Question

In fig, If $△ABE\cong △ACD$, show that $△ADE$ is similar to $△ABC$

Answer 1

Given,

$△ABE\cong △ACD$

Hence,$AB=AC$ by CPCT. .....$\left(1\right)$

and $AE=AD$ by CPCT.

$\Rightarrow AD=AE$ .....$\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$

$\frac{AD}{AB}=\frac{AE}{AC}$ ....$\left(3\right)$

In $△ADE$ and $△ABC$

$\angle A=\angle A$ (common angle)

$\frac{AD}{AB}=\frac{AE}{AC}$ from $\left(3\right)$

$\therefore △ADE$ is similar to $△ABC$ by SAS criterion.

Hence proved.