if PA and PB are two tangents drawn from a point P to a circle wid centre o touching it at A and B , prove that OP is the perpendicular bisector of AB.

We have the following situation-

Let OP intersect AB at C

In ΔPAC and ΔPBC, we have

PA = PB  ( Tangent from an external point are equal)

∠APC = ∠BPC  ( PA and PB are equally inclined to OP)

PC = PC  ( Common)

∴ ΔPAC ΔPBC  (by SAS congurency criteria)

So by corresponding part of congruent triangle,

⇒ AC = BC   ......(1)

∠ACP = ∠BCP  ......(2)

But ∠ACP + BCP = 180°  .......(3)

From (2) and (3)

∠ACP = ∠BCP = = 90°

Hence, from (1) and (2), we can conclude that OP is the perpendicular bisector  of AB

Hence proved..........