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Name ________________________________________ Date __________________ Class__________________ LESSON 12-4 Reteach Inscribed Angles Inscribed Angle Theorem The measure of an inscribed angle is half the measure of its intercepted arc. ∠ABC is an inscribed angle. p is an AC intercepted arc. 1 p m∠ABC = m AC 2 Inscribed Angles An inscribed angle subtends a semicircle if and only if the angle is a right angle. If inscribed angles of a circle intercept the same arc, then the angles are congruent. ∠ABC and ∠ADC intercept p, so ∠ABC ≅ ∠ADC. AC Find each measure. q 1. m∠LMP and mMN p 2. m∠GFJ and mFH _________________________________________ Find each value. 3. x ________________________________________ 4. m∠FJH _________________________________________ ________________________________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 12-30 Holt McDougal Geometry Name ________________________________________ Date __________________ Class__________________ LESSON 12-4 Reteach Inscribed Angles continued Inscribed Angle Theorem ∠A and ∠C are supplementary. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. ∠B and ∠D are supplementary. ABCD is inscribed in :E. Find m∠G. Step 1 Find the value of z. m∠E + m∠G = 180° EFGH is inscribed in a circle. 4z + 3z + 5 = 180 Substitute the given values. 7z = 175 z = 25 Simplify. Divide both sides by 7. Step 2 Find the measure of ∠G. m∠G = 3z + 5 = 3(25) + 5 = 80° Substitute 25 for z. Find the angle measures of each quadrilateral. 5. RSTV 6. ABCD _________________________________________ 7. JKLM ________________________________________ 8. MNPQ _________________________________________ ________________________________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 12-31 Holt McDougal Geometry 11. 90°; 90°; 90°; 90° congruent. Draw any inscribed angle. Use the compass to copy the arc that this angle intercepts. Mark off the same arc from the vertex of the inscribed angle. Connect the points. 12. 68°; 95°; 112°; 85° 13. 59°; 73°; 121°; 107° Practice C 1. Possible answer: It is given that AC ≅ AD. In a circle, congruent chords intercept congruent arcs, so q ≅ AED q. DC p is congruent to itself by ABC the Reflexive Property of Congruence. By the Arc Addition Postulate and the Addition Property of Congruence, q ≅ ADC q . ∠ABC intercepts ADC q , so ACD 1 q . ∠AED intercepts m∠ABC = m ADC 2 q , so m∠AED = 1 m ACD q . By ACD 2 substitution, m∠ABC = m∠AED. Therefore ∠ABC ≅ ∠AED. 2. Possible answer: It is given that p ≅ RS p. By the definition of congruent PQ p = m RS p . ∠PSQ intercepts arcs, m PQ p , and ∠RQS intercepts RS p . So PQ m∠PSQ must equal m∠RQS. Therefore ∠PSQ ≅ ∠RQS. ∠PSQ and ∠RQS are congruent alternate interior angles of QR 6. cannot be inscribed in a circle Reteach q = 96° 1. m∠LMP = 18°; m MN p = 72° 2. m∠GFJ = 55°; m FH 3. 16.4 4. 45° 5. 70°; 88°; 110°; 92° 6. 120°; 75°; 60°; 105° 7. 132°; 90°; 48°; 90° 8. 101°; 86°; 79°; 94° Challenge 1. chord; inscribed 2. a. 45° b. 67.5° and PS. So QR & PS. c. |180 − 11.25n| 3. cannot be inscribed in a circle d. 0 < |n − 16| < 16 4. Can be inscribed in a circle; possible answer: The two congruent angles of the kite are opposite, so they must be right angles. Draw a diameter. Draw segments from opposite ends of the diameter to any point on the circle. Use the compass to copy one of the segments across the diameter. Draw the fourth side. ⎡ ⎛ 2 n ⎞⎤ ⎢180 ⎜ 1 − ⎟⎥ , p ⎠⎦ ⎝ ⎣ p p where 0 < n − n − < 2 2 e. 180 − 360 n , or p f. Answers will vary. Students may choose any values of n and p for which n 5 = . Sample answer: n = 15, p = 36 p 12 Problem Solving 1. 160° 2. 112°; 52°; 68°; 128° 5. Can be inscribed in a circle; possible answer: The pairs of base angles of a trapezoid inscribed in a circle must be 3. C 4. G 5. B 6. G Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. A52 Holt McDougal Geometry