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Theorem 9.11 The sum of either pair of opposite angles of a cyclic quadrilateral is 180°. Given : ABCD is a cyclic quadrilateral. of a circle with centre at O To Prove : ∠ BAD + ∠ BCD = 180° ∠ ABC + ∠ ADC = 180° Proof: Chord AB Angles in same segment are equal. ∠5 = ∠8 Chord BC Angles in same segment are equal. ∠1 = ∠6 Chord CD Angles in same segment are equal. ∠2 = ∠4 Chord AD Angles in same segment are equal. ∠7 = ∠3 ∠1 + ∠2 + ∠3 + ∠4 + ∠7 + ∠8 + ∠5 + ∠6 = 360° (∠1 + ∠2 + ∠7 + ∠8) + (∠3 + ∠4 + ∠5 + ∠6) = 360° ∴ (∠1 + ∠2 + ∠7 + ∠8) + (∠7 + ∠2 + ∠8 + ∠1) = 360° ⇒ 2 (∠1 + ∠2 + ∠7 + ∠8) = 360° ∠1 + ∠2 + ∠7 + ∠8 = 180° (∠1 + ∠2) + (∠7 + ∠8) = 180° ∠BAD + ∠BCD = 180° Similarly, ∠ABC + ∠ADC = 180° Hence, Proved. From (1), (2) , (3), (4) ∠ 3 = ∠ 7 ∠ 4 = ∠ 2 ∠ 6 = ∠ 1 ∠ 5 = ∠ 8

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo