From a point P outside a circle with centre O, tangents PA and PB are drawn to the circle. Prove that OP is the right bisector of the segment AB.

To prove that OP is the right bisector of the segment AB, we need to show two things:

1. OP bisects AB, i.e., AP = BP and,
2. OP is perpendicular to AB, i.e., ∠APB = 90°.

Step 1: Proving AP = BP Since PA and PB are tangents to the circle from an external point P, they are equal in length. This is a property of tangents from an external point to a circle. So, AP = BP.

Step 2: Proving ∠APB = 90° In triangle OPA and OPB, we have:

• OA = OB (Radii of the same circle)
• OP = OP (Common)
• AP = BP (From Step 1)

Therefore, triangle OPA ≅ triangle OPB (By SSS rule of congruency)

So, ∠OPA = ∠OPB (Corresponding parts of congruent triangles are equal)

But, ∠OPA + ∠OPB = 180° (Linear pair)

So, 2∠OPA = 180°

Therefore, ∠OPA = 90°

So, ∠APB = ∠OPA = 90°

Hence, OP is the right bisector of AB.