Question

XY and X'Y' are two parallel tangents to a circle with center Oand another tangent AB with point of contact C intersecting XY at A and X'Y' at B.prove that/_ AOB =90°

Answer 1

This is your answer you can prove the vertex angles to be equal by joining OP and OQ and then proving the two triangles OAP and OCA congruent then using CPCT

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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