Transversals EC2

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LESSON
3-3
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.
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
3-25
Holt Geometry
HJJG
Possible answer: Construct FG through
point C and parallel to AB . ∠3 and ∠4
are a linear pair, so m∠3 + m∠4 = 180°
by the Linear Pair Theorem. But the
Angle Addition Postulate shows that m∠4
= m∠ACF + m∠FCD, so by substitution
m∠3 + m∠ACF + m∠FCD = 180°. m∠1 =
m∠ACF by the Alternate Interior Angles
Theorem and m∠2 = m∠FCD by the
Corresponding Angles Postulate.
Therefore m∠1 + m∠2 + m∠3 = 180° by
substitution.
5.
m∠2 = 8(9)° = 72°
m∠7 = 7(9)° + 9° = 72°
m∠2 = m∠7
∠2 ≅ ∠7
j || k
Substitute 9 for x.
Substitute 9 for x.
Trans. Prop. of =
Def. of ≅ ∠s
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
3. The measures of the segments are equal.
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)
4. Possible answer: If a triangle is isosceles,
then the sides opposite the congruent
angles are congruent.
b. 0 < q < 90; p < q
Reteach
1. ∠2 ≅ ∠4 ∠2 and ∠4 are corr. ∠s .
c || d
Problem Solving
Conv. of Corr. ∠s Post.
2.
m∠1 = 2x°
= 2(31)° = 62°
m∠3 = (3x − 31)°
= 3(31)° − 31° = 62°
m∠1 = m∠3
∠1 ≅ ∠3
c || d
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.
Substitute 31 for x.
2. m∠3 = (3y + 7) = 3(21) + 7 = 70° Replace y with 21.
Substitute 31 for x.
Trans. Prop. of =
Def. of ≅ ∠s
Conv. of Corr. ∠s Post.
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 Same-Side Int.
∠s Thm.
3. ∠4 ≅ ∠5 ∠4 and ∠5 are alt. int. ∠s .
j || k
Conv. of Alt. Int. ∠s Thm.
3. A
4.
4. J
5. m∠1 = 40° and m∠2 = 40°, so the sides
are || by the Conv. of the Alt. Int. ∠s Thm.
m∠3 = 12(6)° = 72° Substitute 6 for x.
m∠5 = 18(6)° = 108° Substitute 6 for x.
6. m∠3 = 90° and m∠4 = 90°, so the sides
are || by the Conv. of the Same-Side Int.
∠s Thm.
m∠3 + m∠5 = 72° + 108° = 180° Add angle measures.
j || k
Conv. of Same-Side
Int. ∠s Thm.
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A23
Holt Geometry

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