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

Let POQ be a diameter, bisecting chord AB at L.

Join OA and OB

In △OLA and △OLB:

OA = OB [Both are radii of the same circle]

AL = LB [PQ bisects the chord AB]

OL = OL [common side]

△OLA ≅ △OLB [By the SSS rule]

Since the triangles are congruent, their corresponding parts must be equal:

∠AOL = ∠BOL

This proves that the diameter PQ bisects ∠AOB, which is the angle subtended by the chord AB at the center O.

Hence, proved that the diameter also bisects the angle subtended by the chord at the centre.