Question

From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB.

Answer 1

In $\Delta PAC$ and $\Delta PBC$

$PA=PB$

[length of tangents drawn from external point are equal ]

$\angle APC=\angle BPC$

[PA and PB are equally inclined to OP]

and $PC=PC$ [Common]

So, by SAS criteria of similarity

$\Delta PAC\cong \Delta PBC$

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

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

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

Hence, $OP\perp AB$