Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at

Given:

PQ is a diameter of circle which bisects the chord AB at C.

To Prove: PQ bisects ∠AOB

Proof:

In ∠AOC and ∠BOC

OA = OB [Radius of circle]

OC = OC [Common]

AC = BC [Given]

Then ΔAOC ≅ ΔBOC [By SSS condition]

∠AOC = ∠BOC   [C.P.C.T]

Hence PQ bisects ∠AOB.