In the below figure, XY 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 A and X?Y? at B. Prove that

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Solution

Q.
In the below figure, XY 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 A and X′Y′ at B. Prove that $∠$ AOB = 90°.

Ans.
Join OC. Since, the tangents drawn to a circle from an external point are equal.

$∴$ AP = AC

In Δ PAO and Δ AOC, we have:

AO = AO [Common]

OP = OC [Radii of the same circle]

AP = AC

$\Rightarrow$ Δ PAO $≅$ Δ AOC [SSS Congruency]

$∴$ $∠$ PAO = $∠$ CAO = $∠$ 1

$∠$ PAC = 2 $∠$ 1 ...(1)

Similarly $∠$ CBQ = 2 $∠$ 2 ...(2)

Again, we know that sum of internal angles on the same side of a transversal is $180^{o}$ .

$∴$ $∠$ PAC + $∠$ CBQ = $180^{o}$

$\Rightarrow$ 2 $∠$ 1 + 2 $∠$ 2 = $180^{o}$ [From (1) and (2)]

$\Rightarrow$ $∠$ 1 + $∠$ 2 = $180^{o} / 2 = 90^{o}$ ...(3)

Also $∠$ 1 + $∠$ 2 + $∠$ AOB = 180° [Sum of angles of a triangle]

$\Rightarrow$ 90° + $∠$ AOB = $180^{o}$

$\Rightarrow$ $∠$ AOB = $180^{o} - 90^{o}$ $\Rightarrow$ $∠$ AOB = $90^{o}$ .

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