28. From an external point P, two tangents PA and PB are drawn to the

1.	28. From an external point P, two tangents PA and PB are drawn to the circle with the centre O.Prove that OP is the perpendicular biscctor of AB.
Answer» 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) and PC = PC ( Common) ∴ ΔPAC~= ΔPBC (by SAS congurency criteria) ⇒ AC = BC ......(1) and ∠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.

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

Answer»

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)

and PC = PC ( Common)

∴ ΔPAC~= ΔPBC (by SAS congurency criteria)

⇒ AC = BC ......(1)

and ∠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.