#### Question

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

(i) `∠`AOP = ∠`BOP

(ii) OP is the ⊥ bisector of chord AB

#### Solution

i) In Δ AOP and Δ BOP

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,

Δ AOP ≅ Δ BOP

The corresponding parts of the congruent triangle are congruent

⇒ `∠`AOP = `∠`BOP [by c.p.c.t]

ii) In Δ OAM and ΔOBM

OA = OB (Radii of the same circle)

`∠`AOM =`∠`BOM (Proved `∠`AOP = `∠` BOP )

OM = OM (Common)

∴ By side – Angle – side criterion of congruence,

Δ OAM ≅ `∠`OBM

The corresponding parts of the congruent triangles are congruent.

⇒AM = MB

And `∠`OMA=`∠`OMB

But,

`∠`OMA +`∠`OMB =180°

∴ `∠`OMA = `∠` OMB = 90°

Hence, OM or OP is the perpendicular bisector of chord AB.