Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

(N/A) Let us consider a circle with centre $O$. Let $AB$ be a tangent which touches the circle at $P$.
We have to prove that the line perpendicular to $AB$ at $P$ passes through the centre $O$. We shall prove this by the contradiction method.
Let us assume that the perpendicular to $AB$ at $P$ does not pass through the centre $O$. Let it pass through another point $O'$.
Join $OP$ and $O'P$.
As the perpendicular to $AB$ at $P$ passes through $O'$,therefore,
$\angle O'PB = 90^{\circ} \dots(1)$
$O$ is the centre of the circle and $P$ is the point of contact. We know that the radius through the point of contact is perpendicular to the tangent.
$\therefore \angle OPB = 90^{\circ} \dots(2)$
Comparing equations $(1)$ and $(2)$,we obtain
$\angle O'PB = \angle OPB \dots(3)$
From the figure,it can be observed that,
$\angle O'PB < \angle OPB \dots(4)$
Therefore,$\angle O'PB = \angle OPB$ is not possible.
It is only possible when the line $O'P$ coincides with $OP$.
Therefore,the perpendicular to $AB$ through $P$ passes through the centre $O$.