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

Given: XY and PQ 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 PQ at B

To prove: ∠AOB = 90°

Construction: Join OC and ROS

In ∆AOR and ∆AOC

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

AO = AO [Common]

OR = OC [Radii of semi circle]

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

⇒ ∠AOR = ∠ AOC

⇒ ∠ROC = ∠AOR + ∠AOC

⇒ ∠ROC = ∠AOC + ∠AOC

⇒ ∠ROC = 2∠AOC … [1]

Now, In ∆BOS and ∆BOC

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

BO = BO [Common]

OS = OC [Radii of semi circle]

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

⇒ ∠BOS = ∠ BOC

⇒ ∠SOC = ∠BOS + ∠BOC

⇒ ∠SOC = ∠BOC + ∠BOC

⇒ ∠SOC = 2∠BOC … [2]

Adding [1] and [2]

∠ROC + ∠SOC = 2∠AOC + 2∠BOC

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

⇒ 90° = ∠AOB