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Prove that the line segment joining the point of contact of two parallel tangents of a circle passes through its center.
Given: Two parallel tangents of a circle and a line segment joining joining the point of contact of the given tangents.
To do: To prove that the this line segment passes through the center.
Solution:
Let XBY and PCQ be two parallel tangents to a circle with center O.
Let us Join OB and OC.
Now, $XB\parallel AO$
We know the sum of adjacent interior angles is $180^{o}$.
$\therefore \ \angle XBO+\angle AOB=180^{o}$
$\angle XBO=90^{o} \ (A\ tangent\ to\ a\ circle\ is\ perpendicular\ to\ the\ radius\ through\ the\ point\ of\ contact)$
$\Rightarrow 90^{o} +\angle AOB=180^{o}$
$\Rightarrow \angle AOB=90^{o}$
Similarly, $\angle AOC=90^{o} $
$\angle BOC=\angle AOB+\angle AOC=90^{o}+90^{o}=180^{o}$
Thus we find that $\angle BOC$ is a straight line passing through the center O.
Or we can say that BC is a straight line passing through the center O.
Hence, it is proved that a line segment joining the points of contact B and C of the tangents of the circle passes through the circle.
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