SIMILARITY, CONGRUENCE, AND PROOFS Answer Key

UNIT 1 • SIMILARITY, CONGRUENCE, AND PROOFS

Answer Key

CCGPS Analytic Geometry Teacher Resource© Walch EducationU1-743

Lesson 1: Investigating Properties of Dilations

Pre-Assessment, pp. 1–31. d 2. c3. b

4. a5. b

Warm-Up 1.1.1, p.61. Reflect the garden over the line y = x: A' (3, –7),

B' (7, –3), C' (3, –3), or rotate the garden about the origin 180° counterclockwise: A' (7, –3), B' (3, –7), C' (3, –3).

2. Reflections and rotations are rigid motions. They preserve angle and length measures. The result of either transformation is a congruent figure.

Practice 1.1.1: Investigating Properties of Parallelism and the Center, pp. 24–28

1. The pentagon has been dilated. The corresponding sides of the pentagon are parallel and the scale factor is consistent (k = 2). Additionally, the preimage points and image points are collinear with the center of dilation.

2. The triangle has been dilated. The corresponding sides of the triangle are parallel and the scale factor is consistent (k = 1/3). Additionally, the preimage points and image points are collinear with the center of dilation.

3. The triangle has not been dilated. The scale factor is inconsistent between corresponding sides.

4. The rectangle has not been dilated. The scale factor is inconsistent between corresponding sides.

5. k = 2.5; enlargement6. k = 1.25; enlargement7. k = 0.4; reduction8. k = 0.5; reduction9. No, it is not a dilation. The scale factors are inconsistent.

The scale factor of the vertical sides is 1.5. The scale factor of the horizontal sides is 4/3.

10. k = 3/5; reduction

Warm-Up 1.1.2, p. 291. 0.852. 85%3. 6/5.14. 118%5. Hideki might have pushed his sister again to cause her to

swing higher.

Practice 1.1.2: Investigating Scale Factors, p. 421. 7.22. 10.23. 12.3

4. 205. H' (–21, –9), J' (–15, –18), K' (–18, –24)6. P' (–3, 2), Q' (2.5, 4.5), R' (–1.5, –2)7. M' (–3.75, 6), N' (5.25, –2.25), O' (–7.5, –3)8. A' (8.4, 7), B' (2.8, 2.8), D' (–4.2, 5.6)9. D" (12, 8), E" (24, 8), F" (–6, 16); k = 4

10. S' (–4.5, 3), T' (4.5, 3), U' (4.5, –3), V' (–4.5, –3); the new countertop is longer by a factor of 1.5 feet.

Progress Assessment, pp. 43–471. c2. d3. a4. b5. b

6. c7. d8. a9. a

10. b11. Answers:

a. Enlargement; the scale factor when converted to a decimal is 1.8. Since 1.8 is greater than 1, the scale factor creates an enlargement.

b. G' (0, 0), H' (0, 14.4), I' (9, 14.4), and J' (9, 0)c. The perimeter of the original book’s front cover is

26 inches. The perimeter of the dilated book is 46.8 inches. The ratio 46.8/26 is equal to the scale factor of the dilation, which is 1.8 or 180%.

d. The area of the original book’s front cover (the preimage) is 40 in2. The area of the dilated book cover (the image) is 129.6 in2. The ratio of the areas is 129.6/40 and is equal to the square of the scale factor of the dilation, 1.82 or 3.24.

Lesson 2: Constructing Lines, Segments, and Angles

Pre-Assessment, pp. 48–50 1. d2. c3. c

4. d5. b

Warm-Up 1.2.1, p. 541. The areas do not overlap. 2. Maryellen could use the leash as a compass to determine

where to place the stake to minimize overlap.

Practice 1.2.1: Copying Segments and Angles, p. 781–10. Check students’ work for accuracy.

Warm-Up 1.2.2, p. 79 1. Check students’ work for accuracy. 2. 90˚

CCGPS Analytic Geometry Teacher ResourceU1-744

© Walch Education

Practice 1.2.2: Bisecting Segments and Angles, pp. 101–1021–10. Check students’ work for accuracy.

Warm-Up 1.2.3, p. 1031. Check students’ work for accuracy. 2. Check students’ work for accuracy.

Practice 1.2.3: Constructing Perpendicular and Parallel Lines, p. 128 1–10. Check students’ work for accuracy.

Progress Assessment, pp. 129–133 1. c2. b3. a4. c5. d

6. b7. c8. b9. d

10. c11. Check students’ work for accuracy.

Lesson 3: Constructing Polygons

Pre-Assessment, p. 134 1. d2. b3. c

4. b5. a

Warm-Up 1.3.1, p. 1371. Copy the given 60˚ angle to construct a triangle. Extend

the sides of the copied angles to determine the center of the audience.

60˚

Speaker 1 Speaker 2

Center of audience

2. No, it is not possible to construct a second non-congruent triangle using the given information. The given information includes two angle measures and a side length between those two angles. If you have two triangles and any two angles and the included side are equal, then the triangles are congruent.

Practice 1.3.1: Constructing Equilateral Triangles Inscribed in Circles, pp. 159–1601–3. Check students’ work for accuracy. Be sure each of the

vertices lies on the circle.4. Check students’ work for accuracy. Be sure each of the

vertices lies on the circle and the radius of the circle is twice the length of the given segment.

5. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment.

6–8. Check students’ work for accuracy. Be sure each of the vertices lies on the circle.

9. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is twice the length of the given segment.


Warm-Up 1.3.2, p. 1611. Each of the new angles measures 45˚.2. The angle bisector intersects the square at the bisected

angle and the angle opposite the bisected angle.3. Antonia will have created 4 triangles.4. Each triangle will have two 45˚ angles and one 90˚ angle.

Practice 1.3.2: Constructing Squares Inscribed in Circles, p. 180


3–6. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to the length of the given segment.

7. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to twice the length of the given segment.


9–10. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment.


Warm-Up 1.3.3, p. 181

1. The length of each side of the square planting bed will be 1

4 the length of the original plank of wood.

2. To determine the length of each side of the square planting

bed, first find the midpoint of the plank of wood by

bisecting the plank. This construction will divide the plank

into two equal pieces. Next, find the midpoint of one of the

halves by bisecting it. The result is a piece of the plank that

is 1

4 the original length of the plank of wood.

Practice 1.3.3: Constructing Regular Hexagons Inscribed in Circles, p. 204

1. Check students’ work for accuracy. Be sure each of the vertices lies on the circle.








Progress Assessment, pp. 205–209 1. a2. b3. a4. c5. d

6. b7. c8. a9. c

10. d11. a. Check students’ work for accuracy.

b. Find the perpendicular bisector of each side of the equilateral triangle created when lines are drawn to connect the vending machines. Extend the bisectors to locate intersection points with the circle. The points of intersection represent the placement of the recycling bins.

Lesson 4: Exploring Congruence

Pre-Assessment, pp. 210–213 1. d2. a3. b

4. d5. c

Warm-Up 1.4.1, p. 2161. The coordinates:

rx-axis

(A (2, 5)) = A' (2, –5)r

x-axis(B (3, 5)) = B' (3, –5)

rx-axis

(C (3, 2)) = C' (3, –2)r

x-axis(D (5, 2)) = D' (5, –2)

rx-axis

(E (5, 1)) = E' (5, –1)r

x-axis(F (2, 1)) = F' (2, –1)

2.

x

y

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

C D

C' D'

EF

E'F'

A B

A' B'( 2, –5 )

( 2, –1 )

( 3, –5 )

( 3, –2 )

( 5, –2 )

( 5, –1 )


© Walch Education

Practice 1.4.1: Describing Rigid Motions and Predicting the Effects, pp. 235–238

1. Translation; right 4 units and up 3 units; the orientation stayed the same.

2. Reflection; the line of reflection is x = –1. The orientation changed, and the preimage and the image are mirror reflections of each other. The line of reflection is the perpendicular bisector of the segments connecting the vertices of the preimage and image.

3. Rotation; the orientation changed, but the figures are not mirror images of each other.

4.

x

y

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

C

A B

R

B'

A'C'

5.

x

y

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

B

C'

A

C

D

B'A'D'

6.

x

y

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

B B'

AC C' A'

7.

T

B'

D'C'A'

D

A

BC

8.

A

DA'

D'

B'

C'

B

C

N

S

W E


9.

A'

B'

C'

B

A C

10. Answers may vary. Method 1: Rotate the paddle 3 times using a 90º angle of rotation and point R. Method 2: Reflect the paddle over the line m to create paddle 2. Rotate paddle 2 90º counterclockwise about point R to create paddle 3. Reflect paddle 3 over line n to create paddle 4.

R

A

m

n

3 2

1 4

Warm-Up 1.4.2, p. 2391.

RTV

TV cabinet

2. Since the drawing is a scaled representation of the cabinet and TV, use the units provided on the grid and draw the diagonal. Use the Pythagorean Theorem to calculate d 65 8.06= ≈ , which is the length of the diagonal of the cabinet. The TV is 8 units. Therefore, the TV should fit, but it will be a tight fit.

Practice 1.4.2: Defining Congruence in Terms of Rigid Motions, pp. 258–261

1. Not congruent; a dilation occurred with a scale factor of 2/3, indicating a reduction. Dilations are non-rigid motions.

2. Congruent; a rotation occurred. Rotations are rigid motions.3. Congruent; a translation occurred by moving the figure

9.5 units to the left and 4.5 units up. Translations are rigid motions.

4. Not congruent; a horizontal stretch has occurred with a scale factor of 1.5. Stretches are non-rigid motions.

5. Not congruent; a vertical stretch has occurred with a scale factor of 1.5. Stretches are non-rigid motions.

6. Not congruent; a horizontal compression has occurred with a scale factor of 2/3. Compressions are non-rigid motions.

7. The speaker has been rotated and translated. Since these are both rigid motions, the figures are congruent.

8. The corner cabinet has been rotated. Since rotations are rigid motions, the figures are congruent.

9. The inner square has been dilated by a scale factor of 5/3. Since dilations are non-rigid motions, the squares are not congruent.

10. Answers may vary. Sample answer: The triangles are congruent. Triangle 1 can be reflected over a vertical line passing through the top right vertex to create triangle 3. Triangle 3 can be reflected over another vertical line passing through the top right vertex to create triangle 5. Triangle 1 can be reflected over a horizontal line passing through the center of the triangle and then translated


© Walch Education

2 units to the right to create triangle 2. Triangle 2 can be reflected over a vertical line passing through the bottom right vertex to create triangle 4. Since all the transformations described are rigid motions, the triangles are congruent.

Progress Assessment, pp. 262–2701. b2. d3. a4. c5. a

6. b7. b8. c9. b

10. c11. a. Answers may vary. Sample answer: Looking at the sugar

and flour containers, the left handle can be reflected over the vertical line passing through the center of each container to create the right handle. The handles are all congruent and can be created from rigid motions.

b. The rectangles of the containers have undergone a non-rigid motion of a dilation with a scale factor of 4/3. The flour container is larger than the sugar container. This means that the flour and sugar containers are not congruent.

Lesson 5: Congruent Triangles

Pre-Assessment, p. 2711. d2. c3. c

4. a5. d

Warm-Up 1.5.1, pp. 274–2751. The position of Piece 2 is the result of a horizontal

translation of 6 units to the right and a vertical translation of 3 units down.

2. The position of Piece 3 is the reflection of Piece 2 over a vertical line.

3. A series of congruency transformations results in a congruent figure, so Piece 3 is congruent to Piece 1.

Practice 1.5.1: Triangle Congruency, pp. 292–2941. Sample answer: FHG BCD≅ 2. Sample answer: JNP RTV≅ 3. Sample answer: BDF PLN≅ 4. Q W∠ ≅∠ , R X∠ ≅∠ , S Y∠ ≅∠ , QR WX≅ , RS XY≅ ,

QS WY≅5. A C∠ ≅∠ , F G∠ ≅∠ , H J∠ ≅∠ , AF CG≅ , FH GJ≅ ,

AH CJ≅6. L H∠ ≅∠ , P J∠ ≅∠ , Q K∠ ≅∠ , LP HJ≅ , PQ JK≅ ,

LQ HK≅7. Yes, the triangles are congruent; ADH JPK≅ .8. No, the triangles are not congruent.9. Yes, the triangles are congruent; EDF IHG≅ .

10. Yes, the triangles are congruent; RST WVU≅ .

Warm-Up 1.5.2, p. 2951. The base of the pyramid is a square; therefore,

AB BC CD DA≅ ≅ ≅ ; the triangles are congruent, so the corresponding parts of the triangles are also congruent: AE BE CE DE≅ ≅ ≅ .

2. The corresponding angles of the congruent triangles are congruent; therefore, EBA EAD EDC ECB∠ ≅∠ ≅∠ ≅∠ and EAB EBC ECD EDA∠ ≅∠ ≅∠ ≅∠ .

Practice 1.5.2: Explaining ASA, SAS, and SSS, pp. 312–315

1. SSS2. Congruency cannot be determined; the identified

congruent parts form SSA, which is not a triangle congruence statement.

3. SAS4. ASA5. SAS6. Congruency cannot be determined; there is not enough

information about both of the triangles to determine if the triangles are congruent.

7. ABC FGE≅ ; SAS or ASA8. There is not enough information to determine if the sails

are congruent. The information provided follows SSA, which is not a triangle congruence statement.

9. Congruency cannot be determined; there is not enough information about both of the triangles to determine if the triangles are congruent.

10. The plots of land are congruent by SAS or SSS.

Progress Assessment, pp. 316–3191. c2. a3. b4. a5. d

6. b7. a8. c9. d

10. b11. a. AB BC≅ , BAD BCD∠ ≅∠

b. There is not enough information to determine if the triangles are congruent. The information provided follows SSA, which is not a triangle congruence statement.

c. FJ HJ≅ , FJK HJE∠ ≅∠ d. FKJ HKJ≅ by SAS

e. The minimum amount of information needed to prove that two triangles are congruent is two angles and the included side, or two sides and the included angle, or three sides. (Students may also recall that given two angles and a side not included, congruency may be proven by AAS.)


Lesson 6: Defining and Applying Similarity

Pre-Assessment, pp. 320–3221. d2. a3. a

4. a5. c

Warm-Up 1.6.1, p. 3241. 2,722.5 feet2. 1,440 feet

3. 1,282.5 feet

Practice 1.6.1: Defining Similarity, pp. 342–3461. ∠C = 133˚, ∠D = 37˚, ∠E = 10˚, EF = 4.5, DE = 5.52. ∠M = 26˚, ∠N = 14˚, ∠P = 140˚, NP = 9.8, PM = 5.6, KL = 43. Similarity transformations preserve angle measures. The

triangles are not similar because the angle measures in each triangle are different.

4. The triangles are similar. ABC can be dilated by a scale factor of 2 with the center at (0, 0) and then reflected over the line x = 6 to obtain DEF .

5. The triangles are similar. ABC can be dilated by a scale factor of 2.5 with center at (0, 0) to obtain DEF .

6. The triangles are similar. ABC can be dilated by a scale factor of 1 with the center at (0, 0), translated horizontally –4 units and vertically –4 units, and then reflected over the line y = –4 to obtain DEF .

7. The triangles are similar. ABC can be dilated by a scale factor of 1/3 with the center at (0, 0) and then rotated 90˚ clockwise about the origin to obtain DEF .

8. The triangles are similar. ABC can be dilated by a scale factor of 1 with the center at (0, 0) and then rotated 180˚ clockwise about the origin to obtain DEF .

9. Similarity transformations preserve angle measures. The triangles are not similar because the angle measures in each triangle are different.

10. The triangles are similar. ABC can be dilated by a scale factor of 1/2 with the center at (0, 0) and then translated vertically –6 units to obtain DEF .

Warm-Up 1.6.2, p. 3471. 1/12 2. 3.875 meters

Practice 1.6.2: Applying Similarity Using the Angle-Angle (AA) Criterion, pp. 359–362

1. Yes, ABC ZYX� ∼� because of the AA Similarity Statement.2. Yes, ABD ECD� ∼� because of the AA Similarity Statement.3. There is not enough information to determine similarity.4. ABC YZX� ∼� ; x = 3 1/35. ABC DEA� ∼� ; x = 4; BC = 3; AE = 66. AED CBD� ∼� ; x = 1; ED = 2; DB = 67. 6 feet 8 inches8. 1063 1/3 feet = 1063 feet 4 inches 9. 36 feet

10. 2.56 feet

Progress Assessment, pp. 363–3691. d2. c3. b4. c5. a

6. d7. b8. c9. a

10. b11. Answers:

a. 96.25 feetb. approximately 2.18 feetc. Triangles can be formed using the height of the object

and the length of the shadow as two sides. The height of the object and the length of the shadow create a 90˚ angle. Connecting these two sides creates the third side of the triangle and an acute angle. This acute angle is congruent to the angle created by a nearby object and its shadow at the same time of day. Because two angles are known to be congruent, the Angle-Angle Similarity Statement affirms the triangles are similar and the side lengths are proportional.

Lesson 7: Proving Similarity

Pre-Assessment, pp. 370–3711. a2. c3. b

4. a5. b

Warm-Up 1.7.1, p. 3741. Two angles in each triangle are congruent, as noted by

the arc marks. According to the Angle-Angle Similarity Statement, if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

2. x = 5.5 cm; y = 8.75 cm; z = 3.5 cm

Practice 1.7.1: Proving Triangle Similarity Using Side-Angle-Side (SAS) and Side-Side-Side (SSS) Similarity, pp. 388–390

1. ABC DEF� ∼� by SAS 2. ABC EDF� ∼� by SSS3. ABD ECD� ∼� by SAS4. ABC FED� ∼� ; SSS5. not similar; corresponding sides are not proportional6. ABC DEF� ∼� ; SAS7. not similar; corresponding sides are not proportional8. x = 99. x = 2 2/3

10. x = 5

Warm-Up 1.7.2, p. 3911. x = 32. 7 units

3. y = 44. 25 units


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Practice 1.7.2: Working with Ratio Segments, pp. 408–411

1. BD = 3.8 units2. BE = 11 2/3 units3. EC = 9 units4. BD = 4 units5. CD = 3.5 units; BD = 2.5 units6. CD = 7.5 units; CB = 9 units7. No; the sides are not proportional.8. Yes; the sides are proportional.9. No; the sides are not proportional.

10. Yes; the sides are proportional.

Warm-Up 1.7.3, p. 412

1. 44,980 212.1 cm≈ 2. No, the diagonals are not congruent.3. To be “square,” the lengths of each diagonal must be

congruent. The lengths are not congruent. The door, as assembled, is not square.

Practice 1.7.3: Proving the Pythagorean Theorem Using Similarity, pp. 428–432

1. x = 6 units2. x = 10 units3. x 8 3 13.9 units= ≈ 4. x = 12 units; f = 28.8 units5. b = 65 units; f = 144 units6. x = 6 units; a = 7.5 units; b = 10 units7. a 4 5 8.9 units= ≈ ; f = 4 units8. x 4 6 9.8 units= ≈ ; a 33 11.5 units= ≈ ;

b 4 22 18.8 units= ≈9. a = 20 units; b = 15 units; c = 25 units

10. Statements Reasons

1. ABC with sides of length a, b, and c, where c2 = a2 + b2

Draw DE equal in length to a on line l. Draw line m perpendicular to DE . Locate a point, F, on m so that DF is equal in length to b. Connect F and E. Call this segment x.

1. Given

2. FED is a right triangle. 2. Line m is perpendicular to line l.

3. a2 + b2 = x2 3. Pythagorean Theorem

4. x2 = c2, or x = c 4. Segments are congruent.

5. ABC FED≅ 5. Side-Side-Side Congruence Statement

6. C D∠ ≅∠ C∠ is a right angle.

6. Corresponding angles of congruent triangles are congruent.

7. ABC is a right triangle 7. C∠ is a right angle.

Warm-Up 1.7.4, p. 4331. ACE ABD� ∼�2. 238 feet

3. 176 sections

Practice 1.7.4: Solving Problems Using Similarity and Congruence, pp. 451–455

1. 6 m2. 3 ft3. 252 ft4. 700 ft5. 54 m

6. 5.7 m7. 27.5 m8. 33.8 ft9. 3.2 m

10. 84 m

Progress Assessment, pp. 456–4611. c2. b3. d4. a5. c

6. b7. c8. b9. a

10. c11. Answers:

a. ABC ADE� ∼� because of the Side-Angle-Side (SAS)

Similarity Statement. BAC DAE∠ ≅∠ . AD

DB

14

21

2

3= =

and AE

EC

17.5

26.25

2

3= = ; therefore, the side lengths are

proportional.

b. DE BC because of the converse to the Triangle Proportionality Theorem, which states that if a line divides two sides proportionally, then the line is parallel to the third side.

Lesson 8: Proving Theorems About Lines and Angles

Pre-Assessment, pp. 462–4641. d2. c3. b

4. a5. d

Warm-Up 1.8.1, p. 4681. 1∠m must be between 75º and 105º.2. 2∠m must be between 75º and 105º.3. 3∠m must be between 75º and 105º.4. 4∠m must be between 75º and 105º.


Practice 1.8.1: Proving the Vertical Angles Theorem, pp. 494–497

1. Answers may vary. Sample answer: Adjacent angles are 2∠ and 3∠ as well as 3∠ and 4∠ . Nonadjacent angles

are 1∠ and 4∠ as well as 5∠ and 2∠ .2. Supplementary angles are 1∠ and 2∠ as well as 1∠ and

5∠ . Statement: 1 2 180; 1 5 180∠ + ∠ = ∠ + ∠ =m m m m . 3. 2∠ and 5∠ are vertical angles. Statement: 2 5∠ ≅∠ . 4. 2∠ and 3∠ are complementary angles. Statement:

2 3 90∠ + ∠ =m m .5. 68º6. 98º7. 30º8. 37º9. Answers may vary. Sample answer:

Given that 1∠ and 2∠ form a linear pair and 2∠ and 3∠ form a linear pair, prove that 1 3∠ ≅∠ .

14

32

l m

1∠ and 2∠ are supplementary and 2∠ and 3∠ are supplementary because of the definition of a linear pair. 1 3∠ ≅∠ because angles supplementary to the same or

congruent angles are congruent.10. Answers may vary. Sample answer:

Statements Reasons

1. DB is the perpendicular bisector of AC .E is a point on DB .

2. B is the midpoint of AC .

3. ≅AB BC4. ∠EBA and ∠EBC are

right angles.5. ≅EB EB 6. ≅ EBA EBC 7. ≅EA EC8. EA = EC

1. Given

2. Definition of perpendicular bisector

3. Definition of midpoint4. Definition of

perpendicular bisector5. Reflexive Property6. SAS7. CPCTC8. Definition of

congruence

Warm-Up 1.8.2, p. 4981. 2 70∠ =m2. Completed table:

Angle MeasureAngle relationship used to determine measure

1 110º Linear pairs are supplementary. 1∠ and 2∠ are a linear pair.

2 70ºThe sum of the interior angles of a triangle equals 180º.

3 110º Vertical angles are congruent. 1∠ and 3∠ are vertical angles.

4 80ºCorresponding angles of similar triangles are congruent. 4∠ and 7∠ are corresponding angles in similar triangles.

5 100º Linear pairs are supplementary. 4∠ and 5∠ are a linear pair.

6 100º Vertical angles are congruent. 5∠ and 6∠ are vertical angles.

7 80º Given

8 30º Given

Practice 1.8.2: Proving Theorems About Angles in Parallel Lines Cut by a Transversal, pp. 522–524

1. 74º, because the given angles are corresponding angles and corresponding angles of a set of parallel lines intersected by a transversal are congruent.

2. 67º because alternate interior angles in a set of parallel lines intersected by a transversal are congruent.

3. 139º because same-side interior angles in a set of parallel lines intersected by a transversal are supplementary.

4. 42º because the given angle is a vertical angle with 7∠ , which is a corresponding angle with 5∠ , and in parallel lines intersected by a transversal, corresponding angles are congruent.

5. 167º because same-side exterior angles in a set of parallel lines intersected by a transversal are supplementary.

6. 115º7. 94º8. 1 79∠ =m , 2 3 101∠ = ∠ =m m , x = 19, and y = 11;

same-side exterior angles are supplementary in a set of parallel lines intersected by a transversal.


© Walch Education

9.

1 2

3 4

l

m

n

Statements Reasons

1. mn and l is the transversal.2. 1∠ and 2∠ are a linear pair.

3∠ and 4∠ are a linear pair.3. 1∠ and 2∠ are supplementary.

3∠ and 4∠ are supplementary.

4. 1 4

2 3

∠ ≅∠∠ ≅∠

5. 1∠ and 3∠ are supplementary. 2∠ and 4∠ are supplementary.

1. Given2. Definition of

linear pair3. If two angles form

a linear pair, then they are supplementary.

4. Alternate interior angles

5. Substitution

10.

1 23 4

l

m

n5 67 8

Line l is a transversal and mn. 1 5∠ ≅∠ and 2 6∠ ≅∠ because they are corresponding angles. 5 8∠ ≅∠ and 6 7∠ ≅∠ because of vertical angles. Therefore, 1 8∠ ≅∠

and 2 7∠ ≅∠ by the Transitive Property.

Progress Assessment, pp. 525–5321. a2. c3. d4. b5. b

6. c7. b8. d9. a

10. c

11. 3∠ =m 92; justifications may vary. Sample answer: First label the vertices and lines in the diagram with points. Then, draw in another line that passes through vertex C so that it is parallel to the two given parallel lines. Determine angle relationships with the angles that were given. 2∠ can now be labeled with ∠ABC. 46∠ =m ABC . ∠ABC is an alternate interior angle with ∠BCF . Therefore, they are congruent. By congruence of angles, the measures are equal. This means that 46∠ =m BCF . From the Angle Addition Postulate, 3∠ = ∠ + ∠m m BCF m FCE. Find

∠m FCE. Using � ��AE as the transversal to parallel lines � ��

AB and � ��DE , 2 134∠ = ∠ =m GAC m and this angle

is an alternate interior angle with ∠JEC . Therefore, these angles are congruent and have a measure of 134º. Now, using

� ��

FC DE and � ��AE as the transversal, ∠FCE

is a same-side interior angle with ∠JEC . Same-side interior angles are supplementary. This means that

180∠ + ∠ =m FCE m JEC . By substituting in 134∠ =m JEC and following with the Subtraction Property, 46∠ =m FCE .

3∠ = ∠ + ∠m m BCF m FCE . Therefore, 3 92∠ =m .

134º

134º

46º

46º

46ºC

A B

D E

F

G H

I J

Lesson 9: Proving Theorems About Triangles

Pre-Assessment, pp. 533–5351. b2. c3. b

4. a5. c

Warm-Up 1.9.1, p. 5391. 70˚2. 20˚3. Sample response: The angle created by the mirror and

the reflected ray is congruent to the angle created by the mirror and the incident ray. This is true because we are told that the angle of incidence is congruent to the angle of reflection. The angle created by the mirror and the flashlight is the complement of the angle of incidence. The angle created by the mirror and the reflected ray is also the complement to the angle of reflection. By subtracting the angle of reflection from 90˚, we are able to determine that the angle measures 20˚.


Practice 1.9.1: Proving the Interior Angle Sum Theorem, pp. 559–561

1. m∠B = 882. m∠C = 373. m∠A = 22; m∠B = 1404. m∠A = m∠B = m∠C = 605. m∠A = 62; m∠B = 376. m∠A = 156; m∠B = 247. m∠CAB = 80; m∠ABC = 468. m∠CAB = 55; m∠ABC = 359. m∠CAB = 14; m∠ABC = 155

10. A line drawn parallel to AB of ABC through C creates two exterior angles, ∠4 and ∠5. The measures of ∠4, ∠3, and ∠5 are equal to 180˚ because of the Angle Addition Postulate and the definition of a straight angle. ∠1 is congruent to ∠4 and ∠2 is congruent to ∠5 because of the Alternate Interior Angles Theorem. It follows that the measures of ∠1 and ∠4 are equal and the measures of ∠2 and ∠5 are equal because of the definition of congruent angles. By substitution, the sum of the measures of ∠1, ∠2, and ∠3 is 180˚.

Warm-Up 1.9.2, p. 5621. ABC has two congruent sides, so by definition, ABC

is an isosceles triangle.2. 47˚3. 94˚

Practice 1.9.2: Proving Theorems About Isosceles Triangles, pp. 582–585

1. m∠B = 80; m∠C = 502. m∠B = 57.5; m∠C = 65; m∠D = 57.53. m∠A = m∠C = 30; m∠B = 1204. m∠A = m∠B = 24; m∠C = 1325. x = 26. x = 90; y = 1357. x = 17; y = 158. ABC is isosceles; ∠ ≅∠A B .9. ABC is not isosceles.

10.

Statements Reasons

1. Draw a line AD perpendicular to BC .

2. ∠ADB and ∠ADC are right angles and are congruent.

3. ADB and ADC are right triangles.

4. ≅AD AD 5. ∠ ≅∠B C 6. ≅ ADB ADC 7. ≅AB AC

8. ABC is isosceles.

1. Given a point and line, there is only one line perpendicular though the point.

2. Perpendicular lines form right angles; right angles are congruent.

3. Definition of right triangles

4. Reflexive Property5. Given6. AAS Congruence Statement7. Congruent Parts of Congruent

Triangles are Congruent8. Definition of isosceles triangle

Warm-Up 1.9.3, p. 5861. approximately 3,780 feet2. (2, 1)

Practice 1.9.3: Proving the Midsegment of a Triangle, pp. 611–616

1. BC = 13.6; XZ = 7.1; m∠BZX = 202. AC = 15; YZ = 6.25; m∠XZY = 433. YZ = 184. BC = 75. (–12, –4), (3, –4), and (7, 8)6. (–2, 3), (2, –2), and (5, 5)

7. Use the slope formula to show that EF and BC have the

same slope equal to −1

2. Therefore, EF BC . Use the

distance formula to find = 17EF and = 2 17BC .

So, =1

2EF BC .

8. Use the slope formula to show that EF and AB have the

same slope equal to −5

2. Therefore, EF AB . Use the


So, =1

2EF AB .

9. The midpoint of AC = (–1, –2); the midpoint of

BC = (7, –1); use the slope formula to show that EF

and AB have the same slope equal to 1

8. Therefore,

EF AB . Use the distance formula to find = 65EF and

= 2 65BC . So, =1

2EF AB .


© Walch Education

10. The midpoint of AC = (0, –5); the midpoint of

BC = (3, 0); use the slope formula to show that EF


3. Therefore,


= 2 34BC . So, =1

2EF AB .

Warm-Up 1.9.4, p. 617

1. = − +1

26y x

2. 178.9 yards

Practice 1.9.4: Proving Centers of Triangles, pp. 656–658

1. (2, –4) is the circumcenter of ABC because the distance from this point to each of the vertices is 2 5 .

2. (8, 6) is the orthocenter of ABC because this point is a

solution to the equation of each altitude: = − +1

3

26

3y x ,

x = 8, and y = x – 2.

3. The midpoints of ABC are T (3, 4.5), U (–7.5, 0), and

V (–4.5, 4.5). The equations of each of the medians of the

triangle are = +2

35y x , y = –x, and = +

1

4

15

4y x . (–3, 3) is

a solution to the equation of each median.

4. The distance from A to U is approximately 16.23 units.

The distance from (–3, 3) to A is 10.82. (–3, 3) is 2

3 the

distance from A. The distance from B to V is approximately

6.36 units. The distance from (–3, 3) to B is approximately

4.24 units. (–3, 3) is 2

3 the distance from B. The distance

from C to U is approximately 18.55 units. The distance

from (–3, 3) to C is approximately 12.37 units. (–3, 3) is 2

3

the distance from C.5. ABC is an obtuse triangle. The incenter is the

intersection of the angle bisectors. The incenter of a triangle is always inside the triangle.

6. ABC is an obtuse triangle. The orthocenter of an obtuse triangle is always outside of the triangle.

7. It is given that ABC has angle bisectors AN , BP , and CM , and that ⊥XR AB , ⊥XS BC , and ⊥XT AC . Because any point on the angle bisector is equidistant from the sides of the angle, XR = XT, XT = XS, and XR = XS. By the Transitive Property, XR = XT = XS.

8. The circumcenter and orthocenter cannot be used to determine the location of the fire station. The triangle created when the towns are connected is obtuse. The circumcenter and orthocenter would fall outside the triangle created.

9. The incenter should be determined to achieve the largest possible pond. Once the incenter is found, a circle could be inscribed for the area of the pond.

10. The incenter should be determined. This center of the triangle is equidistant to the swings, basketball court, and gazebo. All other locations for the fountain would create different distances between each attraction.

Progress Assessment, pp. 659–6641. b2. d3. c4. a5. a

6. d7. b8. c9. d

10. c11. Answers:

a. The midpoint of AB is (–1, 5.5), the midpoint of BC is (0, 3.5), and the midpoint of AC is (–2, 3).

b. The length of each midsegment is one-half the length of the side it is parallel to.

c. The centroid is located at (–1, 4).

Lesson 10: Proving Theorems About Parallelograms

Pre-Assessment, p. 6651. a2. b3. a

4. d5. c

Warm-Up 1.10.1, p. 6681.

∠1m ∠2m

45º 135º

60º 120º

75º 105º

90º 90º

2. 1∠ and 2∠ are same-side interior angles. Same-side interior angles are supplementary. Therefore, use the equation as follows:

m m

m m

1 2 180

2 180 1

∠ + ∠ =∠ = − ∠

Substitute each given measure of 1∠ into the equation in order to solve for the measure of 2∠ .

Practice 1.10.1: Proving Properties of Parallelograms, pp. 693–695

1. Yes, it’s a parallelogram because opposite sides are parallel; m mAB CD 4= = and m mBC DA 0= = .

2. Yes, it’s a parallelogram because opposite sides are parallel;

m mFG HI 1= = and m mGH IF

5

3= = − .


3. No, it’s not a parallelogram because opposite sides are not congruent: JK = 26 , LM = 5, KL = 10 , and MJ = 3.

4. Yes, it’s a parallelogram because opposite sides are congruent: MN OP 37= = and NO PM 2 13= = .

5. No, it’s not a parallelogram because the midpoints of the

diagonals are not the same, indicating that the diagonals

do not bisect each other: MPR 1,7

2=

and MQS 2,3( )= .

6. Yes, it’s a parallelogram because the midpoints of the diagonals are the same, indicating that the diagonals bisect each other: M (6, 4)= .

7. m A m C 104∠ = ∠ = and m B m D 76∠ = ∠ = ; x = 12 and y = 15


9. Given that quadrilateral ABCD is a parallelogram, by definition opposite sides are parallel and opposite sides are congruent. This means that AD BC , with the diagonals acting as transversals and AD BC≅ . By alternate interior angles, BDA DBC∠ ≅∠ and DAC ACB∠ ≅∠ . By ASA, DPA BPC≅ .

10.

Statements Reasons

1. ABCD EBHG FIJG, ,

2. D B

B G

G I

∠ ≅∠∠ ≅∠∠ ≅∠

3. D I∠ ≅∠

1. Given

2. Opposite angles in a parallelogram are congruent.

3. Transitive Property

Warm-Up 1.10.2, p. 6961. The distance between Carrollton and Campton is about

84 miles.2. Atlanta is at (6, 3.25) on the grid.3. Each city is about 42 miles from Atlanta.

4. The slope is 5

16.

5. 16

5−

Practice 1.10.2: Proving Properties of Special Quadrilaterals, p. 729

1. Quadrilateral ABCD is a parallelogram and a rectangle. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals bisect each other, and not all four sides are congruent.

2. Quadrilateral EFGH is a kite. Justification: adjacent sides are congruent and the diagonals intersect at a right angle.

3. Quadrilateral IJKL is a parallelogram, a rectangle, a rhombus, and a square. Justification: opposite sides are

parallel, consecutive sides are perpendicular, the diagonals bisect each other, and all four sides are congruent.

4. Quadrilateral MNOP is a parallelogram, a rhombus, and a square. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals are perpendicular, and all four sides are congruent.

5. Quadrilateral PQRS is a parallelogram. Justification: opposite sides are parallel, plus the diagonals are congruent and bisect each other.

6. Quadrilateral TUVW is an isosceles trapezoid. Justification: one pair of opposite sides is parallel and the other pair of sides is congruent.

7. Quadrilateral WXYZ is a parallelogram and a rectangle. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals bisect each other, and not all four sides are congruent.

8. Quadrilateral ABCD is a parallelogram, a rectangle, a rhombus, and a square. Justification: opposite sides are parallel, consecutive sides are perpendicular, the diagonals bisect each other, and all four sides are congruent.

9. Given that quadrilateral ABCD is a rhombus, the diagonals of a rhombus bisect each other and are perpendicular. Therefore, DP PB≅ and AP PC≅ . The diagonals of a rhombus bisect the opposite pairs of angles in the rhombus so that DAP PAB DCP BCP∠ ≅∠ ≅∠ ≅∠ . The diagonals of a rhombus are perpendicular. By the definition of perpendicular lines, four right and congruent triangles are formed. Therefore, DPA APB BPC CPD∠ ≅∠ ≅∠ ≅∠ . By ASA, APD APB CPB CPD≅ ≅ ≅ .

10. The slopes are opposite reciprocals of each other:

mc

b b cAC 2 2

=+ +

and mc

b c bBD 2 2

=−

+ −. The

product of the slopes is –1, showing that they are opposite reciprocals of each other.

Progress Assessment, pp. 730–7321. a2. b3. a4. d5. d

6. d7. c8. a9. b

10. d

11. Quadrilateral BCDE is a parallelogram. Justifications

may vary. Students should have at least two of the

following justifications: Opposite sides are parallel

m m m mBC DE CD BE1 and1

2= = = =

; opposite sides are

congruent ED BC CD BE4 2 and 2 5( )= = = = ; and/or

diagonals bisect each other because the midpoints of the

diagonals are the same (M = (3, 2)).


© Walch Education

Unit Assessment pp. 733–741

1. b2. b3. b4. b5. d6. a

7. a8. c9. d

10. b11. d12. c

13. a.

B

A

C

b. No, it is not possible to construct a second non-congruent triangle because the parts of the triangle given are two sides and the included angle. The Side-Angle-Side Congruence Statement ensures that two triangles are congruent given this information.

14. Because they are vertical angles, ∠ = ∠m AHB m EHD .

Because AB DE , ∠ = ∠m ABH m EDH . These two pairs of

equal angles indicate that the AA Similarity Postulate can

be used: � ∼�HAB HED . Because they are corresponding

sides of similar triangles, =HA

HE

BH

DH. Using the property

of proportions, we conclude that =BH

HA

DH

HE.

15.

Statements Reasons

1. ZA CH 2. ∠ = ∠m AZO m CHO

3. ZC AH4. ∠ = ∠m CZO m AHO

5. ZH = ZH6. ≅ CZH AHZ 7. AH = ZC8. ∠ = ∠m ZCO m HAO

9. ≅ ZCO HAO 10. ZO = OH

1. Given2. Alternate Interior Angles

Theorem3. Given4. Alternate Interior Angles

Theorem5. Reflexive Property6. ASA Congruence Statement7. CPCTC8. Alternate Interior Angles

Theorem9. ASA Congruence Statement

10. CPCTC


Practice 1.1.1: Investigating Properties of Parallelism and the Center, pp. 14–18

1. The triangle has been dilated. The corresponding sides of the triangle are parallel and the scale factor is consistent (k = 1/2). Additionally, the preimage points and image points are collinear with the center of dilation.

2. The quadrilateral has not been dilated. The scale factor is inconsistent between corresponding sides.

3. The rectangle has not been dilated. The scale factor is inconsistent between corresponding sides.

4. The rectangle has been dilated. The corresponding sides of the rectangle are parallel and the scale factor is consistent (k = 1.6). Additionally, the preimage points and image points are collinear with the center of dilation.

5. k = 1/3; reduction6. k = 0.4; reduction7. k = 1; congruency transformation8. k = 1.25; enlargement; the scale factor is greater than 1;

therefore, it is an enlargement.9. No, because the scale factors of corresponding sides are

inconsistent. 10. k = 1.5; enlargement; the scale factor is greater than 1;

therefore, it is an enlargement.

Practice 1.1.2: Investigating Scale Factors, p. 241. 10.1252. 28.53. 4.64. 3/55. T' (–3, –1), U' (–2, –2), V' (–2/3, –1)6. B' (–2, 0), D' (–10, –12), E' (6, –8)7. N' (–9.6, –3.2), O' (4.8, 8), P' (6.4, –12.8)8. E' (1.2, 2.7), F' (1.5, 0.9), G' (2.7, 3)9. I" (3.375, 2.8125), J" (1.125, 1.125), K" (–1.6875, 2.25);

k = 9/16 or 0.562510. 6 feet by 8 feet

Practice 1.2.1: Copying Segments and Angles, pp. 42–431–10. Check students’ work for accuracy.

Practice 1.2.2: Bisecting Segments and Angles, pp. 59–601–10. Check students’ work for accuracy.

Practice 1.2.3: Constructing Perpendicular and Parallel Lines, p. 78 1–10. Check students’ work for accuracy.

Practice 1.3.1: Constructing Equilateral Triangles Inscribed in Circles, pp. 96–97 1–3. Check students’ work for accuracy. Be sure each of the

vertices lies on the circle.4. Check students’ work for accuracy. Be sure each of the

vertices lies on the circle and the radius of the circle is twice the length of the given segment.



9. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is twice the length of the given segment.


Practice 1.3.2: Constructing Squares Inscribed in Circles, p. 108





9–10. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment.

Practice 1.3.3: Constructing Regular Hexagons Inscribed in Circles, pp. 123–124




UNIT 1 • SIMILARITY, CONGRUENCE, AND PROOFS

Student Book Answer Key


© Walch Education






Practice 1.4.1: Describing Rigid Motions and Predicting the Effects, pp. 141–145

1. Rotation; the orientation changed, but the images are not mirror reflections of each other.

2. Translation; 5 units left and 3 units up; the orientation stayed the same.

3. Reflection; the line of reflection is y = –3; the orientation changed and the preimage and image are mirror reflections of each other; the line of reflection is the perpendicular bisector of the segments connecting the vertices of the preimage and image.

4.

x

y

A'C

B

A-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

B'

C'

5.

x

y

BA

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

R

C

A'

B'

C'

6.

x

y

BC

A

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

A'

B'C'

7.


8.

R

CA B

A'

B'

C'

9. Answers may vary. Sample answer: Translate both chairs 2 units to the right.

Couch

Co�ee table

Chair 2

Chair 1

Fire

plac

e

10. Answers may vary. Sample answer: First, reflect the hexagon over a horizontal line just below the solar panel. Then translate the reflected hexagon about 1 unit to the right and 1 unit down.

A''

F''F'

F

E' D'E D

E''C''

D''

B''C'

A B

C

B'A'

Practice 1.4.2: Defining Congruence in Terms of Rigid Motions, pp. 157–161

1. Congruent; a translation occurred 7 units to the left and 3 units up. Translations are rigid motions.

2. Congruent; a rotation has occurred. Rotations are rigid motions.

3. Not congruent; a vertical compression has occurred with a scale factor of 1/2. Compressions are non-rigid motions.

4. Not congruent; a horizontal compression has occurred with a scale factor of 2/3. Compressions are non-rigid motions.

5. Congruent; a reflection has occurred. Reflections are rigid motions.

6. Not congruent; a vertical stretch has occurred with a scale factor of 5/3. Stretches are non-rigid motions.

7. The outer triangle has been dilated by a scale factor of 2/3. Since dilations are non-rigid motions, the triangles are not congruent.

8. The target has undergone a rotation. Since rotations are rigid motions, the targets are congruent.

9. The art is a reflection. Since reflections are rigid motions, the A+ on top is congruent to the A+ on the bottom.

10. Answers may vary. Sample answer: The windowpanes are congruent. Pane 1 can be translated to the right 2 units to create pane 2. Pane 1 can be translated 4 units to the right to create pane 3. Panes 1, 2, and 3 can be reflected over the horizontal line passing through the bottom of the panes to create panes 4, 5, and 6. Since all the transformations described are rigid motions, the panes are congruent.


© Walch Education

Practice 1.5.1: Triangle Congruency, pp. 172–1751. Sample answer: NLM HIJ≅2. Sample answer: PQR TSV≅3. Sample answer: USW LNQ≅4. H M∠ ≅∠ , I N∠ ≅∠ , J P∠ ≅∠ , HI MN≅ , IJ NP≅ ,

HJ MP≅5. B H∠ ≅∠ , D J∠ ≅∠ , E L∠ ≅∠ , BD HJ≅ , DE JL≅ ,

BE HL≅ 6. N T∠ ≅∠ , P V∠ ≅∠ , R X∠ ≅∠ , NP TV≅ , PR VX≅ ,

NR TX≅ 7. Yes, the triangles are congruent; DGA LJH≅ . 8. Yes, the triangles are congruent; MNP RTS≅ . 9. Yes, the triangles are congruent; BCD FGE≅ .

10. No, the triangles are not congruent.

Practice 1.5.2: Explaining ASA, SAS, and SSS, pp. 185–187

1. SAS2. ASA3. Congruency cannot be determined; the identified


4. ASA5. Congruency cannot be determined; the identified


6. SSS7. DEF TVS≅ ; SAS8. There is not enough information to determine if the pieces

of wood are congruent. The information provided follows AAA, which is not a triangle congruence statement.

9. The pieces of fabric are congruent; SAS10. The pieces of fabric are congruent; SAS or SSS

Practice 1.6.1: Defining Similarity, pp. 200–2041. ∠C = 44˚, ∠D = ∠E = 68˚, CB = 4.5, DF = 2.12. ∠L = ∠M = ∠P = 45˚, MP = 9, NP = 63. The triangles are similar. ABC can be dilated by a scale

factor of 2 with the center at (0, 0) to obtain DEF .4. The triangles are similar. ABC can be dilated by a scale

factor of 1 with center at (0, 0) and then translated –3 units vertically and 2 units horizontally to obtain DEF .


6. The triangles are similar. ABC can be dilated by a scale factor of 2.5 with the center at (0, 0) and then translated 3 units vertically and 5 units horizontally to obtain DEF .

7. The triangles are similar. ABC can be dilated by a scale factor of 2/7 with the center at (0, 0) and then rotated 180˚ clockwise about the origin to obtain DEF .

8. The triangles are similar. ABC can be dilated by a scale factor of 3/2 with the center at (0, 0) and then reflected over the line x = 0 to obtain DEF .


10. The triangles are similar. ABC can be dilated by a scale factor of 5 with the center at (0, 0) and then rotated 90˚ counterclockwise about the origin to obtain DEF .

Practice 1.6.2: Applying Similarity Using the Angle-Angle (AA) Criterion, pp. 211–215

1. There is not enough information to determine similarity.2. Yes, ABC XYZ� ∼� because of the AA Similarity

Statement. 3. Yes, ABC YXZ� ∼� because of the AA Similarity

Statement.4. ABC YXZ� ∼� ; x = 3 1/35. ABC ZYX� ∼� ; x = 26. BCD BDA� ∼� ; x = 10 2/37. 3 feet8. 9 feet9. 2.16 meters

10. 60 feet

Practice 1.7.1: Proving Triangle Similarity Using Side-Angle-Side (SAS) and Side-Side-Side (SSS) Similarity, pp. 226–229

1. ABC FDE� ∼� by SSS2. ACE BCD� ∼� by SAS3. ABC DEF� ∼� by SSS4. not similar; corresponding sides are not proportional5. ABC EFD� ∼� ; SAS6. not similar; corresponding sides are not proportional7. ABC DEF� ∼� ; SSS8. x = 79. x = 5

10. x = 6

Practice 1.7.2: Working with Ratio Segments, pp. 239–242

1. CD = 12 3/8 units2. BC = 3 1/3 units3. DE = 8.4 units4. CD = 6 units5. BC = 10 units; CD = 12 units6. CB = 21 units; CD = 35 units7. Yes; the sides are proportional.8. No; the sides are not proportional.9. No; the sides are not proportional.

10. Yes; the sides are proportional.

Practice 1.7.3: Proving the Pythagorean Theorem Using Similarity, pp. 252–256

1. x 4 2 5.7= ≈ units

2. x3 26

27.6= ≈ units


3. x 3 1.73= ≈ units

4. x5 6

219.4= ≈ units

5. a 2 6 4.9= ≈ units; e = 1/5; f = 4 4/5

6. a = 600; b = 175; x = 168

7. c 4 2 5.7= ≈ units; e f 2 2 2.8= = ≈ units

8. a 6 2.4= ≈ units; b 3 1.7= ≈ units; x 2 1.4= ≈ units9. e = 3.6; f = 6.4

10. A segment, DE , drawn on line l is equal in length to a. Locate a point F on a line m equal in length to b, which is perpendicular to DE . Connect points F and E. Call this segment x. FED is a right triangle, so a2 + b2 = x2. It is also true that a2 + b2 = c2, so x2 = c2, or x = c. ABC FED≅ by the Side-Side-Side Congruence Statement. For this reason, C D∠ ≅∠ and C∠ must be a right angle, making ABC

a right triangle.

Practice 1.7.4: Solving Problems Using Similarity and Congruence, pp. 266–271

1. 33 ft2. 8.75 ft3. 58 1/3 ft4. 649.6 m5. 23 m

6. 3.2 ft7. 33.8 ft8. 22.3 m9. 7.8 m

10. 30 m

Practice 1.8.1: Proving the Vertical Angles Theorem, pp. 292–295

1. Answers may vary. Sample answer: Adjacent angles are 2∠ and 3∠ as well as 3∠ and 4∠ . Nonadjacent angles

are 1∠ and 4∠ as well as 5∠ and 2∠ .2. Supplementary angles are 7∠ , 1∠ , and 2∠ . Statement:

7 1 2 180∠ + ∠ + ∠ =m m m .3. 1∠ and 4∠ are vertical angles. Statement: 1 4∠ ≅∠ . 4. 4∠ and 5∠ are complementary angles. Statement:

4 5 90∠ + ∠ =m m .5. 131º6. 79º7. 102º8. 48º9. Since ⊥

� �� AB CD as this was given, 1∠ and 2∠ , 2∠ and

3∠ , and 1∠ and 4∠ all form linear pairs. This means that those pairs of angles are also supplementary by the Supplement Theorem. Therefore, 1 2 180∠ + ∠ =m m ,

2 3 180∠ + ∠ =m m , and 1 4 180∠ + ∠ =m m . Given that 1∠ is a right angle, by the definition of right angles, 1 90∠ =m . Use substitution so that 90 2 180+ ∠ =m ,

and by the Subtraction Property, 2 90∠ =m . By definition, 2∠ is a right angle. Use substitution so that 90 3 180+ ∠ =m , and by the Subtraction Property

3 90∠ =m . By definition, 3∠ is a right angle. Use substitution so that 90 4 180+ ∠ =m , and by the

Subtraction Property 4 90∠ =m . By definition, 4∠ is a right angle. Therefore, 2∠ , 3∠ , and 4∠ are right angles.

10. x = 13

Practice 1.8.2: Proving Theorems About Angles in Parallel Lines Cut by a Transversal, pp. 312–316

1. 88º, because alternate interior angles in a set of parallel lines intersected by a transversal are congruent.

2. 90º, because same-side interior angles in a set of lines intersected by a transversal are supplementary.

3. 86º, because alternate interior angles in a set of lines intersected by a transversal are congruent.

4. 57º, because same-side exterior angles in a set of parallel lines intersected by a transversal are supplementary.

5. 144º, because corresponding angles in a set of lines intersected by a transversal are congruent.

6. 111º7. 95º8. 1 100∠ =m , 2 3 80∠ = ∠ =m m , x = 11, and y = 12 9.

Statements Reasons

1. mn and l is the transversal.

2. 2 6∠ ≅∠

3. 6∠ and 8∠ are a linear pair.

4. 6∠ and 8∠ are supplementary.

5. 6 8 180∠ + ∠ =m m6. 2 8 180∠ + ∠ =m m7. 2∠ and 8∠ are

supplementary.

1. Given

2. Corresponding Angles Postulate

3. Definition of a linear pair

4. If two angles form a linear pair, then they are supplementary.

5. Supplement Theorem6. Substitution7. Supplement Theorem

10. Since l is a transversal and l ⊥ m, 1∠ is a right angle. 1 90∠ =m because of the definition of a right angle.

Since lines m and n are parallel, 1 2∠ ≅∠ because of the Corresponding Angles Postulate. 1 2∠ = ∠m m because of the definition of congruent angles. 2 90∠ =m by substitution. 2∠ is a right angle. Therefore, l ⊥ m because of the definition of perpendicular lines.

Practice 1.9.1: Proving the Interior Angle Sum Theorem, pp. 333–336

1. m∠B = 202. m∠B = 563. m∠B = 81; m∠C = 214. m∠A = 85; m∠B = 52; m∠C = 435. m∠A = 92; m∠B = 296. m∠A = 128; m∠B = 527. m∠CAB = 26; m∠ABC = 248. m∠CAB = 25; m∠ABC = 659. m∠CAB = 10; m∠ABC = 125


© Walch Education

10. ABC is a triangle. By the Triangle Sum Theorem, the sum of the measures of ∠1, ∠2, and ∠3 is equal to 180˚. If two angles form a linear pair, they are supplementary, so ∠3 and ∠4 are also supplementary. The sum of the measures of ∠3 and ∠4 is equal to 180˚ by the definition of supplementary angles. The sum of the measures of ∠3 and ∠4 is equal to the sum of the measures of ∠1, ∠2, and ∠3 by substitution. The measure of ∠4 is equal to the sum of the measures of ∠1 and ∠2 by the Subtraction Property.

Practice 1.9.2: Proving Theorems About Isosceles Triangles, pp. 350–353

1. m∠A = 40; m∠C = 702. m∠ADB = 108; m∠DCB = 72; m∠DBC = 363. m∠A = 120; m∠B = m∠C = 304. m∠A = m∠C = 45; m∠B = 905. x = 36. x = 78; y = 517. x = 138. ABC is not isosceles. 9. ABC is isosceles; ∠ ≅∠B C .

10. Given ABC is equiangular, prove ABC is equilateral. Since ABC is equiangular, ∠ ≅∠A B and ∠ ≅∠B C . By the converse of the Isosceles Triangle Theorem, ≅AB BC and ≅AC BC , so by the Transitive Property, ≅ ≅AB BC AC . Therefore, ABC is equilateral.

Practice 1.9.3: Proving the Midsegment of a Triangle, pp. 372–378

1. BC = 12; XZ = 7.5; m∠BZX = 552. BC = 21; YZ = 5.625; m∠AXY = 723. XY = 54. AB = 425. (–9, –2), (5, 8), (5, –2)6. (2, 6), (4, 3), (8, 5)

7. Use the slope formula to show that EF and BC have

the same slope equal to −3

4. Therefore, EF BC .

Use the distance formula to find EF = 5 and BC = 10.

So, =1

2EF BC .

8. Use the slope formula to show that EF and BC have

the same slope equal to –1. Therefore, EF BC . Use the


So, =1

2EF BC .

9. The midpoint of AC = (1, 1); the midpoint of

BC = (4, 2); use the slope formula to show that EF


3. Therefore,


= 2 10AB . So, =1

2EF AB .

10. The midpoint of AC = (1, 1); the midpoint of

BC = (–1, 6); use the slope formula to show that EF

and AB have the same slope equal to −5

2. Therefore,


= 2 29AB . So, =1

2EF AB .

Practice 1.9.4: Proving Centers of Triangles, pp. 406–408

1. (–1, –3) is the circumcenter of ABC because the distance from this point to each of the vertices is 10 .

2. (–2, 1) is the orthocenter of ABC because this point is a solution to the equation of each altitude: x = –2 and y = 1. ABC is a right triangle, so the orthocenter is the vertex B.

3. The midpoints of ABC are T (7.5, –0.5), U (4, –3), and

V (3.5, 0.5). The equations of each of the medians of the

triangle are y = 2x – 11, y = –x + 4, and = −1

52y x . (5, –1)

is a solution to the equation of each median.

4. The distance from A to U is approximately 6.71 units. The

distance from (5, –1) to A is 4.47 units. (5, –1) is 2

3 the

distance from A. The distance from B to V is approximately

6.36 units. The distance from (5, –1) to B is approximately

4.24 units. (5, –1) is 2

3 the distance from B. The distance

from C to T is approximately 7.65 units. The distance from

(5, –1) to C is approximately 5.10 units. (5, –1) is 2

3 the

distance from C.5. ABC is an obtuse triangle. The incenter is the

intersection of the angle bisectors. The incenter of a triangle is always inside the triangle.

6. ABC is a right triangle. The orthocenter of a right triangle is always on a vertex of the triangle.

7. It is given that ABC has perpendicular bisectors p, q, and r of AB , BC , and AC . X is on the perpendicular bisector of AB , so it is equidistant from A and B. AX = BX by the definition of equidistant. The perpendicular bisector of BC also has the point X, so BX = CX. AX = CX by the Transitive Property of Equality; therefore, AX = BX = CX.

8. The circumcenter and orthocenter cannot be used to determine the location of the trauma center. The triangle created when the towns are connected is obtuse. The circumcenter and orthocenter would fall outside the triangle created.

9. The incenter should be determined. This center of the triangle is equidistant to the each location within the park. All other locations for the first aid station would create different distances between each location.

10. The incenter should be determined to achieve the maximum amount of space for the dog. Once the incenter is found, the dog’s leash can be staked in the ground.


Practice 1.10.1: Proving Properties of Parallelograms, pp. 428–430

1. No, it’s not a parallelogram because opposite sides are not

parallel: mTU 5= , mVW 2= , mUV

2

3= − , mWT

1

3= .

2. Yes, it’s a parallelogram because opposite sides are parallel:

m mWX YZ 5= = and m mXY ZW

2

5= = .

3. Yes, it’s a parallelogram because opposite sides are congruent: GH IJ 10= = and HI JG 26= = .

4. No, it’s not a parallelogram because opposite sides are not congruent: AB 13= , CD 2= , BC 29= , and DA 3 2= .

5. Yes, it’s a parallelogram because the midpoints of the

diagonals are the same, indicating the diagonals bisect each

other: M 3,3

2= −

.

6. No, the midpoints are not the same, indicating the

diagonals do not bisect each other: MMO (1,1)= and

MNP

3

2,3

2=

.



9. Given that AB is parallel to DE , we can use the Alternate

Interior Angles Theorem to show that m BAH m DEH∠ = ∠ .

Given that AD is parallel to BC , the same theorem

indicates that m ABF m EAD∠ = ∠ . From these pairs of

angles, it can be seen that � ∼�ABF EDA . Using the full

meaning of the triangles being similar, we can conclude

that corresponding sides are in a constant ratio; or, in other

words, AB

ED

BF

DA= .

10. Opposite sides are parallel: m mb

aAB CD= = and

m mBC DA 0= = . The diagonals bisect each other because

they have the same midpoint: Ma c b

2,2

=+

.

Practice 1.10.2: Proving Properties of Special Quadrilaterals, pp. 453–454

1. Quadrilateral ABCD is a parallelogram, a rectangle, a rhombus, and a square. Justification: opposite sides are parallel, consecutive sides are perpendicular, the diagonals bisect each other, and all four sides are congruent.

2. Quadrilateral EFGH is a parallelogram and a rectangle. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals bisect each other, and not all four sides are congruent.

3. Quadrilateral JKLM is an isosceles trapezoid. Justification: one pair of opposite sides is parallel and the other pair of sides is congruent.

4. Quadrilateral NOPQ is a parallelogram. Justification: opposite sides are parallel, plus the diagonals are congruent and bisect each other.

5. Quadrilateral STUV is a parallelogram and a rhombus. Justification: opposite sides are parallel, adjacent sides are not perpendicular, the diagonals are perpendicular, and all four sides are congruent.

6. Quadrilateral WXYZ is a parallelogram, a rhombus, and a square. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals are perpendicular, and all four sides are congruent.

7. Quadrilateral ABCD is a kite. Justification: adjacent sides are congruent and the diagonals intersect at a right angle.

8. Quadrilateral FGHJ is a parallelogram and a rectangle. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals bisect each other, and not all four sides are congruent.

9.

Statements Reasons 1. Quadrilateral ABCD is a square.2. DP PB≅

AP PC≅

3. AC DB⊥

4. APD APB BPC CPD∠ ≅∠ ≅∠ ≅∠ , and they are right angles.

5. APD APB BPC CPD∠ ≅∠ ≅∠ ≅∠

6. APD APB CPB CPD≅ ≅ ≅

1. Given2. The diagonals are

congruent and bisect each other.

3. The diagonals of a square are perpendicular.

4. Definition of perpendicular lines

5. All right angles are congruent.

6. SAS Congruence Statement

10. Given APD APB CPB CPD≅ ≅ ≅ , Corresponding Parts of Congruent Triangles are Congruent. This means that AD AB CB CD≅ ≅ ≅ . Also by CPCTC, ∠ ≅∠PAD PCB. This means that BC AD since these are alternate interior angles. Then, by CPCTC again, ∠ ≅∠BAP PCD. This means that AB CD because these angles are alternate interior angles. Since opposite sides are parallel and the sides are all congruent, by definition, quadrilateral ABCD is a rhombus.

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SIMILARITY, CONGRUENCE, AND PROOFS Answer Key