inetteacher.cominetteacher.com/upload1/104410/docs/dg unit 13 solutions.pdf · symmetric: if figure...

Symmetric: If Figure A is congruent to Figure B,then Figure B is congruent to Figure A.

If �XYZ � �LMN, then �LMN � �XYZ.

4. Reflexive property of equality; reflexive property ofcongruence

5. Transitive property of congruence

6. Subtraction property of equality

7. Division property of equality

8. Distributive; Subtraction; Addition; Division

9. Given; Addition property of equality; Multiplicationproperty of equality; Commutative property ofaddition

10. True; definition of midpoint

11. True; Midpoint Postulate

12. True; definition of angle bisector

13. True; Angle Bisector Postulate

14. False; Line Intersection Postulate

15. False; Line Postulate

16. True; Angle Addition Postulate

17. True; Segment Addition Postulate

18. ● That all men are created equal.

● That they are endowed by their creator withcertain inalienable rights, that among these arelife, liberty, and the pursuit of happiness.

● That to secure these rights, governments are insti-tuted among men, deriving their just powers fromthe consent of the governed.

● That whenever any form of government becomesdestructive to these ends, it is the right of thepeople to alter or to abolish it, and to institutenew government, laying its foundations on suchprinciples and organizing its powers in such formas to them shall seem most likely to effect theirsafety and happiness.

19. 1. AO�� and BO�� are radii; 3. �AOB is isosceles, Defi-nition of isosceles

20. 1. �1 � �2, Given; 2. m � n; 3. �3 � �4,Corresponding Angles Postulate

21. 1. Given; 2. AD�� BC�, Given; 4. �ABC � �BAD,SSS Congruence Postulate; 5. �D � �C

Y Z

XIf ,

then .

M N

L

M N

L

Y Z

X

�

�

6. Both x and y can be related to the circumference ofthe cone’s base, which is the length of the sector’sarc. Label the radius of the arc L, the slant height ofthe cone, so the arc’s length is �36

x0°� � 2�L �

�L��18x0°��. The radius of the cone, then, is this

circumference divided by 2�, or L��36x0°��. The sine of

�2y

� is this radius over L, or �36x0°�, so y � 2 sin�1��36

x0°��.

Or use the Law of Cosines with c � L � �18x0°�, a � L,

and b � L to arrive at the equivalent answer:y � cos�1�1 � �.

7. Possible answer:

If m�A � 20�, (sin A)2 � (cosA)2 � (0.3420)2

� (0.9397)2 � 0.1170 � 0.8830 � 1.

If m�A � 40�, (sin A)2 � (cos A)2 � (0.6428)2

� (0.7660)2 � 0.4132 � 0.5868 � 1.

If m�A � 75�, (sin A)2 � (cos A)2 � (0.9659)2

� (0.2588)2 � 0.9330 � 0.0670 � 1.

Conjecture: For any angle A, (sin A)2 � (cos A)2 � 1.

From the diagram, sin A � �ac� and cosA � �

bc�.

Paragraph Proof: Because sin A � �ac� and cos A � �

bc�,

(sinA)2 � (cos A)2 � , and, by the

Pythagorean Theorem, a 2 � b 2 � c 2. Therefore,

(sin A)2 � (cos A)2 � �a2 �

c2b2

� � �cc

2

2� � 1.

CHAPTER 13

LESSON 13.1

EXERCISES

1. A postulate is a statement accepted as true withoutproof. A theorem is deduced from other theoremsor postulates.

2. Subtraction: Equals minus equals are equal. Multi-plication: Equals times equals are equal. Division:Equals divided by nonzero equals are equal.

3. Reflexive: Any figure is congruent to itself.

�ABC � �ABC

Transitive: If Figure A is congruent to Figure B andFigure B is congruent to Figure C, then Figure A iscongruent to Figure C.

If PQ� � RS� and RS� � XY�, then PQ� � XY�.

P Q R SIf , .

X Y P Qthen

X Y� ��

C

AB

a2 � b2�c2

x2�64,800

Discovering Geometry Solutions Manual CHAPTER 13 279

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©2008 Kendall Hunt Publishing

Surface area: cylinder, cone, sphere

Cone: l (slant height) � �122 ��62� � �180� �

�36 • 5� � 6�5� cm

S � �r2 � �rl � �(6)2 � �(6)�6�5�� 36� � 36�5�� 366.0 cm2

Sphere: S � 4�r 2 � 4�(5)2 � 100� � 314.2 cm2

Cylinder: S � 2�r2 � 2�rH � 2�(5.5)2

� 2�(5.5)(5.5) � 121� � 380.1 cm2

Length of longest rod that will fit inside: cone,cylinder, sphere

Cone: length of longest rod � slant height �6�5� cm � 13.4 cm

Sphere: length of longest rod � diameter � 10 cm

Cylinder: length of longest rod � hypotenuse ofright triangle with one leg of length 11 cm (diam-eter of base) and other leg of length 5.5 cm (heightof cylinder) � �112 ��5.52� � �151.25� � 12.3 cm

28. 30 ft. Let h represent the height at which the guywires cross. Add labels to the figure for reference.

Use similar triangles. From the two right trianglesthat share the angle with measure �, �

hx� � �

46

84� � 0.75,

so h � 0.75x. From the two triangles that share theangle with measure �, �64

h� x� � �

86

04� � 1.25, so h �

1.25(64 � x). Therefore, 0.75x � 1.25(64 � x). Solvethis equation for x.

0.75x � 1.25(64 � x)

0.75x � 80 � 1.25x

2x � 80

x � 40

h � x(tan �) � 40(0.75) � 30 ft

You may want to challenge students to do thisproblem without knowing the distance between thetwo towers (the height of the crossing is independentof this distance). Here is a possible solution.

Add a vertical segment 80 ft tall, going through thepoint of intersection. Also add a horizontal segmentfrom the top of the 80-ft tower to the top of thenew vertical segment, and another from the baseof the 48-ft tower to the base of the new vertical

h

x y

80 ft

��

48 ft

64 ft

22. 2. AB� � CB�; 3. �ABD � �CBD, Definition ofangle bisector; 4. Reflexive property of congruence;5. �BAD � �BCD, SAS Congruence Postulate;6. �A � �C, CPCTC

23. Add the two integers: (2n � 1) � (2m � 1) � 2n �2m � 2 � 2(n � m � 1), which is always even.

24. Multiply the two integers: (2n � 1)(2m � 1) �4nm � 2n � 2m � 1 � 4nm � 2n � 2m � 2 � 1 �2(2nm � n � m � 1) � 1, which is always odd.

25. Let n be any integer. Then the next two consecutiveintegers are n � 1 and n � 2. The sum of thesethree integers is (n) � (n � 1) � (n � 2) �n � n � 1 � n � 2. Combining like terms gives3n � 3 � 3(n � 1), which is divisible by 3.

26. � 299 m. Draw the perpendicular segment from thetop of the mountain to the ground. Labels havebeen added to the resulting figure for referencebelow. In this figure, x represents the distance fromB to the mountain peak, D.

Notice that the figure contains two right triangles.

From �ACD, tan 22° � �220h� y�. From �BCD,

tan 38° � �hy�.

From the first equation, h � (220 � y)(tan 22°), andfrom the second equation, h � y(tan 38°). Therefore,(220 � y)(tan 22°) � y(tan 38°). Solve this equation.

(220 � y)(tan 22°) � y(tan 38°)

220(tan 22°) � y(tan 22°) � y(tan 38°)

220(tan 22°) � y(tan 38°) � y(tan 22°)

220(tan 22°) � y(tan 38° � tan 22°)

y ��tan23280°t�an

t2a2n°22°�

h � y(tan 38°) � ��tan23280°t�an

t2a2n°22°��(tan 38°)

By the Pythagorean Theorem, x2 � h2 � y2, sox � �h2 � y�2� � 299 m.

27. Volume: sphere, cylinder, cone

Cone: V � �13��r2H � �

13��(6)2(12) � 144� � 452.4 cm3

Sphere: V � �43��r 3 � �

43��(5)3 � �

503

0�� 523.6 cm3

Cylinder: V � �r 2H � �(5.5)2(5.5) � 522.7 cm3

A C

D

hx

yB22°

220 m

38°

280 CHAPTER 13 Discovering Geometry Solutions Manual

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measure of that angle is 180° � y � 180° � 126° �54°. (You can also find this angle measure by AIAusing x � 54°.) The measure of the third angle ofthis triangle is 180° � 63° � 54° � 63°, so thetriangle is isosceles. The side opposite one of the 63°angles is 7.3, so the side opposite the other 63° angleis also 7.3 (Converse of the Isosceles TriangleConjecture). Finally, look at the obtuse triangle withangles of measure y and 27°. Because y � 126°, themeasure of the third angle in this triangle is 180° �126° � 27° � 27°, so this triangle is also isosceles.From the 63°-63°-54° triangle, you know that thelength of the side opposite the 27° angle on the farright is 7.3, so the length of the side opposite theother 27° angle must also be 7.3, that is, a � 7.3.

31. a. ��4�. The shaded region is a quarter-circlewith radius 1, so A � �

14��r 2 � �

14��(1)2 � �

�4�.

b. 1 � ��4�. The area of the shaded region is the

difference between the area of the square of sidelength 1 and the area of the shaded region in31a, so A � 12 � �

�4� � 1 � �

�4�.

c. ��2� � 1. By looking at the figures from 31a and31b and using the symmetry of these figures, youcan see that the area of the shaded region is thedifference between the area you found in 31a andthe area you found in 31b. Therefore,

A � ��4� � �1 � �

�4��

�4� � 1 � �

�4� � �

�2� � 1.

IMPROVING YOUR REASONING SKILLS

Because only four moves are allowed and the letter ineach place changes, each letter must change only once.There are several possibilities for each string and fororiginal creations. Here’s one of many possible sets:

1. MATH ⇒ MATE ⇒ RATE ⇒ ROTE ⇒ ROSE

2. MATH ⇒ MATE ⇒ MARE ⇒ MORE ⇒ CORE

3. MATH ⇒ MASH ⇒ MAST ⇒ MOST ⇒ HOST

4. MATH ⇒ MASH ⇒ MASS ⇒ MESS ⇒ LESS

5. MATH ⇒ LATH ⇒ LATE ⇒ LAVE ⇒ LIVE

Sample answer for original creation: MATH ⇒ PATH ⇒PITH ⇒ PITA ⇒ PICA

LESSON 13.2

EXERCISES

1. Linear Pair Postulate

2. Parallel Postulate, Angle Addition Postulate, LinearPair Postulate, CA Postulate (through the AIATheorem)

3. Parallel Postulate

4. Perpendicular Postulate

segment. Mark all congruent angles, by theCorresponding Angles Conjecture.

By the AA Similarity Conjecture, the two triangleshaving a tower as one side are similar, so �

xy� � �

48

80� �

�35�. Similarly, the two right triangles are similar, so�80

h� h� � �

xy� � �

35�. Solve this equation for h.

�80h� h� � �

35�

5h � 3(80 � h)

5h � 240 � 3h

8h � 240

h � 30

Note that we never once referred to the distancebetween the two towers. This same argument canbe used to show that if the two towers haveheights a and b, then the height of the intersectionof the guy wires is h � �a

a�b

b�, which in our caseis h � �

8800

�

. 4488

� � �3182480

� �30.

29. FG � 2�6� and DG � 2�3� because ABGF andBCDG are parallelograms. Triangle FGD is right(m�FGD � 90°) by the Converse of thePythagorean Theorem because (2�6�)2 � (2�3�)2 �62. But m�FGB � 128° by the ParallelogramConsecutive Angles Conjecture and m�DGB � 140°by the Parallelogram Opposite Angles Conjecture.So, m�FGD � 92° because the sum of the anglesaround G is 360°. So, m�FGD is both 90° and 92°.

30. x � 54°, y � 126°, a � 7.3 m. First look at the righttriangle that contains the 27° angle. The measure ofthe third angle in this triangle is 90° � 27° � 63°.This angle and the angle marked as congruent to itform a linear pair with the angle of measure x. Thus,x � 2(63°) � 180°, so x � 54°. Now look at the righttriangle in which the hypotenuse has length 7.3. Inthis triangle, the measure of the unmarked angle is90° � 54° � 36°. Next look at the triangle thatcontains the side of length 6.8. By the AIA Conjec-ture, the angle opposite this side is a right angle. Theangle whose measure is the sum of this right angleand the adjacent 36° angle forms a vertical anglewith the angle of measure y, so y � 90° � 36° �126°. Now look at the triangle in the center of thefigure that contains a 63° angle and an angle thatforms a linear pair with the angle of measure y. The

xh

y

80 ft

80 ft � h

48 ft


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8. Given: Two lines cut by a transversal to formcongruent alternate interior angles

Show: The two lines are parallel

Given: Lines �1 and �2 cut by transversal �3;�1 � �2

Show: �1 � �2

Flowchart Proof

9. Given: Two parallel lines cut by a transversal to formalternate exterior angles

Show: The alternate exte-rior angles are congruent

Given: Lines �1 and �2 cut by transversal �3; �1 � �2

Show: �1 � �2

Flowchart Proof

10. Given: Two lines cut by a transversal to formcongruent alternate exterior angles

Show: The lines are parallel

Given: Lines �1 and �2 cut by transversal �3;�1 �2

Show: �1 � �2

Flowchart Proof

�1 � �2

Converseof AIATheorem

�3 � �4

Transitiveproperty�1 � �2

Given

�1 � �3�4 � �2

VA Theorem

�1 and �2 cutby transversal �3

Given

2

4

1

3

�3

�1

�2

�1 � �2

Given

�1 � �3

VA Theorem�1 � �2

Transitiveproperty ofcongruence

�3 � �2

CA Postulate

2

1

3

�3

�1

�2

�1 � �2

Given

�2 � �3

VA Theorem

�1 � �3

Transitiveproperty ofcongruence

�1 � �2

CA Postulate

21

3

�3

�1

�2

5. Given: Two angles are both congruent andsupplementary

Show: Each angle is a right angle

(These angles may form a linear pair, as in thefigure below, but they do not have to for the proofto work.)

Given: �1 � �2 and �1 and �2 are supple-mentary

Show: �1 and �2 are right angles

Paragraph Proof: By the definition of supplemen-tary angles, m�1 � m�2 � 180°. By the definitionof congruence, m�1 � m�2. Therefore, by substitu-tion, m�1 � m�1 � 180°, or 2m�1 � 180°. By thedivision property, m�1 � 90°. Then, by the transi-tive property, m�2 � 90°. Thus, by the definition ofa right angle, �1 and �2 are both right angles.

6. Given: Two angles are congruent

Show: The supplements of the two angles arecongruent

(The figure below shows supplementary angles aslinear pairs, but they do not have to be for theproof to work.)

Given: �2 � �3, �1 and �2 are supplementary,�4 and �3 are supplementary

Show: �1 � �4

Paragraph Proof: By the definition of supplemen-tary angles, m�1 � m�2 � 180° and m�4 �m�3 � 180°. Then, by the transitive property,m�1 � m�2 � m�4 � m�3. By the definition ofcongruence, m�2 � m�3, so, by substitution,m�1 � m�2 � m�4 � m�2. By the subtractionproperty, m�1 � m�4. Therefore, by the definitionof congruence, �1 � �4.

7. Given: Two angles are right angles

Show: The two angles are congruent

Given: �1 and �2 are right angles

Show: �1 � �2

Paragraph Proof: By the definition of a rightangle, m�1 � 90° and m�2 � 90°. Then, by thetransitive property, m�1 � m�2. Therefore, by thedefinition of congruence, �1 � �2.

1 2

1 2 43

12


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Show: �1 � �3

Paragraph Proof: By the Line Postulate, constructAB�� with A on �1 and B on �2. This transversal willintersect �3 by the Parallel Postulate (otherwise therewould be 2 lines through point B parallel to �3). Bythe Interior Supplements Theorem, �1 and �2 aresupplementary, so, by the definition of supplemen-tary angles, m�1 � m�2 � 180°. By the CAPostulate, �2 � �3, so, by the definition ofcongruence, m�2 � m�3. Then, by the substitutionproperty, m�1 � m�3 � 180° and �1 and �3 aresupplementary by definition. Therefore, �1 � �3 bythe Converse of the Interior Supplements Theorem.

14. Given: Two lines in the same plane are eachperpendicular to a third line

Show: The two lines are parallel to each other

Given: Coplanar lines �1, �2, and �3 with �1 � �3and �2 � �3

Show: �1 � �2

Flowchart Proof

�1 � �2

Right Angles AreCongruent Theorem

�1 is a right angle;�2 is a right angle

Definition ofperpendicular

�1 � �3, �2 � �3

Given

�1 � �2

Converse ofAIA Theorem

21 �1

�2

�3

11. Given: Two parallel lines cut by a transversal toform interior angles on the same side of thetransversal

Show: The two angles are supplementary

Given: Lines �1 and �2 cut by transversal �3; �1 � �2

Show: �1 and �2 are supplementary

Flowchart Proof (See proof at bottom of page)

12. Given: Two lines cut by a transversal to formsupplementary interior angles on the same side ofthe transversal

Show: The two lines are parallel

Given: Lines �1 and �2 cut by transversal �3;�1 and �2 are supplementary

Show: �1 � �2

Paragraph Proof: By the definition of supplemen-tary angles, m�1 � m�2 � 180°. By the Linear PairPostulate, �1 and �3 are supplementary. Then, bythe definition of supplementary angles, m�1 �m�3 � 180°. Therefore, by the substitution prop-erty, m�1 � m�3 � m�1 � m�2. By the subtrac-tion property, m�3 � m�2. Angles 3 and 2 arealternate interior angles, so, by the Converse of theAIA Theorem, �1 � �2.

13. Given: Two lines in the same plane, each parallelto a third line

Show: The two lines areparallel to each other

Given: Coplanar lines �1,�2, and �3 with �1 � �2 and �3 � �2

21

3

�1

�2

�3

A

B

2

13

�3

�1

�2

2

13

�3

�1

�2


�2 � �3

CA Postulate

�1 and �2 cutby transversal �3

�1 and �2 aresupplementary

Definition ofsupplementary

�1 and �3 aresupplementary

Linear PairPostulate

m�1 � m�3 �180°

Definition ofsupplementary

m�1 � m�2 �180°

Substitutionpropertym�2 � m�3

Definition ofcongruence

Lesson 13.2, Exercise 11.

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The area is less than that of one sheet of plywood.However, it is impossible to cut the correct sizepieces from one piece, because so many of thedimensions add to 3 ft, leaving a one-foot strip thatis too small for any pieces but the gables. Twosheets would be enough.

19. A�(6, �10), B�(�2, �6), C�(0, 0); mapping rule:(x, y) → (2x � 8, 2y � 2). Draw �ABC on graphpaper and locate the center of the dilation, (8, 2).Label (8, 2) as P. To find the vertices of the imagetriangle, �A�B�C�, draw PA�, PB�, and PC�. Extendeach of these segments to locate A�, B�, and C�such that A�A � PA, B�B � PB, and C�C � PC.

Use the figure to find the coordinates of the imagetriangle: A�(6, �10), B�(�2, �6), and C�(0, 0).Because the center of dilation is (8, 2) rather than(0, 0) and the scale factor is 2, the mapping ruleis (x, y) → (8 � 2(x � 8), 2 � 2(y � 2)) �(2x � 8, 2y � 2). You can verify that this rulegives the same results as the graph for the imageof each point of �ABC.

IMPROVING YOUR VISUAL THINKING SKILLS

EXTENSION

Given: �1 � �2, �1 and �2 cut by transversal �3

Show: �2 � �6

1 24 3

5 67 8

�3

�1

�2

FYJG

BK

x

y

C (4, 1)P (8, 2)

B (3, –2) A (7, –4)

A� (6, –10)

B� (–2, –6)

C� (0, 0)

gable

front right roof

roofleft

gable

back

4 ft

8 ft

15. Given: Right triangle with �1 as the right angleand �2 and �3 as the acute angles

Show: �2 and �3 are complementary

Paragraph Proof: By the TriangleSum Theorem, m�1 � m�2 �m�3 � 180°. By the definition of right angle,m�1 � 90°. Using the subtraction property, m�2 �m�3 � 90°, so by the definition of complementaryangles, �2 and �3 are complementary.

16.

17. 1066 cm3. The volume of the truncated pyramid isthe difference between the volume of the completepyramid (before the top was sliced off) and thevolume of the top that was sliced off, which is asmaller pyramid. These two pyramids will bereferred to as large pyramid and small pyramid inthis solution.

The cut is two-thirds of the distance from the baseto the vertex, or one-third of the distance from thevertex to the base. This means that the height of thesmall pyramid is one-third the height of the largeone. The two pyramids are similar, so each sideof the base of the small triangle is one-third thelength of the corresponding side of the largetriangle. By the Proportional Volumes Conjecture,

� ��13��3

� �217�

Then the volume of the truncated pyramid is

1 � �217� � �

22

67� of the volume of the large pyramid, or

�22

67�(1107) � 1066 cm3.

18. No, assuming the bottom is included. (If the bottomis not included and the gables can be cut separatelyfrom the rectangular parts of the back and front,then yes.) Make a table showing the pieces needed.

Area (ft)2

Bottom (2-by-3 rectangle) 6

2 sides �3-by-1�12� rectangle� 9

Back and front �2-by-1�12� rectangle� 6

2 rooftops �3-by-�2� rectangle� � 8�12�

2 gable ends (2-by-1 triangle) 2

Total 31�12�

Plywood (4-by-8 rectangle) 32

Vsmall pyramid��Vlarge pyramid

Linear PairPostulate

Converse ofAIA Theorem

Converse ofAEA Theorem

VA Theorem CA Postulate

1

2

3


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2. Given: A segment and a point that is equidistantfrom the endpoints of the segment

Show: The point is on the perpendicular bisector ofthe segment

Let AB� be the segment and P be the point. In order toprove the theorem, two cases need to be considered.

Case 1: A, P, and B are collinear. (P is on AB�between A and B.)

Given: A, P, and B are collinear; PA � PB

Show: P is on the perpendicular bisector of AB�

Paragraph Proof: PA � PB is given, so PA� � PB�by the definition of congruence. Because P is onAB�, this means that P is the midpoint of AB� bythe definition of midpoint. Then P is on theperpendicular bisector of AB� by the definition ofperpendicular bisector.

Case 2: A, P, and B are not collinear. (P is noton AB�.)

Given: A, P, and B are not collinear; PA � PB

Show: P is on the perpendicular bisector of AB�

Paragraph Proof: Let E be the midpoint of AB�(Midpoint Postulate). Use the Line Postulate todraw PE�. PA � PB is given, so PA� � PB� by thedefinition of congruence. EA� � EB� by the defini-tion of midpoint. PE� � PE� by the reflexive propertyof congruence. Therefore, �AEP � �BEP by theSSS Congruence Postulate. Then �AEP �BEP byCPCTC and �AEP and �BEP are a linear pair byconstruction and are thus supplementary by theLinear Pair Postulate. So, �AEP and �BEP are bothright angles by the Congruent and SupplementaryTheorem. Therefore, PE� is the perpendicular bisectorof AB� by the definition of perpendicular bisector.

3. Given: Isosceles triangle

Show: The base angles of the triangle are congruent

Given: Isosceles triangle ABC

Show: �A � �B

Plan: Use the reflexive property and the SSSCongruence Postulate to get �ABC � �BAC.Therefore, �A � �B by CPCTC.

A B

C

P

E BA

A P B

Paragraph Proof: �2 � �7 by AEA. �7 � �6 by theVA Theorem, so �2 � �6 by the transitive property ofcongruence.

LESSON 13.3

EXERCISES

1. Given: A point on the perpendicular bisector of asegment

Show: The point is equidistant from the endpointsof the segment

Given: AB� with perpendicular bisector CD��;CD�� intersects AB� at E; P is a point on CD��

Show: AP � BP

Case 1: A, B, and P are collinear.

Paragraph Proof: P is on AB� and CD�� (given), soP � E by the Line Intersection Postulate. P is themidpoint of AB� by the definition of perpendicularbisector, so AP � BP by the definition of midpoint.

Case 2: A, B, and P are not collinear.

Plan: Use the SAS Congruence Postulate to get�AEP � �BEP. Then use CPCTC to get AP� � BP�.

Proof:

Statement Reason

1. CD�� is the perpen- 1. Givendicular bisector ofAB�

2. E is the midpoint 2. Definition of perpen-of AB� dicular bisector

3. AE� � BE� 3. Definition ofmidpoint

4. PE� � AB� 4. Definition of perpen-dicular bisector

5. �PEA and �PEB 5. Definition ofare right angles perpendicular

6. �PEA � �PEB 6. Right Angles AreCongruent Theorem

7. PE� � PE� 7. Reflexive property ofcongruence

8. �AEP � �BEP 8. SAS CongruencePostulate

9. AP� � BP� 9. CPCTC

10. AP � BP 10. Definition ofcongruence

C

P

E

D

BA


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Proof:

Statement Reason

1. AC� � BC� 1. Definition of isosceles

2. BC� � AC� 2. Symmetric property ofcongruence

3. AB� � AB� 3. Reflexive property ofcongruence

4. �ABC � �BAC 4. SSS CongruencePostulate

5. �A � �B 5. CPCTC

4. Given: Triangle with two congruent angles

Show: The triangle is isosceles

Given: �ABC with �A � �B

Show: �ABC is isosceles

Plan: Use the reflexive property and the ASACongruence Postulate to get �ABC � �BAC. Thenuse CPCTC and the definition of isosceles triangle.

Proof:

Statement Reason

1. �A � �B 1. Given

2. �B � �A 2. Symmetric property ofcongruence

3. AB� � BA� 3. Reflexive property ofcongruence

4. �ABC � �BAC 4. ASA CongruencePostulate

5. AC� � BC� 5. CPCTC

6. �ABC is isosceles 6. Definition of isoscelestriangle

5. Given: An angle with a point that is equidistantfrom the sides of the angle

Show: The point is on the bisector of the angle

Given: �BCA with P in the interior of the angle;P is equidistant from CA�� and CB��.

Show: P is on the angle bisector of �BCA

Paragraph Proof: Use the Line Postulate andPerpendicular Postulate to draw perpendicularsegments PB� and PA� from P to the sides of theangle and also to draw BA�. By the definition ofdistance from a point to a line and the givenequidistance, PB � PA, so PB� � PA� by the defini-tion of congruence. Thus, �PAB is isosceles withlegs PB� and PA�. Therefore, by the Isosceles TriangleTheorem, �PAB � �PBA. By the Angle AdditionPostulate, m�CAP � m�BAC � m�PAB, and

C A

B

P

A B

C


m�CBP � m�CBA � m�PBA. m�CAP �m�CBP � 90° by the definition of perpendicular.Thus, by the subtraction property, �BAC � �ABC.Then, by the Converse of the Isosceles TriangleTheorem, CB� � CA�. By the reflexive property ofcongruence, CP� � CP�. Therefore, �ACP � �BCPby the SSS Congruence Postulate, and thus�ACP � �BCP by CPCTC. Therefore, CP�� is theangle bisector of �BCA by the definition of anglebisector, and P is on the angle bisector.

6. Given: A triangle with the three perpendicularbisectors of the sides

Show: The three perpendicular bisectors areconcurrent

Given: �ABC with lines �, m, and n as perpen-dicular bisectors of sides AB�, BC�, and AC�,respectively

Show: �, m, and n are concurrent

Paragraph Proof: By the Line Intersection Postu-late, lines � and m intersect in exactly one point;call that point P. Because P is on the perpendicularbisectors of both AB� and BC�, by the PerpendicularBisector Theorem, AP � BP and BP � CP. Then,by the transitive property, AP � CP. Then, by theConverse of the Perpendicular Bisector Theorem,point P is on line n, the perpendicular bisector ofAC�. Therefore �, m, and n are concurrent by thedefinition of concurrent.

7. Given: A triangle with the three angle bisectors

Show: The three angle bisectors are concurrent

Given: �ABC with lines �, m, and n as the anglebisectors of �A, �B, and �C, respectively

Show: �, m, and n are concurrent

Paragraph Proof: By the Line Intersection Postu-late, lines � and m intersect in exactly one point;call that point Q. Because Q is on the angle bisec-tors of both �A and �B, by the Angle BisectorTheorem, Q is equally distant from AB�� and AC�� andalso from AB�� and BC��. Then, by the transitive prop-erty, Q is equally distant from AC�� and BC��. Then, bythe Converse of the Angle Bisector Theorem, pointQ is on n. Therefore �, m, and n are concurrent bythe definition of concurrent.

�

A B

C

Q

nm

A

�

B

P

mn C

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8. Given: A triangle with one exterior angle drawn

Show: The measure of the exterior angle is equal to thesum of the measures of its two remote interior angles

Given: �ABC with interior angles �1, �2, and �3at vertices A, B, and C, respectively; �4 an exterior angle at vertex C

Show: m�4 � m�1 � m�2

Plan: Use the Linear Pair Postulate and the definition of supplementary angles to getm�3 � m�4 � 180°. Then use the Triangle Sum Theorem and the transitiveproperty to get m�1 � m�2 � m�3 � m�3 � m�4. Therefore, m�1 � m�2 �m�4 by the subtraction property.

Proof: Statement Reason

1. �4 is an exterior angle at vertex C 1. Given

2. �3 and �4 are supplementary 2. Linear Pair Postulate

3. m�3 � m�4 � 180° 3. Definition of supplementary

4. m�1 � m�2 � m�3 � 180° 4. Triangle Sum Theorem

5. m�1 � m�2 � m�3 � m�3 � m�4 5. Transitive property

6. m�1 � m�2 � m�4 6. Subtraction property

9. Given: Any quadrilateral

Show: The sum of the measures of the four angles is 360°

Given: Quadrilateral ABCD

Show: m�A � m�ABC � m�C � m�CDA � 360°

Plan: Use the Triangle Sum Theorem and the addition propertyto get m�A � m�1 � m�3 � m�C � m�4 � m�2 � 360°. Then use the AngleAddition Postulate and the substitution property to get m�A � m�ABC � m�C �m�CDA � 360°.


1. Construct DB� 1. Line Postulate

2. m�A � m�1 � m�3 � 180° 2. Triangle Sum Theorem

3. m�C � m�4 � m�2 � 180° 3. Triangle Sum Theorem

4. m�A � m�1 � m�3 � m�C � 4. Addition propertym�4 � m�2 � 360°

5. m�A � (m�1 � m�2) � m�C � 5. Commutative and associative (m�3 � m�4) � 360° properties of addition

6. m�1 � m�2 � m�ABC 6. Angle Addition Postulate

7. m�3 � m�4 � m�ADC 7. Angle Addition Postulate

8. m�A � m�ABC � m�C � 8. Substitution propertym�ADC � 360°


Show: The medians to the congruent sides are congruent

Given: Isosceles triangle ABC with AC� � BC�; BN� is the median to AC� and AM�� is the median to BC�

Show: BN� � AM��

Plan: Use the definitions of median and midpoint to get BM � �12�BC and AN � �

12�AC.

Then use the multiplication property and the substitution property to get AN�� BM��.By the reflexive property, the Isosceles Triangle Theorem, and the SAS CongruencePostulate, �ABN � �BAM. Therefore, BN� � AM�� by CPCTC.

A B

N M

C

A

D

B

C

3 4

21

A

B

C3 4

2

1


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1. BN� is the median to AC�; 1. GivenAM�� is the median to BC�

2. N is the midpoint of AC�; 2. Definition of medianM is the midpoint of BC�

3. AC� � BC� 3. Given

4. AC � BC 4. Definition of congruence

5. �12�AC � �

12�BC 5. Multiplication property

6. AN � �12�AC; BM � �

12�BC 6. Definition of midpoint

7. AN � BM 7. Substitution property

8. AN�� BM�� 8. Definition of congruence

9. AB� � BA� 9. Reflexive property of congruence

10. �CAB � �CBA 10. Isosceles Triangle Theorem

11. �ABN � �BAM 11. SAS Congruence Postulate

12. BN� � AM�� 12. CPCTC


Show: The angle bisectors to congruent sides are congruent

Given: Isosceles triangle ABC with AC� � BC�; BQ� is the anglebisector to AC� and AP� is the angle bisector to BC�

Show: AP� � BQ�

Plan: Use the definition of angle bisector to get m�PAB � �12�m�CAB and m�QBA �

�12�m�CBA. Then use the Isosceles Triangle Theorem, the multiplication property, andthe substitution property to get �PAB �QBA. By the reflexive property and theASA Congruence Postulate, �ABP �BAQ. Therefore, AP� BQ� by CPCTC.


1. BQ� is the angle bisector to AC�; 1. GivenAP� is the angle bisector to BC�

2. BQ� is the angle bisector of �CBA; 2. AC� is opposite �ABC; BC� is opposite �CABAP� is the angle bisector of �CAB

3. m�PAB � �12�m�CAB; m�QBA � 3. Definition of angle bisector

�12�m�CBA

4. AC� � BC� 4. Given

5. �CAB �CBA 5. Isosceles Triangle Theorem

6. m�CAB � m�CBA 6. Definition of congruence

7. �12�m�CAB � �

12�m�CBA 7. Multiplication property

8. m�PAB � m�QBA 8. Substitution

9. �PAB �QBA 9. Definition of congruence

10. AB� BA� 10. Reflexive property of congruence

11. �ABP �BAQ 11. ASA Congruence Postulate

12. AP� BQ� 12. CPCTC


Show: The altitudes to congruent sides are congruent

Given: Isosceles triangle ABC with AC� � BC�; AS� is the altitude toBC�, and BT� is the altitude to AC�

Show: AS� BT�A B

T S

C

A B

Q P

C


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Plan: Use the Isosceles Triangle Theorem, the Right Angles Are Congruent Theorem,and the SAA Theorem to get �ASB � �BTA. Then AS� � BT� by CPCTC.


1. AS� is the altitude to BC�; 1. GivenBT� is the altitude to AC�

2. AS� � BC�; BT� � AC� 2. Definition of altitude

3. �ASB is a right angle; 3. Definition of perpendicular�BTA is a right angle

4. �ASB �BTA 4. Right Angles Are Congruent Theorem

5. AC� BC� 5. Given

6. �ABC �CAB 6. Isosceles Triangle Theorem


8. �ASB �BTA 8. SAA Theorem

9. AS� BT� 9. CPCTC


13. Theorem A: median → angle bisector

Given: Isosceles triangle ACB with AC� BC�; CD�� isthe median to AB�

Show: CD�� is the angle bisector of �ACB

Paragraph Proof: By the definition of median,AD�� BD�. AC� BC� is given. CD�� CD�� by thereflexive property of congruence. Therefore,�ADC �BDC by the SSS Congruence Postulate.�ACD �BCD by CPCTC, and CD�� is the anglebisector of �ACB by the definition of angle bisector.

Theorem B: angle bisector → altitude

Given: Isosceles triangle ACB with AC� BC�; CD�� isthe angle bisector of �ACB

Show: CD�� is the altitude from C to AB�

Paragraph Proof: By the definition of anglebisector, �ACD �BCD. AC� BC� is given.CD�� CD�� by the reflexive property of congruence.Therefore, �ADC � �BDC by the SAS CongruencePostulate. �CDA �CDB by CPCTC. By theLinear Pair Postulate and the Congruent andSupplementary Theorem, �ADC and �BDC areboth right angles. Therefore, CD�� is the altitude bythe definitions of perpendicular and altitude.

A B

C

D

A B

C

D

Theorem C: altitude → median

Given: Isosceles triangle ACB with AC� BC�; CD�� isthe altitude from C to AB�

Show: CD�� is the median to AB�

Paragraph Proof: By the definitions of altitude andperpendicular, �CDA and �CDB are right angles.Then �CDA � �CDB by the Right Angles AreCongruent Theorem. AC� BC� is given, andtherefore �A � �B by the Isosceles TriangleTheorem. Thus, �ADC � �BDC by the SAATheorem, and AD�� BD� by CPCTC. Therefore, CD��is the median to AB� by the definition of median.

14. x � 6, y � 3. Label points in the figure forreference.

�ABD �ACB by the AA Similarity Conjecture.(�A is common to both triangles; �ADB and�ABC are both right angles.) Corresponding sidesof similar triangles are proportional, so �A

ACB� � �

AA

DB�.

Use this proportion to find y.

�

y(y � 12) � �3�5��2

y 2 � 12y � 45

y�3�5�

3�5��y � 12

53

y

x

A

B C

D

12

A B

C

D

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TU � TQ � UQ � 6.1 � 3.6 � 2.5. Apply thePythagorean Theorem to �STU to find d � ST.

d 2 � ��471.87��2� (2.5)2

d � ��471�.87��2�� (2.5)2� � �478.12� � 21.9

The distance between the two nests is about 21.9 m.

16. First image: (0, 3); second image: (6, 1). Let B repre-sent the first image of A under the glide reflectionand let C represent the second image. Use a graphto find the images.

To find the images of A and B under the reflection,draw the line x � y � 5, or y � 5 � x.

First glide reflection: The image of A(�2, 9) underthe reflection across the line x � y � 5 is (�4, 7);the image of (�4, 7) under the translation (x, y) →(x � 4, y � 4) is (0, 3), so the coordinates of Bare (0, 3).

Second glide reflection: Start with (0, 3), the imageof A under the first glide reflection. The image of(0, 3) under the reflection is (2, 5); the image of(2, 5) under the translation is (6, 1), so thecoordinates of C are (6, 1).

17. BC � FC makes ABCF a rhombus, so the diagonals,AC� and BF�, are perpendicular. m�FGC � 90°, som�CFG � m�FCG � 90°. FD� � GE�, so m�2 �m�CFG � 90°. By subtraction and transitivity,m�2 � m�FCG. �1 � �FCG by AIA, so �1 � �2 by transitivity.

18. a. . The arcs show that the shaded region is an

equilateral triangle with side length 1. Any alti-

tude of this triangle divides the triangle into two

congruent 30°-60°-90° triangles with shorter leg

of length �12�, so the length of the longer leg, which

is the height of the triangle, is �12�(�3�) � .

Therefore, A � �12�bh � �

12�(1)� � � .

b. ��6�. The shaded region is a sector of a circle withradius 1. The central angle of this sector is the

�3��4

�3��2

�3��2

�3��4

B (0, 3)

x � y � 5

A (–2, 9)

(–4, 7)

(2, 5)

C (6, 1)x

y

10–5

5

–5

10

5

y 2 � 12y � 45 � 0

(y � 15)(y � 3) � 0

y � 15 � 0 or y � 3 � 0

y � �15 or y � 3

Reject �15 because y represents a side length andtherefore must be positive, so y � 3.

Now apply the Pythagorean Theorem to �ABD tofind x:

x2 � y 2 � �3�5��2

x 2 � 9 � 45

x 2 � 36

x � 6

15. � 21.9 m. Let S be the location of the nest on thetop of the shorter tree and T be the location of thenest on the taller tree.

Triangle PQR is a right triangle with hypotenuseRQ�. Apply the Pythagorean Theorem to this triangleto find PQ.

16.72 � (PQ)2 � 27.42

PQ � �27.42 �� 16.72� � �471.87�

(This length is approximately 21.72 m, but do notround this intermediate result.)

Because both trees are vertical, SP� � TQ�. ST� and PQ�are not parallel. Thus, PSTQ is a trapezoid. DrawSU�, the perpendicular segment from S to TQ�, toform parallelogram PSUQ (which is also arectangle). Opposite sides of a parallelogram arecongruent, so SU � PQ � �471.87�. Now look at�STU. Because SU� � TU�, this is a right trianglewith hypotenuse ST�. Because SP� and UQ�� are oppo-site sides of a parallelogram, UQ � SP � 3.6, so

16.7 m

27

.4 m

3.6 m

P

R

Q

d

T

S

U6.1 m


DG4SM_890_13.qxd 12/21/06 11:21 AM Page 290


same angle as the angle in the lower left of theshaded triangle in 18a. Because that triangle isequilateral, the measure of each of its angles is60°, so the sector has a 60° central angle. There-fore, A � �3

6600°°��r 2 � �

16��(1)2 � �

�6�.

c. ��3� � . The difference between the areas of the

shaded regions in 18b and 18a is ��6� � . To

find the area of the shaded region, add twice this

difference to the area of the triangle from 18a:

A � � 2� � � � � �

� �

19. One possible sequence:

1. Fold A onto B and crease. Label the crease �1.Label the midpoint of the arc M.

2. Fold line �1 onto itself so that M is on the crease.Label this crease �2.

M is the midpoint of AB�, and �2 is the desiredtangent.

20. m�BAC � 13°. AB� � BC�, so �ABC is a right

triangle with hypotenuse AC�. By the Pythagorean

Theorem, AB � �52 � 1�22� � 13. Thus,

tan �BAC � �BA

CB� � �1

33�, and m�BAC �

tan�1��133�� 13°.

21. In each figure, draw YW�� and WX�� to form �WXY,�XOW, and �YOW.

a. B. In each figure, OW�� is a radius and YX� is adiameter.

In Figure A, OW � 10 cm, so YX � 20 cm.Because m�WOX � 40° and �WOX is a centralangle of the circle, mWX� � 40°. Because �WYXis an inscribed angle that intercepts WX�,m�WYX � �

12�(40°) � 20°. Because it is inscribed

in a semicircle, �YWX is a right angle and�WYX is a right triangle with hypotenuse YX�.

Therefore, sin 20° � �WYX

X� � �

W20

X�, so

WX � 20 sin 20°, and cos 20° � �YYWX� � �

Y2W0�, so

BA

M

�1

�2

�3��

4��3

�3��

2��3

�3��

4�3��

4��6

�3��

4

�3��4

�3��4


YW � 20 cos 20°. Thus, perimeter of �WXY �

YX � YW � WX � 20 � 20 sin 20° + 20 cos 20°

� 45.6 cm.

In Figure B, OW � 20 cm, so YX � 40 cm. Usethe same reasoning as above to find the othertwo side lengths in �WXY. First m�WOX � 10°,so mWX� � 10°, and m�WYX � �

12�(10°) � 5°.

Then WX � 40 sin 5° and YW � 40 cos 5°. Thus,perimeter of �WXY � YX � YW � WX � 40� 40 sin 5° + 40 cos 5° � 83.3 cm.

Therefore, the perimeter of �WXY is greater inFigure B.

b. B. In both figures, �XOW is an isosceles trianglewhose legs are radii of the circle. Apply the SASTriangle Area Conjecture to �XOW in bothfigures.

Figure A: A � �12�(OW)(OX)sin �WOX

� �12�(10)(10)sin 40° � 32.1 cm2

Figure B: A � �12�(OW)(OX)sin �WOX

� �12�(20)(20)sin 10° � 34.7 cm2

Therefore, the area of �XOW is greater inFigure B.

22. a. x � �12�(a � c); y � �

12�(a � b); z � �

12�(b � c)

b. w � �12�(a � b); x � �

12�(b � e); y � �

12�(e � d);

z � �12�(d � a)


The reasoning used for solving these puzzles will vary.Here is a possible approach to solving the first puzzle.Focus on the three squares of nine boxes at the top.The upper left square has 1 and 9 in the middle row,and the upper middle square has 9 and 1 in the bottomrow. The top right square also needs a 1 and a 9, andboth digits must fall in the top row to avoid repetition.Because the right column contains a 1, the 9 must goin the top right box, leaving the 1 in the adjacent box.

The upper right square is now missing 3, 6, and 7. The3 must go in the middle row because there is already a3 in the bottom row of the upper left square. Becausethere is a 6 in the second column from the right, the6 must go in the lower right square of the upper right

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box, and the 7 must be adjacent to it. Note: In the completed puzzles below, the bold numbers are the ones that aregiven in puzzles in the textbook.

LESSON 13.4

EXERCISES

1. Given: A quadrilateral with two pairs of congruent oppositeangles

Show: The quadrilateral is a parallelogram

Given: Quadrilateral ABCD with �A � �C and �B � �D

Show: ABCD is a parallelogram

Plan: Use x to represent the measures of one pair of congruent angles and y to repre-sent the other pair. Use the Quadrilateral Sum Theorem and the division property toget x � y � 180°. Therefore, the opposite sides are parallel by the Converse of theInterior Supplements Theorem.


1. �A � �C; �B � �D 1. Given

2. m�A � m�C � x; m�B � m�D � y 2. Definition of congruence

3. x � x � y � y � 360° 3. Quadrilateral Sum Theorem

4. 2x � 2y � 360° 4. Combine like terms

5. 2(x � y) � 360° 5. Distributive property

6. x � y � 180° 6. Division property

7. �A and �D are supplementary 7. Definition of supplementary

8. DC�� AB� 8. Converse of the Interior Supplements Theorem

9. �A and �B are supplementary 9. Definition of supplementary

10. AD�� BC� 10. Converse of the Interior Supplements Theorem

11. ABCD is a parallelogram 11. Definition of parallelogram

2. Given: A quadrilateral with one pair of opposite sides that are both parallel and congruent

Show: The quadrilateral is a parallelogram

Given: Quadrilateral ABCD with AD�� BC� and AD�� BC�

Show: ABCD is a parallelogram

Plan: Use the AIA Theorem, the reflexive property, and the SAS Congruence Postu-late to get �ADC � �CBA. Then use CPCTC and the Converse of the AIA Theoremto get AB� � DC��.

D C

BA

3 2

41

D C

BA

8

7 6

5

1

7 6 5

1

5 2

8 3

5

4 7 2

3

6 9

5

9 2 8

7

6

1

3 2

5

4 8

7 9

2

4

3

1 4

6

1

2 1 9

9

38

5

7

1 8

9

6

7

3

6 2

4

9 8

6 7

3 5

4

8

1

9

4 3

4

8

9

24

32

7

1

6

5

4

91

3 9 1

2

5 8

4

2 7

7

8 7

6

6

5 8

8 4

1

3 6 1

6

4 5

5 6 4

7 9 2 4 5

3

4

2

9

8 3 5

6 4

7 6

1 9

2

7 3

8 2

8

7

9

3 1

3

79

1 9 5

2

85

25

6

72

8

1 3 9

3 1

6

4


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1. Construct AC� 1. Line Postulate

2. AD�� BC� 2. Given

3. �1 � �2 3. AIA Theorem


5. AC� � CA� 5. Reflexive property of congruence

6. �ADC � �CBA 6. SAS Congruence Postulate

7. �3 � �4 7. CPCTC

8. DC�� AB� 8. Converse of the AIA Theorem


3. Given: Rhombus

Show: Each diagonal of the rhombus bisects two opposite angles

Given: Rhombus ABCD with diagonals AC� and BD�

Show: AC� bisects �DAB and �BCD; BD� bisects �ADC and �CBA

Plan: Use the definition of rhombus, the reflexive property, and the SSS CongruencePostulate to get �ABC � �ADC. Then use CPCTC and the definition of anglebisector to prove that AC� bisects �DAB and �BCD. Repeat using diagonal BD�.


1. ABCD is a rhombus 1. Given

2. AB� � AD�� 2. Definition of rhombus

3. BC� � DC�� 3. Definition of rhombus

4. AC� � AC� 4. Reflexive property of congruence

5. �ABC � �ADC 5. SSS Congruence Postulate

6. �1 � �2; �6 � �5 6. CPCTC

7. AC� bisects �DAB and �BCD 7. Definition of angle bisector

8. AD�� CD�� 8. Definition of rhombus

9. AB� � CB� 9. Definition of rhombus

10. BD� � BD� 10. Reflexive property of congruence

11. �ADB � �CDB 11. SSS Congruence Postulate

12. �3 � �4; �8 � �7 12. CPCTC

13. BD� bisects �ABC and �ADC 13. Definition of angle bisector

4. Given: Parallelogram

Show: The consecutive angles are supplementary

Given: Parallelogram ABCD

Show: �A and �B are supplementary; �B and �C aresupplementary; �C and �D are supplementary; �D and �A are supplementary

Plan: Use the definition of a parallelogram to get AD�� BC� and AB� � DC��. Then usethe Interior Supplements Theorem.

Flowchart Proof

Definition ofparallelogram

AD � BC

Definition ofparallelogram

AB � DC

ABCD is aparallelogram

Given

�A and �B are supplementary�C and �D are supplementary

Interior Supplements Theorem

�B and �C are supplementary�D and �A are supplementary


D C

BA

D C

BA

65

43

78

21


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5. Given: A quadrilateral with four congruent sides

Show: The quadrilateral is a rhombus

Given: Quadrilateral ABCD with AB� BC� CD�� DA�

Show: ABCD is a rhombus

Plan: Use the reflexive property and the SSS Congruence Postulate to get �ABD �CDB. Then use CPCTC and the Converse of the AIA Theorem to get AB� � CD�� andAD�� CB�. Therefore, ABCD is a rhombus by the definitions of parallelogram andrhombus.


1. Construct BD� 1. Line Postulate

2. AB� CD�� 2. Given

3. AD�� CB� 3. Given

4. BD� DB� 4. Reflexive property of congruence

5. �ABD �CDB 5. SSS Congruence Postulate

6. �1 �2 6. CPCTC

7. AB� � CD�� 7. Converse of the AIA Theorem

8. �3 �4 8. CPCTC

9. AD�� CB� 9. Converse of the AIA Theorem


11. ABCD is a rhombus 11. Definition of rhombus

6. Given: A quadrilateral with four congruent angles

Show: The quadrilateral is a rectangle

Given: Quadrilateral ABCD with �A � �B � �C � �D

Show: ABCD is a rectangle

Plan: Use the Converse of the Opposites Angles Theorem to prove that ABCD is aparallelogram. Then use the definition of rectangle.


1. �A � �C; �B � �D 1. Given

2. ABCD is a parallelogram 2. Converse of Opposite Angles Theorem

3. �A � �B � �C � �D 3. Given

4. m�A � m�B � m�C � m�D 4. Definition of congruence

5. m�A � m�B � m�C � m�D � 360° 5. Quadrilateral Sum Theorem

6. 4m�A � 360° 6. Substitution property

7. m�A � 90° 7. Division property

8. m�A � m�B � m�C � m�D � 90° 8. Substitution property

9. ABCD has four right angles 9. Definition of right angle

10. ABCD is a rectangle 10. Definition of rectangle

7. Given: Rectangle

Show: The diagonals of the rectangle are congruent

Given: Rectangle ABCD with diagonals BD� and AC�

Show: BD� � AC�

Plan: Use the definition of a rectangle to prove that ABCD is a parallelogram and�DAB � �CBA. Then use the Opposite Sides Theorem, the reflexive property, andthe SAS Congruence Postulate to get �DAB � �CBA. Finish with CPCTC.

CD

BA

CD

BA

D C

BA

41

23


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1. ABCD is a rectangle 1. Given

2. ABCD is a parallelogram 2. Definition of rectangle

3. DA� CB� 3. Opposite Sides Theorem

4. �DAB and �CBA are right angles 4. Definition of rectangle

5. �DAB � �CBA 5. Right Angles Are Congruent Theorem


7. �DAB � �CBA 7. SAS Congruence Postulate

8. BD� AC� 8. CPCTC

8. Given: A parallelogram with congruent diagonals

Show: The parallelogram is a rectangle

Given: Parallelogram ABCD with DB� AC�

Show: ABCD is a rectangle

Plan: Use the Opposite Sides Theorem, the reflexive property of congruence, and theSSS Congruence Postulate to get �DAB �CBA. Repeat the above steps to get�ADC �CBA and �DAB � �BCD. Then use CPCTC and the transitive propertyto get �DAB �CBA �BCD �ADC. Finish with the Four Congruent AnglesRectangle Theorem.

Flowchart Proof

9. Given: Isosceles trapezoid

Show: The base angles are congruent

Given: Isosceles trapezoid ABCD with DC�� AB�, AD�� BC�

Show: �A � �B

Plan: Use the Parallel Postulate to construct DE� � CB�. Then use the Opposite SidesTheorem and the transitive property to prove that �AED is isosceles. Therefore,�A � �B by the Isosceles Triangle Theorem and the CA Postulate.

CD

BA E

ABCD is aparallelogram

Given

ABCD is arectangle

Four CongruentAngles RectangleTheorem

�DAB � �CBA ��ADC � �BCD

Transitive property

SSS CongruencePostulate

�DAB � �CBA


Given

DB � AC

AD � BCDC � AB

Opposite SidesTheorem

Reflexiveproperty

AB � BA

Reflexiveproperty

AC � CA

�DAB � �CBA

CPCTC


�ADC � �CBA

Reflexiveproperty

DB � BD �DAB � �BCD

CPCTC

�ADC � �CBA

�DAB � �BCD

CPCTC

CD

BA


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1. DC�� AB� 1. Given

2. Construct DE� � CB� 2. Parallel Postulate

3. DCBE is a parallelogram 3. Definition of parallelogram

4. DE� BC� 4. Opposite Sides Theorem


6. AD�� DE� 6. Transitive property of congruence

7. �ADE is isosceles 7. Definition of isosceles triangle

8. �A �DEA 8. Isosceles Triangle Theorem

9. �DEA �B 9. CA Postulate

10. �A � �B 10. Transitive property of congruence

10. Given: Isosceles trapezoid

Show: The diagonals are congruent

Given: Isosceles trapezoid ABCD with DC�� AB�, AD�� BC�

Show: AC� BD�

Plan: Use the Isosceles Trapezoid Theorem, the reflexive property, and the SASCongruence Postulate to get �DAB �CBA. Then AC� BD� by CPCTC.

Flowchart Proof

11. Given: Parallelogram with a diagonal that bisects two opposite angles

Show: The parallelogram is a rhombus

Given: Parallelogram ABCD; AC� bisects �DAB and �BCD

Show: ABCD is a rhombus

Plan: Use the Opposite Angles Theorem, the multiplication property, and thedefinition of angle bisector to get �1 � �3. Then use the Converse of theIsosceles Triangle Theorem, the definition of isosceles triangle, and the OppositeSides Theorem to get AB� BC� DC�� AD��.


1. AC� bisects �DAB and �BCD 1. Given

2. �1 �2; �3 �4 2. Definition of angle bisector

3. m�1 � m�2; m�3 � m�4 3. Definition of congruence

4. m�1 � m�2 � m�DAB; 4. Angle Addition Postulatem�3 � m�4 � m�BCD

5. 2m�1 � m�DAB; 2m�3 � m�BCD 5. Substitution property

6. m�1 � �12�(m�DAB); m�3 � �

12�(m�BCD) 6. Multiplication property

7. ABCD is a parallelogram 7. Given

8. �DAB � �BCD 8. Opposite Angles Theorem

9. m�DAB � m�BCD 9. Definition of congruence

CD

BA

34

21

Isosceles trapezoidABCD with DC � AB �DAB � �CBA

Isosceles TrapezoidTheorem

�DAB � �CBA

SAS CongruencePostulate

Given

AD � BC

Reflexive property

AB � BA

CPCTC

BD � AC

CD

BA

(Proof continued)

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10. �12�(m�DAB) � �

12�(m�BCD) 10. Multiplication property

11. m�1 � m�3 11. Substitution property

12. AB� � BC� 12. Converse of the Isosceles Triangle Theorem

13. AB� DC��; AD�� BC� 13. Opposite Sides Theorem

14. AB� BC� DC�� AD�� 14. Transitive property of congruence

15. ABCD is a rhombus 15. Definition of rhombus

12. Given: A pair of parallel lines intersected by a second pair of parallel lines; the distance between each pair oflines is the same

Show: The parallelogram formed is a rhombus

Given: WZ�� XY��, XW�� YZ��, distance between WZ�� andXY�� equals distance between XW�� and YZ��

Show: XWZY is a rhombus

Plan: Use the Converse of the Angle Bisector Theorem to prove that WY�� is the anglebisector of �Y and of �W. Therefore, WXYZ is a rhombus by the Converse of theRhombus Angles Theorem.


1. WZ�� XY��, XW�� YZ�� 1. Given

2. WXYZ is a parallelogram 2. Definition of parallelogram

3. Distance from W to XY�� 3. Givendistance from W to YZ��

4. WY�� is the angle bisector of �Y 4. Converse of the Angle Bisector Theorem

5. Distance from Y to XW�� 5. Givendistance from Y to WZ��

6. WY�� is the angle bisector of �W 6. Converse of the Angle Bisector Theorem

7. XWZY is a rhombus 7. Converse of the Rhombus Angles Theorem

W Z

X YP

Q


mentary (Parallelogram Consecutive AnglesConjecture).

b. Sometimes. �ADM and �MAD are complemen-tary if m�AMD � 90°, that is, if AC� � BD�. Butthe diagonals of a parallelogram are perpendic-ular only if the parallelogram is a rhombus. Thisparallelogram may or may not be a rhombus;you don’t have enough information to tell.

c. Sometimes. A diagonal of a parallelogram couldbe longer than the sum of the lengths of twoopposite sides, or it could be shorter or the samelength. There is not enough information to tell.

d. Never. By the Triangle Inequality Conjecture,AD � CD � AC.

16. 2386 ft2. Divide the grazing area into fivesections and find thearea of each.

Notice that eachsection is a sector ofa circle, but the radii

E A

B

25 1560°

30

12

12

18

1010

8 8

C

D

13.

14.

15. a. Always. Because its diagonals bisect each other,ABCD is a parallelogram (converse ofParallelogram Diagonals Conjecture). Then�BAD and �ADC are supplementary becauseconsecutive angles of a parallelogram are supple-

ASA Congruence Postulate

Parallelogram Diagonal Lemma

Opposite Sides Theorem

Opposite AnglesTheorem

Converse of the Rhombus Angles Theorem

Converse of the AngleBisector Theorem

Double-Edged Straightedge Theorem

Converse of the IT Theorem

IT Theorem

Line Postulate Angle AdditionPostulate


CA Postulate


Consecutive Angles Theorem

Linear Pair Postulate

(Exercise 11 Proof continued)

DG4SM_890_13.qxd 12/21/06 11:21 AM Page 297


supplementary, m�ADC � 180° � 50° � 130°.Because opposite sides of a parallelogram arecongruent, CD � 5. Use the Law of Cosines to findAC, the length of the resultant vector, V��1 � V��2.

(AC)2 � (AD)2 � (CD)2 � 2(AD)(CD)(cos 130°)

(AC)2 � 92 � 52 � 2(9)(5)(cos 130°)

AC � �92 � 5�2 � 2(�9)(5)(c�os 130�°)� � 12.8

(Keep all digits of the value of AC in your calculatorfor use in the next calculation.)

Let � � m�CAD. Use the Law of Sines to find �.

�sin

5�

� �

sin � �

� � sin�1� � � 17.4°

The bearing of the resultant vector, V��1 � V��2, is90° � 17.4° � 72.6°.

19. a. B. Look at �XYZ in both figures and find theunknown side lengths in each of these triangles.In both figures, let � � m�XYZ.

Figure A: Apply the Pythagorean Theorem toobtain XZ � �42 � 5�2� � �41�. �XYZ is a righttriangle, with right angle Z. Use the tangent ratioto find �.

tan � � �XYZ

Z� �

� � tan�1 � 35.4°

Figure B: Follow the steps applied to Figure A.

XZ � �62 � 5�2� � �61�

tan � � �XYZ

Z� �

� � tan�1 � 52.5°

Thus, m�XYZ is greater in Figure B.

b. A. Remember that the shortest distance betweentwo points is a straight line. To find the straightline from X to Y on the surface of Figure A, drawa net of the prism. The path has to cross twofaces, and there are six ways to do so, so there aresix paths to consider. But each path is the mirrorimage of the path opposite it, so there are onlythree different paths.

�42 � (�5 � 9)�2� � �212� � 14.6

X

Z Y95

4

�61��6

�61��6

�41��9

�41��9

5 sin 130°��AC

5 sin 130°��AC

sin 130°�AC

of the circles differ depending on the positions ofthe sectors in the diagram.

Sector A: r � 15 ft; central angle � 60°, so sector is�16� of circle.

Asector A � �16��r 2 � �

16��(15)2 � �

16�� 225 � 37.5� ft2

Sector B: r � 40 ft; central angle � 90°, so sector is�14� of circle.

Asector B � �14��r2 � �

14��(40)2 � �

14�� 1600 � 400� ft2

Sector C: r � 30 ft; central angle � 90°, so sector is�14� of circle.

Asector C � �14��r2 � �

14��(30)2 � �

14�� 900 � 225� ft2

Sector D: r � 18 ft; central angle � 90°, so sector is�14� of circle.

Asector D � �14��r2 � �

14��(18)2 � �

14�� 324 � 81� ft2

Sector E: r � 8 ft; central angle � 90°, so sector is�14� of circle.

Asector E � �14��r2 � �

14��(8)2 � �

14�� 64 � 16� ft2

To find the total grazing area, add the areas of thefive sectors:

37.5� + 400� + 225� + 81� + 16� � 759.5� �2386 ft2

17.

18. �V1 � V��2 has length 12.8 and bearing 72.6°. Label thevertices in the figure, and use the given bearingsand vector lengths to find angle measures and sidelengths in parallelogram ABCD and �ACD.

Because the bearing of V�1 is 40° and the bearingof V�2 is 90°, m�BAD � 90° � 40° � 50°. Becauseconsecutive angles of a parallelogram are

V2�

V1�

V1 � V240°

N

A D

CB

50°�

Lines of RotationalName symmetry symmetry

parallelogram none 2-fold

trapezoid none none

kite 1 diagonal none

square 2 diagonals 4-fold2 � bisectors of sides

rectangle 2 � bisectors of sides 2-fold

rhombus 2 diagonals 2-fold

isosceles trapezoid 1 � bisector of sides none


DG4SM_890_13.qxd 12/21/06 11:21 AM Page 298


same angle and �COD intercepts DC�, mDC� �38°. From 20c, m�BOC � 52°, so mCB� � 52°.Therefore, mEAB� � 360° � 38° � 38° � 52° �232°.

e. 19°. Notice that �HED is the same angle as�CED, which is an inscribed angle thatintercepts DC�. From 20d, mDC� � 38°, som�HED � m�CED � �

12�mDC� � �

12�(38°) � 19°.

IMPROVING YOUR VISUAL THINKING SKILLS

Look at each net and see if adjacent faces are related asthey are in the cube. A good problem-solving techniquehere is to eliminate choices. For example, the net in 1Acan be eliminated because the base of the green face istouching the red face, unlike in the cube.

1. D

2. C

EXTENSION

Answers will vary.

EXPLORATION • PROOF AS CHALLENGE AND DISCOVERY

IMPROVING YOUR ALGEBRA SKILLS

There is a flaw in the proof. If h � n � p, then h � n �p � 0. The step from h(h � n � p) � n(h � n � p) toh � n is not valid, because when you divide both sidesby h � n � p, you are dividing by zero.

LESSON 13.5

EXERCISES

1. D. Paris is in France; Tucson is in the United States;London is in England. Therefore, Bamako must bethe capital of Mali.

2. C. The “Sir” in 2A shows that Halley was English;Julius Caesar was an emperor, not a scientist;Madonna is a singer. Therefore, Galileo Galilei mustbe the answer.

3. No, the proof is claiming only that if two particularangles are not congruent, then the two particularsides opposite them are not congruent. A differentpair of sides might still be congruent.

4. Yes, this statement is the contrapositive of theconjecture proved in Example B, so they are logi-cally equivalent.

5. 1. Assume the opposite of the conclusion;2. Triangle Sum Theorem; 3. Substitution propertyof equality; 4. m�C � 0, Subtraction property ofequality.

6. Paragraph Proof: Assume ZOID is equiangular.Then, by the definition of equiangular, �Z ��O � �I � �D. Then, by the Four Congruent

�(4 � 9�)2 � 5�2� � �194� � 13.9

�(5 � 4�)2 � 9�2� � �162� � 12.7

Notice that the expressions for Figure A are thethree possible ways to group the three dimen-sions into a pair and a “singleton,” so you canfind the corresponding distances for Figure Bwithout drawing.

�52 � (�6 � 6)�2� � �169� � 13

�62 � (�5 � 6)�2� � �157� � 12.5

�62 � (�6 � 5)�2� � �157� � 12.5

The shortest distance is �162� � 12.7 for Figure Aand �157� � 12.5 for Figure B. So the value isgreater in Figure A.

20. a. 19°. �EOD is a central angle of circle O, somED� � 38°. �EAD is an inscribed angle thatalso intercepts ED�, so m�A � �

12�(38°) � 19°. OA�

and OE� are both radii of the circle, so �AOE isan isosceles triangle with OA � OE. Therefore, bythe Isosceles Triangle Theorem, m�AEO �m�A � 19°.

b. 52°. Look at �DGO. Because GF�� is tangentto circle O at D, OD�� GF�� by the TangentConjecture. Thus, m�ODG � 90°. �GOD and�EOD are the same angle, so m�GOD � 38°.Therefore, m�DGO � 180° � 90° � 38° � 52°.

c. 52°. Because GF�� EC��, �HEO � �DGO by theCA Postulate. �OEC and �OEH are the sameangle, so m�OEC � 52°. �OEC is isoscelesbecause OE� and OC�� are both radii of the circle.Therefore, �OCE � �OEC by the IsoscelesTriangle Theorem, so m�OCE � 52°. Finally,because OB� � EC��, �BOC � �OCE by AIA, som�BOC � 52°.

d. 232°. mEAB� � 360° � mED� � mDC� � mCB�. In20a, you found that mED� � 38°. Look at thecentral angles for ED�, DC�, and CB�. Because OH��is the altitude to the base of isosceles triangleEOC, it is also the angle bisector, so m�COH �38°. Then, because �COH and �COD are the

X

Z Y9

5

4

X

Z

Y

9

5

4


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Angles Rectangle Theorem, ZOID is a rectangle.Therefore, ZOID is a parallelogram, which creates acontradiction because, by definition, a trapezoid hasexactly one pair of parallel sides and thus cannot bea parallelogram. Therefore, the assumption thatZOID is equiangular is false and the conjectureis true.

7. Paragraph Proof: Assume CD�� is the altitude toAB�. Then, by the definition of altitude, CD�� AB�,and by the definition of perpendicular, �CDA and�CDB are right angles. Thus, �CDA � �CDB bythe Right Angles Are Congruent Theorem. CD�� CD�� by the reflexive property of congruence. It isgiven that CD�� is a median, so D is the midpoint ofAB�, and AD�� BD� by the definition of midpoint.Then �ADC � �BDC by the SAS CongruencePostulate. Therefore, AC� BC� by CPCTC, so�ABC is isosceles. It is given that �ABC is scalene.Therefore, the assumption that CD�� is the altitude toAB� is false and the conjecture is true.

8. Paragraph Proof: Assume ZO � ID. Then ZO�ID� by the definition of congruence. ZO� � ID� isgiven. Therefore, by the Opposite Sides Parallel andCongruent Theorem, ZOID is a parallelogram. Thiscreates a contradiction because ZOID is a trapezoid,and a trapezoid cannot be a parallelogram. Thus,the assumption that ZO � ID is false, and theconjecture is true.

9. Given: A chord of a circle

Show: The perpendicular bisector of the chordpasses through the center of the circle

Given: Circle O with chord AB� and perpendicularbisector CD��

Show: CD�� passes through O

Paragraph Proof: It is given that CD�� is theperpendicular bisector of AB�. Then, by the defini-tion of perpendicular bisector, D is the midpoint ofAB�. Assume CD�� does not pass through O. Use theLine Postulate to construct OB� and OA� and thePerpendicular Postulate to construct OE�, where OE�is the perpendicular segment from O to AB�.Because OE� � AB�, �OEB and �OEA are rightangles, so �OEB � �OEA by the Right Angles AreCongruent Theorem. OB� OA� because they areradii of the same circle, so �B � �A by theIsosceles Triangle Theorem. OE� OE� by thereflexive property of congruence. Therefore,�OEB �OEA by the SAA Congruence Theorem,

C

O

A

B

D

E

and EB� EA� by CPCTC. Then E is the midpointof AB�, and OE�� is the perpendicular bisector of AB�.This contradicts the assumption that the perpen-dicular bisector of AB� does not pass through O.(A segment can have only one midpoint and oneperpendicular bisector.) Thus, the assumption isfalse and the conjecture is true.

10. a � 75°, b � 47°, c � 58°. By the Interior Supple-ments Theorem, a � 180° � 105° � 75°. b � a �58° � 180°, or b � 75° � 58° � 180°, so b � 180° �75° � 58° � 47°. By the CA Postulate, c � 58°.

11. 42� ft3 � 132 ft3. The volume of the container is thesum of the volumes of a cylinder with radius 3 ft andheight 3 ft and a cone with radius 3 ft and height 5 ft.

Vcylinder � �r2H � �(3)2(3) � 27� ft3

Vcone � �13��r2H � �

13��(3)2(5) � 15� ft3

Vcontainer � 27� + 15� � 42� ft3 � 132 ft3

12. a. � �1�2�. The shaded region is the difference

between the shaded area in Lesson 13.1, Exer-cise 31a and the shaded area in Lesson 13.3,Exercise 18c, so subtract the areas you found inthose exercises:

A � ��4� � ��

�3� � � � �

�4� � �

�3� �

� � �1�2�

b. 1 � �3� � ��3�. To find the area of the shaded

region, subtract four times the shaded area from12a from the area of the square.

A � 1 � 4� � �1�2�� 1 � �3� � �

�3�

13. a. Always. This is the Angles Inscribed in a Semi-circle Conjecture.

b. Never. An angle inscribed in a major arc inter-cepts a minor arc, where the sum of the majorarc and the minor arc is the complete circle. Bythe Inscribed Angle Conjecture, the measure ofan angle that intercepts a minor arc will be halfof an arc measure that is greater than 0° and lessthan 180°, so the angle measure will be greaterthan 0° and less than 90°. Thus, the angle willalways be acute.

c. Sometimes. Because your book defines angles tohave measures that are greater than 0° and lessthan 180°, this statement is true only for minorarcs. If the arc is a semicircle, its measure is 180°.The measure of a major arc is found bysubtracting the measure of the correspondingminor arc from 360°.

d. Sometimes. The measure of an angle formed bytwo intersecting chords equals the average of the

�3��4

�3��4

�3��4

�3��4

�3��4

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measures of the two intercepted arcs, so thisstatement is true only when the two interceptedarcs are congruent. That happens when thechords are diameters.

e. Always. Make a sketch.Draw radii to the two points of tangency.

Because the two radii are congruent and thetwo tangent segments are congruent (TangentSegments Conjecture), the quadrilateral is akite. By the Tangent Conjecture, the twononvertex angles are right angles. Therefore, thetwo vertex angles must be supplementary. Thesetwo angles are the angle formed by the tangentsand the central angle of the minor intercepted arc.


Possible derivation:

V � �13�BH � �

13��

12�h(a � b)�H � �

13��

12�(2x)(2b � b)�H

� �13�[x(3b)]H � bxH � x 2H � x 2 � �

1x2� � 12x

EXTENSIONS

A. Conjecture: The largest angle of a triangle musthave a measure of at least 60°.


Given: Triangle ABC with �A the largest angle

Show: m�A � 60°

Paragraph Proof: Assume m�A 60°. Because�A is the largest angle of the triangle, m�B m�A and m�C m�A. Therefore, m�B 60°and m�C 60°. Then m�A � m�B � m�C 60° � 60° � 60° � 180°, or m�A � m�B �m�C 180°. This creates a contradiction because,by the Triangle Sum Theorem, m�A � m�B �m�C � 180°. Thus the assumption that m�A 60° is false, and the conjecture is true. (If thetriangle is isosceles or equilateral so that m�B �m�A and/or m�C � m�A, the proof is still valid.)

B. Conjecture: The largest side of a triangle must beopposite an angle with a measure greater than orequal to 60°.

Given: �ABC with BC� the longest side

Show: m�A � 60°

Paragraph Proof: Because �A is opposite BC�, andBC� is the longest side of �ABC, �A is the largestangle of the triangle by the Side-Angle InequalityConjecture. By the theorem that was proved inExtension A, m�A � 60°.

LESSON 13.6

EXERCISES

1. Given: Two inscribed angles in a circle that intercept the same or congruent arcs

Show: The inscribed angles are congruent

Case 1: Two angles intercepting the same arc

Given: Circle with inscribed angles XAW and XBWintercepting XW�

Show: �XAW � �XBW

Paragraph Proof: By the Inscribed Angle Theorem, m�XAW � �12�(mXW�) and

m�XBW � �12�(mXW�). Therefore, by the transitive property, m�XAW � m�XBW, so

�XAW � �XBW by the definition of congruence.

Case 2: Two angles intercepting congruent arcs

Given: Circle with inscribed angle ZAY intercepting ZY� andinscribed angle XBW intercepting XW�; ZY� � XW�

Show: �ZAY � �XBW

Paragraph Proof: By the Inscribed Angle Theorem,

m�ZAY � �12�(mZY�) and m�XBW � �

12�(mXW�). Because

ZY� � XW� is given, �12�(mZY�) � �

12�(mXW�) by the definition of

congruence and the multiplication property. Therefore, m�ZAY � m�XBW by the

transitive property, so �ZAY �XBW by the definition of congruence.

A Z

Y

XB

W

A

B

W

X

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2. Given: Quadrilateral inscribed in a circle

Show: The opposite angles of the quadrilateral are supplementary

Given: Circle with inscribed quadrilateral ABCD

Show: �A and �C are supplementary; �B and �D aresupplementary

Plan: Use the Inscribed Angle Theorem, the addition property, andthe distributive property to get m�A � m�C � �

12�(mBCD� � mDAB�). Then use the

definition of degrees in a circle, the substitution property, and the definition ofsupplementary to get �A and �C are supplementary. Repeat the above steps, usingdifferent angles and arcs, to get �B and �D are supplementary.


1. ABCD is inscribed in the circle 1. Given

2. m�A � �12�(mBCD�); m�C � �

12�(mDAB�) 2. Inscribed Angle Theorem

m�B � �12�(mACD�); m�D � �

12�(mABC�)

3. m�A � m�C � �12�(mBCD�) � �

12�(mDAB�) 3. Addition property

m�B � m�D � �12�(mACD�) � �

12�(mABC�)

4. m�A � m�C � �12�(mBCD� � mDAB�) 4. Distributive property

m�B � m�D � �12�(mACD� � mABC�)

5. m�A � m�C � �12�(arc measure of circle) 5. Arc Addition Postulate

m�B � m�D � �12�(arc measure of circle)

6. m�A � m�C � �12�(360°) 6. Arc measure of full circle � 360°

m�B � m�D � �12�(360°)

7. m�A � m�C � 180° 7. Multiplication property

m�B � m�D � 180°

8. �A and �C are supplementary 8. Definition of supplementary

�B and �D are supplementary

3. Given: Circle with parallel secants

Show: The secants intercept congruent arcs

Given: Circle with secants AB�� and CD��; AB�� CD��

Show: AC� BD�

Plan: Use the Line Postulate to construct AD��. Then usethe AIA Theorem and the Inscribed Angle Theorem toget AC� BD�.


1. Construct AD�� 1. Line Postulate

2. �CDA �BAD 2. AIA Theorem

3. m�CDA � m�BAD 3. Definition of congruence

4. m�CDA � �12�(mAC�); 4. Inscribed Angle Theorem

m�BAD � �12�(mBD�)

5. �12�(mAC�) � �

12�(mBD�) 5. Transitive property

6. mAC� � mBD� 6. Multiplication property

7. AC� BD� 7. Definition of congruence

D

B

C

A

D

AB

C


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4. Given: Parallelogram inscribed in a circle

Show: The parallelogram is a rectangle

Given: Parallelogram ACER inscribed in a circle

Show: ACER is a rectangle

Plan: Use the Cyclic Quadrilateral Theorem, the OppositeAngles Theorem, and the Congruent and SupplementaryTheorem to show that all four angles are right angles.


1. ACER is inscribed in a circle 1. Given

2. �A and �E are supplementary; 2. Cyclic Quadrilateral Theorem�C and �R are supplementary

3. ACER is a parallelogram 3. Given

4. �A �E; �C �R 4. Opposite Angles Theorem

5. �A and �E are right angles; 5. Congruent and Supplementary Theorem�C and �R are right angles

6. ACER is a rectangle 6. Definition of rectangle

5. Given: A circle and a point outside the circle; two segmentstangent to the circle with the point as one endpoint andthe point of tangency as the other endpoint

Show: The segments are congruent

Given: Circle O with point P outside the circle; PS��and PT�� tangent to the circle at S and T, respectively

Show: PS� � PT�

Plan: Use the Line Postulate to construct OS�, OT�, and OP�.Then use the Tangent Theorem, the Converse of the Angle Bisector Theorem, and the SAA Congruence Postulate to get �OSP �OTP. Use CPCTC to show PS� PT�.


1. Construct OS�, OT�, and OP� 1. Line Postulate

2. OS� OT� 2. Definition of circle

3. OS�� PS�; OT�� PT� 3. Tangent Theorem

4. �OSP is a right angle; 4. Definition of perpendicular�OTP is a right angle

5. �OSP �OTP 5. Right Angles Are Congruent Theorem

6. O is on the angle bisector of �SPT 6. Converse of the Angle Bisector Theorem

7. �1 �2 7. Definition of angle bisector

8. �OSP �OTP 8. SAA Theorem

9. PS� PT� 9. CPCTC

6. Given: Circle with two intersecting chords

Show: The measure of the angle formed by the chords ishalf the sum of the measures of the two intercepted arcs

Given: Circle with chords AB� and CD�� intersecting at E

Show: m�1 � �12�(mAC� � mBD�)

Paragraph Proof: Use the Line Postulate to construct AD��,forming �2 and �3. By the Inscribed Angle Theorem,m�2 � �

12�(mAC�) and m�3 � �

12�(mBD�). Then, by the addition and distributive

properties, m�2 � m�3 � �12�(mAC�) � �

12�(mBD�) � �

12�(mAC� � mBD�). By the

Triangle Exterior Angle Theorem, m�1 � m�2 � m�3, so, by the transitive property, m�1 � �

12�(mAC� � mBD�).

A

B

D

CE1

23

S

O

P

T

12

A

R

C

E

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11. A(4.0, 2.9), P(�1.1, �2.8). To find the coordinatesof A, draw the perpendicular segment from A to thex-axis, forming a right triangle with horizontal legof length a and vertical leg of length b.

�5a

� � cos 36°, so a � 5 cos 36° � 4.0.

�5b

� � sin 36°, so b � 5 sin 36° � 2.9.

To find the coordinates of P, draw a perpendicular segment from P to they-axis, forming a right triangle withhorizontal leg of length ⏐p⏐ and verticalleg of length ⏐q⏐. (Absolute-value barsare used here because the side lengths ofthe triangle are positive, but p and q, the coordinatesof P, are negative because P is in Quadrant III.)

� sin 22°, so ⏐p⏐ � 3 sin 22°, ⏐p⏐ � 1.1, and

p � �1.1.

� cos 22°, so ⏐q⏐ � 3 cos 22°, ⏐q⏐ � 2.8, and

q � �2.8.

12. BY � 3, YC � 1, AZ � 9. First look at �ABC with

XZ�� BC�. By the Parallel/Proportionality Conjecture,

�AX

XB� � �

AZC

Z�. Then �

62� � �

A3Z�, so 3 � �

62� � 3 � �

A3Z� and

AZ � 9. Now look at �ABC again, this time with

ZY�� AB�. Apply the Parallel/Proportionality

⏐q⏐�3

⏐p⏐�3

3 � q �

� p �P

5

36°a

b

A

9.

10.

SSS Congruence

Postulate

Linear Pair Postulate

CA Postulate

Parallel Postulate

Angle Addition Postulate

ASA CongruencePostulate

ITTheorem

VA Theorem

Inscribed Angle Theorem

Cyclic Quadrilateral Theorem

Parallelogram Inscribedin a Circle Theorem

ParallelogramDiagonal Lemma

Opposite AnglesTheorem

Right Angles Are Congruent

Theorem

Exterior Angle Sum Theorem

Arc Addition Postulate

AIA Theorem

Triangle Sum Theorem

Congruent and Supplementary Theorem

ASA Congruence Postulate

Line Postulate

Converse of the Angle Bisector Theorem


Converse of the IT Theorem

SAA Theorem

Tangent Theorem

IT Theorem

Angle AdditionPostulate


SAS CongruencePostulate

PerpendicularPostulate

MidpointPostulate

Tangent Segments Theorem


7. Intersecting Secants Theorem: The measure of an angle formed by twosecants intersecting outside a circle is half the difference between the measure of thelarger intercepted arc and the measure of the smaller intercepted arc.

Given: Circle O with point P outside the circle; secant PB�� intersects the circle at Aand B; secant PD�� intersects the circle at C and D

Show: x � �12�(mBD� � mAC�)

Paragraph Proof: Using the Line Postulate, construct AD��. By the Triangle ExteriorAngle Theorem, b � a � x, so, by the subtraction property, x � b � a. By theInscribed Angle Theorem, b � �

12�(mBD�) and a � �

12�(mAC�). Then, by the substitution

and distributive properties, x � �12�(mBD�) � �

12�(mAC�) � �

12�(mBD� � mAC�).

8. Given: Circle with an angle inscribed in a semicircle

Show: The inscribed angle is a right angle

Given: Circle A with �BDC inscribed in semicircle BDC

Show: �BDC is a right angle

Paragraph Proof: By the Inscribed Angle Theorem, m�BDC � �12�mBEC�. By the

definition of semicircle, mBEC� � 180°, so by the division property, m�BDC � 90°.By the definition of right angle, �BDC is a right angle. Because only definitions,properties, and the Inscribed Angle Theorem are needed to prove this conjecture,it is a corollary of the Inscribed Angle Theorem.

E

A C

D

B

B

A

D

C

x

P

b

aO

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Conjecture again to get the proportion � �AZC

Z�.

BC � 4, BY � 4 � YC, so �4 �YC

YC� � �39�. Then 9(YC)

� 3(4 � YC), or 9(YC) � 12 � 3(YC), so 12(YC) �

12, and YC � 1. Therefore, BY � 4 � YC � 3.

13. BC�, AB�, AC�, CD��, AD�. First use the Triangle SumTheorem to find the angle measures that are notgiven. In �ABC, m�ABC � 180° � 60° � 59° �61°. In �ADC, m�ACD � 180° � 57° � 61° � 62°.Now apply the Side-Angle Inequality Conjecture toeach triangle to find the relative lengths of the sides.

In �ABC, BC� is opposite the smallest angle, AB� isopposite the middle-size angle, and AC� is oppositethe largest angle, so BC � AB � AC.

In �ADC, AC� is opposite the smallest angle, CD�� isopposite the middle-size angle, and AD�� is oppositethe largest angle, so AC � CD � AD.

Because CD�� is common to both triangles and is thelongest side of one triangle but the shortest side ofthe other, you can arrange all five segments in orderfrom shortest to longest: BC�, AB�, AC�, CD��, AD�.

14. 32.5. Let P and Q be the centers of the smaller andlarger circles, respectively. Draw radii PA� and QB�.Also draw a segment AR� so that AR� � PQ� and R ison QB�.

By the Tangent Theorem, PA�� AB� and QB�� AB�,so PA� � QB�. AR� � PQ� by construction. Therefore,PARQ is a parallelogram. Because opposite sides ofa rectangle are congruent, QR � PA, so QR � 10, soBR � 20 � 10 � 10, and AR � PQ, so AR � 34.Apply the Pythagorean Theorem to right triangleABR to find AB: (AB)2 � (BR)2 � (AR)2, so AB �

�(AR)2�� (BR�)2� � �342 ��102� � �1056� � 32.5.

15. As long as P is inside the triangle, a � b � c � h.

Proof: Let x be the length of a side. The areas of

the three small triangles are �12�xa, �

12�xb, and �

12�xc.

The area of the large triangle is �12�xh. So, �

12�xa �

�12�xb � �

12�xc � �

12�xh. Divide by �

12�x on both sides to

get a � b � c � h.

ha

c

b

x

P

3434

2010

10

10

10

A

P Q

R

B

YC�BY 16. a. 27°. �P is formed by two secants intersecting

outside the circle, so apply the IntersectingSecants Theorem: m�P � �

12�(mNQ� � mOM�) �

�12�(94° � 40°) � �

12�(54°) � 27°.

b. 47°. �QON is an inscribed angle that interceptsNQ�, so, by the Inscribed Angle Theorem,m�QON � �

12�(94°) � 47°.

c. 67°. �QRN is formed by two intersecting chords,QM�� and ON��, so, by the Intersecting ChordsTheorem, m�QRN � �

12�(mNQ� � mOM�) �

�12�(94° � 40°) � �

12�(134°) � 67°.

d. 133°. Look at �QMP. From 16a, m�P � 27°.�PQM is the same angle as �OQM, which is aninscribed angle that intercepts MO�, so, by theInscribed Angle Theorem, the measure of thisangle is �

12�(40°) � 20°. Therefore, by the Triangle

Sum Theorem, m�QMP � 180° � 27° � 20° �133°. Another way to find m�QMP is to noticethat �QMN is an inscribed angle that interceptsNQ�, so m�QMN � 47°. Then, because �QMNand �QMP form a linear pair, m�QMP �180° � 47° � 133°.

e. Cannot be determined. m�ONF depends onmOMN�, or mMO� � mMN�. You know thatmOQ� � mMN� � 360° � 94° � 40° � 226°,but there is not enough information todetermine mOQ� and mMN� individually.

f. Cannot be determined. As in 16e, there is notenough information to determine mMN�.

17. First construct OP�, then bisect it. Label themidpoint M. Then construct a circle with center Mand radius PM. Label the intersection of the twocircles A and B. Finally, construct PA� and PB�, whichare the required tangents.

�OAP and �OBP are right angles; since they areinscribed in semicircles.


The color of a spot is determined by the colors of thetwo spots above it. If they are the same, the spot isyellow; otherwise, the spot is green. (The last spot in arow is always green because there can’t be two same-colored spots above it.) A row could never be all yellow

A

B

O M P

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LESSON 13.7

EXERCISES

1. Given: Two similar triangles with corresponding altitudes

Show: The altitudes are proportional tothe corresponding sides

Given: �BTA �GVL; TH�� is the altitude to BA�; VE� is the altitude to GL�

Show: �

Plan: Use the Right Angles Are Congruent Theorem, CASTC, and the AA SimilarityPostulate to get �BHT �GEV. Therefore, �G

BTV� � �

TV

HE� by CSSTP.


1. �BTA �GVL 1. Given

2. �B �G 2. CASTC

3. TH�� is the altitude to BA�; 3. GivenVE� is the altitude to GL�

4. TH�� BA�; VE� � GL� 4. Definition of altitude

5. �THB is a right angle; 5. Definition of perpendicular�VEG is a right angle

6. �THB �VEG 6. Right Angles Are Congruent Theorem

7. �BHT �GEV 7. AA Similarity Postulate

8. � 8. CSSTP

2. Given: Two similar triangles with corresponding medians

Show: The medians are proportional to thecorresponding sides

Given: �BGI �SAM; GY� is the median toBI�; AL� is the median to SM��

Show: �BSAG� � �

GA

YL�

YB I

G

LS M

A

Corresponding AltitudesTheorem


AA Similarity Postulate

TH�VE

BT�GV

TH�VE

BT�GV

EG L

V

HB A

T


because the last spot is always green. Because there areonly 25 � 32 possible ways to color a row (two choicesfor each of the first five spots), some row must berepeated after at most 32 rows, and then the whole cyclewill repeat. For example, the first row is repeated as theninth row, so the cycle is eight rows long. The cyclebased on this starting pattern does not include anall-green row. However, some other starting patterncan lead to an all-green row. In fact, starting with an

all-green row gives an eight-row cycle, which after theninth row is all green, and any row in the cycle wouldlead to all green. The figure below shows rows 7 through11. The shaded spots represent yellow.

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Paragraph Proof: By the definitions of median and midpoint, BY � YI andSL � LM. Then, by the Segment Addition Postulate and the substitution property,BI � 2BY and SM � 2SL, or BY � �

12�BI and SL � �

12�SM. �BGI �SAM is given,

so, by CSSTP,

�BSAG� � �

SBMI� � � �

BSLY�

Also, �B �S by CASTC. Therefore, �BYG �SLA by the SAS Similarity

Theorem. Therefore, �BSAG� � �

GA

YL� by CSSTP.

3. Given: Two similar triangles with corresponding angle bisectors

Show: The angle bisectors are proportional to thecorresponding sides

Given: �ACB �DFE; CP� is the angle bisectorof �ACB; FQ� is the angle bisector of �DFE

Show: �DAC

F� � �FC

QP�

Plan: Use the definition of angle bisector, the Angle Addition Postulate, and the substi-tution property to get m�ACB � 2m�1 and m�DFE � 2m�2. Then use CASTC andthe AA Similarity Postulate to get �APC �DQF. Therefore, �D

ACF� � �F

CQP� by CSSTP.


1. CP� is the angle bisector of �ACB; 1. GivenFQ� is the angle bisector of �DFE

2. �1 �3; �2 �4 2. Definition of angle bisector

3. m�1 � m�3; m�2 � m�4 3. Definition of congruence

4. m�ACB � m�1 � m�3; 4. Angle Addition Postulatem�DFB � m�2 � m�4

5. m�ACB � 2m�1; m�DFB � 2m�2 5. Substitution property

6. �ACB �DFE 6. Given

7. �A �D 7. CASTC

8. �ACB �DFE 8. CASTC

9. m�ACB � m�DFE 9. Definition of congruence

10. 2m�1 � 2m�2 10. Transitive property

11. m�1 � m�2 11. Division property

12. �1 �2 12. Definition of congruence

13. �APC �DQF 13. AA Similarity Postulate

14. �DAC

F� � �FC

QP� 14. CSSTP

A BP

C

1 3

D EQ

F

2 4

SAS SimilarityTheorem

SAS SimilarityTheorem

Corresponding MediansTheorem

SegmentAdditionPostulate

SegmentDuplicationPostulate

ParallelPostulate

CAPostulate

SASCongruencePostulate

AASimilarityPostulate

�12�BI��12�SM


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4. Given: Triangle cut by a line passing through two sides and parallel to the third side

Show: The line divides the two sides proportionally

Given: �ACB with DE�� AB�

Show: �CD

DA� � �C

EBE�

Plan: Use the CA Postulate and the AA Similarity Postulate to get �CDE �CAB.Then use CSSTP and the Segment Addition Postulate to get �CD

C�D

DA� � �

CEC�E

EB�.

Therefore, �CD

DA� � �C

EBE� by algebra and the subtraction property.


1. DE�� AB� 1. Given

2. �1 �2 2. CA Postulate

3. �DCE �ACB 3. Reflexive property

4. �CDE �CAB 4. AA Similarity Postulate

5. �CC

DA� � �

CC

BE� 5. CSSTP

6. CA � CD � DA; CB � CE � EB 6. Segment Addition Postulate

7. �CDC�D

DA� � �

CEC�E

EB� 7. Substitution property

8. �CC

DD� � �C

DDA� � �

CC

EE� � �C

EBE� 8. Algebra

9. 1 � �CD

DA� � 1 � �C

EBE� 9. Algebra

10. �CD

DA� � �C

EBE� 10. Subtraction property

5. Given: Triangle cut by a line passing through two sides,dividing them proportionally

Show: The line is parallel to the third side

Given: �ACB with DE�� intersecting CA� and CB� such that

�CD

DA� � �C

EBE�

Show: DE�� AB�

Plan: Use the addition property to get �CD

DA� � 1� �C

EBE� � 1. Then use algebra and

the Segment Addition Postulate to get �CC

DA� � �

CC

BE�. Therefore, �ABC �DEC by the

SAS Similarity Theorem, �1 �2 by CASTC, and DE�� AB� by the CA Postulate.


1. �CD

DA� � �C

EBE� 1. Given

2. 1 � �CD

DA� � 1 � �C

EBE� 2. Addition property

3. �CC

DD� � �C

DDA� � �

CC

EE� � �C

EBE� 3. Algebra

4. �CDC�D

DA� � �

CEC�E

EB� 4. Algebra

5. CA � CD � DA; CB � CE � EB 5. Addition property

6. �CC

DA� � �

CC

BE� 6. Substitution property

7. �ACB �DCE 7. Reflexive property

8. �ABC �DEC 8. SAS Similarity Theorem

9. �1 �2 9. CASTC

10. DE�� AB� 10. CA Postulate

A B

C

D E1

2

A B

C

D E1

2


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6. Given: Right triangle with altitude to the hypotenuse

Show: The two smaller triangles formed are similar to each other and to the original triangle

Given: Right triangle ABC with right angle ACB; CD�� is the altitude to hypotenuse AB�

Show: �ADC �ACB �CDB

Plan: Use the Right Angles Are Congruent Theorem, the reflexive property, and the AA Similarity Postulate to get �ADC �ACB and �ACB �CDB.Therefore, �ADC �ACB �CDB by the transitive property of similarity.


1. �ACB is a right angle 1. Given

2. CD�� is the altitude to hypotenuse AB� 2. Given

3. CD�� AB� 3. Definition of altitude

4. �CDA is a right angle 4. Definition of perpendicular

5. �CDA �ACB 5. Right Angles Are Congruent Theorem

6. �1 �1 6. Reflexive property

7. �ADC �ACB 7. AA Similarity Postulate

8. �CDB is a right angle 8. Definition of perpendicular

9. �ACB �CDB 9. Right Angles Are Congruent Theorem

10. �2 �2 10. Reflexive property

11. �ACB �CDB 11. AA Similarity Postulate

12. �ADC �ACB �CDB 12. Transitive property of similarity

7. Given: Right triangle with altitude to the hypotenuse

Show: The length of the altitude is the geometric mean of the lengths of the two segments on the hypotenuse

Given: Right triangle ABC with right angle ACB;CD�� is the altitude to hypotenuse AB�

Show: CD is the geometric mean of AD and DB

Paragraph Proof: By the Three Similar Right Triangles, �ADC �CDB. Then, byCSSTP, �C

ADD� � �

CD

DB�, or �h

x� � �

hy�. This is equivalent to h2 � xy, or h � �xy� (h must be

positive). Thus, by the definition of geometric mean, h is the geometric mean ofx and y, or CD is the geometric mean of AD and DB.

8. Given: Right triangle

Show: The square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two legs

Given: Right triangle with legs of lengths a and b andhypotenuse of length c

Show: a2 � b2 � c2

Paragraph Proof: Use the Perpendicular Postulate to draw the altitude to thehypotenuse. By the Three Similar Right Triangles Theorem, each of the small righttriangles is similar to the large right triangle. Because the small triangle on the rightis similar to the large triangle, �

ac� � �

c �a

d� by CSSTP. This proportion is equivalent to

�c �a

d� � �ac

�. (The ratio of the length of the shorter leg to the length of the hypotenuseis the same in both triangles.) By algebra, either of these proportions is equivalent toa2 � c(c � d). By the distributive property, a2 � c2 � cd. Now look at the othersmall triangle, which is also similar to the large triangle by the Three Similar Right

dc

c – d

b a

BA

C

Dx y

h

BA

C

D1 2

DG4SM_890_13.qxd 12/21/06 11:21 AM Page 309


Triangles Theorem, so �bc� � �

db� by CSSTP. This proportion is equivalent to �d

b� � �b

c�. (The

ratio of the length of the hypotenuse to the length of the longer leg is the same inboth triangles.) By algebra, either of these proportions is equivalent to b2 � cd.Substitute b2 for cd in the equation a2 � c2 � cd to obtain a2 � c2 � b2. Then, bythe addition property, a2 � b2 � c2.

9. Given: Triangle with the square of one side equal to the sum of the squares ofthe other two sides

Show: The triangle is a right triangle

Given: �ABC with side lengths a, b,and c such that a2 � b2 � c2

Show: �ABC is a right triangle with right angle C

Paragraph Proof: Construct right triangle DEF with �E a right angle, and with legsof lengths a and b. Let x represent the length of the hypotenuse of this triangle. Bythe Pythagorean Theorem, a2 � b2 � x2. Because a2 � b2 � c2 is given, x2 � c2

by the transitive property of equality. Because x and c both represent side lengthsand therefore must be positive numbers, x � �c2� � c. Therefore, �DEF �ABCby the SSS Congruence Postulate. Then �E �C by CPCTC. �E is a right angle byconstruction and �E �C, so �C must also be a right angle. Therefore, �ABC is aright triangle with right angle C.

10. Given: Two right triangles with the hypotenuse and one leg of one triangle congruent to thehypotenuse and one leg of the other triangle

Show: The triangles are congruent

Given: Right triangles HYP and LEG with right angles Y and E; HP� LG�; YP� EG�

Show: �HYP �LEG

Plan: Use the Pythagorean Theorem to write expressions for the lengths of theunknown legs. Show that the expressions are equivalent. The triangles are congruentby SSS or SAS.


1. �HYP is a right triangle with right 1. Givenangle Y; �LEG is a right triangle with right angle E

2. HP� LG�; YP� EG� 2. Given

3. HP � LG; YP � EG 3. Definition of congruence

4. (HY)2 � (YP)2 � (HP)2; 4. Pythagorean Theorem(LE)2 � (EG)2 � (LG)2

5. HY � �(HP)2�� (YP�)2�; 5. Algebra

LE � �(LG)2�� (EG�)2�6. HY � LE 6. Substitution and transitive properties

7. HY� LE� 7. Definition of congruence

8. �HYP �LEG 8. SSS Congruence Postulate

Y

H

P E

L

G

B A

C

c

ba

D F

E

x

ba


DG4SM_890_13.qxd 12/21/06 11:21 AM Page 310


triangles into two congruent right triangles in whichthe length of one leg is half the length of the chord,and the hypotenuse is a radius of the circle.

Apply the Pythagorean Theorem to each of the righttriangles shown to find x and y.

x � �(4.8)2�� (2.6�)2�y � �(4.8)2�� (4.1�)2�The distance between the two chords is x � y �

�(4.8)2�� (2.6�)2� � �(4.8)2�� (4.1�)2� � 1.5 cm.

Now look at the case where the chords are onopposite sides of the diameter.

Notice that the distances of the two chords from thecenter, x and y, are the same as before, but now thedistance between the chords is x � y �

�(4.8)2�� (2.6�)2� � �(4.8)2�� (4.1�)2� � 6.5 cm.

15. a. C. The perimeter of the hexagon is 6(10) �60 cm, and the perimeter of the pentagon is5(12) � 60 cm, so the perimeters are equal.

b. A. Divide each polygon into congruent isoscelestriangles by drawing congruent segments from thecenters of the polygons to each of their vertices.In each case, the apothem is an altitude of one ofthese triangles, so this segment divides the isoscelestriangle into two congruent right triangles.

Regular hexagon:

m�RST � �16�(360°) � 60°,

so �RST is an equilateraltriangle and �RSU is a 30°-60°-90° triangle. Inthis triangle, RU� is theshorter leg and SU� is thelonger leg. Because RU ��12�(10) � 5 cm, a � SU � 5�3� cm � 8.66 cm.

R TU60°

30°

S

a

5.2 cm

x

y

8.2 cm

2.6 cm

4.1 cm

4.8 cm4.8 cm

x y

11.

12.

13.

14. Approximately 1.5 cm or 6.5 cm. There are twopossible answers because the chords can beeither on the same side or on opposite sides of thediameter they are parallel to. First look at the casewhere the chords are on the same side of thediameter. The longer chord will be closer to thecenter of the circle.

In each case, the distance from the center to thechord is the altitude of an isosceles triangle whoselegs are radii and therefore have length �

12�(9.6) �

4.8 cm. These altitudes divide each of the isosceles

5.2 cm

x

y8.2 cm

9.6 cm

Right AnglesAre Congruent

Theorem

Three SimilarRight Triangles

Theorem

Perpendicular Postulate

PythagoreanTheorem


SAS Congruence

PostulateSSS

Congruence Postulate

Segment DuplicationPostulate

CA Postulate

Parallel Postulate


SAS Similarity Theorem

SSS Similarity Theorem

CAPostulate

AA SimilarityPostulate

Segment AdditionPostulate

Parallel/ProportionalityTheorem


DG4SM_890_13.qxd 12/21/06 11:21 AM Page 311


Look at �PSQ. Because PS� is tangent to circle Q,PS� � SQ� by the Tangent Conjecture, som�PSQ � 90°. Therefore, m�SQP � 180° �57° � 90° � 33°. Because �SQZ is a centralangle in circle Q and �SQZ and �SQP are thesame angle, mSZ� � 33°. Because �PSQ �PTQ by the SSS Congruence Postulate (usingthe two pairs of congruent radii and thecommon side), m�TQZ � m�SQZ � 33°,so m�SQT � 66°. In circle Q, �SLT is aninscribed angle that intercepts SZT�, so m�SLT ��12�(mSZT�) � �

12�(66°) � 33°. �GLT is the same

angle as �SLT, so m�GLT � m�SLT � 33°.

b. 66°. m�SQT was found in the solution for 18a.

c. 57°. PS� PT� because they are both radii ofcircle P, and QS� QT� because they are bothradii of circle Q. Therefore, PSQT is a kite.Then PQ� � ST� because the diagonals of a kiteare perpendicular, so �SRQ is a right triangle.Look at the angles in this triangle. From 18a,m�SQR � 33°, so m�RSQ � 90° � 33° �57°. �RSQ is the same angle as �TSQ, som�TSQ � 57°.

d. 62°. Look at �GTL. �SGT is an inscribed anglein circle P, so m�SGT � �

12�(mSWT�). mSWT� �

m�SPT by the definition of arc measure, andm�SPT � 2(57°) � 114° by the Kite DiagonalConjecture. So, m�SGT � �

12�(114°) � 57°. This

gives one angle measure in �GTL because �SGTand �LGT are the same angle. From 18a,m�GLT � 33° (�GLT is the same angle as�SLT ). Then, by the Triangle Sum Conjecture,m�GTL � 180° � 57° � 33° � 90°. By theAngle Addition Postulate, m�GTS � m�STL �m�GTL. Because �GTS is an inscribed angle incircle P, m�GTS � �

12�mGS� � �

12�(118°) � 59°.

Then m�STL � m�GTL � m�GTS � 90° �59° � 31°. Because �STL is an inscribed anglein circle Q, m�STL � �

12�(mSL�), or mSL� �

2(m�STL) � 2(31°) � 62°.

e. �PSQ and �PTQ are both right angles by theTangent Theorem and thus are supplementary.Therefore, �SPT and �SQT must also besupplementary by the Quadrilateral SumTheorem. Therefore, PSQT is cyclic by theConverse of the Cyclic Quadrilateral Theorem.

NAQ

P

L

G

S

Z

WR

T

57°

118°

Regular pentagon:

m�WXY � �15�(360°) � 72°. Because �WXY is

isosceles, m�XWY � �12�(180° � 72°) � 54°.

Now look at right triangle WXZ. Here WZ ��12�(WY) � 6 cm. Because �XWZ and �XWYare the same angle, m�XWZ � 54°. Then �6

a� �

tan 54°, so a � 6 tan 54° � 8.26 cm.

Thus, the apothem is greater in the hexagon.

c. A. Recall that you can find the area of a regularpolygon by using the formula A � �

12�aP, where

a is the length of the apothem and P is theperimeter (Regular Polygon Area Conjecture).Because both of the regular polygons have thesame perimeter (from 15a), the polygon with thegreater apothem will have the greater area. From15b, you know that this is the hexagon.

d. A. Recall that the Polygon Sum Conjecture statesthat the sum of the measures of the n interiorangles of an n-gon is 180°(n � 2). It follows thatas the number of sides of an n-gon increases, thesum of the measures of the interior angles willincrease. Therefore, because the hexagon hasmore sides than the pentagon, the sum of themeasures of the interior angles will be greater inthe hexagon.

e. C. Recall that the Exterior Angle Sum Conjecturestates that for any polygon, the sum of the mea-sures of a set of exterior angles (one exteriorangle at each vertex) is 360°. Therefore, the sumis the same for both polygons.

16. a. The vectors are diagonals of your quadrilateral.

b. A 180° rotation about the midpoint of thecommon side; the entire tessellation maps ontoitself.

17. a. A 180° rotation about the midpoint of any side

b. Possible answer: A vector running from eachvertex of the quadrilateral to the opposite vertex(or any multiple of that vector)

18. a. 33°. Label the intersection point of PQ� and ST� asR and the intersection points of the two circleswith PQ� as Z and W, as shown in the figurebelow. Also construct GT�.

W YZ54°

36°

X


DG4SM_890_13.qxd 12/21/06 11:21 AM Page 312


�x �x

32� � �23

72�

32x � 27(x � 32)

32x � 27x � 864

5x � 864

x � 172.8

Then, by the Pythagorean Theorem, r �

�(172.8�)2 � (1�3.5)2� � 172.3 � 172. The rise, whichis the radius of the semicircular arch, is approxi-mately 172 cm. The span is the diameter of thearch, so it is twice the rise. Therefore, the span isapproximately 344.6 � 345 cm. Use the inverse sineto find the central angle measure: � = sin�1��11732.5.8�� 4.5°, so the central angle is approximately 2(4.5°) �9°. Divide into 180° to find the number ofvoussoirs: �

189

0°

°� � 20.

IMPROVING YOUR ALGEBRA SKILLS

1. Add; 5x � y � 29

2. Add; 10x � 2y � 50

3. Subtract; x � y � 4

EXTENSIONS

A.

Given: �ABC �DEF with �DAB

E� � �BE

CF� � �D

ACF� � s;

BD� is the altitude to AC�; EQ� is the altitude to DF�

Show: � s2

Paragraph Proof: Choose AC� as the base of �ABC

and DF� as the base of �DEF. Let b � DF and

h � EQ. �DAC

F� � s, so �AbC� � s, and AC � sb. BD� is the

altitude to AC� and EQ� is the altitude to DF�, so, by

the Corresponding Altitudes Theorem, �BEQ

D� � �D

ABE�.

Thus, by the transitive property, �BEQ

D� � s, or �

BhD� � s,

so BD � hs. Now find the area of each triangle:

Area of �DEF � �12�(DF)(EQ) � �

12�bh

Area of �ABC � �12�(AC)(BD) � �

12�(sb)(sh)

� �12�s2bh = s2��

12�bh�

�AA

rreeaa

oo

ff

��

AD

BECF� � � s2

s2��12�bh�

��12�bh

Area of �ABC��Area of �DEF

D

E

B

F CAQ D

hsh

b sb

f. Because PS� is tangent to circle Q, m�PSQ � 90°(Tangent Theorem). Because m�PSQ � 90°, SQ�must be tangent to circle P (Converse of TangentTheorem).

19. a.�2� � � 1

Substitute the area expressions you found in Lesson 13.1, Exercise 31c and in Lesson 13.5,Exercise 12b.

A �

�

� � �1�2� � � 1

b. ��6� � � 1

Substitute the area expressions you found in Lesson13.5, Exercise 12a and in 19a above.

A � � � �1�2�� 1

�2� � � 1�

� � �1�2� � �1

�2� � � 1 � ��

�6� � � 1

20. a. �CDG �CFG by SAA; �GEA �GEB bySAS; �DGA �FGB by the Hypotenuse LegTheorem.

b. CD�� CF� and DA� FB� by CPCTC; CD �DA � CF � FB (addition property). Therefore,CA� CB�, and �ABC is isosceles.

c. The figure is inaccurate.

d. The angle bisector does not intersect the perpen-dicular bisector inside the triangle as shown,except in the special case of an isosceles triangle,when they coincide.

21. 173 cm, 345 cm, 20 stones. Draw the trapezoid, andextend the legs until they meet to form a triangle.Let r represent the rise. Use proportional sides insimilar triangles to find x.

32 cm

27 cm

13.5 cm

32 cm 32 cm

x r�

�3��4�3��2

�3��4

�3��2�3��4

�

�3��4

�3��2

��6� � �3� � 2��2

��2� � 1 � 1 � �3� � �

�3�

��2

��2� � 1� � �1 � �3� � �

�3��

��2

�

2

�3��2


DG4SM_890_13.qxd 12/21/06 11:21 AM Page 313


The shorter leg of the right triangle has length b.Because AD��, the hypotenuse of this triangle, is aside of the rhombus, its length is a. Therefore, bythe Pythagorean Theorem, the length of the longerleg is �a2 � b�2�. This is the y-coordinate of D andtherefore also the y-coordinate of C. Thus, thecoordinates of C are �a � b, �a2 � b�2�� andthe coordinates of D are �b, �a2 � b�2��.

4. Possible answer:

Slope RE�� 0a �

�00

� � �0a� � 0

Slope CE�� ba

��

0a� � �0

b� (undefined)

Slope TC� � �ab

��

0b

� � �0a� � 0

Slope TR� � �0b �

�00� � �0

b� (undefined)

Opposite sides have the same slope and are there-fore parallel by the parallel slope property. (In thiscase, two sides have undefined slope.) Two sides arehorizontal and two sides are vertical, so the anglesare all right angles. Because RECT is a parallelogramwith four right angles, it is a rectangle.

5. Possible answer:

Coordinates of X � ��a �2

0�, �

0 �2

0�� 2

a�, 0�

Coordinates of Y � ��a �2

b�, �

0 �2

c�� a �

2b

�, �2c

��Coordinates of Z � ��b �

20

�, �c �

20

�� 2b

�, �2c

��X, Y, and Z are the midpoints of TR�, RI�, and TI�,respectively, by the coordinate midpoint property.Therefore XY�, YZ�, and ZX� are midsegments bydefinition.

I

x

y

(b, c)

R (a, 0)T (0, 0) 0

Z ( ),b2

c2

Y ( ),a + b2

c2

X ( ),a2

C

x

y

(a, b)

E (a, 0)R (0, 0)

T (0, b)

B.

Given: Two similar polygons with scale factor s

Show: The area of one polygon is s2 times the areaof the other polygon

Paragraph Proof: Divide each polygon into (n � 2)triangles by drawing the diagonals from one vertex.(The process is shown for hexagons.) Show that�AFE �GLK and �ABC �GHI by the SASSimilarity Theorem. Then use CSSTP and CASTCand the subtraction property to show that�AED �GKJ and �ACD �GIJ.

(Although this process is shown only for hexagons,it can be generalized to any polygons.) Then,because the polygons are made up of similar trian-gles, you can use the result from Extension A toshow that their areas are similar, with ratio s2.

USING YOUR ALGEBRA SKILLS 13

EXERCISES

1. B(a, 0). The y-axis passes through the midpoint ofAB�, the base of the isosceles triangle. B lies on thex-axis. Because (0, 0) is the midpoint of AB�, B mustbe the same distance from the origin as A, so thecoordinates of B are (a, 0).

2. C(a � b, c). Because ABCD is a parallelogram,DC�� AB�. AB� lies along the x-axis, so DC�� must alsobe a horizontal segment. Therefore, the y-coordinateof C must be the same as the y-coordinate of D,which is c. Because opposite sides of a parallelogramare congruent, DC � AB. Because AB � a � 0 � a,the x-coordinate of C must be a � b to get CD �(a � b) � b � a.

3. C�a � b, �a2 � b�2��, D�b, �a2 � b�2��. Thelength of AB� is a � 0 � a. All sides of a rhombusare congruent, so the length of each of the otherthree sides must also be a. A rhombus is aparallelogram, so DC�� AB�. Therefore, DC�� must bea horizontal segment, and C and D will have the samey-coordinate. The x-coordinate of C must be a � bto get CD � (a � b) � b � a. Draw a perpen-dicular segment from D to the x-axis to form aright triangle.

D (b, ?) C (?, ?)

B (a, 0)A (0, 0)

y

x

a2 � b2

A

E D

B

CF

G

K J

H

IL


DG4SM_890_13.qxd 12/21/06 11:21 AM Page 314


Given: Rectangle RECT with diagonals RC� and TE�

Show: RC� TE�

Proof: To show that two segments are congruent,use the distance formula to show that they have thesame length.

TE � �(a � 0�)2 � (0� � b)2� � �a2 � b�2�RC � �(0 � a�)2 � (0� � b)2� � �a2 � b�2�RC� TE� because both segments have the samelength. Therefore, the diagonals of a rectangle arecongruent.

9. Given: A triangle with one midsegment

Show: The midsegment is parallel to the third sideand half the length of the third side

Given: �TRI and midsegment ZY�

Show: ZY� � TR� and ZY � �12�TR

Proof: To show that two segments are parallel, usethe parallel slope property. To compare lengths, usethe distance formula.

Slope TR�� 0a �

�00

� � �0a� � 0

Slope ZY�� 0

The slopes are the same, so the segments areparallel by the parallel slope property. TR� and ZY�are horizontal lines, so to find their lengths, subtracttheir x-coordinates.

TR � a � 0 � a

ZY � �a �

2c

� � �2c

� � �2a

� � �12�a � �

12�TR

Thus, the midsegment is half the length of the thirdside. Therefore, the midsegment of a triangle isparallel to the third side and half the length of thethird side.

0��2a

�

�d2� � �

d2�

��a �

2c

� � �2c

�

I

x

y

(c, d)

R (a, 0)T (0, 0)

Z ( ),c2

d2

Y ( ),a + c2

d2

C

x

y

(a, b)

E (a, 0)R (0, 0)

T (0, b)

6.


�00

� � �0a� � 0

Slope RA�� a �d �

c �0

a� � ��dc�

Slope AP�� ad–

�c –

dc� � �a �

02c� � 0

Slope PT�� dc �

�00

� � �dc�

TR� and AP� have the same slope and are parallelby the parallel slope property. PT� and RA� haveunequal slopes, so they are not parallel. Thus, TRAPhas only one pair of parallel sides and is a trapezoidby definition.

Use the distance formula to find the lengths of thenonparallel sides.

PT � �(c � 0�)2 � (d� � 0)2� � �c2 � d�2�RA � �(a � c� � a)2� � (d �� 0)2�

� �(�c)2�� d2� � �c2 � d�2�The nonparallel sides of the trapezoid have thesame length, so trapezoid TRAP is isosceles bydefinition.

7.

The y-axis splits �EQU into two 30°-60°-90° trian-gles, and the altitude of the equilateral triangle isthe longer leg of each of the 30°-60°-90° triangles,so the y-coordinate of U will be a�3�.

EQ � 2a

EU � �(�a � 0�)2 � �0 �� a�3��2� � �a2 � 3�a2�� 4a2� � 2a

UQ � �(a � 0)2� � �0 �� a�3��2� � �a2 � 3�a2�� 4a2� � 2a

Therefore, EQ � EU � UQ, so EQ� EU� UQ��,and �EQU is equilateral by definition.

8. Given: A rectangle with both diagonals

Show: The diagonals are congruent

x

y

U

E Q

(a, 0)

(0, a 3)

(–a, 0)

P

x

y

(c, d) A (a – c, d)

R (a, 0)T (0, 0)


DG4SM_890_13.qxd 12/21/06 11:21 AM Page 315


length to each other. Therefore, KITE is a kite bydefinition. Therefore, if only one diagonal of aquadrilateral is the perpendicular bisector of theother diagonal, then the quadrilateral is a kite.

12. Given: A quadrilateral with midpoints connected toform a second quadrilateral

Show: The second quadrilateral is a parallelogram

Given: Quadrilateral QUAD with midpoints P, R, L,and G

Show: PRLG is a parallelogram

Proof: To show that a quadrilateral is a parallelo-gram, show that opposite sides have the same slope.

Slope PR�� bc

�

Slope RL�� d �e

a�

Slope LG�� bc

�

Slope GP�� d �e

a�

PR� and LG� have the same slope, and RL� and GP�have the same slope, so the opposite sides areparallel by the parallel slope property, and PRLG isa parallelogram by definition. Therefore, the figureformed by connecting the midpoints of the sides ofa quadrilateral is a parallelogram.

13. Given: An isosceles triangle with the midpoint ofthe base connected to the midpoint of each leg toform a quadrilateral

Show: The quadrilateral is a rhombus

x

y

C (2a, 0)

A (a, h)

B (0, 0) E (a, 0)

D ( ),a2

h2

F ( ),3a2

h2

�2e

��d �

2a

�

�2e

� � 0��d2� � �2

a�

��2

c�

��2b�

�2e

� � �c �

2e

��d2� � �

b �2

d�

�2e

��d �

2a

�

�c �

2e

� � �2c

��b �

2d

� � �a �

2b

�

�2c

��2b

�

�2c

� � 0��a �

2b

� � �2a

�

D

x

y

(d, e)

A (b, c)

U (a, 0)Q (0, 0)

G( ),d

2

c2

e2

P ( ), 0a2

L ( ),b + d2

c + e2

R ( ),a + b2

10. Given: Trapezoid

Show: The midsegment is parallel to the bases

Given: Trapezoid TRAP with midsegment MN��

Show: MN�� TR�

Proof: By the coordinate midpoint property, thecoordinates of M are ��2

b�, �2

c�� and the coordinates of

N are ��a �2

d�, �2

c��. To show that the midsegment and

bases are parallel, find their slopes.

Slope MN�� 0


�00

� � �0a� � 0

Slope PA�� dc �

�cb� � �d �

0b� � 0

The slopes are equal. Therefore, the lines are parallel.

11. Given: A quadrilateral in which only one diagonalis the perpendicular bisector of the other

Show: The quadrilateral is a kite

Given: Quadrilateral KITE with diagonals IE� andTK�. IE� is the perpendicular bisector of TK�, but TK�is not the perpendicular bisector of IE� �⏐b⏐ � ⏐c⏐�Show: KITE is a kite

Proof: To show that a quadrilateral is a kite, use thedistance formula to show that only two pairs ofadjacent sides have the same length.

KI � �(0 � a�)2 � (b� � 0)2� � �a2 � b�2�IT � �(0 � (��a))2�� (b �� 0)2� � �a2 � b�2�TE � �(0 � (��a))2�� (c �� 0)2� � �a2 � c�2�EK � �(a � 0�)2 � (0� � c)2� � �a2 � c�2�Adjacent sides KI� and IT� have the same length andadjacent sides TE� and EK� have the same length, andbecause ⏐b⏐ � ⏐c⏐, the pairs are not equal in

I

x

y

(0, b)

K(a, 0)T (–a, 0)

E (0, c)

M(0, 0)

0��a �

2d � b�

�2c

� � �2c

��a �

2d

� � �2b

�

P

x

y

(b, c) A (d, c)

R

M N

(a, 0)T (0, 0)


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leave no gap, and three of any polygon with morethan six sides overlap. (See the Platonic Solids video.)

2. If each vertex is an endpoint of an even number ofedges, then whenever a traveler enters that vertex onone edge, at least one unused edge is available fordeparture. (The vertex at which the traveler beginswill have one edge left for ending there.) If all buttwo vertices have an even number of edges, then thesame argument applies to the nonspecial vertices;moreover, one of those special vertices can be usedas a beginning and the other as an ending, so thenetwork can be traveled. If more than two verticeshave an odd number of edges, the network cannotbe traveled, because one edge of one of thosevertices can be used as a beginning and one edge ofanother of those vertices can be used as an ending,but the third vertex will have a leftover edge whichcan be neither a beginning nor an ending.

3. The n-gon can be divided into n triangles with acommon vertex by drawing a segment from aninterior point to each of the n vertices. (Eachsegment drawn to one of the n vertices is the sideof two triangles, and each side of the n-gon is theside of one triangle; this gives 2n � n � 3n sidesand therefore n triangles.) Each triangle has interiorangles whose measures sum to 180°. Because thesum of the measures of the interior angles of then-gon is the sum of the measures of the trianglesminus the 360° around the central point, it equalsn � 180° � 360° � (n � 2)180°.

4. To consider PQ� reflected across two lines, OM�� andOL��, you must consider three cases: PQ� is outsidethe angle and is reflected across the nearer line first;PQ� is inside �MOL; and PQ� is outside the angleand is reflected across the farther line first. Considerthe second case:

Because reflection is an isometry, PO� P�O�� andP�O�� PO��, so PO� P O��. Likewise, QO�� Q�O�� and Q�O�� Q O��, so QO�� Q O��. By angleaddition, m�POP � (m�POM � m�MOP�) �(m�P�OL � m�LOP). Also, because reflection is anisometry, �P OM �MOP� and �P�OL �LOP.So, by substitution, m�POP � 2(m�MOP� �m�P�OL). But m�MOP� � m�P�OL � m�MOL.

OM

Q

Q'

Q''

P

LP'

P''

Given: Isosceles triangle ABC with midpoint ofbase, E, and midpoints of legs, D and F, connectedto form quadrilateral ADEF

Show: ADEF is a rhombus

Proof: Use the distance formula to show that all thesides of ADEF have the same length.

AD � �a � �2a��2

� ��h � �h2��2� � ��2

a��2

��h2��2�

�

AF � ��32a� �� a�2

��h2� ��h�2� � ��2

a��2

��h2��2�

�

DE � �a � �2a��2

� ��0 � �h2��2� � ��2

a��2

��h2��2�

�

EF � ��32a� �� a�2

��h2� ��0�2� � ��2

a��2

��h2��2�

�

AD � AF � DE � EF by the transitive property.Therefore, ADEF is a rhombus by the definition of arhombus.

PROJECT

Project should satisfy the following criteria:

● Proofs, which might not match the sample proofs, arelogically valid and clearly written. Students need notcreate the proofs themselves; they might research theproofs and write them up to demonstrate under-standing.

● Students use the Triangle Sum Theorem for proof 3.To prove it formally for any n-gon requires mathemat-ical induction.

● To illustrate proof 4, students draw rays that start atthe point of intersection of the two intersecting linesand that pass through corresponding points on theoriginal figure, and then draw its first and secondimages.

● In proof 5, students find the ordered pairs of the threemidpoints and the equations of the three medians.Then they solve the equations simultaneously.

Sample proofs:

1. Three, four, or five triangles fitted about a pointleave a gap that can be closed by folding thetriangles to start a tetrahedron, an octahedron, oran icosahedron, respectively. Likewise, three squaresor three pentagons leave a gap and can be foldedto start a cube or a dodecahedron. Three hexagons

�a2 � h�2��2

�a2 � h�2��2

�a2 � h�2��2

�a2 � h�2��2


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� + · · · � � 2

There are 17 groups of positive integers that satisfythis equation. Therefore, there are 17 choices ofpolygons that can be fitted around a vertex withoutgaps or overlaps. In four cases, these polygons canbe arranged in two ways (such as 32.4.12 and3.4.3.12). Thus, there are 21 possible types ofvertices: 36, 34.6, 33.42, 32.4.3.4, 32.4.12, 3.4.3.12,3.6.3.6, 32.62, 3.4.2.6, 3.4.6.4, 3.7.42, 3.8.24, 3.9.18,3.10.15, 3.122, 44, 4.5.20, 4.6.12, 4.82, 52.10, and 63.After checking each of these types, you will find 11that are edge-to-edge tilings by regular polygonswith all vertices of the same type: 36, 34.6, 33.42,32.4.3.4, 3.6.3.6, 3.4.6.4, 3.122, 44, 4.6.12, 4.82, and63. (See Tilings and Patterns by Gruenbaum andShephard.)

EXTENSION

Given: Triangle with side lengths a, b, and c;altitude with length h to side of length a

Show: Area of triangle � �s(s � a�)(s � b�)(s � c�)�,where s � �

a �2b � c�

Paragraph Proof: The altitude divides the triangleinto two right triangles and divides the side oflength a into two parts. By the PythagoreanTheorem, the lengths of these parts are �b2 � h�2�and �c2 � h�2�. Therefore, a � �b2 � h�2� �

�c2 � h�2�. To eliminate radicals from this equation,square both sides twice.

a � �b2 � h�2� � �c2 � h�2�a2 � ��b2 � h�2� � �c2 � h�2��2

a2 � b2 � h2 � 2�b2 � h�2��c 2 � h�2� + c 2 � h2

a2 � b2 � c 2 � 2h2 � 2�b2 � h�2��c 2 � h�2�(a2 � b2 � c 2 � 2h2)2 � �2�b2 � h�2��c 2 � h�2��2

[(a2 � b2 � c 2) � 2h2]2 � �2�b2 � h�2��c 2 � h�2��2

(a2 � b2 � c 2)2 � 4h2(a2 � b2 � c 2) � 4h4

� 4(b2 � h2)(c2 � h2)

(a2 � b2 � c 2)2 � 4h2a2 � 4h2b2 � 4h2c2 � 4h4

� 4(b2c2 � b2h2 � c2h2 � h4)

(a2 � b2 � c 2)2 � 4h2a2 � 4h2b2 � 4h2c2 � 4h4

� 4b2c2 � 4b2h2 � 4c2h2 � 4h4

(a2 � b2 � c 2)2 � 4h2a2 � 4b2c2

4h2a2 � 4b2c2 � (a2 � b2 � c 2)2

b ch

a

nk � 2�nk

n2 � 2�n2

n1 � 2�n1

Therefore, m�POP � 2m�MOL. The same proce-dure applies to �QOQ . Because PQ� is the samedistance from point O as is P Q �� and because bothimages move the same number of degrees, therefore,by the definition of rotation, the image P Q �� is arotation of PQ� about point O a distance of 2m�MOL.Similar proofs can be given for the other cases.

5. The coordinates of the midpoints are

D��c �2

e�, �

d �2

f� �, E��a �

2e

�, �b �

2f

��, and F��a �2

c�, �

b �2

d��.

The equations of the medians are

AD��: y � �dc �

�ef

��

22

ab

�x �

BE��: y � x �

CF��: y � x �

Solving any two of these equations simultaneouslyyields the coordinates of the centroid, (x, y) �

� , �b �

3d � f� �.

6. Here is one of many proofs that �2� is irrational:Assume that �2� is rational and then arrive at acontradiction. If �2� is rational, then it can bewritten as a fraction, �

ab�, in reduced form. Because

�ab� � �2�, �b

a2

2� � 2. Therefore a2 � 2b2, which showsthat 2 is a factor of a2. The only way that 2 can be a factor of the square of a is if 2 is a factor of a, soa � 2c for some integer c. Substitute this into theequation a2 � 2b2 to get 4c2 � 2b2, so 2c2 � b2.This shows that 2 is a factor of b2, so it must alsobe a factor of b. This is a contradiction because weassumed that a and b had no common factors, yetwe showed that they have a common factor of 2.This contradiction arose from the assumption that�2� is rational, so �2� is irrational.

7. An Archimedean tiling is a distinct edge-to-edgetiling of regular polygons with all vertices of the sametype. For these shapes to fill the plane from edge toedge without gaps or overlaps, their angles, whenarranged around a point, must have measures thatadd exactly to 360°. An Archimedean tiling can beconstructed using only equilateral triangles, squares,or regular hexagons. Any other regular polygonwould create either a gap or an overlap in the singleshape tiling. Hence, there are three monohedraltessellations, which are Archimedean tilings.

To find other possible edge-to-edge tilings, use thefact that the measure of an interior angle of aregular n-gon is �

180°(nn

� 2)� degrees. If the n1-gon,

n2-gon, . . . , and nk-gon meet at a vertex, youknow that

� � · · · �

� 360°

180°(nk � 2)��nk

180°(n2 � 2)��n2

180°(n1 � 2)��n1

a � c � e��3

f(a � c) � e(b � d)��a � c � 2e

b � d � 2f��a � c � 2e

d(a � e) � c(b � f )��a � e � 2c

b � f � 2d��a � e � 2c

b(c � e) � a(d � f )��c � e � 2a


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CHAPTER 13 REVIEW

EXERCISES

1. False. The quadrilateral could be an isoscelestrapezoid.

2. True

3. False. The figure could be an isosceles trapezoidor a kite.

4. True

5. False. The angles are supplementary but not neces-sarily congruent.

6. False. See Lesson 13.5, Example B.

7. True

8. Perpendicular

9. Congruent

10. The center of the circle

11. Four congruent triangles that are similar to theoriginal triangle

12. An auxiliary theorem proven specifically to helpprove other theorems

13. If a segment joins the midpoints of the diagonals ofa trapezoid, then it is parallel to the bases.

14. Angle Bisector Postulate

15. Perpendicular Postulate

16. Assume the opposite of what you want to prove,then use valid reasoning to derive a contradiction.

17. a. That smoking is not glamorous

b. If smoking were glamorous, then this smokerwould look glamorous. This smoker does notlook glamorous, therefore smoking is notglamorous.

18. False. The parallelogram is a rhombus. (It could bea square, but it might not be.)

19. False. In fact, the measure of the angle betweenthem equals the measure of one of the other baseangles. The figure below shows a possible counter-example.

50° 50°50°

80°

80°80°

100° 100°80°

50°

Let A represent the area of the triangle. Using a asthe length of the base, the standard formula for thearea of a triangle gives A � �

12�ah. The last equation

above contains 4h2a2 on one side, so rewriteA � �

12�ah to get an expression for 4h2a2.

A � �12�ah

2A � ah

4A2 � a2h2

16A2 � 4a2h2 � 4h2a2

Because 16A2 � 4h2a2 and 4h2a2 = 4b2c2 �(a2 � b2 � c 2)2, by the transitive property 16A2 �4b2c2 � (a2 � b2 � c 2)2. Rearrange terms in thisequation to write A2 as a product of four fractions.Some of the steps below involve factoring the differ-ence of two squares using the factoring patternx2 � y 2 � (x � y)(x � y).

16A2 � 4b2c2 � (a2 � b2 � c 2)2

16A2 � (2bc)2 � (a2 � b2 � c 2)2

16A2 � [(2bc) � (a2 � b2 � c 2)]

� [(2bc) � (a2 � b2 � c 2)]

16A2 � (2bc � a2 � b2 � c 2)(2bc � a2 � b2 � c 2)

16A2 � [a2 �(b2 � 2bc � c 2)]

� [(b2 � 2bc � c 2) � a2]

16A2 � [a2 �(b � c)2][(b � c)2 � a2]

16A2 � [a �(b � c)][a � (b � c)]

� [(b � c) � a][(b � c) � a]

16A2 � (a �b � c)(a � b � c)� (b � c � a)(b � a � c)

A2 �

A2 � ��a �b2

� c��a �

2b � c��b �

2c � a��

� ��b �2a � c��

In Hero’s formula, s represents the semiperimeter of

the triangle: s � �a �

2b � c�. Then s � a � �

a �2b � c� �

�22a� � �

b �2c � a�, s � b � �

a �2b � c� � �

22b� � �

a �2b � c�,

and s � c � �a �

2b � c� � �

22c� � �

a �2b � c�.

Because A2 � ��a �2b � c��a �

2b � c��b �

2c � a��

� ��b �2a � c�� a �

2b � c��b �

2c � a��a �

2b � c��

� ��a �2b � c��, A2 � s(s � a)(s � b)(s � c).

Therefore, A � �s(s � a�)(s � b�)(s � c�)�.

(a � b � c)(a � b � c)(b � c � a)(b � a � c)��16


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20. False. The sum of the measures of the interiorangles of the pentagon formed by the perpendicularbisectors is (5 � 2)(180°) � 540°, so x � 540° �(180° � 2b) � 360° � 2b. The perpendicular bisec-tors of the sides of congruent sides of the isoscelestrapezoid will be perpendicular if 360° � 2b � 90°.Solving this equation gives b � 135°. Therefore, thestatement is true only if the measures of one pair ofbase angles of the trapezoid are 135°.

21. True (except in the special case of an isosceles righttriangle, in which the segment is not definedbecause the feet coincide).

Given: Isosceles �ABC with AC� BC�; altitudesAD�� and BE�; ED�

Show: ED� � AB�

Paragraph Proof: AC� BC� is given, so �EAB �DBA by the Isosceles Triangle Theorem. �AEB �BDA by the definition of altitude and the Right

Angles Are Congruent Theorem. AB� AB� by the

reflexive property of congruence. �AEB �BDA

by SAA. Therefore AE� BD� by CPCTC. By the

definition of congruence, AE � BD and AC � BC,

so AC � AE � BC � BD by the substitution and

subtraction properties. But, by the Segment Addi-

tion Postulate, AC � AE � EC and BC � BD � DC,

so EC � DC by the transitive property. Then �AEC

C� �

�DBC

C� by substitution and the division property. Thus

ED� divides AC� and BC� proportionately. Therefore

ED� � AB� by the Converse of the Parallel/Propor-

tionality Theorem.

C

B

DE

AX

a a

bb

x

x � 540° � (180° � 2b)

x � 360° � 2b


22. True

Given: Rhombus ROME with diagonals RM�� andEO� intersecting at B

Show: RM�� EO�

Proof:

Statement Reason

1. �1 �2 1. Rhombus AnglesTheorem

2. RO � RE 2. Definition of rhombus

3. RO� RE� 3. Definition ofcongruence

4. RB� RB� 4. Reflexive property ofcongruence

5. �ROB �REB 5. SAS CongruencePostulate

6. �3 �4 6. CPCTC

7. �3 and �4 are 7. Linear Pair Postulatesupplementary

8. �3 and �4 are right 8. Congruent andangles Supplementary

Theorem

9. RM�� EO� 9. Definition ofperpendicular

23. True

Given: Parallelogram ABCD with AE� bisecting�DAB and CF� bisecting �DCB

Show: AE� � CF�

Paragraph Proof: m�1 � �12�m�BAD and m�2 �

�12�m�DCB by the definition of angle bisector.�BAD �DCB by the Opposite Angles Theorem.Therefore, �

12�m�BAD � �

12�m�DCB by the multipli-

cation property, so m�1 � m�2 by the transitiveproperty. AB� � DC�� by the definition of parallelo-gram, so �2 and �3 are supplementary by the Inte-rior Supplements Theorem, that is, m�2 � m�3 �180°. Then, by substitution, m�1 � m�3 � 180°,so, by definition and the substitution property, �2and �3 are supplementary. Therefore, AE� � CF� bythe Converse of the Interior Supplements Theorem.

A BF

CD E

1 3

2

ME

R O

B

12

34

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24. Plan: Use the Inscribed Angle Theorem, the addition property, and the distributiveproperty to get m�P � m�E � m�N � m�T � m�A � �

12�(mTN� � mAT� �

mPA� � mEP� � mNE�). Because there are 360° in a circle, m�P � m�E � m�N �

m�T � m�A � 180°.


1. m�P � �12�mTN�; m�E � �

12�mAT�; 1. Inscribed Angle Theorem

m�N � �12�mPA�; m�T � �

12�mEP�;

m�A � �12�mNE�

2. m�P � m�E � m�N � m�T � 2. Addition property

m�A � �12�mTN� � �

12�mAT� � �

12�mPA� �

�12�mEP� � �

12�mNE�

3. m�P � m�E � m�N � m�T � 3. Distributive property

m�A � �12�(mTN� � mAT� � mPA� �

mEP� � mNE�)

4. m�P � m�E � m�N � m�T � 4. There are 360° in a circle,

m�A � �12�(360°) � 180° substitution property

25. Given: Right triangle RTH

Show: At least one nonright angle (�H or �T) has measure less than or equal to 45°

Indirect Paragraph Proof: Assume m�H � 45° and m�T � 45°.Then m�H � m�T � 45° � 45°, so m�H � m�T � 90°.Because m�R � 90°, this would give m�H � m�T � m�R � 90° � 90°, orm�H � m�T � m�R � 180°. But this creates a contradiction because m�H �m�T � m�R � 180° by the Triangle Sum Theorem. Therefore, the assumption thatm�H � 45° and m�T � 45° is false, and the conclusion that either m�H � 45° orm�T � 45° (or possibly both) is true.

26. Given: �TYR with midsegment MS��

Show: MS�� TR� and MS � �12�(TR)

Plan: Use the definition of midpoint, the Segment AdditionPostulate, the substitution property, and the division propertyto get �

MTY

Y� � �

12� and �R

SYY� � �

12�. Then use the reflexive property and

the SAS Similarity Theorem to get �MSY �TRY. Therefore,MS � �

12�(TR) by CSSTP and the multiplication property, and MS�� TR� by CASTC and

the CA Postulate.


1. MS�� is a midsegment of �TYR 1. Given

2. M is the midpoint of YT�; S is the 2. Definition of midsegmentmidpoint of YR�

3. MY � MT; YS � SR 3. Definition of midpoint

4. TY � MY � MT; RY � SY � SR 4. Segment Addition Postulate

5. TY � 2(MY ); RY � 2(SY ) 5. Substitution

6. �MTY

Y� � �

12�; �R

SYY� � �

12� 6. Division property

7. �MTY

Y� � 7. Transitive property

8. �MYS �TYR 8. Reflexive property

9. �MSY �TRY 9. SAS Similarity Theorem

SY�RY

T R

Y

SM

R

T

H


(Proof continued)

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Statement Reason

10. �MTR

S� � �

MTY

Y� 10. CSSTP

11. �MTR

S� � �

12� 11. Transitive property

12. MS � �12�(TR) 12. Multiplication property

13. �YMS �YTR 13. CASTC

14. MS�� TR� 14. CA Postulate

27. Given: Trapezoid ZDYO with midsegment TR�

Show: TR� � ZO�; TR � �12�(ZO � DY )

Plan: Use the Line Postulate to extend ZO� and DR�.Then use the Line Intersection Postulate to label Pas the intersection of ZO�� and DR��. �DYR �PORby the SAA Congruence Theorem; thus, DY� OP� by CPCTC. Use the Triangle Midsegment Theorem and the substitution property to getTR � �

12�(ZO � DY ). Also, apply the Triangle Midsegment Theorem to get TR� � ZO�.


1. Extend ZO� and DR� 1. Line Postulate

2. ZO�� and DR�� intersect at point P 2. Line Intersection Postulate

3. TR� is the midsegment of trapezoid 3. GivenZDYO

4. R is the midpoint of YO�� 4. Definition of midsegment

5. YR � OR 5. Definition of midpoint

6. YR� OR�� 6. Definition of congruence

7. �DRY �PRO (�3 �4) 7. VA Theorem

8. DY� � ZO� 8. Definition of trapezoid

9. DY� � ZP� 9. Z, O, and P are collinear

10. �YDR �OPR 10. AIA Theorem

11. �DYR �POR 11. SAA Theorem

12. DY� PO� 12. CPCTC

13. DY � OP 13. Definition of congruence

14. TR � �12�(ZP) 14. Triangle Midsegment Theorem (�ZDP)

15. ZO � OP � ZP 15. Segment Addition Postulate

16. TR � �12�(ZO � OP) 16. Transitive property

17. TR � �12�(ZO � DY) 17. Substitution property

18. TR� � ZP� 18. Triangle Midsegment Theorem (�ZDP)

19. TR� � ZO� 19. Z, O, and P are collinear

28. a. Conjecture: The quadrilateral formed when the midpoints of the sides of arectangle are connected is a rhombus.

b. C

B

D G

E

FH

A

Z O P

D Y

RT

3

4 2

1


(Exercise 26 Proof continued)

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Given: Rectangle ABCD with E, F, G, and H the midpoints of AB�, BC�, CD��, and DA�,respectively

Show: EFGH is a rhombus

Plan: Use the Right Angles Are Congruent Theorem and the SAS Congruence Postu-late to get �AEH �BEF �DGH �CGF. Then use CPCTC.


1. ABCD is a rectangle 1. Given

2. ABCD is a parallelogram 2. Definition of rectangle

3. BC� DA�; AB� CD�� 3. Opposite Sides Theorem

4. BC � DA; AB � CD 4. Definition of congruence

5. F is the midpoint of BC�; H is the 5. Givenmidpoint of DA�; E is the midpoint of AB�; G is the midpoint of CD��

6. BF � FC; DH � HA; 6. Definition of midpointAE � EB; CG � GD

7. BF � FC � BC; DH � HA � DA; 7. Segment Addition PostulateAE � EB � AB; CG � DG � CD

8. 2(BF) � 2(FC) � BC; 2(DH) � 8. Substitution property2(HA) � DA; 2(AE) � 2(EB) � AB;2(GC) � 2(GD) � CD

9. BF � FC � �12�(BC); DH � HA � �

12�(DA); 9. Multiplication property

AE � BE � �12�(AB); CG � DG � �

12�(CD)

10. BF � CF � DH � AH; AE � BE � 10. Transitive propertyCG � DG

11. BF� CF� DH�� AH��; AE� BE� 11. Definition of congruenceCG� DG��

12. �A, �B, �C, and �D are right angles 12. Definition of rectangle

13. �A �B �C �D 13. Right Triangles Are Congruent Theorem

14. �AEH �BEF �DGH �CGF 14. SAS Congruence Postulate

15. EH� EF� GH�� GF� 15. CPCTC

16. EFGH is a rhombus 16. Four Congruent Sides Rhombus Theorem

29. a. Conjecture: The quadrilateral formed when the midpoints of the sides of arhombus are connected is a rectangle.

b.

Given: Rhombus IJKL with M, N, O, and P the midpoints of IJ�, JK�, KL�, and LI�,respectively

Show: MNOP is a rectangle

Paragraph Proof: Apply the Triangle Midsegment Theorem to �LIJ to get PM�� LJ�and to �LKJ to get ON�� LJ�. Then, by the Parallel Transitivity Theorem, PM�� ON��.Similarly, apply the Triangle Midsegment Theorem to �ILK to get PO� � IK� and to�IJK to get MN�� IK�. Then, by the Parallel Transitivity Theorem, PO� � MN��. Thus,MNOP has two pairs of opposite parallel sides and is by definition a parallelogram.Because IJKL is a rhombus, LJ� � IK� by the Rhombus Diagonals Conjecture. Then�IYL, �KYL, �IYJ, and �KYJ are right angles by the definition of perpendicular.

M

N

P

Y X

O

I

K

JL


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Using the parallel sides found earlier from the Triangle Midsegment Theorem, MNOPis made up of four small parallelograms, so �IYL �MPO, �KYL �PON,�IYJ �PMN, and �KYJ �MNO by the Opposite Angles Theorem. Thus, thefour angles of MNOP are all right angles. Thus, MNOP is a parallelogram with fourright angles, so, by definition, it is a rectangle.

30. a. Conjecture: The quadrilateral formed when the midpoints of the sides of a kiteare connected is a rectangle.

b.

Given: Kite IJKL with M, N, O, and P the midpoints of IJ�, JK�, KL�, and LI�,respectively

Show: MNOP is a rectangle

Paragraph Proof: Apply the Triangle Midsegment Theorem to �LIJ to get PM�� LJ�and to �LKJ to get ON�� LJ�. Then, by the Parallel Transitivity Theorem, PM�� ON��.Also, by the Triangle Midsegment Theorem, PM � �

12�(LJ) and ON � �

12�(LJ), so PM �

ON by the transitive property. Then PM�� ON�� by the definition of congruence.Therefore, MNOP is a parallelogram by the Opposite Sides Parallel and CongruentTheorem. Because IJKL is a kite, LJ� � IK� by the Kite Diagonals Conjecture. Then, bythe CA Postulate and Interior Supplements Theorem, the four angles of MNOP areall right angles. Thus, MNOP is a parallelogram with four right angles, so it is arectangle by definition.

31. Given: Circle O with chords AB� and CD�� intersecting at P

Show: AP � PB � DP � PC

Plan: Use the Line Postulate to construct chords DB� andAC�. Then use the Inscribed Angles Intercepting ArcsTheorem and the AA Similarity Postulate to get �APC �DPB. Therefore, by CSSTP and the multiplication property,AP � PB � DP � PC.


1. Construct DB� and AC� 1. Line Postulate

2. �CAB �CDB; �ACD �DBA 2. Inscribed Angles Intercepting Arcs Theorem

3. �APC �DPB 3. AA Similarity Postulate

4. �DAP

P� � �PP

CB� 4. CSSTP

5. (DP)(PB) � �DAP

P� � (DP)(PB) � �PP

CB� 5. Multiplication property

6. AP � PB � DP � PC 6. Algebra

A

B

C

PO

D

I

K

L J

M

N

P

O


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