Tangents from P, a point in the exterior of ⨀(O, r) touch the circle at A and B. Prove that and bisects.

Let PA and PB be tangents to the circle drawn from point P which is in the exterior of circle (O, r). Given A and B are on the circle.

We have to prove that OP ⊥ AB and OP bisects AB.

Proof:

Here PA and PB are tangents drawn to the circle from an exterior point P.

∴ OP intersects AB at C.

Also given that A and B are on the circle.

We know that the tangents drawn to a circle from a point in the exterior of the circle are congruent.

∴ PA = PB

⇒ OP = OP [common]

⇒ OA = OB [radii of circle]

∴ By SSS theorem,

ΔOAP ≅ ΔOBP

Then, ∠AOP = ∠BOP

⇒ ∠AOC = ∠BOC [C ∈ OP]

⇒ OA = OB [radii of circle]

⇒ OC = OC [common]

∴ By SAS theorem,

ΔAOC ≅ ΔBOC

∴ AC = BC and ∠ACO = ∠BCO = 90°

Now, as C ∈ OP,

⇒ OP bisects AB.

Also AC ⊥ OC and BC ⊥ OC

⇒ OP ⊥ AB [∵ A – C – B]

∴ OP ⊥ AB and OP bisects AB.

Hence proved.