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In fig, l and m are two parallel tangents to a circle with center O, touching the circle at A and B respectively. Another tangent at C intersects the line l at D and m at E. prove that$\angle DOE=90^{o}$
Given: A circle with center O, two parallel tangents l and m touching the circle at A and B. an another tangent touching the circle at C and intersecting l and m at D and E respectively.
To do: To prove $\angle DOE=90^{o}$
Solution:
l and m are two parallel tangents to the circle with centre O touching the circle at A and B respectively.
DE is a tangent at the point C, which intersects l at D and m at E.
Follow the steps:
Join OC.
In $\vartriangle ODA$ and $\vartriangle ODC$,
$OA=OC\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( Radii\ of\ the\ same\ circle)$
$AD=DC\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( Length\ of\ tangents\ drawn\ from\ an\ external\ point\ to\ a\ circle\ are\ equal)$
$DO=OD\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( Common\ side)$
$\vartriangle ODA\cong \vartriangle ODC\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( SSS\ congruence\ criterion)$
$\angle DOA=\angle COD\ \ \ .................( 1)$
Similarly, $\vartriangle OEB\cong \vartriangle OEC$
$\angle EOB=\angle COE$
Now, AOB is a diameter of the circle. Hence, it is a straight line.
$\angle DOA\ +\angle COD+\angle COE+\angle EOB\ =\ 180^{o}$
From $( 1)$ and $( 2)$,
We have: $2\angle COD+2\angle COE\ =\ 180^{o}$
$\angle COD+\angle COE\ =\frac{180^{o}}{2}$
$=90^{o}$
And we know $\angle COD+\angle COE\ =\angle DOE$
$\Rightarrow \angle DOE=90^{o}$
Hence, proved that $\angle DOE=90^{o}$.
Updated on: 10-Oct-2022
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