
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Theorems
Theorem 9.2 Important
Theorem 9.3 Important
Theorem 9.4 You are here
Theorem 9.5 Important
Theorem 9.6
Theorem 9.7 Important
Theorem 9.8
Theorem 9.9 Important
Theorem 9.10
Theorem 9.11 Important
Angle in a semicircle is a right angle Important
Only 1 circle passing through 3 non-collinear points Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
Theorem 9.4 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.