Theorems

Theorem 9.1

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 April 16, 2024 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.