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.

Explanation

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

To prove: PQ bisects `angleAOB`

Proof: In `triangleBOC` and `triangleAOC`

`OA=OB` (radius of the circle)

`OC=OC` (common)

`AC=BC` (given)

`triangleAOC~=triangleBOC` (By SSS)

So, `angleAOC=angleBOC` (By CPCT)

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