Last updated at Aug. 13, 2018 by Teachoo

Transcript

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

Theorems

Theorem 10.1

Theorem 10.2 Important

Theorem 10.3 Important

Theorem 10.4

Theorem 10.5 Deleted for CBSE Board 2022 Exams

Theorem 10.6 Important

Theorem 10.7

Theorem 10.8 Important

Theorem 10.9

Theorem 10.10 Important

Theorem 10.11 You are here

Theorem 10.12 Important

Angle in a semicircle is a right angle Important

Chapter 10 Class 9 Circles (Term 2)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.