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

"

**Given: **

$XY$ and $X’Y’$ are two parallel tangents to a circle, $x$ with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X’Y’$ at $B$.

**To do: **

We have to prove that $\angle AOB=90^{o}$.

**Solution:**

$XY$ and $X\'Y\'$ are two parallel tangents to the circle with centre $O$ touching the circle at $P$ and $Q$ respectively.

$AB$ is a tangent at the point $C$, which intersects $XY$ at $A$ and $X\'Y\'$ at $B$.

Follow the steps:

Join $OC$.

In $\vartriangle OAP$ and $\vartriangle OAC$,

$OP=OC\ \ \ \ \ \ \ \ \ ( Radii\ of\ the\ same\ circle)$

$AP =AC\ \ \ \ \ \ \ \ ( Length\ of\ tangents\ drawn\ from\ an\ external\ point\ to\ a\ circle\ are\ equal)$

$OA=OA\ \ \ \ \ \ \ \ \ \ ( Common\ side)$

$\vartriangle OAP\cong \vartriangle OAC\ \ \ \ \ \ \ \ \ \ \ \ ( SSS\ congruence\ criterion)$

$\angle AOP=\angle AOC\ \ .................( 1)$

Similarly, $\vartriangle OBC\cong \vartriangle OBQ$

$\angle BOQ=\angle BOC\ \ \ \ .....................( 2)$

Now, AOB is a diameter of the circle.

Hence, it is a straight line.

$\angle AOP\ +\angle AOC+\angle BOQ+\angle BOC\ =\ 180^{o}$

From $( 1)$ and $( 2)$,

We have: $2\angle AOC+2\angle BOC\ =\ 180^{o}$

$\angle AOC+\angle BOC\ =\frac{180^{o} }{2}$

$=90^{o}$

We know,

$\angle AOC+\angle BOC=\angle AOB=90^{o}$

$\Rightarrow \angle AOB=90^{o}$

Hence proved.

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