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In the given figure, XY and X'Y' are two parallel tangents to a circle with centre and another tangent AB with point of contact C, is intersecting XY at A and X'Y' at B. Prove that $\angle AOB = 90^{o}$.
Given: A circle with center O, two parallel tangents XY and X'Y' touching the circle an another tangent touching the circle at C and intersecting l and m at A and B respectively.
To do: To prove $\angle AOB=90^{o}$.
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: $2\angle AOC+2\angle BOC\ =\ 180^{o}$
$\angle AOC+\angle BOC\ =\frac{180^{o} }{2}$
$=90^{o}$
And we know $\angle AOC+\angle BOC=\angle AOB=90^{o}$
$\Rightarrow \angle AOB=90^{o}$
Hence proved that $\angle AOB=90^{o}$.
Updated on: 10-Oct-2022
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