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

Solution: let us assume $\perp$ through P passes through $O^{\prime }$
$O^{\prime }P\perp AP$ so, $\angle AP{O}^{\prime }={90}^{\circ }$
now, join O & P, if $OP\perp AP$
then, $\angle APO={90}^{\circ }$
$\angle APO$ Should be greater than $\angle AP{O}^{\prime }$
but both the angle are equal
so, $\perp$from P must passes through O
hence proved