Q. 94.5( 437 Votes )

In Fig. 10.13, 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 :

Let us join point O to C.


In ΔOPA and ΔOCA,


OP = OC (Radius of the same circle)


AP = AC (Tangents from point A)


AO = AO (Common side)


ΔOPA ΔOCA (SSS congruence criterion)

∠POA = COA … (i)

Similarly, ΔOQB ΔOCB

QOB = COB … (ii)


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


Therefore, POA + COA + COB + QOB = 180°


From equations (i) and (ii), it can be observed that 2COA + 2 COB = 180°

∠ COA + ∠ COB = 180° / 2

COA + COB = 90°

AOB = 90°

Hence Proved.

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