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Theorem 10.3 - Chapter 10 Class 9 Circles (Term 2)
Last updated at Aug. 25, 2021 by Teachoo
Chapter 10 Class 9 Circles
Serial order wise
Transcript
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.
