Question

From any external point $P$, two tangent $PA$ and $PB$ are drown to the circle with centre $O$. Prove that $OP$ is the perpendicular bisector of $AB$.

Answer 1

Suppose $OP$ intersects $AB$ at $C$.

In triangles $PAC$ and $PBC$, we have

$PA=PB$ [$\because$ tangents from an external point are equal]

$\angle APC=BPC$ [$\because PA$ and $PB$ are equally inclined to $OP$]

and, $PC=PC$

So, by $SAS-$criterion of congruence, we have

$△PAC\cong △PBC$

$\Rightarrow AC=BC$ and $\angle ACP=\angle BCP$

But, $\angle ACP+\angle BCP={180}^o$

$\therefore \angle ACP+\angle BCP={90}^o$

Hence, $OP\perp AB$.