# In the given figu

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 = 2AOC [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 = 2BOC [2]

POC + QOC = 2AOC + BOC

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

90° = AOB

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