Question

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

Answer 1

drawn a circle of centre O(0,r) .tangents PB , PA are drawn from an exterior point P on circle as shown in figure.

we have to prove that OP is perpendicular on 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.
$\textbf{As we know that the tangents drawn to}\\\textbf{a circle from a point in the exterior}\\\textbf{of the circle are congruent}$
∴ 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.

Answer 2

Suppose OP intersects AB at C

In ∆PAC and ∆PBC ,

PA = PB

[ Tangents from an external point are

equal ]

<APC = <BPC

[ PA and PB are equally inclined to OP ]

PC = CP ( common side )

∆PAC congruent to ∆PBC

[ SAS criterion ]

AC = BC

<ACP = <BCP

<ACP + <BCP = 180°

<ACP = <BCP = 90°

Therefore ,

OP perpendicular to AB

I hope this helps you.

: )