# Theorem 10.11

Last updated at July 11, 2018 by Teachoo

Last updated at July 11, 2018 by Teachoo

Transcript

Theorem 10.11 The sum of either pair of opposite angles of a cyclic quadrilateral is 180º. Given : ABCD is a cyclic quadrilateral. of a circle with centre at O To Prove : ∠ BAD + ∠ BCD = 180° ∠ ABC + ∠ ADC = 180° Proof : By angle sum property of 𝑎 quadrilateral ∠A + ∠B + ∠C + ∠D = 360° ∠1 + ∠2 + ∠3 + ∠4 + ∠7 + ∠8 + ∠5 + ∠6 = 360° (∠1 + ∠2 + ∠7 + ∠8) + (∠3 + ∠4 + ∠5 + ∠6) = 360° ∴ (∠1 + ∠2 + 27 + 28) + (∠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.

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.