Theorem 10.3 : The perpendicular from the center of a circle to a chord bisects the chord Class 9

Given: a circle with center O. AB is a chord of a circle and OD bisects AB

To prove :  AD = BD

Proof : In △AOD and △BOD

⇒ OA = OB (Both Radius)

⇒∠ODA = ∠ODB (Both 90°, given)

⇒ OD = OD (Common)

∴ △AOD ⩭ △BOD (by SAS congruency)

⇒ AD = ∠BD ( By CPCT )

Hence proved