In Figure, $X Y$ and $X^{'} Y^{'}$ are two parallel tangents to a circle with centre $O$ and another tangent $A B$ with point of contact $C$ intersecting $X Y$ at $A$ and $X^{'} Y^{'}$ at $B$ . Prove that $∠$ $A O B$ $= 90^{o}$ .

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Solution

Join $O$ and $C$ .

In $△ O P A$ and $△ O C A$

1) $O P = O C$ [Radius of the circle]

2) $A P = A C$ [Tangents from point A]

3) $A O = A O$ [Common side]

$△ O P A ≅ △ C O A$ [sss congruence criterior]

$∠ P O A = ∠ C O A$ [By CPCT]

Similarly, $△ O Q B ≅ △ O C B$ [sss congruence criterior]

$∠ Q O B = ∠ O Q B$ [By CPCT]

$P O Q$ is a diameter of the circle.

$∠ P O A + ∠ C O A + ∠ Q O B = 180^{o}$

$2 ∠ C O A + 2 ∠ C O B = 180^{o}$

$\Rightarrow ∠ C O A + ∠ C O B = 90^{o}$

$∠ A O B = 90^{o}$

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In 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=90o.

In Figure, $X Y$ and $X^{'} Y^{'}$ are two parallel tangents to a circle with centre $O$ and another tangent $A B$ with point of contact $C$ intersecting $X Y$ at $A$ and $X^{'} Y^{'}$ at $B$ . Prove that $∠$ $A O B$ $= 90^{o}$ .