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Transcript
```UNIT 1 • SIMILARITY, CONGRUENCE, AND PROOFS
Lesson 1: Investigating Properties of Dilations
Pre-Assessment, pp. 1–3
1. d
2. c
3. b
4. a
5. b
Warm-Up 1.1.1, p.6
1. 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; enlargement
6. k = 1.25; enlargement
7. k = 0.4; reduction
8. k = 0.5; reduction
9. 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. 29
1. 0.85
2. 85%
3. 6/5.1
4. 118%
5. Hideki might have pushed his sister again to cause her to
swing higher.
Practice 1.1.2: Investigating Scale Factors, p. 42
1. 7.2
2. 10.2
3. 12.3
4. 20
5. 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. 43–47
1. c
6. c
2. d
7. d
3. a
8. a
4. b
9. a
5. b
10. b
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. d
2. c
3. c
4. d
5. b
Warm-Up 1.2.1, p. 54
1. 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. 78
1–10. Check students’ work for accuracy.
Warm-Up 1.2.2, p. 79
1. Check students’ work for accuracy.
2. 90˚
U1-743
CCGPS Analytic Geometry Teacher Resource
Practice 1.2.2: Bisecting Segments and Angles,
pp. 101–102
1–10. Check students’ work for accuracy.
Warm-Up 1.2.3, p. 103
1. 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. c
6. b
2. b
7. c
3. a
8. b
4. c
9. d
5. d
10. c
11. Check students’ work for accuracy.
Lesson 3: Constructing Polygons
Pre-Assessment, p. 134
1. d
2. b
3. c
4. b
5. a
Warm-Up 1.3.1, p. 137
1. Copy the given 60˚ angle to construct a triangle. Extend
the sides of the copied angles to determine the center of
the audience.
Speaker 1
Speaker 2
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. 159–160
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.
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.
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 1.3.2, p. 161
1. 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
Center of audience
1–2. Check students’ work for accuracy. Be sure each of the
vertices lies on the circle.
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.
8. 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.
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.
60˚
U1-744
CCGPS Analytic Geometry Teacher Resource
Warm-Up 1.3.3, p. 181
1. The length of each side of the square planting bed will be
1
the length of the original plank of wood.
4
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
1
is
the original length of the plank of wood.
4
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.
2–3. 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.
4. 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.
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. Check students’ work for accuracy. Be sure each of the
vertices lies on the circle.
7–8. 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.
9. 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.
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.
Progress Assessment, pp. 205–209
1. a
6. b
2. b
7. c
3. a
8. a
4. c
9. c
5. d
10. d
11. a. Check students’ work for accuracy.
b. F ind 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. d
2. a
3. b
4. d
5. c
Warm-Up 1.4.1, p. 216
1. The coordinates:
rx-axis(A (2, 5)) = A' (2, –5)
rx-axis(B (3, 5)) = B' (3, –5)
rx-axis(C (3, 2)) = C' (3, –2)
rx-axis(D (5, 2)) = D' (5, –2)
rx-axis(E (5, 1)) = E' (5, –1)
rx-axis(F (2, 1)) = F' (2, –1)
2.
10
9
y
8
7
6
5
4
A
3
2
1
B
C
F
D
E
x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
( 5, –1 )
-2 ( 2, –1 )
( 5, –2 )
-3
F'
-4
-5
-6
-7
C'
( 3, –2 )
A'
E'
D'
B'
( 2, –5 ) ( 3, –5 )
-8
-9
-10
U1-745
CCGPS Analytic Geometry Teacher Resource
Practice 1.4.1: Describing Rigid Motions and
Predicting the Effects, pp. 235–238
6.
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.
y
10
9
8
7
6
5
4
3
x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
A
-2
-3
-4
-5
-6
8
7
-7
-8
-9
B'
4
3
B'
2
1
B
10
9
6
5
y
C'
C
A'
-10
7.
2
1
R C'
A'
x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
-2
C
-3
-4
A
B
-5
-6
-7
-8
-9
B
-10
5.
10
9
T
C
D
y
A
8
D'
C'
A'
B'
7
6
5
4
3
A'
B'
2
1
A
B
D
8.
C
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
D'
-2
C'
-3
-4
N
x
W
E
S
-5
-6
-7
-8
-9
-10
A'
D'
A
B
D
C
B'
C'
U1-746
CCGPS Analytic Geometry Teacher Resource
Warm-Up 1.4.2, p. 239
9.
1.
B
A
C
A'
C'
TV cabinet
R
TV
B'
using a 90º angle of rotation and point R. Method 2: Reflect
m
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
3
2
R
1
A
n
4
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.
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
U1-747
CCGPS Analytic Geometry Teacher Resource
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. 262–270
1. b
6. b
2. d
7. b
3. a
8. c
4. c
9. b
5. a
10. c
11. a. A
nswers 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 nonrigid 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. 271
1. d
2. c
3. c
4. a
5. d
Warm-Up 1.5.1, pp. 274–275
1. 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. 292–294
1. 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. 295
1. 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. SSS
2. Congruency cannot be determined; the identified
congruent parts form SSA, which is not a triangle
congruence statement.
3. SAS
4. ASA
5. SAS
6. 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 ASA
8. 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. 316–319
1. c
6. b
2. a
7. a
3. b
8. c
4. a
9. d
5. d
10. b
11. 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.)
U1-748
CCGPS Analytic Geometry Teacher Resource
Lesson 6: Defining and Applying Similarity
Pre-Assessment, pp. 320–322
1. d
2. a
3. a
4. a
5. c
Warm-Up 1.6.1, p. 324
1. 2,722.5 feet
2. 1,440 feet
3. 1,282.5 feet
Practice 1.6.1: Defining Similarity, pp. 342–346
1. ∠C = 133˚, ∠D = 37˚, ∠E = 10˚, EF = 4.5, DE = 5.5
2. ∠M = 26˚, ∠N = 14˚, ∠P = 140˚, NP = 9.8, PM = 5.6, KL = 4
3. 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. 347
1. 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 1/3
5. ABC ∼ DEA ; x = 4; BC = 3; AE = 6
6. AED ∼CBD ; x = 1; ED = 2; DB = 6
7. 6 feet 8 inches
8. 1063 1/3 feet = 1063 feet 4 inches
9. 36 feet
10. 2.56 feet
Progress Assessment, pp. 363–369
1. d
6. d
2. c
7. b
3. b
8. c
4. c
9. a
5. a
10. b
a. 96.25 feet
b. approximately 2.18 feet
c. 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. 370–371
1. a
2. c
3. b
4. a
5. b
Warm-Up 1.7.1, p. 374
1. 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 SSS
3. ABD ∼ECD by SAS
4. ABC ∼FED ; SSS
5. not similar; corresponding sides are not proportional
6. ABC ∼ DEF ; SAS
7. not similar; corresponding sides are not proportional
8. x = 9
9. x = 2 2/3
10. x = 5
Warm-Up 1.7.2, p. 391
1. x = 3
2. 7 units
3. y = 4
4. 25 units
U1-749
CCGPS Analytic Geometry Teacher Resource
Practice 1.7.2: Working with Ratio Segments,
pp. 408–411
1. BD = 3.8 units
2. BE = 11 2/3 units
3. EC = 9 units
4. BD = 4 units
5. CD = 3.5 units; BD = 2.5 units
6. CD = 7.5 units; CB = 9 units
7. 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.
6. ∠
C ≅ ∠D
∠
C is a right angle.
6. Corresponding angles
of congruent triangles
are congruent.
Warm-Up 1.7.4, p. 433
1. ACE ∼ ABD
2. 238 feet
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 units
2. x = 10 units
3. x = 8 3 ≈ 13.9 units
4. x = 12 units; f = 28.8 units
5. b = 65 units; f = 144 units
6. x = 6 units; a = 7.5 units; b = 10 units
7. a = 4 5 ≈ 8.9 units ; f = 4 units
8. 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. 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.
5. Side-Side-Side
Congruence
Statement
7. ABC
is a right triangle 7. ∠C is a right angle.
Warm-Up 1.7.3, p. 412
Statements
5. 
ABC ≅FED
3. 176 sections
Practice 1.7.4: Solving Problems Using Similarity and
Congruence, pp. 451–455
1. 6 m
2. 3 ft
3. 252 ft
4. 700 ft
5. 54 m
6. 5.7 m
7. 27.5 m
8. 33.8 ft
9. 3.2 m
10. 84 m
Progress Assessment, pp. 456–461
1. c
2. b
3. d
4. a
5. c
6. b
7. c
8. b
9. a
10. c
a. ABC ∼ ADE because of the Side-Angle-Side (SAS)
= =
Similarity Statement. ∠BAC ≅ ∠DAE .
DB 21 3
AE 17.5 2
=
= ; therefore, the side lengths are
and
EC 26.25 3
proportional.
Reasons
1. Given
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. 462–464
1. d
2. c
3. b
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.
4. a
5. d
Warm-Up 1.8.1, p. 468
1.
2.
3.
4.
m∠1
m∠2
m∠3
m∠ 4
must be between 75º and 105º.
must be between 75º and 105º.
must be between 75º and 105º.
must be between 75º and 105º.
U1-750
CCGPS Analytic Geometry Teacher Resource
Practice 1.8.1: Proving the Vertical Angles Theorem,
pp. 494–497
∠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: m∠1 + m∠2 = 180; m∠1 + m∠5 = 180 .
3. ∠2 and ∠5 are vertical angles. Statement: ∠2 ≅ ∠5 .
4. ∠2 and ∠3 are complementary angles. Statement:
m∠2 + m∠3 = 90 .
5. 68º
6. 98º
7. 30º
8. 37º
Given that ∠1 and ∠2 form a linear pair and ∠2 and
∠3 form a linear pair, prove that ∠1 ≅ ∠3.
l
m
3
4
∠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.
Statements
1. DB is the
perpendicular bisector
of AC .
E is a point on DB .
2. B is the midpoint of
AC .
3. AB ≅ BC
4. ∠
EBA and ∠EBC are
right angles.
5. EB ≅ EB
6. EBA ≅EBC
7. EA ≅ EC
8. EA = EC
1. m∠2 = 70
2. Completed table:
Angle relationship used
to determine measure
Angle Measure
1
110º
Linear pairs are supplementary. ∠1 and
∠2 are a linear pair.
2
70º
The sum of the interior angles of a triangle
equals 180º.
3
110º
Vertical angles are congruent. ∠1 and ∠3
are vertical angles.
4
80º
Corresponding 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
2
1
Warm-Up 1.8.2, p. 498
Reasons
1. Given
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. 115º
7. 94º
8. m
∠1 = 79, m∠2 = m∠3 = 101 , x = 19, and y = 11;
same-side exterior angles are supplementary in a set of
parallel lines intersected by a transversal.
2. Definition of
perpendicular bisector
3. Definition of midpoint
4. Definition of
perpendicular bisector
5. Reflexive Property
6. SAS
7. CPCTC
8. Definition of
congruence
U1-751
CCGPS Analytic Geometry Teacher Resource
9.
l
m
1
2
n
3 4
Statements
Reasons
1. m  n 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.
1. Given
2. Definition of
linear pair
3. If two angles form
a linear pair,
then they are
supplementary.
4. Alternate interior
angles
5. Substitution
4. ∠1 ≅ ∠ 4
∠2 ≅ ∠3
5. ∠1 and ∠3 are supplementary.
∠
2 and ∠ 4 are supplementary.
11. m
∠3 = 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. m∠ABC = 46 . ∠ABC
is an alternate interior angle with ∠BCF . Therefore, they
are congruent. By congruence of angles, the measures are
equal. This means that m∠BCF = 46 . From the Angle
m∠3 = m∠BCF + m∠FCE. Find
m
∠
FCE. Using
AE as the transversal to parallel lines
AB and DE , m∠GAC = m∠2 = 134 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
m∠FCE + m∠JEC = 180. By substituting in m∠JEC = 134
and following with the Subtraction Property, m∠FCE = 46.
m∠3 = m∠BCF + m∠FCE. Therefore, m∠3 = 92.
G
A
B
H
46º
134º
46º
C
46º
F
l
10.
1
3
2
I
m
4
134º
E
J
D
Lesson 9: Proving Theorems About Triangles
Pre-Assessment, pp. 533–535
n
5 6
7 8
L ine l is a transversal and m  n. ∠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. 525–532
1. a
2. c
3. d
4. b
5. b
6. c
7. b
8. d
9. a
10. c
1. b
2. c
3. b
4. a
5. c
Warm-Up 1.9.1, p. 539
1. 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˚.
U1-752
CCGPS Analytic Geometry Teacher Resource
Practice 1.9.1: Proving the Interior Angle Sum
Theorem, pp. 559–561
1. m∠B = 88
2. m∠C = 37
3. m∠A = 22; m∠B = 140
4. m∠A = m∠B = m∠C = 60
5. m∠A = 62; m∠B = 37
6. m∠A = 156; m∠B = 24
7. m∠CAB = 80; m∠ABC = 46
8. m∠CAB = 55; m∠ABC = 35
9. 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. 562
1. 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 = 50
2. m∠B = 57.5; m∠C = 65; m∠D = 57.5
3. m∠A = m∠C = 30; m∠B = 120
4. m∠A = m∠B = 24; m∠C = 132
5. x = 2
6. x = 90; y = 135
7. x = 17; y = 15
8. 
ABC is isosceles; ∠A ≅ ∠B .
9. 
ABC is not isosceles.
10.
Statements
perpendicular to
BC .
are right angles and
are congruent.
3. 
are right triangles.
5. ∠B ≅ ∠C
7. AB ≅ AC
8. ABC is isosceles.
Reasons
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 Property
5. Given
6. AAS Congruence Statement
7. Congruent Parts of Congruent
Triangles are Congruent
8. Definition of isosceles triangle
Warm-Up 1.9.3, p. 586
1. approximately 3,780 feet
2. (2, 1)
Practice 1.9.3: Proving the Midsegment of a Triangle,
pp. 611–616
1. BC = 13.6; XZ = 7.1; m∠BZX = 20
2. AC = 15; YZ = 6.25; m∠XZY = 43
3. YZ = 18
4. BC = 7
5. (–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
1
same slope equal to − . Therefore, EF  BC . Use the
2
distance formula to find EF = 17 and BC = 2 17 .
1
So, EF = BC .
2
8. Use the slope formula to show that EF and AB have the
5
same slope equal to − . Therefore, EF  AB . Use the
2
distance formula to find EF = 29 and BC = 2 29 .
1
So, EF = AB .
2
9. The midpoint of AC = (–1, –2); the midpoint of
BC = (7, –1); use the slope formula to show that EF
1
and AB have the same slope equal to . Therefore,
8
EF  AB . Use the distance formula to find EF = 65 and
1
BC = 2 65 . So, EF = AB .
2
U1-753
CCGPS Analytic Geometry Teacher Resource
10. The midpoint of AC = (0, –5); the midpoint of
BC = (3, 0); use the slope formula to show that EF
5
and AB have the same slope equal to . Therefore,
3
EF  AB . Use the distance formula to find EF = 34 and
1
BC = 2 34 . So, EF = AB .
2
Warm-Up 1.9.4, p. 617
1
1. y = − x + 6
2
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
1 26
solution to the equation of each altitude: y = − x + ,
3
3
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
2
1 15
triangle are y = x + 5 , y = –x, and y = x + . (–3, 3) is
3
4
4
a solution to the equation of each median.
4. The distance from A to U is approximately 16.23 units.
2
The distance from (–3, 3) to A is 10.82. (–3, 3) is the
3
distance from A. The distance from B to V is approximately
6.36 units. The distance from (–3, 3) to B is approximately
2
4.24 units. (–3, 3) is the distance from B. The distance
3
from C to U is approximately 18.55 units. The distance
2
from (–3, 3) to C is approximately 12.37 units. (–3, 3) is
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. 659–664
1. b
6. d
2. d
7. b
3. c
8. c
4. a
9. d
5. a
10. c
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).
Parallelograms
Pre-Assessment, p. 665
1. a
2. b
3. a
4. d
5. c
Warm-Up 1.10.1, p. 668
1.
m∠1
m∠2
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∠1 + m∠2 = 180
m∠2 = 180 − m∠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;
mAB = mCD = 4 and mBC = mDA = 0 .
2. Yes, it’s a parallelogram because opposite sides are parallel;
5
mFG = mHI = 1 and mGH = mIF = − .
3
U1-754
CCGPS Analytic Geometry Teacher Resource
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
 7
do not bisect each other: M PR =  1,  and M QS = ( 2,3) .
 2
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
8. m∠A = m∠C = 82 and m∠B = m∠D = 98 ; x = 14 and
y = 20
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
1. Given
2. ∠D ≅ ∠B
∠B ≅ ∠G
∠G ≅ ∠I
3. ∠D ≅ ∠I
2. Opposite angles in
a parallelogram
are congruent.
3. Transitive Property
Warm-Up 1.10.2, p. 696
1. 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.
5
4. The slope is
.
16
16
5. −
5
Practice 1.10.2: Proving Properties of Special
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.
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:
c
−c
mAC =
and mBD =
. The
2
2
b+ b +c
b2 + c 2 − b
product of the slopes is –1, showing that they are opposite
reciprocals of each other.
Progress Assessment, pp. 730–732
1. a
2. b
3. a
4. d
5. d
6. d
7. c
8. a
9. 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
1

 mBC = mDE = 1 and mCD = mBE = 2  ; opposite sides are
congruent ED = BC = 4 2 and CD = BE = 2 5 ; and/or
(
)
diagonals bisect each other because the midpoints of the
diagonals are the same (M = (3, 2)).
U1-755
CCGPS Analytic Geometry Teacher Resource
Unit Assessment
14. Because they are vertical angles, m∠AHB = m∠EHD .
pp. 733–741
1. b
2. b
3. b
4. b
5. d
6. a
13. a.
Because AB  DE , m∠ABH = m∠EDH . These two pairs of
7. a
8. c
9. d
10. b
11. d
12. c
equal angles indicate that the AA Similarity Postulate can
be used: HAB ∼HED . Because they are corresponding
sides of similar triangles,
HA
=
BH
. Using the property
DH
BH DH
of proportions, we conclude that
=
.
HA HE
C
HE
15.
A
Statements
B
b. N
o, it is not possible to construct a second noncongruent triangle because the parts of the triangle
given are two sides and the included angle. The SideAngle-Side Congruence Statement ensures that two
triangles are congruent given this information.
Reasons
1. ZA CH
2. m∠AZO = m∠CHO
1. Given
2. Alternate Interior Angles
Theorem
3. Given
3. ZC  AH
4. m∠CZO = m∠AHO 4. Alternate Interior Angles
Theorem
5. Reflexive Property
5. ZH = ZH
6. ASA Congruence Statement
6. CZH ≅  AHZ
7. CPCTC
7. AH = ZC
8. m∠ZCO = m∠HAO 8. Alternate Interior Angles
Theorem
9. ASA Congruence Statement
9.  ZCO ≅ HAO
10. CPCTC
10. ZO = OH
U1-756
CCGPS Analytic Geometry Teacher Resource
UNIT 1 • SIMILARITY, CONGRUENCE, AND PROOFS
Practice 1.1.1: Investigating Properties of Parallelism
and the Center, pp. 14–18
Practice 1.3.1: Constructing Equilateral Triangles
Inscribed in Circles, pp. 96–97
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; reduction
6. k = 0.4; reduction
7. k = 1; congruency transformation
8. 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.
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.
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.
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 1.3.2: Constructing Squares Inscribed in
Circles, p. 108
1. 10.125
2. 28.5
3. 4.6
4. 3/5
5. 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 0.5625
10. 6 feet by 8 feet
1–2. Check students’ work for accuracy. Be sure each of the
vertices lies on the circle.
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.
8. 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.
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 1.2.1: Copying Segments and Angles,
pp. 42–43
Practice 1.3.3: Constructing Regular Hexagons
Inscribed in Circles, pp. 123–124
1–10. Check students’ work for accuracy.
1. Check students’ work for accuracy. Be sure each of the
vertices lies on the circle.
2–3. 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.
4. 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.
Practice 1.1.2: Investigating Scale Factors, p. 24
Practice 1.2.2: Bisecting Segments and Angles,
pp. 59–60
1–10. Check students’ work for accuracy.
Practice 1.2.3: Constructing Perpendicular and
Parallel Lines, p. 78
1–10. Check students’ work for accuracy.
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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. Check students’ work for accuracy. Be sure each of the
vertices lies on the circle.
7–8. 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.
9. 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.
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.
5.
10
9
8
4
3
2
1
C C'
A'
C'
A'
A
-3
-4
-5
-6
x
B
C
-7
-8
-9
-10
6.
10
9
8
7
6
5
4
3
2
1
A'
y
A
C
B
x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
C'
B B'
A
R
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
-2
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.
y
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
-2
-3
B'
7
6
5
Practice 1.4.1: Describing Rigid Motions and
Predicting the Effects, pp. 141–145
10
9
8
7
6
5
4
3
2
1
y
B'
x
7.
-4
-5
-6
-7
-8
-9
-10
U1-758
CCGPS Analytic Geometry Teacher Resource
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.
8.
B'
A
A
B
C' C
R
A'
B
F
C
E'
F'
A'
E D
E''
F''
A''
D'
B'
D''
C' C''
B''
2 units to the right.
Practice 1.4.2: Defining Congruence in Terms of
Rigid Motions, pp. 157–161
Fireplace
Chair 1
Coffee
table
Couch
Chair 2
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.
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.
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Practice 1.5.1: Triangle Congruency, pp. 172–175
1. 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. SAS
2. ASA
3. Congruency cannot be determined; the identified
congruent parts form SSA, which is not a triangle
congruence statement.
4. ASA
5. Congruency cannot be determined; the identified
congruent parts form SSA, which is not a triangle
congruence statement.
6. SSS
7.  DEF ≅ TVS ; SAS
8. 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; SAS
10. The pieces of fabric are congruent; SAS or SSS
Practice 1.6.1: Defining Similarity, pp. 200–204
1. ∠C = 44˚, ∠D = ∠E = 68˚, CB = 4.5, DF = 2.1
2. ∠L = ∠M = ∠P = 45˚, MP = 9, NP = 6
3. 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 .
5. Similarity transformations preserve angle measures. The
triangles are not similar because the angle measures in each
triangle are different.
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 .
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 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 1/3
5. ABC ∼ ZYX ; x = 2
6. BCD ∼BDA ; x = 10 2/3
7. 3 feet
8. 9 feet
9. 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 SSS
2. ACE ∼BCD by SAS
3. ABC ∼ DEF by SSS
4. not similar; corresponding sides are not proportional
5. ABC ∼EFD ; SAS
6. not similar; corresponding sides are not proportional
7. ABC ∼ DEF ; SSS
8. x = 7
9. x = 5
10. x = 6
Practice 1.7.2: Working with Ratio Segments,
pp. 239–242
1. CD = 12 3/8 units
2. BC = 3 1/3 units
3. DE = 8.4 units
4. CD = 6 units
5. BC = 10 units; CD = 12 units
6. CB = 21 units; CD = 35 units
7. 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. x =
3 26
2
≈ 7.6 units
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CCGPS Analytic Geometry Teacher Resource
Subtraction Property m∠ 4 = 90 . By definition, ∠ 4 is a
right angle. Therefore, ∠2 , ∠3 , and ∠ 4 are right angles.
10. x = 13
3. x = 3 ≈ 1.73 units
4. x =
5 6
2
≈ 19.4 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 units
9. 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 ft
2. 8.75 ft
3. 58 1/3 ft
4. 649.6 m
5. 23 m
6. 3.2 ft
7. 33.8 ft
8. 22.3 m
9. 7.8 m
10. 30 m
Practice 1.8.1: Proving the Vertical Angles Theorem,
pp. 292–295
∠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:
m∠7 + m∠1 + m∠2 = 180 .
3. ∠1 and ∠ 4 are vertical angles. Statement: ∠1 ≅ ∠ 4 .
4. ∠ 4 and ∠5 are complementary angles. Statement:
m∠ 4 + m∠5 = 90 .
5. 131º
6. 79º
7. 102º
8. 48º 9. 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, m∠1 + m∠2 = 180 ,
m∠2 + m∠3 = 180 , and m∠1 + m∠ 4 = 180 . Given that
∠1 is a right angle, by the definition of right angles,
m∠1 = 90 . Use substitution so that 90 + m∠2 = 180 ,
and by the Subtraction Property, m∠2 = 90 . By
definition, ∠2 is a right angle. Use substitution so
that 90 + m∠3 = 180 , and by the Subtraction Property
m∠3 = 90 . By definition, ∠3 is a right angle. Use
substitution so that 90 + m∠ 4 = 180 , and by the
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. 111º
7. 95º
8. m∠1 = 100, m∠2 = m∠3 = 80 , x = 11, and y = 12
9.
Statements
1. m  n and l is the
transversal.
2. ∠2 ≅ ∠6
3. ∠
6 and ∠8 are a linear
pair.
4. ∠
6 and ∠8 are
supplementary.
5. m∠6 + m∠8 = 180
6. m∠2 + m∠8 = 180
7. ∠2 and ∠8 are
supplementary.
Reasons
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 Theorem
6. Substitution
7. Supplement Theorem
10. Since l is a transversal and l ⊥ m, ∠1 is a right angle.
m∠1 = 90 because of the definition of a right angle.
Since lines m and n are parallel, ∠1 ≅ ∠2 because of the
Corresponding Angles Postulate. m∠1 = m∠2 because
of the definition of congruent angles. m∠2 = 90 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 = 20
2. m∠B = 56
3. m∠B = 81; m∠C = 21
4. m∠A = 85; m∠B = 52; m∠C = 43
5. m∠A = 92; m∠B = 29
6. m∠A = 128; m∠B = 52
7. m∠CAB = 26; m∠ABC = 24
8. m∠CAB = 25; m∠ABC = 65
9. m∠CAB = 10; m∠ABC = 125
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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 = 70
2. m∠ADB = 108; m∠DCB = 72; m∠DBC = 36
3. m∠A = 120; m∠B = m∠C = 30
4. m∠A = m∠C = 45; m∠B = 90
5. x = 3
6. x = 78; y = 51
7. x = 13
8. 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 = 55
2. BC = 21; YZ = 5.625; m∠AXY = 72
3. XY = 5
4. AB = 42
5. (–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
3
the same slope equal to − . Therefore, EF  BC .
4
Use the distance formula to find EF = 5 and BC = 10.
1
So, EF = BC .
2
8. Use the slope formula to show that EF and BC have
the same slope equal to –1. Therefore, EF  BC . Use the
distance formula to find EF = 2 and BC = 2 2 .
1
So, EF = BC .
2
9. The midpoint of AC = (1, 1); the midpoint of
BC = (4, 2); use the slope formula to show that EF
1
and AB have the same slope equal to . Therefore,
3
EF  AB . Use the distance formula to find EF = 10 and
1
AB = 2 10 . So, EF = AB .
2
10. The midpoint of AC = (1, 1); the midpoint of
BC = (–1, 6); use the slope formula to show that EF
5
and AB have the same slope equal to − . Therefore,
2
EF  AB . Use the distance formula to find EF = 29 and
1
AB = 2 29 . So, EF = AB .
2
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
1
triangle are y = 2x – 11, y = –x + 4, and y = x − 2 . (5, –1)
5
is a solution to the equation of each median.
4. The distance from A to U is approximately 6.71 units. The
2
distance from (5, –1) to A is 4.47 units. (5, –1) is the
3
distance from A. The distance from B to V is approximately
6.36 units. The distance from (5, –1) to B is approximately
2
4.24 units. (5, –1) is the distance from B. The distance
3
from C to T is approximately 7.65 units. The distance from
2
(5, –1) to C is approximately 5.10 units. (5, –1) is the
3
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.
U1-762
CCGPS Analytic Geometry Teacher Resource
Practice 1.10.1: Proving Properties of
Parallelograms, pp. 428–430
1. No, it’s not a parallelogram because opposite sides are not
2
1
parallel: mTU = 5 , mVW = 2 , mUV = − , mWT = .
3
3
2. Yes, it’s a parallelogram because opposite sides are parallel:
2
mWX = mYZ = 5 and mXY = mZW = .
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
3

other: M =  3, −  .
2

6. No, the midpoints are not the same, indicating the
diagonals do not bisect each other: M MO = (1,1) and
 3 3
M NP =  ,  .
 2 2
7. m∠A = m∠C = 165 and m∠B = m∠D = 15 ; x = 15 and
y = 21
8. m∠A = m∠C = 92 and m∠B = m∠D = 88; x = 18 and
y = 24
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
AB BF
=
words,
.
ED DA
b
10. Opposite sides are parallel: mAB = mCD = and
a
mBC = mDA = 0 . The diagonals bisect each other because
 a+c b
they have the same midpoint: M = 
, .
 2 2 
Practice 1.10.2: Proving Properties of Special
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.
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
1. Given
2. The diagonals are
congruent and
bisect each other.
3. The diagonals
3. AC ⊥ DB
of a square are
perpendicular.
4.
Definition of
4. ∠APD ≅ ∠APB ≅ ∠BPC ≅ ∠CPD,
perpendicular lines
and they are right angles.
5. ∠APD ≅ ∠APB ≅ ∠BPC ≅ ∠CPD 5. All right angles are
congruent.
6.
SAS Congruence
6.  APD ≅ APB ≅CPB ≅CPD
Statement
10. Given  APD ≅ APB ≅CPB ≅CPD , Corresponding
Parts of Congruent Triangles are Congruent. This
means that AD ≅ AB ≅ CB ≅ CD . Also by CPCTC,