Q. 254.1( 8 Votes )

In the given figu

Answer :

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]


Adding [1] and [2]


POC + QOC = 2AOC + BOC


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


90° = AOB


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