Question

From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:

∠AOP = ∠BOP

Answer 1

In ΔAOP and ΔBOP

AP = BP   ...(Tangents from P to the circle)

OP = OP   ...(Common)

OA = OB   ...(Radii of the same circle)

∴ By Side – Side – Side criterion of congruence,

ΔAOP ≅ ΔBOP

The corresponding parts of the congruent triangle are congruent

`=>` ∠AOP = ∠BOP  ...[By c.p.c.t]