Theorem 9.3 - Chapter 9 Class 9 Circles
Last updated at Feb. 14, 2025 by Teachoo
Last updated at Feb. 14, 2025 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.
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
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo