Question

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

Answer 1

We have,

$△FEC\cong △GBD$

$\Rightarrow$ $EC=BD$..........(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

$\frac{AE}{EC}=\frac{AD}{BD}$

$\Rightarrow$ $DE||BC$ [By the converse of basic proportionality theorem]

$\Rightarrow$ $\angle 1=\angle 3$ and $\angle 2=\angle 4$

Thus, in ${△}^{\prime }sADE$ and $ABC$, we have

$\angle A=\angle A$

$\angle 1=\angle 3$

$\angle 2=\angle 4$

So, by AAA-criterion of similarity, we have

$△ADE\sim △ABC$ $\left[Henceproved\right]$