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Theorem 10.3 Class 9 - Perpendicular from center to a chord bisects it

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Theorem 10.3 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.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.