Question

In the Figure, $l$ and $m$ are two parallel tangent at $A$ and $B$. The tangent at $C$ makes an intercept $DE$ between $l$ and $m$. Prove that $\angle DFE={90}^o$.

Answer 1

Since triangles drawn from an external point to a circle are equal.

Therefore,$DA=DC$

Thus, in triangles $AFD$ and $DFC$, we have

$DA=DC$ and $DF=DF$ and $AF=CF$ [Radii of the same dircle]

So, bt $SSS-$criterion of congruence, we have
$△ADF\cong DFC$
$\Rightarrow \angle ADF=\angle CDF$
$\Rightarrow \angle ADC=2\angle CDF$

Similarly, we can prove that
$\angle BEF=\angle CEF$
$\Rightarrow \angle CEB=2\angle CEF$

Now, $\angle ADC+2\angle CEB={180}^o$ [sum of the interior angles on the same sides of transversal is ${180}^o$]
$\Rightarrow 2\angle CDF+2\angle CEF={180}^o$ [using equation $\left(i\right)$ and $\left(ii\right)$]
$\Rightarrow \angle CDF+\angle CEF={90}^o$

[$\because \angle CDF,\angle CEF$ and $\angle DEF$ are angles of a triangle$\because \angle CDF+\angle CEF+\angle DEF={180}^o$]

$\Rightarrow {180}^o-\angle DEF={90}^o$

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