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

Solution

Since tangents drawn from an external point to a circle are equal.

Therefore, AP = AC.

Thus, in triangles AOP and AOC, we have

AP = AC

AO = AO ..........[Common side]

OP = OC ..........[ Radii of the same circle ]

So, by SSS- criterion of congruence,

We have,

$∆ AOP ≅ ∆ AOC \Rightarrow ∠ PAO = ∠ CAO \Rightarrow ∠ PAC = 2 ∠ CAO . . . . . . . . . (i)$

Similarly, we can prove that $∠ QBO = ∠ CBO$

$\Rightarrow ∠ CBQ = 2 ∠ CBO nOW, ∠ PAC + ∠ CBQ = 180 ° . . . . . . . . . . . . (ii) [Sum of the interior angle on the same side of transvarsal is 180 °] \Rightarrow 2 ∠ CAO + 2 ∠ CBO = 180 ° . . . . . . . [Using (i) and (ii)] \Rightarrow ∠ CAO + ∠ CBO = 90 ° \Rightarrow 180 ° - ∠ AOB = 90 ° . . . . . . . [Since ∠ CAO, ∠ CBO and ∠ AOB are angles of a triangle, ∠ CAO + ∠ CBO + ∠ AOB = 180 °] \Rightarrow ∠ AOB = 90 °$

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Circles

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

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