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

(i) ∠AOP = ∠BOP

(ii) OP is the ⊥ bisector of chord AB.

The circle with centre O and tangents PA and PB drawn from point P outside the circle is shown in the below figure:

(i) In ∆AOP and ∆BOP, we have

⇒ AP = BP [Tangents from an exterior point P, are equal in length.]

⇒ OA = OB [Radius of circle]

⇒ OP = OP [Common]

∴ ΔAOP ≅ ΔBOP by SSS axiom.

∴ ∠AOP = ∠BOP [By C.P.C.T.]

Hence, proved that ∠AOP = ∠BOP.

(ii) In ∆OAM and ∆OBM, we have

OA = OB [Radii of the same circle]

∠AOM = ∠BOM [Proved ∠AOP = ∠BOP]

OM = OM [Common]

∴ ΔOAM ≅ ΔOBM [By SAS axiom]

By C.P.C.T.

⇒ AM = MB and ∠OMA = ∠OMB = x

From figure,

⇒ ∠OMA + ∠OMB = 180° [As, AB is a straight line]

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$

⇒ x = 90°.

Thus, ∠OMA = ∠OMB = 90°.

Hence proved, that OP is the ⊥ bisector of chord AB.