Question

In fig., $l$ and $m$ are two parallel tangents to a circle with centre $O$, touching the circle at $A$ and $B$ respectively. Another tangent at $C$ intersects the line $l$ at $D$ and $m$ at $E$. Prove that $\angle DOE={90}^o$

Answer 1

Join $OC$

In triangle $ODA$ and triangle $ODC$

$OA=OC$ (Radii of the same circle )

$AD=DC$ (Length of tangent drawn from an external point to a circle are equal)

$DO=OD$ (common side)

$\Delta DOA\cong \Delta ODC$

$\therefore \angle DOA=\angle COD$

$\Delta DOA\cong \Delta ODC$

$\therefore \angle DOA=\angle COD$

Similarly$\Delta OEB\cong \Delta OEC$

$\therefore \angle EOB=\angle COE$

$AOB$ is a diameter of the circle.

Hence, it is a straight line.

$\therefore \angle DOA+\angle COD+\angle COE+\angle EOB=180$

$\Rightarrow 2\angle COD+2\angle COE=180$

$\Rightarrow \angle COD+\angle COE=90$

$\Rightarrow \angle DOE={90}^0$

Hence proved.