Question

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

(i) $\angle AOP=\angle BOP$,

(ii) OP is the $\perp$ bisector of chord AB.

Answer 1

In ∆POA and ∆POB

PA = PB (tangents drawn from an external point are equal)

OA = OB (radii of same circle)

OP = OP(common)

∆POA ≅ ∆POB (SSS)

⇒∠AOP = ∠BOP (CPCT)⇒∠APO = ∠BPO (CPCT)

In ∆APM and ∆BPM, OP = OP (Common)

∠APM = ∠BPM (Proved above)

PA = PB (tangents drawn from an external point are equal)

∆APM ≅∆BPM (SAS)⇒∠AMP = ∠BMP (CPCT)⇒AM = BM (CPCT) ........(1)now, ∠AMP + ∠BMP = 180° [Linear pair]⇒∠AMP + ∠AMP = 180°⇒2∠AMP = 180°⇒∠AMP = 90° = ∠BMP ......(2)So, from (1) and (2), we say that OP is the perpendicular bisector of AB.