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Theorem 10.11 - Chapter 10 Class 9 Circles (Term 2)

Last updated at Aug. 13, 2018 by

Theorem 10.11 - Sum of opposite angles in cyclic quadrilateral is 180

Theorem 10.11 - Chapter 10 Class 9 Circles - Part 2
Theorem 10.11 - Chapter 10 Class 9 Circles - Part 3

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Theorem 10.11 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.