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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Exe
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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
Linear Pair Conjecture
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°
9. See flowchart below.
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
13. possible answer:
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
ANSWERS TO EXERCISES 63
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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
64 ANSWERS TO EXERCISES
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)
12. possible answer:
13. See flowchart below.
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
ANSWERS TO EXERCISES 65
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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)
66 ANSWERS TO EXERCISES
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)
ANSWERS TO EXERCISES 67
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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�
68 ANSWERS TO EXERCISES
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
ANSWERS TO EXERCISES 69
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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)
70 ANSWERS TO EXERCISES
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LESSON 5.7
1. See flowchart below. 2. See flowchart below.
3. 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)
ANSWERS TO EXERCISES 71
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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
72 ANSWERS TO EXERCISES
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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)
ANSWERS TO EXERCISES 73
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
27. See flowchart below.
28. possible answer:
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
27. (Chapter 5 Review)
28. (Chapter 5 Review)