Q. 184.9( 10 Votes )

In given figure X

Answer :

Given that: XY is tangent to the circle at point P

X’Y’ is tangent to the circle at point Q


AB is tangent to the circle at point C


XY X’Y’


To prove: AOB = 90°


Proof: First join OC such that OC is perpendicular to AB.



ACO = 90° & BCO = 90°


We know already that OPA = OQB = 90° (as XY and X’Y’ are tangents at P and Q respectively; and tangents are always perpendicular to any part of the circle)


Now taking ∆AOP and ∆AOC, observe that


OP = OC [ they are radius of the same circle; radius of a circle have equal lengths]


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


OA = OA [ they are same sides of the triangles]


Thus, by SSS-congruency ∆AOP ∆AOC.


POA = AOC [ corresponding parts of congruent triangles are equal] …(i)


Taking ∆COB and ∆BOQ, observe that


OQ = OC [ they are radius of the same circle and hence, are equal]


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


OB = OB [ they are same sides of the triangle]


Thus, by SSS-congruency ∆COB ∆BOQ.


COB = BOQ [ corresponding parts of congruent triangles are equal] …(ii)


Breaking up line PQ into angles, we get


POA + AOC + COB + BOQ = 180°


AOC + AOC + COB + COB = 180° [By equations (i) & (ii)]


2 AOC + 2 COB = 180°


2(AOC + COB) = 180°


AOC + COB = 180°/2


AOB = 90°


Hence, proved.


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