In the given figure, △ FEC = △ GDB and ∠ 1 = ∠ 2 . Prove that △ ADE △ ABC .

$\triangle FEC = \triangle GDB$
Therefore, $\angle 3 = \angle 4$
In $\triangle ABC$,
$\angle 3 = \angle 4$
$\therefore AB = AC ............(i)$
In $\triangle ADE$, $\angle 1 = \angle 2$
$AD = AE .............(ii)$
Dividing (ii) by (i)
$\frac{AD}{AB} = \frac{AE}{AC}$
$\Rightarrow DE \parallel BC$
$\angle 1 = \angle 3$ and $\angle 2 = \angle 4$
$\therefore \triangle ADE \sim \triangle ABC$