#### Question In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C, is intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

#### Video Explanation

#### Explanatory Answer

Because the radius and tanget meet at right angles at the point of tangency, ∠OPA = 90°; ∠OQB = 90° and ∠OCA = 90°

AP and AC are equal tangents from A. OP and OC are equal as they are radii to the circle. And OA is a side common to triangles OPA and OCA.

By SSS rule ΔOPA \\equiv ) ΔOCA (Trianlges OPA and OCA are congruent)

Therefore, ∠PAO = ∠CAO. Let it be ‘x’.

And ∠AOP = ∠AOC. Let it be ‘a’.

BQ and BC are equal tangents from an external point B.

OQ and OC are radii and are equal.

OB is a side common to OQB and OCB.

By SSS rule ΔOQB \\equiv ) ΔOCB (Trianlges OQB and OCB are congruent)

Therefore, ∠OBQ = ∠OBC. Let it be ‘y’.

And ∠QOB = ∠COB. Let it be ‘b’.

∠AOB = ∠AOC + ∠COB = a + b

In Δ OPA, x + a + 90° = 180°

Or a = 90° - x

Similarly in ΔOCB, y + b + 90° = 180°

Or b = 90° - y

Therefore, ∠AOB = a + b = (90 – x) + (90 – y) = 180 – (x + y)

Lines xy and x’y’ are parallel

Therefore, ∠CBy’ = 2x (Alternate interior angles are equal)

Because x’y’ is a straight line, ∠QBC + ∠CBy’ = 180°

I.e., 2y + 2x = 180°

Or x + y = 90°

Therefore, ∠AOB = 180 – (x + y) = 180 – 90 = 90°

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