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

Explanation

Let us support PQ does not pass through O.
Since, `PQ _|_`bisector of `AB`
`/_QOP=90^@`⟶(1)
P is the mid-point of AB.
`QP` passes through center `OP_|_AB ,/_OPA=90^@`-(2)
Comparing 1 & 2,`/_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.