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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