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Name ________________________________________ Date __________________ Class__________________ LESSON 4-9 Reteach Isosceles and Equilateral Triangles Theorem Examples Isosceles Triangle Theorem If two sides of a triangle are congruent, then the angles opposite the sides are congruent. If RT ≅ RS, then �T � �S. Converse of Isosceles Triangle Theorem If two angles of a triangle are congruent, then the sides opposite those angles are congruent. If �N � �M, then LN ≅ LM. You can use these theorems to find angle measures in isosceles triangles. Find m∠E in UDEF. m∠D = m∠E 5x� = (3x + 14)� 2x = 14 x=7 Isosc. U Thm. Substitute the given values. Subtract 3x from both sides. Divide both sides by 2. Thus m∠E = 3(7) + 14 = 35�. Find each angle measure. 1. m∠C = _____________________ 2. m∠Q = _____________________ 3. m∠H = _____________________ 4. m∠M = _____________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 4-70 Holt McDougal Geometry Name ________________________________________ Date __________________ Class__________________ LESSON 4-9 Reteach Isosceles and Equilateral Triangles continued Equilateral Triangle Corollary If a triangle is equilateral, then it is equiangular. (equilateral U → equiangular U) Equiangular Triangle Corollary If a triangle is equiangular, then it is equilateral. (equiangular U → equilateral U) If �A � �B � �C, then AB ≅ BC ≅ CA . You can use these theorems to find values in equilateral triangles. Find x in USTV. USTV is equiangular. Equilateral U → equiangular U (7x + 4)� = 60° The measure of each ∠ of an equiangular U is 60°. 7x = 56 x=8 Subtract 4 from both sides. Divide both sides by 7. Find each value. 5. n = _____________________ 6. x = _____________________ 7. VT = _____________________ 8. MN = _____________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 4-71 Holt McDougal Geometry 9. 693 ft 11. 6.3 13. 4 1 yd 2 10. 50° 2. x = 36; y = 72; z = 36 12. 60° 3. 2A 14. 65° 4. Possible answer: The coordinates of Y are (a, 3a ). The Midpoint Formula shows that the midpoints of the sides of ⎛1 3 ⎞ ⎛3 3 ⎞ a ⎟ , B ⎜ a, a⎟ , UXYZ are A ⎜ a, ⎜2 ⎟ ⎜ 2 ⎠ ⎝2 2 ⎟⎠ ⎝ and C(a, 0). The Distance Formula gives these distances: AX = AY = AC = AB = BC = XC = BY = BZ = CZ = a. Thus by SSS, UABC ≅ UXAC ≅ UYAB ≅ UCBZ, and the triangles are equilateral. 15. 8 Practice B 1. Possible answer: It is given that HI is congruent to HJ , so ∠I must be congruent to ∠J by the Isosceles Triangle Theorem. ∠IKH and ∠JKH are both right angles by the definition of perpendicular lines, and all right angles are congruent. Thus by AAS, UHKI is congruent to UHKJ. IK is congruent to KJ by CPCTC, so HK bisects IJ by the definition of segment bisector. 2. 58.1 ft 4. 2 Reteach 1. 51° 2. 47° 3. 72° 4. 60° 3. 45° 5. 12 6. 33 5. 36 or 9 7. 18 8. 13 4 3 6. 76° 7. 8. 10 9. 30° Challenge 1. x = 10; By Isosc. U Thm. and U Sum Thm., m∠TSV = 54°. UQSV ≅ UTSV by SSS, so m∠VSQ = 54° by CPCTC. m∠QSR = 72° by Lin. Pair Thm. and by ∠ Add. Post. By Isosc. U Thm. and U Sum Thm., m∠SQR = 36°. Solve (3x + 6)° = 36°. x = 10 10. 89 Practice C 1. Possible answer: UABC is an isosceles triangle with vertices A(0, b), B(a, 0), and C(−a, 0). D is the midpoint of BC , so D has coordinates (0,0). The slope of AD b−0 b = , so the slope is undefined. A is 0−0 0 line with an undefined slope is a vertical line. The slope of BC is 0−0 0 = = 0 . A line with a zero a − ( −a ) 2a 2. 42.2; UFJH is an equilat. U, so HJ = JF = FH = 15 and x = 3. So GJ = 12.2. FG ≅ FJ since the sides opp. the ∠s are ≅. So FG = FJ = 15. UEFH ≅ UGFJ by AAS, so the corr. sides are ≅ by CPCTC. P = 12.2 + 15 + 15 = 42.2 3. 108°; UFJH is equiangular, so m∠JFH = m∠FJH = 60°. By Lin. Pair Thm. and ∠ Add. Post., m∠FJG = 66°. By Isosc. U Thm. and U Sum Thm., m∠GFJ = 48°. m∠GFH = 48° + 60° = 108° slope is a horizontal line. Because AD is vertical and BC is horizontal, AD ⊥ BC . Problem Solving 1. 14 in. 2. 40° 3. 11 ft; m∠GJH = 72° − 36° = 36°. m∠GHJ = 36° by Alt. Int. ∠s Thm. By Converse of Isosceles Triangle Theorem, GJ = GH = 11 ft. Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. A46 Holt McDougal Geometry