Question

in the given figure X Y 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 ab and x'y' at b. <br />prove that angle AOB =90 degree

Answer 1

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Answer 2

Given: 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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