Question

In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C, intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Answer 1

Given: XY and X’Y’ are two parallel tangents to the circle with centre O touching the circle at P and Q, respectively. AB is a tangent at the point C, which intersects XY at A and X’Y’ at B.

To prove: ∠AOB = 90^°

Construction: Let us join point O to C.

Proof: In ΔOAP and ΔOAC,

OP = OC ...(Radii of the same circle)

AP = AC  ...(The lengths of tangents drawn from an external point to a circle are equal.)

AO = OA  ....(Common side)

ΔOAP ≅ ΔOAC  ...(SSS congruence criterion)

∴ ∠AOP = ∠COA   ....(1) (C.P.C.T)

Similarly, ΔOBQ ≅ ΔOBC

∴ ∠BOQ = ∠COB   ....(2)

POQ is a diameter of the circle. Hence, it is a straight line.

∴ ∠AOP + ∠COA + ∠BOQ + ∠COB = 180º

2∠COA + 2∠COB = 180º   ...[From (1) and (2)]

⇒ ∠COA + ∠COB = 90º

⇒ ∠AOB = 90°

Hence proved.