Question

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

OP is the ⊥ bisector of chord AB.

Answer 1

In ΔOAM and ΔOBM

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

∠AOM = ∠BOM  ...(Proved ∠AOP = ∠BOP)

OM = OM  ...(Common)

∴ By Side – Angle – Side criterion of congruence,

ΔOAM ≅ ΔOBM

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.