1. Chapter 10 Class 9 Circles
  2. Serial order wise
  3. Theorems

Theorem 10.3

Last updated at Nov. 23, 2017 by

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

  1. Chapter 10 Class 9 Circles
  2. Serial order wise
    3. Theorems
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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.

Chapter 10 Class 9 Circles
Serial order wise

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