Download 3.3 practice

Name ________________________________________ Date __________________ Class__________________
LESSON
3-3
Practice B
Proving Lines Parallel
Use the figure for Exercises 1–8. Tell whether lines m and n
must be parallel from the given information. If they are, state
your reasoning. (Hint: The angle measures may change for
each exercise, and the figure is for reference only.)
1. ∠7 ≅ ∠3
2. m∠3 = (15x + 22)°, m∠1 = (19x − 10)°,
x=8
_________________________________________
3. ∠7 ≅ ∠6
__________________________________________
4. m∠2 = (5x + 3)°, m∠3 = (8x − 5)°,
x = 14
_________________________________________
__________________________________________
_________________________________________
__________________________________________
5. m∠8 = (6x − 1)°, m∠4 = (5x + 3)°, x = 9
6. ∠5 ≅ ∠7
_________________________________________
7. ∠1 ≅ ∠5
__________________________________________
8. m∠6 = (x + 10)°, m∠2 = (x + 15)°
_________________________________________
__________________________________________
9. Look at some of the printed letters in a textbook. The small horizontal and
vertical segments attached to the ends of the letters are called serifs. Most of the
letters in a textbook are in a serif typeface. The letters on this page do not have
serifs, so these letters are in a sans-serif typeface. (Sans means “without” in French.)
The figure shows a capital letter A with serifs. Use the given information to write a
paragraph proof that the serif, segment HI, is parallel to segment JK .
Given: ∠1 and ∠3 are supplementary.
Prove: HI || JK
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
3-20
Holt McDougal Geometry
3. m∠1 + m∠4 + m∠ABE +
m∠DEB = 360°
3. Add. Prop. of =
7. ∠1 ≅ ∠7
8. ∠4 ≅ ∠6
9. m∠2 + m∠5 = 180°
4. m∠3 + m∠CEB + m∠CBE
= 180°
4. Given
10. m∠3 + m∠8 = 180°
5. m∠DEB + m∠CEB = 180°
5. Lin. Pair Thm.
6. m∠3 + m∠CEB + m∠CBE
= m∠DEB + m∠CEB
6. Subst. (Steps 4,
5)
11. m∠6 = 47° by the Corresponding Angles
Postulate
7. m∠3 + m∠CBE = m∠DEB
8. m∠1 + m∠3 + m∠4 +
m∠ABE + m∠CBE = 360°
9. m∠2 = m∠ABE + m∠CBE
10. m∠1 + m∠2 + m∠3 +
m∠4 = 360°
12. m∠3 = 133° by the Same-Side Interior
Angles Theorem
7. Subtr. Prop. of
=
8. Subst. (Steps 3,
7)
3-3 PROVING LINES PARALLEL
Practice A
9. Angle Add.
Post.
10. Subst. (Steps
8, 9)
1. parallel
2. Conv. of Corr. ∠s Post.
3. m∠7 = 68°, ∠3 ≅ ∠7, Conv. of Corr. ∠s
Post.
Reteach
1. no
2. yes
4. transversal; congruent
3. 67°
4. 142°
5. supplementary
5. 92°
6. 125°
7.
7. 111°
8. 90°
9. 138°
10. 56°
11. 130°
12. 118°
6. parallel
Statements
1. ∠1 and ∠3 are
supplementary.
Challenge
1. Justifications may vary. All lines directed
due north are parallel. A heading that is
read off the compass is the same as the
ship’s heading.
2. about 102°
3. about 38°
4. about 170°
5. about 256°
Reasons
1. a. Given
2. b. ∠2 and ∠3 are
supplementary.
2. Linear Pair Thm.
3. ∠1 ≅ ∠2
3. c. ≅ Supps. Thm.
4. d. m || n
4. Conv. of Corr. ∠s Post.
Practice B
1. m || n; Conv. of Alt. Int. ∠s Thm.
2. m || n; Conv. of Corr. ∠s Post.
Problem Solving
1. 17; Alt. Int. ∠s Thm.
3. m and n are parallel if and only if
m∠7 = 90°.
2. 102°; Alt. Ext. ∠s Thm.
3. x = 10; y = 3; (12x + 2y)° = 126° by the Corr.
∠s Post. and (3x + 2y)° = 36° by the Alt. Int.
∠s Thm.
4. D
5. H
4. m || n; Conv. of Same-Side Int. ∠s Thm.
5. m and n are not parallel.
6. m || n; Conv. of Corr. ∠s Post.
7. m || n; Conv. of Alt. Ext. ∠s Thm.
Reading Strategies
8. m and n are not parallel.
1. ∠1 ≅ ∠5
2. ∠2 ≅ ∠6
3. ∠3 ≅ ∠7
4. ∠4 ≅ ∠8
5. ∠2 ≅ ∠8
6. ∠3 ≅ ∠5
9. Sample answer: The given information
states that ∠1 and ∠3 are
supplementary. ∠1 and ∠2 are also
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A23
Holt McDougal Geometry
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
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A24
Holt McDougal Geometry