Question

In given Figure, If $△ABE\cong △ACD$, then prove that $△ADE\sim △ABC$

Answer 1

It is given that

$△ABE\cong △ACD$

$\therefore$ $AB=AC$ [$\therefore$ Corresponding parts of congruent triangles are equal]

and, $AE=AD$

$\Rightarrow$ $\frac{AB}{AD}=\frac{AC}{AE}$

$\Rightarrow$ $\frac{AB}{AC}=\frac{AD}{AE}$.......(i)

Thus, in triangles ADE and ABC, we have

$\frac{AB}{AC}=\frac{AD}{AE}$ [from (i)]

and, $\angle BAC=\angle DAE$ [Common]

Hence, by SAS-criterion of similarity, we have

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