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LESSON
3-3
Reteach
Proving Lines Parallel
Converse of the Corresponding
Angles Postulate
If two coplanar lines are cut by a transversal
so that a pair of corresponding angles are
congruent, then the two lines are parallel.
You can use the Converse of the
Corresponding Angles Postulate
to show that two lines are parallel.
Given: ∠1 ≅ ∠3
∠1 ≅ ∠3
q || r
∠1 ≅ ∠3 are corresponding angles.
Converse of the Corresponding Angles Postulate
Given: m∠2 = 3x°, m∠4 = (x + 50)°, x = 25
m∠2 = 3(25)° = 75°
Substitute 25 for x.
m∠4 = (25 + 50)° = 75°
Substitute 25 for x.
m∠2 = m∠4
Transitive Property of Equality
∠2 ≅ ∠4
Definition of congruent angles
q || r
Converse of the Corresponding Angles Postulate
For Exercises 1 and 2, use the Converse of the Corresponding Angles
Postulate and the given information to show that c || d.
1. Given: ∠2 ≅ ∠4
2. Given: m∠1 = 2x°, m∠3 = (3x − 31)°, x = 31
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Name ________________________________________ Date __________________ Class__________________
LESSON
3-3
Reteach
Proving Lines Parallel continued
You can also prove that two lines are parallel by using the converse of any of
the other theorems that you learned in Lesson 3-2.
Theorem
Hypothesis
Converse of the Alternate
Interior Angles Theorem
Conclusion
a || b
∠2 ≅ ∠3
Converse of the Alternate
Exterior Angles Theorem
f || g
∠1 ≅ ∠4
Converse of the Same-Side
Interior Angles Theorem
s || t
m∠1 + m∠2 = 180°
For Exercises 3–5, use the theorems and the given information to
show that j || k.
3. Given: ∠4 ≅ ∠5
4. Given: m∠3 = 12x°, m∠5 = 18x°, x = 6
5. Given: m∠2 = 8x°, m∠7 = (7x + 9)°, x = 9
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supplementary by the Linear Pair
Theorem. Therefore ∠3 and ∠2 must be
congruent by the Congruent Supplements
Theorem. Since ∠3 and ∠2 are
congruent, HI and JK are parallel by the
Converse of the Corresponding Angles
Postulate.
Practice C
1. x = 11; y = −5; m∠1 = 57°; m∠2 = 57°;
m∠3 = 123°
Reteach
1. ∠2 ≅ ∠4 ∠2 and ∠4 are corr. ∠s .
c || d
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
Substitute 31 for x.
Substitute 31 for x.
Trans. Prop. of =
Def. of ≅ ∠s
Conv. of Corr. ∠s Post.
3. ∠4 ≅ ∠5 ∠4 and ∠5 are alt. int. ∠s .
2.
j || k
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.
3. The measures of the segments are equal.
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 Same-Side
Int. ∠s Thm.
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
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)
b. 0 < q < 90; p < q
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