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Answers to Exercises13. 17

14. 15

15. the twelfth century

16. The angles of the trapezoid measure 67.5° and

112.5°; 67.5° is half the value of each angle of a regular

octagon, and 112.5° is half the value of 360° � 135°.

17. Answers will vary; see the answer for

Developing Proof on page 259. Using the Triangle

Sum Conjecture, a � b � j � c � d � k � e � f �l � g � h � i � 4(180°), or 720°. The four angles in

the center sum to 360°, so j � k � l � i � 360°.

Subtract to get a � b � c � d � e � f � g �h � 360°.

18. x � 120°

19. The segments joining the opposite midpoints

of a quadrilateral always bisect each other.

20. D

21. Counterexample: The base angles of an

isosceles right triangle measure 45°; thus they are

complementary.

67.5�

135�

ANSWERS TO EXERCISES 61

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CHAPTER 5 • CHAPTER CHAPTER 5 • CHAPTER

LESSON 5.1

1. See table below.

2. See table below.

3. 122°

4. 136°

5. 108°; 36°

6. 108°; 106°

7. 105°; 82°

8. 120°; 38°

9. The sum of the interior angle measures of the

quadrilateral is 358°. It should be 360°.

10. The measures of the interior angles shown sum

to 554°. However, the figure is a pentagon, so the

measures of its interior angles should sum to 540°.

11. 18

12. a � 116°, b � 64°, c � 90°, d � 82°, e � 99°,

f � 88°, g � 150°, h � 56°, j � 106°, k � 74°,

m � 136°, n � 118°, p � 99°; Possible explanation:

The sum of the angles of a quadrilateral is 360°, so

a � b � 98° � d � 360°. Substituting 116° for a

and 64° for b gives d � 82°. Using the larger

quadrilateral, e � p � 64° � 98° � 360°.

Substituting e for p, the equation simplifies to

2e � 198°, so e � 99°. The sum of the angles of a

pentagon is 540°, so e � p � f � 138° � 116° �540°. Substituting 99° for e and p gives f � 88°.

5

Number of sides of polygon 7 8 9 10 11 20 55 100

Sum of measures of angles 900° 1080° 1260° 1440° 1620° 3240° 9540° 17640°

Number of sides of 5 6 7 8 9 10 12 16 100equiangular polygon

Measure of each angle 108° 120° 128�47

�° 135° 140° 144° 150° 157�12

�° 176�25

�°of equiangular polygon

1. (Lesson 5.1)

2. (Lesson 5.1)

62 ANSWERS TO EXERCISES

LESSON 5.2

1. 360°

2. 72°; 60°

3. 15

4. 43

5. a � 108°

6. b � 45�13

�°

7. c � 51�37

�°, d � 115�57

�°

8. e � 72°, f � 45°, g � 117°, h � 126°

9. a � 30°, b � 30°, c � 106°, d � 136°

10. a � 162°, b � 83°, c � 102°, d � 39°, e � 129°,

f � 51°, g � 55°, h � 97°, k � 83°

11. See flowchart below.

12. Yes. The maximum is three. The minimum is

zero. A polygon might have no acute interior

angles.

13. Answers will vary. Possible proof using

the diagram on the left: a � b � i � 180°, c � d �h � 180°, and e � f � g � 180° by the Triangle

Sum Conjecture. a � b � c � d � e � f � g �h � i � 540° by the addition property of equality.

Therefore, the sum of the measures of the angles of

a pentagon is 540°. To use the other diagram,

students must remember to subtract 360° to

account for angle measures k through o.

14. regular polygons: equilateral triangle and

regular dodecagon; angle measures: 60°, 150°,

and 150°

15. regular polygons: square, regular hexagon, and

regular dodecagon; angle measures: 90°, 120°,

and 150°

16. Yes. �RAC � �DCA by SAS. AD�� CR� by

CPCTC.

17. Yes. �DAT � �RAT by SSS. �D � �R by

CPCTC.

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1 a � b � 180�

2 c � d � 180�

4 a + b + c + d + e + f =

3 e � f � 180�Subtraction propertyof equality

Addition property of equality

6 b + d + f = �

�?

�?

�?

�?

�? �

�? �

�?

5 a + c + e =

Linear Pair Conjecture



540°

180°

360°

Triangle Sum Conjecture

11. (Lesson 5.2)

LESSON 5.3

1. 64 cm 2. 21°; 146° 3. 52°; 128°

4. 15 cm 5. 72°; 61° 6. 99°; 38 cm

7. w � 120°, x � 45°, y � 30°

8. w � 1.6 cm, x � 48°, y � 42°


10. Answers may vary. This proof uses the Kite

Angle Bisector Conjecture.

Given: Kite BENY with vertex angles �B and �N

Show: Diagonal BN� is the perpendicular bisector

of diagonal YE�.

From the definition of kite, BE�� BY�. From the

Kite Angle Bisector Conjecture, �1 � �2. BX��BX� because they are the same segment. By SAS,

�BXY � �BXE. So by CPCTC, XY�� XE�.

Because �YXB and �EXB form a linear pair, they

are supplementary, so m�YXB � m�EXB �180°. By CPCTC, �YXB � �EXB, or m�YXB �m�EXB, so by substitution, 2m�YXB � 180°, or

m�YXB � 90°. So m�YXB � m�EXB � 90°.

Because XY�� XE� and �YXB and �EXB are right

angles, BN� is the perpendicular bisector of YE�.

11. possible answer: �E � �I

12. possible answer:

The other base is ZI�. �Q and �U are a pair of base

angles. �Z and �I are a pair of base angles.

Z I

Q U

E

KT

I

B N

Y

E

21 X


OW�� is the other base. �S and �H are a pair of

base angles. �O and �W are a pair of base angles.

SW�� HO��.

14. Only one kite is possible

because three sides determine

a triangle.

15.

16. infinitely many, possible construction:

17. 80°, 80°, 100°, 100°

18. Because ABCD is an isosceles trapezoid, �A

� �B. �AGF � �BHE by SAA. Thus, AG�� BH��by CPCTC.

19. a � 80°, b � 20°, c � 160°, d � 20°, e � 80°,

f � 80°, g � 110°, h � 70°, m � 110°,n � 100°;

Possible explanation: Because d forms a linear pair

with e and its congruent adjacent angle,d � 2e �180°.Substituting d � 20° gives 2e � 160°,so e � 80°.

Using theVerticalAngles Conjecture and d � 20°, the

unlabeled angle in the small right triangle measures

20°, which means h � 70°. Because g and h are a

linear pair, they are supplementary, so g � 110°.

B O

NE

S H

WI

F

E

BN

W O

S H


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9. (Lesson 5.3)

�? and �?

6�1 � and5

�3 ��? Congruence

shortcutDefinition ofangle bisector

4

�?

Given

1 BE � BY

Given

2 EN � YN

Same segment

3�? � �?

�BEN � �? �

?

�?

BN�� BN�

�BYN �2 BN� bisects �B, BN� bisects �N

�4

CPCTCSSS


LESSON 5.4

1. three; one 2. 28

3. 60°; 140° 4. 65°

5. 23 6. 129°; 73°; 42 cm

7. 35 8. See flowchart below.

9. Parallelogram. Draw a diagonal of the original

quadrilateral. The diagonal forms two tri-angles.

Each of the two midsegments is parallel to the

diagonal, and thus the midsegments are parallel to

each other. Now draw the other diagonal of the

original quadrilateral. By the same reasoning, the

second pair of midsegments is parallel. Therefore,

the quadrilateral formed by joining the midpoints is

a parallelogram.

10. The length of the edge of the top base

measures 30 m. We know this by the Trapezoid

Midsegment Conjecture.

11. Ladie drives a stake into the ground to create a

triangle for which the trees are the other two

vertices. She finds the midpoint from the stake to

each tree. The distance between these midpoints is

half the distance between the trees.

12. Explanations will vary.

80

40 60

60 cm

Cabin

13. If a quadrilateral is a kite, then exactly one

diagonal bisects a pair of opposite angles. Both the

original and converse statements are true.

14. a � 54°, b � 72°, c � 108°, d � 72°, e � 162°,

f � 18°, g � 81°, h � 49.5°, i � 130.5°, k � 49.5°,

m � 162°, n � 99°; Possible explanation: The third

angle of the triangle containing f and g measures

81°, so using the Vertical Angles Conjecture, the

vertex angle of the triangle containing h also

measures 81°. Subtract 81° from 180° and divide by

2 to get h � 49.5°. The other base angle must also

measure 49.5°. By the Corresponding Angles

Conjecture, k � 49.5°.

15. (3, 8)

16. (0, �8)

17. coordinates: E(2, 3.5), Z(6, 5); the slope of

EZ�� 38

�, and the slope of YT�� 38

�

18.

There is only one kite, but more than one way to

construct it.

K

R

NF

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�FOA withmidsegment LN

Given

LN � OA3

4

1

Triangle MidsegmentConjecture

�IOA withmidsegment RD

Given

2

5

Two lines parallel to the same line are parallel

�

?�

?

�

?

OA� � RD��

LN� � RD��Triangle Midsegment Conjecture

8. (Lesson 5.4)

LESSON 5.5

1. 34 cm; 27 cm

2. 132°; 48°

3. 16 in.; 14 in.

4. 63 m

5. 80

6. 63°; 78°

7.

8.

9. � Vh

� Vw

D

P

P

O

R

R

T S

AL

10.

11. (b � a, c)



14. The parallelogram linkage is used for the

sewing box so that the drawers remain parallel to

each other (and to the ground) so that the contents

cannot fall out.

15. a � 135°, b � 90°

16. a � 120°, b � 108°, c � 90°, d � 42°, e � 69°

a

a

b

b

c

c

d

d

� Vb

� Vc


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ET � NT

EA � LN

2

LEAN is a parallelogram 7

CPCTC

1

AIA Conjecture

3 �AEN � �LNE

�?

�?

4

AIA Conjecture

�? 6

ASA

�?

8

CPCTC

�?

9

Definition of segment bisector

�?

5

Opposite sides congruent

�?

Given �EAL � �NLA �AET � �LNT

AE�� LN�AT�� LT�

EN� and LA� bisect

each other

Definition of

parallelogram

13. (Lesson 5.5)


17. x � 104°, y � 98°. The quadrilaterals on the

left and right sides are kites. Nonvertex angles are

congruent. The quadrilateral at the bottom is an

isosceles trapezoid. Base angles are congruent, and

consecutive angles between the bases are

supplementary.

18. a � 84°, b � 96°

19. No. The congruent angles and side do not

correspond.

20.

21. Parallelogram. Because the triangles are

congruent by SAS, �1 � �2. So, the segments are

parallel. Use a similar argument to show that the

other pair of opposite sides is parallel.

22. Kite or dart. Radii of the same circle are

congruent. If the circles have equal radii, a rhombus

is formed.

1

2

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USING YOUR ALGEBRA SKILLS 5

1.

2.

3. y

x

(2, 9)

(0, 6)

y

x

(3, 8)

(0, 4)

y

x(0, 1)

(1, –1)


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4. y � �x � 2

5. y � ��163� x � �

71

43�

6. y � x � 1

7. y � �3x � 5

8. y � �25

� x � �85

�

9. y � 80 � 4x

10. y � �3x � 26

11. y � ��14

� x � 3

12. y � �65

� x

13. y � x � 1

14. y � � �29

� x � �493�


LESSON 5.6

1. Sometimes true; it is true only if the

parallelogram is a rectangle.

2. Always true; by the definition of rectangle, all

the angles are congruent. By the Quadrilateral Sum

Conjecture and division, they all measure 90°, so

any two angles in a rectangle, including consecutive

angles, are supplementary.

3. Always true by the Rectangle Diagonals

Conjecture.

4. Sometimes true; it is true only if the rectangle is

a square.

5. Always true by the Square Diagonals Conjecture.

6. Sometimes true; it is true only if the rhombus is

equiangular.

7. Always true; all squares fit the definition of

rectangle.

8. Always true; all sides of a square are congruent

and form right angles, so the sides become the legs

of the isosceles right triangle and the diagonal is

the hypotenuse.

9. Always true by the Parallelogram Opposite

Angles Conjecture.

10. Sometimes true; it is true only if the

parallelogram is a rectangle. Consecutive angles of

a parallelogram are always supplementary, but are

congruent only if they are right angles.

11. 20

12. 37°

13. 45°, 90°

14. DIAM is not a rhombus because it is not

equilateral and opposite sides are not parallel.

15. BOXY is a rectangle because its adjacent sides

are perpendicular.

16. Yes. TILE is a rhombus, and a rhombus is a

parallelogram.

false

true

falsetrue

false

true

17.

18. Constructions will vary.

19. one possible construction:

20. Converse: If the diagonals of a quadrilateral

are congruent and bisect each other, then the

quadrilateral is a rectangle.

Given: Quadrilateral ABCD with diagonals

AC�� BD�. AC� and BD� bisect each other

Show: ABCD is a rectangle

Because the diagonals are congruent and bisect

each other, AE�� BE�� DE�� EC�. Using the

Vertical Angles Conjecture, �AEB � �CED and

�BEC � �DEA. So �AEB � �CED and �AED

� �CEB by SAS. Using the Isosceles Triangle

Conjecture and CPCTC, �1 � �2 � �5 � �6,

and �3 � �4 � �7 � �8. Each angle of the

quadrilateral is the sum of two angles, one from

each set, so for example, m�DAB � m�1 � m�8.

By the addition property of equality, m�1 �m�8 � m�2 � m�3 � m�5 � m�4 � m�6 �m�7. So m�DAB � m�ABC � m�BCD �m�CDA. So the quadrilateral is equiangular. Using

�1 � �5 and the Converse of AIA, AB� � CD�.

Using �3 � �7 and the Converse of AIA,

BC� � AD��. Therefore ABCD is an equiangular

parallelogram, so it is a rectangle.

21. If the diagonals are congruent and bisect each

other, then the room is rectangular (converse of the

Rectangle Diagonals Conjecture).

A

E

B

CD

81 2

3

456

7

I E

P S

A

B E

K A

B

K

E

E V

L O

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22. The platform stays parallel to the floor

because opposite sides of a rectangle are parallel

(a rectangle is a parallelogram).

23. The crosswalks form a parallelogram: The

streets are of different widths, so the crosswalks are

of different lengths. The streets would have to meet

at right angles for the crosswalks to form a rectangle.

The corners would have to be right angles and the

streets would also have to be of the same width for

the crosswalk to form a square.

24. Place one side of the ruler along one side of

the angle. Draw a line with the other side of the

ruler. Repeat with the other side of the angle. Draw

a line from the vertex of the angle to the point

where the two lines meet.

25. Rotate your ruler so that each endpoint of the

segment barely shows on each side of the ruler.

Draw the parallel lines on each side of your ruler.

Now rotate your ruler the other way and repeat the

process to get a rhombus. The original segment is

one diagonal of the rhombus. The other diagonal

will be the perpendicular bisector of the original

segment.

26. See f lowchart below.

27. Yes, it is true for rectangles.

Given: �1 � �2 � �3 � �4

Show: ABCD is a rectangle

By the Quadrilateral Sum Conjecture, m�1 �m�2 � m�3 � m�4 � 360°. It is given that all

four angles are congruent, so each angle measures


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90°. Because �4 and �5 form a linear pair,

m�4 � m�5 � 180°. Substitute 90° for m�4 and

solve to get m�5 � 90°. By definition of congruent

angles, �5 � �3, and �5 and �3 are alternate

interior angles, so AD�� BC� by the Converse of the

Parallel Lines Conjecture. Similarly, �1 and �5 are

congruent corresponding angles, so AB� � CD� by the

Converse of the Parallel Lines Conjecture. Thus,

ABCD is a parallelogram by the definition of

parallelogram. Because it is an equiangular

parallelogram, ABCD is a rectangle.

28. a � 54°, b � 36°, c � 72°, d � 108°, e � 36°,

f � 144°, g � 18°, h � 48°, j � 48°, k � 84°

29. possible answers: (1, 0); (0, 1); (�1, 2); (�2, 3)

30. y � �89

�x � �896� or 8x � 9y � �86

31. y � ��170�x � �

152� or 7x � 10y � �24

32. velocity � 1.8 mi/h; angle of path � 106.1°

clockwise from the north

2 mi/h

1.5 mi/h

60�

QU � AD QD � AU

QU � AD

1

4

Given

Given

DU � DU

3

2

Same segment

�QUD � �ADU 5

�1 � �2

�3 � �4

Converse of the Parallel Lines Conjecture

6 QUAD is a parallelogram

Definition of parallelogram

7

QUAD is a rhombus

9 QU � UA � AD � DQ

8

�?

�?

�?

�?

QD � AU

Given SSS

CPCTC

Definition of

rhombus

26. (Lesson 5.6)


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LESSON 5.7

1. See flowchart below. 2. See flowchart below.


4.Given

1 SP � OA

Converse of AIA Conjecture

7 PA � SO

Definition of parallelogram

8 SOAP is a parallelogram

Given

2 SP � OA

SAS

5�SOP � �APO

Same segment

4

PO � PO

AIA Conjecture

3�1 � �2

CPCTC

6�3 � �4

5. parallelogram

6. sample flowchart proof:

Opposite sides ofrectangle are congruent

Definition of rectangle

SAS

Same segment

4

CPCTC

5

31 2 �YOG � �OYI YO � OY IY � GO

YG � IO

�YOG � �OYI

�3 � �4

�1 � �2

SO � KA

AIADefinition ofparallelogram

1�SOA �7

2

Given

3

4

5SOAK is aparallelogram

6

�?

�?

OA � �?

SA � �?

�?

�?

�?

65

GivenConjecture proved in Exercise 1

21�BAT � �THB

CPCTC

3�BAT � �THB

�BAH = � �HBA � �

�?

�? �

?

�?

�?

Parallelogram BATH with diagonal BT

4 Parallelogram BATH with diagonal HA

9

8

65

7

�1 � � �?

4� �

? � � �?

WATR is aparallelogram

�4 � � �?

�? � �?

�? � �?

�?

�? �

?

�?

�?

�?

�?

�?

�?

�? � �?

1

�? � �?

2

�? � �?

3

�AKS

ASA

Same segment

AIA

SA�

Definition of

parallelogram

SK�

CPCTC

�ATH�THA

Given Conjecture proved

in Exercise 1

WA�� RT�

WR�� AT� �WRT � �TAW

WT�� WT��

Given�2 RT� � WA��

RW�� TA�

CPCTC Converse of the

Parallel Lines Conjecture

Definition of

parallelogram

CPCTC

�3Given SSS

Same segment

by Converse of the Parallel

Lines Conjecture

1. (Lesson 5.7)

2. (Lesson 5.7)

3. (Lesson 5.7)


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7.

8. Because AR� is parallel to ZT�, corresponding

�3 and �2 are congruent. Opposite sides of

parallelogram ZART are congruent so AR � TZ.

Because the trapezoid is isosceles, AR � PT, and

substituting gives ZT � PT making �PTZ

isosceles and �1 and �2 congruent. By

substitution, �1 and �3 are congruent.

Given Parallelogram Opposite Sides Conjecture

Parallelogram Opposite Sides Conjecture

SSS

5

CPCTC

6

4 2 �� AR � EB 3 �� BR � EA

Given

1 BEAR is a parallelogram

�� RE � AB

Definition of rectangle

7 BEAR is a rectangle

�EBR � �ARB � �RAE � �BEA

�EBR � �ARB � �RAE � �BEA

9. See sample flowchart below.

10. If the fabric is pulled along the warp or the weft,

nothing happens.However, if the fabric is pulled

along the bias, it can be stretched because the

rectangles are pulled into parallelograms.

11. 30° angles in 4-pointed star, 30° angles in

6-pointed star; yes

12. He should measure the alternate interior

angles to see whether they’re congruent.If they are,

the edges are parallel.

13. ES��: y � �2x � 3; QI��: y � �12

�x � 2

14. (12, 7)

15. �13

�

16.

6 inches

9

12 3

6

9. (Lesson 5.7)

GivenSame segment

2 �� GR � TH

Given

3 5 �RGT � �HTG

SAS CPCTC

6

4�RGT ��HTG

Isosceles TrapezoidConjecture

�� GH � TR 1 Isosceles trapezoid

GTHR�� GT � GT


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CHAPTER 5 REVIEW

1. 360° divided by the number of sides

2. Sample answers: Using an interior angle, set the

interior angle measure formula equal to the angle

and solve for n. Using an exterior angle, divide into

360°. Or find the interior angle measure and go

from there.

3. Trace both sides of the ruler as shown at right.

4. Make a rhombus using the double-edged

straightedge, and draw a diagonal connecting the

angle vertex to the opposite vertex.

5. Sample answer: Measure the diagonals with

string to see if they are congruent and bisect each

other.

6. Sample answer: Draw a third point and connect

it with each of the two points to form two sides of a

triangle. Find the midpoints of the two sides and

connect them to construct the midsegment. The

distance between the two points is twice the length

of the midsegment.

7. x � 10°, y � 40° 8. x � 60 cm

9. a � 116°, c � 64° 10. 100

11. x � 38 cm

12. y � 34 cm, z � 51 cm

13. See table below.

14. a � 72°, b � 108°

15. a � 120°, b � 60°, c � 60°, d � 120°, e � 60°,

f � 30°, g � 108°, m � 24°, p � 84°; Possible

explanation: Because c � 60°, the angle that forms

a linear pair with e and its congruent adjacent

angle measures 60°. So 60° � 2e � 180°, and

e � 60°. The triangle containing f has a 60° angle.

The other angle is a right angle because it forms a

linear pair with a right angle. So f � 30° by the

Triangle Sum Conjecture. Because g is an interior

angle in an equiangular pentagon, divide 540° by 5

to get g � 108°.

16. 15 stones

17. (1, 0)

18. When the swing is motionless, the seat, the bar

at the top, and the chains form a rectangle. When

you swing left to right, the rectangle changes to a

parallelogram. The opposite sides stay equal in

length, so they stay parallel. The seat and the bar

at the top are also parallel to the ground.

19. a = 60°, b = 120°

IsoscelesKite trapezoid Parallelogram Rhombus Rectangle Square

Opposite sidesare parallel No No Yes Yes Yes Yes

Opposite sides are congruent No No Yes Yes Yes Yes

Opposite anglesare congruent No No Yes Yes Yes Yes

Diagonals bisecteach other No No Yes Yes Yes Yes

Diagonals are perpendicular Yes No No Yes No Yes

Diagonals are congruent No Yes No No Yes Yes

Exactly one lineYesof symmetry Yes No No No No

Exactly two linesof symmetry No No No Yes Yes No

13. (Chapter 5 Review)


An

swe

rs to E

xercise

s

20.

Speed: � 901.4 km/h. Direction: slightly west of

north. Figure is approximate.

21.

22. possible answers:

23.

PL z

yN E

x

FL

Y

R

FLx

Y

R

x12

E

RS

Q

x12

900 km/h

50 km/h

Resultantvector

24. possible answers:

25. 20 sides

26. 12 cm



Given: Parallelogram ABCD

Show: AB�� CD� and AD�� CB�See flowchart below.

12

43

A B

D C

Fx

z

R

D

Y

Fx

z

R

D

Y

CPCTC

8 AB � CD

Definition ofparallelogram

2

1

AD �� CB

Definition ofparallelogram

3 AB �� CD

ABCD is a parallelogram

Same segment

5 BD � BD

AIA

4 �1 � �3

AIA

6 �2 � �4

CPCTC

9 AD � CB

ASA

7 �ABD � �CDB

Given

� �? � � �

? 3 5

2 DE � �?

NE � �?

4 DN � �?

DENI isa rhombus

1

6 �1 � � �?

�3 � � �?

7DN bisects�IDE and �INE

�?

�?

�?

�?

�?

�?

�?

Given

Definition of

rhombus

Same segment

SSS

CPCTC Definition of

angle bisector

DI�

NI� �DEN � �DIN �2, �4

DN��

Definition of rhombus



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