Theorems
Theorem 9.2 Important
Theorem 9.3 Important You are here
Theorem 9.4
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 2025 Exams
Last updated at April 16, 2024 by Teachoo
Theorem 9.3 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.