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Name _______________________________________ Date __________________ Class __________________
Problem Solving
Proving Lines Parallel
1. A bedroom has sloping ceilings as shown. Marcel is hanging
a shelf below a rafter. If m1(8x  1), m2(6x7),
and x4, show that the shelf is parallel to the rafter above it.
2. In the sign, m3(3y7), m4(5y5), and y21.
Show that the sign posts are parallel.
Choose the best answer.
3. In the bench, mEFG(4n16), mFJL(3n40),
mGKL(3n22), and n24. Which is a true statement?
A FG || HK by the Converse of the Corr. s Post.
B FG || HK by the Converse of the Alt. Int. s Thm.
C EJ || GK by the Converse of the Corr. s Post.
D EJ || GK by the Converse of the Alt. Int. s Thm.
4. In the windsurfing sail, m5(7c1), m6(9c  1),
m717c, and c6. Which is a true statement?
F RV is parallel to SW .
G SW is parallel to TX .
H RT is parallel to VX .
J Cannot conclude that two segments are parallel
The figure shows Natalia’s initials, which are
monogrammed on her duffel bag. Use the
figure for Exercises5and 6.
5. If m1(4x  24), m2(2x8),
and x16, show that the sides of
the letter N are parallel.
6. If m3(7x13), m4(5x35),
and x11, show that the sides of the
letter H are parallel.
© Houghton Mifflin Harcourt Publishing Company
Holt McDougal Analytic Geometry
Name _______________________________________ Date __________________ Class __________________
4.
Possible answer: If a triangle is
isosceles, then the sides opposite the
congruent angles are congruent.
6. a. m1  m2  m6  (180  p)°
m3  m4  m5  (180  q)°
m7  p°; m8  q°
m9  (180  p–q)
Reteach
1. 2  4 2 and 4 are corr. s .
c || d
Conv. of Corr. s Post.
2.
m1  2x°
 2(31)°  62°
Substitute 31 for x.
m3  (3x  31)°
 3(31)°  31°  62° Substitute 31 for x.
m1  m3
Trans. Prop. of 
1  3
Def. of  s
c || d
Conv. of Corr. s
Post.
3. 4  5 4 and 5 are alt. int. s .
j || k
Conv. of Alt. Int. s Thm.
4.
m3  12(6)°  72° Substitute 6
for x.
m5  18(6)°  108°
Substitute 6 for
x.
m3  m5  72°  108°  180°
Add
angle
measures.
j || k
Conv. of SameSide Int. s
Thm.
5.
m2  8(9)°  72° Substitute 9 for x.
m7  7(9)°  9°  72° Substitute 9 for x.
m2  m7 Trans. Prop. of 
2  7 Def. of  s
j  k
Conv. of Alt. Ext. s
Thm.
Challenge
1. a  22.5
2. a  13
3. a  22
4. Explanations may vary.
5. a. Explanations may vary.
b. 0  c  20; 0  d  100
b. 0  q  90; p  q
Problem Solving
1. m1  (8x  1)  8(4)  1  31°
Replace x
with 4.
m2  (6x  7)  6(4)  7  31°
Replace x
with 4.
1 and 2 are corr. s and they are
congruent, so the shelf is parallel to the
rafter by the Conv. of Corr. s Post.
2. m3  (3y  7)  3(21)  7  70° Replace
y with 21.
m4  (5y  5)  5(21)  5  110°
Replace y with 21.
m3  m4  70°  110°  180°
3 and 4 are supp. s , so the sign
posts are parallel by the Conv. of SameSide Int. s Thm.
3. A
4. J
5. m1  40° and m2  40°, so the sides
are  by the Conv. of the Alt. Int. s
Thm.
6. m3  90° and m4  90°, so the sides
are  by the Conv. of the Same-Side Int.
s Thm.
Reading Strategies
1. Converse of the Alternate Exterior
Angles Theorem
2. Converse of the Same-Side Interior
Angles Theorem
3. Converse of the Alternate Interior
Angles Theorem
4. Converse of the Corresponding Angles
Postulate
5. No; 1  5.
6. 61°
© Houghton Mifflin Harcourt Publishing Company
Holt McDougal Analytic Geometry