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

Let us join point O to C.

In ΔOPA and ΔOCA,

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

AP = AC           ......(Tangents from point A)

AO = AO              ....(Common side)

ΔOPA ≅ ΔOCA (SSS congruence criterion)

Therefore, P ↔ C, A ↔ A, and O ↔ O

∠POA = ∠COA … (i)

Similarly, ΔOQB ≅ ΔOCB

∠QOB = ∠COB … (ii)

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

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180º

From equations (i) and (ii), it can be observed that

2∠COA + 2 ∠COB = 180º

∠COA + ∠COB = 90º

∠AOB = 90°

Answer 2

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: Join OC.

In ΔOAP and ΔOAC,

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

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

AO = OA          ....(Common side)

ΔOAP ≅ ΔOAC   .....(SSS congruence criterion)

∴ ∠AOP = ∠COA               ....(C.P.C.T)        .....(1) 

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.