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

Given: XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C, is intersecting XY at A and X’Y’ at B

To prove: ∠AOB = 90°

In ∆AOP and ∆AOC

AP = AC [tangents drawn from an external point to a circle are equal]

AO = AO [Common]

OP = OC [Radii of semi-circle]

∆AOP ≅ ∆AOC [By Side-Side-Side Criterion]

⇒ ∠AOP = ∠ AOC

⇒ ∠POC = ∠AOP + ∠AOC

⇒ ∠POC = ∠AOC + ∠AOC

⇒ ∠POC = 2∠AOC [1]

Now, In ∆BOQ and ∆BOC

BQ = BC [tangents drawn from an external point to a circle are equal]

BO = BO [Common]

OQ = OC [Radii of semi-circle]

∆BOQ ≅ ∆BOC [By Side-Side-Side Criterion]

⇒ ∠BOQ = ∠ BOC

⇒ ∠QOC = ∠BOQ + ∠BOC

⇒ ∠QOC = ∠BOC + ∠BOC

⇒ ∠QOC = 2∠BOC [2]

Adding [1] and [2]

∠POC + ∠QOC = 2∠AOC + ∠BOC

⇒ 180° = 2(∠AOC + ∠BOC) [∠POC + ∠QOC = 180° , linear pair]

⇒ 90° = ∠AOB