Prove that the perpendicular bisector of a chord of a circle always passes through the centre.

Explanation

To prove: The perpendicular bisector of a chord of a circle always passes through the centre.
Proof:
Let us support `PQ` does not pass through `O`.
Since `PQ` is perpendicular bisector of `AB`
`therefore∠QOP=90°` ................(1)
`P` is the mid-point of `AB`.
`QP` passes through center from this, we can say that
`OP⊥AB` i.e., `∠OPA=90°` ................(2)
Comparing 1 & 2, we can say
`∴∠QOP=∠OPA`, this can be possible only when `O`, `P` and `Q` are on the same line.
`∴` It is a contradiction `PQ` passes through `O`.
Hence Proved.