Question

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.

Answer 1

Given : In circle with center $O,CD$ is the diameter and $AB$ is the chord which is bisected by diameter at $E,OA$ and $OB$ are joined

To prove : $\angle AOE=\angle BOE$

Proof: In $\Delta OAEand\Delta OBE$

$OA=OB$ (Radii of the circle)

$OE=OE$ (Common)

$AE=BE$ (Given)

$\therefore \Delta OAE\cong \Delta OBE$ (By SSS congruency criterian)

$\therefore \angle AOE=\angle BOE$ (CPCT)

Hence, diameter bisects the angle subtended by the chord $AB$ at the center of the circle.