ans are concurrent.

5. Sample measures are given:

Yes,

6. No, the quotient does not change.

Extension

For any triangle in which the angle bisector is containedin the same line as the median, the line will also containan altitude and perpendicular bisector of the triangle.

—CONTINUED—

ADAF

AD �23 AF

ADAF

�72

108�

23

AF � 108 mm

AD � 72 mm

CGCGABAG � BG,

CGAG � BG.

�BAC.m�BAF � m�CAF,

�BCA�ABC

�BAC.→AF

m�CAFm�BAF

VT � VS � 10

10 � VT

100 � �VT�2

36 � 64 � �VT�2

62 � 82 � �VT�2 This will occur for the bisector of the vertex angle of anisosceles triangle and for the bisector of each angle of anequilateral triangle.

Given: ADxxxxy bisects �BAC andADxxxx is a median.

Prove: ADxxxx is an altitude andADxxxx is a perpendicularbisector of BCxxxx.

In this drawing is the angle bisector of andis the median of the triangle. So Sinceis the angle bisector, or

Then by the HL Congruence Theorem.Therefore, or because corresponding parts of congruent triangles arecongruent. Also by the HL CongruenceTheorem. So, or because corresponding parts of congruent triangles arecongruent. Then

So which are linear pairs.Therefore, and This wouldmean that would be an altitude and a perpendicularbisector, also.

Lesson 5.4


1. In if M is the midpoint of N is the midpointof and P is the midpoint of then and

are midsegments of triangle

2. It is convenient to position one of the sides of the trianglealong the x-axis because some of the coordinates of twoof the points will be zero and one side will be horizontaland have a slope of 0.

3. 4.

5.

7.

9.

� 30.6

� 10.6 � 8 � 12

Perimeter � GJ � JH � GH

� 10.6

� 12 � 21.2

GJ �12 � EF

� 16

� 2 � 8

DF � 2 � DG

� 21.2

� 2 � 10.6

EF � 2 � EH

GH � DEJH � DF

�ABC.PNNP,MN,BC,AC,

AB,�ABC,

ADAD�CD.�ADC � �ADB

m�ADC � m�ADBm�ADE.m�CDF � m�ADF � m�BDE �

m�CDF � m�BDE�CDF � �BDE�CFD � �BED

m�ADE � m�ADF�ADE � �ADF�ADE � �ADF

AD � AD.DE � DF.DE � DF

→AD

DC � DB.AD�BAC

→AD

A E B

D

CF

6.

8.

� 8

� 12 � 16

JH �12 � DF

� 12

� 12 � 24

GH �12 � DE


Chapter 5 continued



10.

11.

12. 13.

14.

15.

16.

17.

19. and are congruent by theCorresponding Angles Postulate, as are and

and bythe Alternate Interior Angles Theorem.

So, by the Transitive Property of Congruence,and are congruent, as are and Then

and all congruent by the Third AnglesTheorem and the Transitive Property of Congruence.

20. Sample answer:

Use the construction of the perpendicular bisectors of thesides to find the midpoints D, E, and F.

are the midsegments of �ABC.DE, EF, and DF

A B

E

C

F

D

�MNC are�B, �ALM, �LMN,�NLM.�LMA,�C,

�BNL,�LNM�A, �NMC,�BLN,

�NMC � �LNM�LMA � �NLM�LMA.�BNL, �C,

�NMC�BLN, �A,

� 31

� 24 � 7

� 3 � 8 � 7

LM � 3x � 7

x � 8

�12 x � �4

3x �72 x � 4

3x � 7 �72 x � 3

3x � 7 �12 �7x � 6�

LM �12 � BC

� 18

� 2 � 9

BC � 2 � NC

� 14� 2 � 7AB � 2 � MN

� 10�12 � 20LN �

12 � AC

AB � MNLM � BC

AB � 5.4 � 10 yd � 54 yd

� �29 � 5.4

� �25 � 4

� �52 � 22

AB � ��6 � 1�2 � �6 � 4�2

� �1, 4�

� 22

, 82

A � 0 � 22

, 0 � 8

2 21.

22. Slope of

Slope of

Since the slopes of and are equal,

So and

23.

24.

—CONTINUED—

�b

a � c

�2b

2�a � c�

�2b

2a � 2c

Slope of CB �2b � 02a � 2c

�b

a � c

Slope of DF �b � 0a � c

� �c, 0�� 2c2

, 02F � 0 � 2c

2,

0 � 02

y

xA(0, 0) F (c, 0) B(2c, 0)

E(a � c, b)

C (2a, 2b)

D (a, b)

DF �12

BC.BC � DF

DF �12

BC

�12

� �89��89

2�

�89�4

��894

DF � �22.25

� �89

� �25 � 64

� �52 � 82

BC � ��10 � 5�2 � �6 � ��2��2

� �22.25

� �6.25 � 16

� �2.52 � 42

DF ��5 �52

2

� �4 � 0�2

BC � DF.DFBC

�85

� 4 �25

�452

DF �4 � 0

5 �52

�85

BC �6 � ��2�

10 � 5

� �5, 4�� 102

, 82F � 0 � 10

2,

2 � 62

� 152

, 2� 152

, 42E � 5 � 10

2,

�2 � 62

� 52

, 0� 52

, 02D � 0 � 5

2,

2 � ��2�2

� �6, 6�

� 122

, 122

B � 2 � 102

, 8 � 4

2

18.

� 6

� 24 � 18

� 6 � 4 � 18

AB � 6x � 18

x � 4

�2x � �8

x � 3x � 8

x � 1 � 3x � 9

x � 1 �12 �6x � 18�

MN �12 � AB

� 9

� NC

LM �12 � BC




Chapter 5 continued

24. —CONTINUED—



25.

26.

Slope of

Draw a line through M with slope

Slope of

Draw a line through L with slope 5.

MN �9 � 45 � 4

�51

� 5

13

.

LN �4 � 34 � 1

�13

2

2

y

x

M(5, 9)

N(4, 4)L(1, 3)

C B

A

�12 CA

�12 � 2�a2 � b2

EF � �a2 � b2

�12 CB

�12 �2��a � c�2 � b2�

DF � ��a � c�2 � b2

� 2�a2 � b2

� �4�a2 � b2

� �4�a2 � b2�

� �4a2 � 4b2

� ��2a�2 � �2b�2

CA � ��2a � 0�2 � �2b � 0�2

� 2��a � c�2 � b2

� �4��a � c�2 � b2

� �4��a � c�2 � b2�

� �4�a � c�2 � 4b2

� �22�a � c�2 � �2b�2

CB � ��2a � 2c�2 � �2b � 0�2

� �a2 � b2

EF � ��a � c � c�2 � �b � 0�2

� ��a � c�2 � b2

DF � ��a � c�2 � �b � 0�2

EF � CA.CAEF

�ba

�2b2a

Slope of CA �2b � 02a � 0

�ba

Slope of EF �b � 0

a � c � c

DF � CB.CBDF

Draw a line through N with slope

The lines intersect at and

27.

Draw a line through M with slope

Draw a line through L with slope

Draw a line through N with slope

The lines intersect at and

28.

because

29.

� 31 units� 12 � 10 � 9P � SU � TU � ST

� 9

� QU

ST �12�QR�

� 10

� 12�20�

TU �12 PQ

� 40 units� 14 � 16 � 10P � CD � BD � BC

� 10

� 2 � 5

� 2�GC�

BG � GC. � GC � GC

BC � BG � GC

16 � BD

8 �12 BD

GF �12 BD

C�7, 9�.B�11, 3�,A�3, �1�,

52

.

Slope of LM �6 � 19 � 7

�52

12

.

�24

�12

Slope of MN �6 � 49 � 5

�32

.

�3

�2� �

32

Slope of LN �4 � 15 � 7

�2

2

y

x

M(9, 6)

L(7, 1)

N(5, 4)

A

C

B

C�2, 8�.A�0, �2�, B�8, 10�,

32

.

Slope of LM �9 � 35 � 1

�64

�32


Chapter 5 continued



30.

Perimeter of is 4 times the perimeter of

31. The perimeter of the shaded triangle in Stage 1 is because the length of each side of the shaded triangle is the length of a side in the original triangle.

The total perimeter of the shaded triangle in Stage 2 is

The total perimeter of the shaded triangles in Stage 3 is

32. The bottoms of the legs will be 60 inches apart. Since thecross bar attaches at the midpoints of the legs, the crossbar is a midsegment of the triangle formed by the twolegs and the ground. Since the length of the midsegmentis half of the length of third side, the length of the crossbar, 30 inches, is half of the distance between the bottomsof the legs, 60 inches.

33. is a midsegment of so D is the midpoint ofand By the Midsegment Theorem,

and But F is the midpoint of soThen by the transitive property of equality

and the definition of congruent segments,Corresponding angles and are congruent,so by the SAS Congruence Postulate.�ADE � �DBF

�ABC�ADEDE � BF.

BF �12 BC.

BC,DE �12 BC.DE � BC

AD � DB.AB�ABC,DE

� 238.

12 �

14 �

14 �

14 �

18 �

18 �

18 �

18 �

18 �

18 �

18 �

18 �

18

12 �

14 �

14 �

14 � 11

4.

12

12

Perimeter of �GHI �14 of perimeter of �ABC

� 14 �BC � AC � AB�

� 14 BC �

14 AC �

14 AB

Perimeter of �GHI � GH � HI � GI

� 14 AB

� 12 �1

2 AB� GI �

12 FE

� 14 AC

� 12 �1

2 AC� HI �

12 DE

� 14 BC

� 12 �1

2 BC� GH �

12 FD

�GHI.�ABC

A B

E

C

F H

I G

D

34. In Exercise 33, it was shown that andSo all that is left to show is that

This can be done in the same manner that it was shownthat By using the fact that is a midsegmentof �ABC, By using the fact that E is themidpoint of we can get Therefore,

or and the triangles are congruent bythe SSS Congruence Postulate.

35. Since is closer to than it must be longer thanSince is the midsegment, or

So cannot be 10 or 12 feetlong since it must be longer than , which is 12 feetlong. could be 14 feet long but not 24 feet long sincePQ cannot equal RS. So, or

36. a.

b.

c. The line containing has slope �2 and passesthrough F(4, 5).

An equation of CBvxxy is y � 5 � �2(x � 4).

d.

The line containing has slope 3 and passes through

An equation of is

The line containing has slope and passes through

An equation of is

e. :

—CONTINUED—

y �12 x �

12

y � 2 �12 x �

32

y � 2 �12 �x � 3�

↔AB :

y � 3x � 2 y � �2x � 13

y � 4 � 3x � 6y � 5 � �2x � 8

y � 4 � 3�x � 2�↔ACy � 5 � �2�x � 4�

↔BC :

y � 2 �12 �x � 3�.

↔AB

y � 2 �12 �x � 3�

E�3, 2�.

12AB

y � 4 � 3�x � 2�.↔AC

y � 4 � 3�x � 2�

D�2, 4�.AC

�12

slope of FD � m3 �5 � 44 � 2

� 3�31

slope of EF � m2 �5 � 24 � 3

y � 5 � �2�x � 4�

CB

� �2�2

�1slope of DE � m1 �

4 � 22 � 3

1

1

y

x

F (4, 5)

E(3, 2)A(1, 1)

C (3, 7)

B(5, 3)(2, 4)

D

12 < PQ < 24.MN < PQ < RS,

PQMN

PQMN �12 � 24 � 12 feet.

MN �12 RSMNMN.

MN,RSPQ

AE � DFAE � DFAE �

12 AC.AC,

DF �12 AC.

DEDE � BF.

AE � DF.DE � BF.AD � DB




Chapter 5 continued

f. To find A, find the point of intersection of and

has the equation

has the equation

By substitution,

So,

To find B, find the point of intersection of and

has the equation

has the equation

By substitution,

So,

To find C, find the point of intersection of and

has the equation

has the equation

By substitution,

So,

37.

C�3, 7�.

y � 9 � 2 � 7

y � 3 � 3 � 2

y � 3x � 2

x � 3

5x � 15

3x � �2x � 15

3x � 2 � �2x � 13

y � �2x � 13.↔BC

y � 3x � 2.↔AC

↔BC.

↔AC

B�5, 3�.

y � 3

y � �10 � 13

y � �2 � 5 � 13

y � �2x � 13

x � 5

52 x �252

12 x � �2x �255

12 x �

12 � �2x � 13

y � �2x � 13.↔BC

y �12 x �

12.

↔AB

↔BC.

↔AB

A�1, 1�.

y � 3 � 2 � 1

y � 3 � 1 � 2

y � 3x � 2

x � 1

�52 x � �

52

12 x � 3x �52

12 x �

12 � 3x � 2

y � 3x � 2.↔AC

y �12 x �

12.

↔AB

↔AC.

↔AB 38. is the function

that gives the length of themidsection at Stage n. Fromone stage to the next, thelength is multiplied by


39.

Addition property of equality

40.

Subtraction property of equality

Division property of equality

41.




42.




43.




44.




45.




46.

Distributive property

Simplify


Division property of equality x � 6

5x � 30

5x � 10 � 40

3x � 2x � 10 � 40

3x � 2�x � 5� � 40

x � �11

x � 1 � �10

�2�x � 1� � 20

�2�x � 1� � 3 � 23

x � �73

3x � �7

3x � 10 � 3

9�3x � 10� � 27

x � 2

4x � 8

4x � 1 � 7

2�4x � 1� � 14

x � 4

�4x � �16

5x � 9x � 16

5x � 12 � 9x � 4

x � 3

6x � 18

8x � 2x � 18

8x � 1 � 2x � 17

x � 11

3x � 33

3x � 13 � 46

x � 14

x � 3 � 11

12.

y � 24 � �12�n

Stage

8

4

0

12

16

20

24

2 4 61 3 5

Mid

seg

men

t

len

gth

y

n

Stage n 0 1 2 3 4 5

Midsegment length 24 12 6 3 1.5 0.75


Chapter 5 continued

47.

48.

49.

50. and because and are angle bisectors and angle bisectors divide anangle into two congruent angles.

51. Point D is the incenter of because it is the inter-section of the angle bisectors of

52. because D is the point of intersection ofthe angle bisectors of and D is equidistant fromthe sides of the triangle.

53.

But because D is equidistant from the sides of

So,

Lesson 5.5

Technology Activity 5.5 (p. 294)

Investigate

1. The longest side is opposite the largest angle.

2. The shortest side is opposite the smallest angle.

3. The answers are the same.

The longest side is opposite the largest angle.

The shortest side is opposite the smallest angle.

4. The longest side will always be opposite the largestangle. The shortest side will always be opposite thesmallest angle. The side with the middle length will beopposite the angle with the middle measure.

DF � DE � 6.

�ABC.DE � DF

DE � 6

�DE�2 � 36

�DE�2 � 64 � 100

�DE�2 � 82 � 102

�DE�2 � �EC�2 � �CD�2

�ABCDE � DG � DF

�ABC.�ABC

→CD

→BD,

→AD,�BCD � �ACD�CAD � �BAD

x � 18

�3x � �54

4x � 7x � 54

4x� � 61� � �7x � 7��

x � 7

17x � 119

17x � 61 � 180

�10x � 22�� 7x � 1�� 38� � 180�

x � 23

2x � 46

2x � 134 � 180

�x � 2�� 132� � x� � 180� Extension

Sample answer:

The statement is false because the above example is acounterexample.


1. The 1 inch side is opposite the smallest angle of theinch side is opposite the middle angle of and the

inch side is opposite the largest angle of

2. No, it is not possible to draw a triangle with side lengthsof 5 inches, 2 inches, and 8 inches because the sum of the lengths of any two sides of a triangle must be greaterthan the length of the third side. But is not greaterthan 8.

3.

The smallest angle is and the largest angle is

4. The shortest side is and the longest side is

5. The distance between Guiuan and Masbate has to begreater than miles and less than

miles.


6.

is the shortest side because it is opposite the smallestangle.

is the longest side because it is opposite the largestangle.

7.

is the shortest side. and are the longest sides

8.

is the shortest side. is the longest side.HJJK

m�J � 55�

m�J � 35� � 90�

m�J � m�H � 90�

�RS � ST�.STRSRT

m�R � 65�

m�R � 115� � 180�

m�R � 50� � 65� � 180�

m�R � m�S � m�T � 180�

AB

AC

m�A � 67�

m�A � 113� � 180�

m�A � 42� � 71� � 180�

m�A � m�B � m�C � 180�

165 � 99 � 264165 � 99 � 66

DE.EF

�F.�D

m�E � 45�

m�E � 135� � 180�

32� � m�E � 103� � 180�

m�D � m�E � m�F � 180�

5 � 2

90�.218

62�,178

28�,

length of shortest sidelength of longest side

�63 mm96 mm

� 0.656

measure of smallest anglemeasure of largest angle

�41�

82�� 0.5






Chapter 5 continued

9. is the smallest angle. is the largest angle.



12. 13.

14. and

15.

and

16.

and

17. and 18. and

19. and

20.–23. Answers may vary; sample answers are given.

20.

21. 22.

23. The following combinations of lengths will not producetriangles: 4 inches, 4 inches, and 10 inches; 3 inches, 5inches, and 10 inches; and, 2 inches, 7 inches, and 9inches.

24.

x < 7

�x > �7

2x > 3x � 7

2x � 5 > 3x � 2

x � 2 � x � 3 > 3x � 2

AB � AC > BC

5.5 in.

8.5 in.4 in.

7 in.

6 in.5 in.

2 in.4 in.

8 in. 6 in.

5 in.

8 in.7 in. 7 in. 8 in.

6 in. 6 in.5 in.

�R�T, �S,

�P�N, �Q,�M�L, �K,

HGHJ, JG,

m�G � 25�

m�G � 155� � 180�

m�HGJ � 120� � 35� � 180�

m�G � m�J � m�H � 180�

EFDF, DE,

m�F � 60�

120� � m�F � 180�

90� � 30� � m�F � 180�

m�D � m�E � m�F � 180�

ACAB, BC,

x� > z�

x� > y�y� � z� � x�

�F�H

�Q�R

�B�C 25.

26. It is shorter to cut across the empty lot because the sumof the lengths of the two sidewalks is greater than thelength of the diagonal across the lot. If the corner ofPleasant Street and Pine Street were labeled point A, thecorner of Pine Street and Union Street were labeled pointB, and the corner of Union Street and Oak Hill Avenuewere labeled point C, could be formed. By theTriangle Inequality Theorem, that is,walking around the sidewalks is longer than walkingthrough the lot.

27. The sides and angles could not be positioned as they arelabeled; for example, the longest side is not opposite thelargest angle.

28. No, a kitchen triangle cannot have side lengths of 9 feet,3 feet, and 5 feet because 3 � 5 � 8 and 8 is not greaterthan 9.

29. The boom is raised when the boom lines are shortened.

30. AB must be less than feet.

31. Yes, when the boom is lowered and length of the boomlines, AB, is greater than 100 feet, then will belarger than

32. The third inequality would be and this isnot helpful because since x is positive, x � 14 > 10 forall values of x.

33. so is a right triangle. The largest anglein a right triangle is the right angle, so

so (If one angle of atriangle is larger than another then the side oppositethe larger angle is longer than the side opposite the smaller angle.)

�,MN > MJ.m�MJN > m�MNJ,

�MJNMJ�JN,

x � 14 > 10

�BAC.�ABC

100 � 50 � 150

AB � BC > AC;�ABC

x < 7

�x > �7

2x > 3x � 7

2x � 6 > 3x � 1

x � 2 � x � 4 > 3x � 1

AB � AC > BC


Chapter 5 continued

34.

Statements Reasons

1. 1. Given

2. Extend to D such 2. Ruler Postulatethat

3. 3. Segment Addition Postulate

4. 4. Base Angles Theorem

5. 5. Protractor Postulate

6. 6. Substitution property of equality

7. 7. If one angle of a triangle islarger than another angle, thenthe side opposite the largerangle is longer than the sideopposite the smaller angle.



35. since the side opposite the angle of is longerthan the side opposite the angle of A

36. is the measure of the exterior angle and B

37. D

38.

Statements Reasons

1. 1. Given

2. Let D be a point on 2. A plane contains at least plane M distinct from C. three noncollinear points.

3. 3. Definition of a line perpendicular to a plane

4. is a right angle. 4. If two lines are perpendic-ular, then they intersect toform four right angles.

5. is a right triangle. 5. Definition of right triangle

6. is an acute angle. 6. The non-right angles in aright triangle are acuteangles.

7. 7. Definition of an acuteangle

8. 8. Definition of a right angle

9. m�PDC < m�PCD 9. Substitution property ofequality

10. 10. If one angle of a triangle islarger than another angle,then the side opposite thelarger angle is longer thanthe side opposite thesmaller angle.

PD > PC

m�PCD � 90�

m�PDC < 90�

�PDC

�PCD

�PCD

↔CD�PC

PC�plane M

x� � y� � z�.z�

y� �n � 3 > n�.x�x� > y�

AB � AC > BC

AD � AC > BC

DC > BC

m�DBC > m�1

m�DBC > m�2

�1 � �2

AD � AC � DC

AB � AD.AC

�ABC


39.–41. Answers may vary. Sample answers are given.

39. The proof for Example 2 on page 230 is a two-columnproof.

40. The proof for Example 1 on page 229 is a paragraphproof.

41. The proof for Example 3 on page 158 is a flow proof.

42. and are corresponding angles. So are and

43. and are vertical angles.

44. and are alternate interior angles. So are and

45. and are alternate exterior angles. So are and

46.




—CONTINUED—

y �12

x �52

y � 1 �12

x �32

y � ��1� �12

�x � 3�

N�3, �1�.

12

,BC

y � �25

x �195

y � 3 � �25

x �45

y � 3 � �25

�x � 2�

M�2, 3�.

�25

AC

y � �4x � 7

y � 1 � �4x � 8

y � 1 � �4�x � 2�

y � 1 � �4�x � ��2��

L ��2, 1�.�4AB

� �25

�2

�5slope of LN � m3 �

1 � ��1��2 � 3

� �4�4

�1slope of MN � m2 �

3 � ��1�2 � 3

�12

�24

slope of LM � m1 �3 � 1

2 � ��2�

�10.�7�2�7

�11.�6�3�6

�9�12

�9.�5�1�5






Chapter 5 continued

46. —CONTINUED—

To find A, find the point of intersection of and

The point A has coordinates


The coordinates of B are

To find C, find the point of intersection of and ↔BC.

↔AC

��1, �3�.

y � �3

y � 4 � 7

y � �4��1� � 7

y � �4x � 7

x � �1

�92

x �92

�4x �12

x �92

�4x � 7 �12

x �52

y �12

x �52

y � �4x � 7

↔BC.

↔AB

��3, 5�.

y � 5

y � 12 � 7

y � �4��3� � 7

y � �4x � 7

x � �3

�185

x �545

�4x � �25

x �545

�4x � 7 � �25

x �195

y � �25

x �195

y � �4x � 7

↔AC.

↔AB

The coordinates of C are

47.



—CONTINUED—

y �53

x �163

y � 2 �53

x �103

y � 2 �53

�x � 2�

y � 2 �53

�x � ��2��

M��2, 2�.

53

BC

y �12

x �132

y � 5 �12

x �32

y � 5 �12

�x � 3�

y � 5 �12

�x � ��3��

L��3, 5�.

12

AB

�53

slope of LN � m3 �5 � 0

�3 � ��6��

�12

�24

slope of MN � m2 �2 � 0

�2 � ��6�

� �3�3

�1slope of LM � m1 �

5 � 2�3 � ��2�

�7, 1�.

y � 1

y �72

�52

y �12

� 7 �52

y �12

x �52

x � 7

9x � 63

5x � �4x � 63

5x � 25 � �4x � 38

12

x �52

��25

x �195

y �12

x �52

y � �25

x �195


Chapter 5 continued

47. —CONTINUED—


To find A, find the intersection of and

The coordinates of A are


The coordinates of B are �1, 7�.

y � 7

y �142

y �12

�132

y �12

� 1 �132

y �12

x �132

x � 1

�7x � �7

3x � 10x � 7

3x � 39 � 10x � 32

12

x �132

�53

x �163

y �53

x �163

y �12

x �132

↔BC.

↔AB

��7, 3�.

y � 3

y � 21 � 18

y � �3��7� � 18

y � �3x � 18

x � �7

72

x ��49

2

12

x � �3x �492

12

x �132

� �3x � 18

y � �3x � 18

y �12

x �132

↔AC.

↔AB

y � �3x � 18

y � �3�x � 6�

y � 0 � �3�x � ��6��

N��6, 0�.�3AC

To find C, find the intersection of and


48.

The line containing has slope 4 and passes through



—CONTINUED—

y � �16

x �73

y � 1 � �16

x �86

y � 1 � �16

�x � 8�

N�8, 1�.

�16

AC

y � �x � 14

y � 5 � �x � 9

y � 5 � �1�x � 9�

M�9, 5�.�1BC

y � 4x � 6

y � 6 � 4x � 12

y � 6 � 4�x � 3�

L�3, 6�.AB

� �1�5

�5Slope of LN � m3 �

6 � 13 � 8

� 4�41

Slope of MN � m2 �5 � 19 � 8

� �16

�1

�6Slope of LM � m1 �

6 � 53 � 9

��5, �3�.

y � �3

y � 15 � 18

y � �3��5� � 18

y � �3x � 18

x � �5

�14

3x �

703

�3x �53

x �703

�3x � 18 �53

x �163

y �53

x �163

y � �3x � 18

↔BC.

↔AC






Chapter 5 continued

48. —CONTINUED—






The coordinates of C are �14, 0�.

y � 0

y � �14 � 14

y � �x � 14

x � 14

56

x �353

�16

x � �x �353

�16

x �73

� �x � 14

y � �x � 14

y � �16

x �73

↔BC.

↔AC

�4, 10�.

y � 10

y � 16 � 6

y � 4 � 4 � 6

y � 4x � 6

x � 4

5x � 20

4x � �x � 20

4x � 6 � �x � 14

y � �x � 14

y � 4x � 6

↔BC.

↔AB

�2, 2�.

y � 2

y � 8 � 6

y � 4 � 2 � 6

y � 4x � 6

x � 2

256

x �253

4x � �16

x �253

4x � 6 � �16

x �73

y � �16

x �73

y � 4x � 6

↔AC.

↔AB

49.

undefined

is vertical.


The line containing is a vertical line passing through




—CONTINUED—

�6, �4�.

y � �4

y � �23

� 6

y � �23

x

x � 6

�43

x � �8

�23

x �23

x � 8

y �23

x � 8

y � �23

x

↔AC.

↔AB

y �23

x � 8

y � 6 �23

x � 2

y � ��6� �23

�x � 3�

N�3, �6�.

23

AC

x � 0

M�0, �4�.BC

y � �23

x

y � 2 � �23

x � 2

y � ��2� � �23

�x � 3�

L�3, �2�.

�23

AB

LN

�40

�slope of LN � m3 ��2 � ��6�

3 � 3

� �23

�2

�3slope of MN � m2 �

�4 � ��6�0 � 3

�23

slope of LM � m1 ��2 � ��4�

3 � 0


Chapter 5 continued

49. —CONTINUED—





Lesson 5.6


1. An indirect proof might also be called a proof by contra-diction because in an indirect proof, you prove that astatement is true by first assuming that its opposite istrue. If this assumption leads to a contradiction, then youhave proved that the original statement is true.

2. To use an indirect proof to show that two lines m and nare parallel, you would first make the assumption thatlines m and n are not parallel.

3. 4. 5.

6. In if you wanted to prove that youwould use the two cases and in anindirect proof.


7. 8. 9.

10. 11. 12.

13. 14. 15.

16. The correct answer is C because and so by the Converse of the Hinge Theorem

17. The correct answer is B because and so by the Hinge Theorem AC > BD.m�3 < m�5

AD � AD,AB � DC,

m�4 < m�5.AC > AB

AD � AD,BD � CD,

m�1 > m�2UT > SVAB > CB

m�1 < m�2m�1 � m�2XY > ZY

m�1 > m�2m�1 � m�2RS < TU

BC � ACBC < ACBC > AC,�ABC,

DC < FEKL < NQm�1 > m�2

�0, �8�.

y � �8

y � 0 � 8

y �23

� 0 � 8

y �23

x � 8

x � 0

y �23

x � 8

↔BC.

↔AC

�0, 0�.

y � 0

y � �23

� 0

y � �23

x

x � 0

y � �23

x

↔BC.

↔AB

18. because

19. because

20. because

21. Given that and assume that

22. Given with Q the midpoint of assume isnot a median.

23. Given with assume

24. C Assume that there are two points, P and Q, where mand n intersect.

B Then there are two lines (m and n) through points Pand Q.

A But this contradicts Postulate 5, which states that thereis exactly one line through any two points.

D It is false that m and n can intersect in two points, sothey must intersect in exactly one point.

25. Case l: Assume that If one side of a triangleis longer than another side, then the angle oppo-site the longer side is larger than the angle oppo-site the shorter side, so But thiscontradicts the given information that

Case 2: Assume that EF � DF. By the Converse of theBase Angles Theorem, But thiscontradicts the given information that

Since both cases produce a contradiction, the assumptionthat EF is not greater than DF must be incorrect and

26. Assume Then m and n intersect in a point and thetriangle shown in the diagram is formed.

by the Triangle SumTheorem. Then by theSubtraction property of equality. But

because and are supple-mentary. So by the Substitutionproperty of equality. Then by simplifying bothsides. But this is not possible; angle measures in a trian-gle cannot be zero.

So the assumption that must be false. Therefore,m � n.

m � n

m�3 � 0180� � 180� � m�3

�2�1m�1 � m�2 � 180�

� m�3m�1 � m�2 � 180�m�1 � m�2 � m�3 � 180�

m � n.

EF > DF.

m�D > m�E.

m�E � m�D.

m�D > m�E.

m�D < m�E.

EF < DF.

m�C � 90�.m�A � m�B � 90�,�ABC

MQNP,�MNP

RS � 7 in.ST � 5 in.,RS � ST � 12 in.

x < 17.5

4x < 70

2 < 4.�4x � 5�� < 65�

x > 1

2x > 2

3x > x � 2

115� > 45�.3x � 1 > x � 3

70� > 60�.x > 9






Chapter 5 continued

27. Case 1: Assume that Then bythe Converse of the Hinge Theorem. But

by the ASA CongruencePostulate, so �S � �T or This is a contradiction. So RS ≤ RT.

Case 2: Assume Then by the Converse of the Hinge Theorem. But

by the ASA CongruencePostulate, so or This is a contradiction. So

Therefore and is an isosceles triangle.

28. The paths are described by two triangles in which twosides of one triangle are congruent to two sides of anoth-er triangle, but the included angle in your friend’s triangleis larger than the included angle of your triangle, so theside representing the distance from the airport is longer inyour friend’s triangle.

29. The paths are described by two triangles in which twosides of one triangle are congruent to two sides of anoth-er triangle, but the included angle in your friend’s triangleis larger than the included angle in your triangle, so theside representing the distance to the airport is longer inyour friend’s triangle.

30. a. As ED increases, increases because ∠EBD isthe angle opposite

As ED increases, decreases because increases and and are supplementary.

b. As ED increases, AD decreases because as EDincreases increases and decreasesmaking AD decrease.

c. The cleaning arm illustrates the Hinge Theorembecause the lengths of and remain constantwhile and ED change. In �EBD and �ABD,

The included angle in �EBD,�EBD, is larger than the included angle in �ABD,�ABD. So is longer than .ADED

BE � BD � BA.m�EBD

BDBE

m�ABDm�EBD

�EBD�DBAm�EBDm�DBA

ED.m�EBD

�RST RS � RT

RS ≥ RT.m�S � m�T.�S � �T

�RUS � �RUT

m�T < m�SRS < RT.

m�S � m�T.�RUS � �RUT

m�T > m�SRS > RT. 31.

Statements Reasons

1. 1. Given

2. Construct a ray from B 2. Protractor Postulateto construct an angle in the interior of that is congruent to

.

3. Locate P on the con- 3. Ruler Postulate structed ray such that

.


5. 5. Corresponding parts of congruent triangles arecongruent

6. Locate H on so 6. Protractor Postulatethat BHxxxy bisects .

7. 7. Definition of angle bisec-tor

8. 8. Transitive property of congruence (Steps 1, 3)

9. 9. Reflexive property of congruence


11. 11. Corresponding parts ofcongruent triangles arecongruent

12. 12. Definition of congruentsegments

13. 13. Segment AdditionPostulate

14. 14. Substitution property ofequality

15. 15. Triangle Inequality


17. 17. Definition of congruentsegments (Step 5)


AC > DF

PC � DF

AC > PC

PH � HC > PC

AC � PH � HC

AC � AH � HC

AH � PH

AH � PH

�ABH � �PBH

BH � BH

PB � AB

�PBH � �ABH

�PBAAC

PC � DF

�PBC � �DEF

BP � ED

�DEF

�ABC

m�ABC > m�DEFAB � DE, BC � EF,


Chapter 5 continued


32. isosceles 33. equilateral, equiangular, and isosceles

34. scalene 35. isosceles

36. equiangular, equilateral, and isosceles

37. isosceles

38.

39.

41.

42. is a median, an altitude, anangle bisector, and a perpendicular bisector. divides intotwo congruent triangles. It is short-er than each side of .

5.6 Quiz 2 (p. 308)

1. 2. If then

3. If the perimeter of then the perimeter of

4.

5.

6.

MP, NP, MN

m�M � 57�

m�M � 123� � 180�

m�M � 48� � 75� � 180�

m�M � m�N � m�P � 180�

MQ, MP, PQ

m�M � 81�

m�M � 99� � 180�

m�M � 49� � 50� � 180�

m�M � m�P � m�Q � 180�

LQ, LM, MQ

m�M � 31�

m�M � 149� � 180�

75� � m�M � 74� � 180�

m�L � m�M � m�Q � 180�

�GHF � 21�CDE � 42,

CE � 16.FG � 8,FG � CE

�RST

�RSTRU

RU

R S

T

U

m�BAC � 84�

m�BAC � 96� � 180�

m�BAC � 51� � 45� � 180�

m�BAC � m�B � m�C � 180�

m�B � 51�

m�B � 32� � 19�

m�B � �x � 19��

32 � x

2x � 32 � 3x

�x � 13�� x � 19�� 3x�

7. is longer than because two sides of arecongruent to two sides of and the included angle

is larger than included angle so

8. The 2nd Group is farther from the camp because thegroups’ paths form two triangles with 2 pairs of congruent sides and the included angle for the 2nd groupis larger than the included angle for the 1st group.

Review (pp. 310–312)

1. If a point is on the perpendicular bisector of a segment,then it is equidistant from the endpoints of the segment.

2. If then U must be on the perpendicular

bisector of

3. If Q is equidistant from and then Q is on thebisector of

4. Let X be the midpoint of Then oror

But because K is equidistant from R, S,and T.

5.

Since W is equidistant from the sides of

6. The special segments are angle bisectors and the point of concurrency is the incenter

7. The special segments are perpendicular bisectors and the point of concurrency is the circumcenter.

8. The special segments are medians and the point of concurrency is the centroid.

9. The special segments are altitudes and the point of concurrency is the orthocenter.

WB � WA � 6.�XYZ,

WA � 6

�WA�2 � 36

�WA�2 � 64 � 100

�WA�2 � 82 � 102

�WA�2 � �AY�2 � �WY�2

KR � KT � 20

20 � KT

400 � KT 2

144 � 256 � KT 2

122 � 162 � KT 2

�KX�2 � �XT�2 � �KT�2

XT � 16.XT �12 � 32

XT �12 STST.

�RST.

→ST,

→SR

RT.↔SQ

UR � UT,

DE > AB.�BCA�DFE�DEF

�ABCABDE



40.

m�C � 45�

m�C � 32� � 13�

m�C � �x � 13��




Chapter 5 continued

10. midpoint of

midpoint of

midpoint of

An equation of the median to XZ___

is

.

An equation of the median to YZ___

is

.

The centroid is the point of intersection of the two lines.

The coordinates of the centroid of are ��43

, 2�.�XYZ

y � 2

y � �32 ��

43�

y � �32

x

�43

� x

3 ��94

x

34

x � 3 ��32

x

y � �32

x

y �34

x � 3

y � �32

x

y � 0 � �32

�x � 0�

m2 �3 � 0

�2 � 0�

3�2

y �34

x � 3

y �34

�x � 4�

y � 0 �34

�x � ��4��

m1 �3 � 0

0 � ��4� �34

� ��2, 3�

� ��42

, 62�

YZ � ��4 � 02

, 0 � 6

2 �

� �0, 3�� 02

, 62�XZ � �0 � 0

2,

6 � 02 �

� ��2, 0�

� ��42

, 02�

XY � ��4 � 02

, 0 � 0

2 � 11. The coordinates of the orthocenter of are since is a right triangle and the two legs are alsoaltitudes of

12. The slope of

The slope of

The slope of

The line containing has slope and passes through.

The line containing has slope 1 and passes through.

The line containing has slope 0 and passes through.

H is the point of intersection of and

The coordinates of H are

J is the point of intersection of and

The coordinates of J are

—CONTINUED—

�10, 1�.

10 � x

�10 � �x

1 � �x � 11

y � 1

y � �x � 11

↔JK.

↔HJ

�6, 5�

y � 5

y � �6 � 11

y � �x � 11

x � 6

�2x � �12

�x � x � 12

�x � 11 � x � 1

y � x � 1

y � �x � 11

↔HK.

↔HJ

y � 1.

y � 1 � 0

y � 1 � 0�x � 6�

N�6, 1�JK

y � x � 1.

y � 3 � x � 4

y � 3 � 1�x � 4�

L�4, 3�HK

y � �x � 11.

y � 3 � �x � 8

y � 3 � �1�x � 8�

M�8, 3��1HJ

LN � m3 �3 � 14 � 6

�2

�2� �1

MN � m2 �3 � 18 � 6

�22

� 1

LM � m1 �3 � 38 � 4

�04

� 0

�XYZ.�XYZ

�0, 0��XYZ



Chapter 5 continued

12. —CONTINUED—

K is the point of intersection of and

The coordinates of K are

13. Let L be the midpoint of M be the midpoint of and N be the midpoint of

The slope of the slope of so



14.

16. The angle measurements in order from least to greatestare and The side measurements inorder from least to greatest are AB, BC, and AC.

17. The angle measurements in order from least to greatestare and The side measurements inorder from least to greatest are EF, DF, and DE.

18.

The angle measurements in order from least to greatestare and The side measurements inorder from least to greatest are GJ, GH, and HJ.

19.

The angle measurements in order from least to greatestare and The side measurements inorder from least to greatest are KM, LM, and KL.

m�M.m�L, m�K,

m�L � 35�

m�L � 55� � 90�

m�L � m�K � 90�

m�G.m�H, m�J,

m�J � 60�

m�J � 120� � 180�

m�J � 50� � 70� � 180�

m�J � m�H � m�G � 180�

m�F.m�D, m�E,

m�B.m�C, m�A,

P � 64

P � 18 � 22 � 24

P � BC � CD � BD

24 � BD

12 �12 BD

GF �12 BD

18 � BC

9 � 9 � BC

BG � GC � BC

BG � GC � 9

LN � JK.JK,LN � �1 �

MN � HJ.HJ,MN � 1 �

LM � HK.HK,LM � 0 �

HK.JK,HJ,

�2, 1�.

2 � x

1 � x � 1

y � 1

y � x � 1

↔JK.

↔HK

20. The length of the third side must be less than the sum ofthe lengths of the other two sides. So the length of thethird side must be less than 300 feet So themaximum length of fencing needed is 600 feet

of fencing.

21. 22. 23.

24. In if then .

25. Assume has two right angles at and Thenand, since

This contradicts theTriangle Sum Theorem. Then the assumption that there issuch a must be incorrect and no triangle has tworight angles.

Chapter 5 Test (p. 313)

1. If P is the circumcenter of then PR, PS, and PTare always equal.

2. If bisects then and are sometimescongruent.

3. The incenter of a triangle never lies outside the triangle.

4. The length of a median of a triangle is sometimes equal tothe length of a midsegment.

5. If is the altitude to side of then isalways shorter than

6. a.

b.

c.

d.

BC � 19.8

BC � 9.9 � 9.9

BC � CF � FB

HE � 5

23 HE �103

HE �13 HE �

103

HE �13 �HE � 10�

HE �13 �HE � HB�

HE �13 EB

10 � HB

100 � �HB�2

36 � 64 � �HB�2

62 � 82 � �HB�2

�HG�2 � �GB�2 � �HB�2

HC � 12

13 HC = 4

HC �23 HC � 4

HC �23 �HC � 6�

HC �23 �HC � HG�

HC �23 CG

AB.AM�ABC,BCAM

CDAD�ABC,→BD

�RST,

�ABC

m�A � m�B � m�C > 180�.m�C > 0�,m�A � m�B � 180�

�B.�A�ABC

MP ≠ QP�M � �Q,�MPQ,

TU � VSm�1 < m�2AB < CB

�100 � 200 � 300�

�100 � 200�.



15.

P � 31

P � 9 � 12 � 10

P � ST � TU � SU

TU � 12

TU �12 � 24

TU �12 PQ

ST � 9

ST �12 � 18

ST �12 RQ

18 � RQ

9 � 9 � RQ

RU � UQ � RQ

RU � UQ � 9




Chapter 5 continued

7. Point H is the centroid of the triangle.

8. is a(n) median, perpendicular bisector, altitude, andangle bisector of

9. and by the Midsegment Theorem.

10. because the side opposite islonger than the side opposite

11. To locate the pool so that its center is equidistant fromthe sidewalks, find the incenter of the triangle by con-structing angle bisectors of two angles of the triangle andlocating the point of intersection of the bisectors. Thispoint will be equidistant from each sidewalk.

12. The converse of the Hinge Theorem guarantees that theangles between the legs get larger as the legs are spreadapart.

13. The maximum distance between the end of two legs is 10 feet because the length of the third side of the trianglemust be less than the sum of the lengths of the other twosides.

14.

If then is longer than because two sides of one triangle are congruent to twosides in another triangle and the measure of the includedangle of one triangle is larger than the measure of theincluded angle of the other triangle (Hinge Thm.).

15.

Statements Reasons

1. 1. Given

2. 2. Segment Addition Post.


4. 4. Triangle Inequality Theorem


16. Assume Then because iftwo angles of a triangle are congruent, then the sidesopposite them are congruent. So by the defini-tion of congruent segments. But this contradicts the givenstatement that Therefore, the assumption mustbe false. So m�D � m�ABC.

AD � AB.

AD � AB

AD � ABm�D � m�ABC.

BE < AE

BE < BC � CE

BC � CE � AE

AC � CE � AE

AC � BC

BCACm�AOC > m�BOC,

A BC

O

�ACB.�BACm�BAC > m�ACB

EF � ABEF �12 AB

�ABC.CG

Chapter 5 Standarized Test (pp. 314–315)

1. 2. D 3. B

B

4. The midpoint of is

The centroid is two thirds of the distance from each vertex to the midpoint of the opposite side.

But

So

The coordinates of C are or

E

5. so

so

So

C

6. A

7. The length of the side has to be greater than or12 inches and less than inches. A

8. A

9. because the sum of the measures of the acuteangles of a right triangle is 90�.x � y � 90

28 � 16 � 4428 � 16

P � 48

P � 12 � 16 � 20

P � NP � PL � NL

20 � NL

400 � �NL�2

144 � 256 � �NL�2

122 � 162 � �NL�2

�NP�2 � �PL�2 � �NL�2

NP � 12MK � 12,

PL � 16KP � 16,

��7, �3�.��7, �11 � 8�

CH � 8

CH �23 � 12

CH �23 MH

MH � 1 � ��11� � 12.

CH �23 MH

� ��7, 1�.

� ��142

, 22�

M � ��12 � ��2�2

, 1 � 1

2 �FG

y � 4

�12 y � �2

92 y � 5y � 2

92 y � 4 � 5y � 6

x � 9

103 x � 30

4x �23 x � 30

4x � 9 �23 x � 21


Chapter 5 continued

10. because the side opposite is longer than theside opposite

11. If then But so C

12. The location of the point of intersection of the perpendic-ular bisectors is the midpoint of because is aright triangle.

13. Let M be the midpoint of

Let N be the midpoint of

Let P be the midpoint of

The slope of

An equation of is

The slope of

An equation of is

The slope of

An equation of is

The centroid is the point of intersection of and

The coordinates of the centroid are �10, 2�.

y � 2

y �15

� 10

y �15

x

x � 10

�95

x � �18

15

x � 2x � 18

y � 2x � 18

y �15

x

↔CM.

↔AN,

↔PB,

y � �14

x �92

y � 0 � �14

�x � 18�↔CM

� �14

�3

�12↔CM � m3 �

3 � 06 � 18

y � 2x � 18

y � 0 � 2�x � 9�↔BP

↔BP � m2 �

6 � 012 � 9

�63

� 2

y �15

x

y � 0 �15

�x � 0�↔AN

�15

�3

15↔AN � m1 �

3 � 015 � 0

� �9, 0�� 182

, 02�P � �0 � 18

2,

0 � 02 �

AC.

� �15, 3�� 302

, 62�N � �12 � 18

2,

6 � 02 �

BC.

� �6, 3�� 122

, 62�M � �0 � 12

2,

0 � 62 �

AB.

�GHJGH

x > 45.x > yx � 45.x � y,

�H.�Gx� > y�

14. The slope of

The slope of the line perpendicular to is undefined.

So the line perpendicular to that passes through B isthe line

The slope of

The slope of the line perpendicular to is because

An equation of the line perpendicular to and passingthrough is

The slope of

The slope of the line perpendicular to is 1 because

An equation of the line parallel to and passingthrough is

The orthocenter is the point of intersection of

and

The coordinates of the orthocenter are

15. a. The coordinates of the centroid are The coor-dinates of the orthocenter are Find the equa-tion of the line passing through the centroid and the orthocenter then show that the cir-cumcenter is also on the line.

An equation of the line passing through the centroidand the orthocenter is

Substitute the coordinates of the circumcenter intothis equation.

—CONTINUED—

y � 5x � 48

y � 2 � 5x � 50

y � 2 � 5�x � 10�

slope � m �12 � 2

12 � 10�

102

� 5

�9, �3��12, 12�,

�10, 2��12, 12�.

�10, 2�.

�12, 12�.

y � 12

y � �24 � 36

y � �2 � 12 � 36

y � �2x � 36

x � 12

↔CP.

↔BM,

↔AN,

y � x.

y � 0 � 1�x � 0�A�0, 0�

↔BC

1 � ��1� � �1.

↔BC

�6

�6� �1.

↔BC � m3 �

6 � 012 � 18

y � �2x � 36.

y � 0 � �2�x � 18�C�18, 0�

↔AB

12

� ��2� � �1.

�2↔AB

�6

12�

12

↔AB � m2 �

6 � 012 � 0

x � 12.

↔AC

↔AC

↔AC � m1 �

0 � 018 � 0

�0

18� 0






Chapter 5 continued

15. —CONTINUED—

Since is true, the circumcenter is on thesame line as the centroid and the orthocenter.Therefore, they are all collinear.

b. The distance from the circumcenter C to the centroidD is CD.

The distance from the circumcenter C to the orthocenter P is CP.

So the distance from the circumcenter to the centroidis one third the distance from the circumcenter to theorthocenter.

Project Chapters 4–5 (pp. 316–317)

Investigation

1. The lines are medians because they are the lines that con-tain the line segments whose endpoints are a vertex of thetriangle and the midpoint of the opposite side.

2. The balancing point of the triangle is the centroid becauseit is the point of intersection of the medians.

3. Answers will vary.

4. Conjecture: The balancing point of a square, a rectangle, a parallelogram, or a rhombus is the point ofintersection of its diagonals.

5. Answers will vary.

Sample answer: I tested the conjecture by making moreexample shapes of each kind. The results were the sameeach time. The balancing point was the point of intersec-tion of the diagonals.

�26 � �26

�26 �13

�3�26�

CD �13

CP

� 3�26

� �9�26

� �234

� �9 � 225

� �32 � 152

CP � ��12 � 9�2 � �12 � ��3��2

� �26

� �1 � 25

� �12 � 52

CD � ��10 � 9�2 � �2 � ��3��2

�3 � �3

�3 � �3

�3 � 45 � 48

�3 � 5 � 9 � 48

Present Your Results

Projects may vary.

Extension

The conjecture does not work for all four–sided shapes. Thefollowing is an example for which it was not true.


Documents

CHAPTER 5 - Mr Jaffemrjaffe.net/GeomRef/unit5solutions.pdfAnswers may vary. Sample answer: At position B, the goalkeeper is closer to both sides of the imaginary triangle formed by