Last updated at Aug. 13, 2018 by Teachoo

Transcript

Theorem 10.4 The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Given : A circle with center at O. AB is chord of circle & OX bisects AB i.e. AX = BX To Prove : OX ⊥ AB Proof : In ∆AOX & ∆BOX OA = OB OX = OX AX = BX ∴ ∆AOX ≅ ∆BOX ∠ AXO = BXO In line AB, Hence, ∠AXO and ∠BXO form linear Pair ∠AXO + ∠BXO = 180° ∠AXO + ∠AXO = 180° 2 ∠AXO = 180° ∠AXO = (180°)/2 ∠AXO = 90° ∴ ∠AXO = ∠BXO = 90° ⇒ OX ⊥ AB Hence, Proved.

Theorems

Theorem 10.1

Theorem 10.2 Important

Theorem 10.3 Important

Theorem 10.4 You are here

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.