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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