Question

In figure, $△FEC\cong △GBD$ and $\angle 1=\angle 2$. Prove that $△ADE\sim △ABC$.

Answer 1

We have,
$△FEC△GBD$
$\Rightarrow EC=BD$ ....(i)
It is given that
$\angle 1=\angle 2$
$\Rightarrow AD=AE$ [sides opposite to equal angles are equal] ....(ii)
From (i) and (ii), we have
$\frac{AE}{EC}=\frac{AD}{ED}$
$\Rightarrow$ DE || BC [By the converse of basic proportionality theorem]
$\Rightarrow \angle 1=\angle 3and\angle 2=\angle 4$
Thus, in ${△}^{\prime }s$ ADE and ABC, we have
$\angle A=\angle A$
$\angle 1=\angle 3$
$\angle 2=\angle 4$
So, by AAA-criterion of similarity, we have
$△ADE\sim △ABC$