Last updated at Aug. 25, 2021 by

Transcript

Theorem 10.3 The perpendicular from the center of a circle to a chord bisects the chord. Given : C is a circle with center at O. AB is a chord such that OX ⊥ AB To Prove : OX bisect chord AB i.e. AX = BX Proof : In ∆OAX & ∆OBX ∠OXA = ∠OXB OA = OB OX = OX ∴ ∆OAX ≅ ∆OBX AX = BX Hence, Proved.

Theorems

Theorem 10.1

Theorem 10.2 Important

Theorem 10.3 Important You are here

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

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.