Tangents drawn to a circle from an external point are

Let ’O’ be the centre of the circle PA,PB are tangents drawn from P we know that, tangent is $\perp$ ar to radius at point of contact.

$\therefore \angle OAP=\angle OBP=90°$

In $\Delta OAP$ and $\Delta OBP$

$OA=OB$ (Radius)

$\angle OAP=\angle OBP$ (Right angle)

$OP=OP$ (common side)

$\therefore$ By R.H.S. Congruency,

$\Delta POA\cong \Delta POB$

$\therefore PA=PB$ (CPCT)

Hence proved.