Question

In figure, if $△ABE\simeq △ACD$, prove that $△ADE\sim △ABC$.

Answer 1

It is given that
$△ABE\simeq △ACD$
$\therefore AB=AC[\therefore Correspondingpartsofcongruenttriangleareequal]$
and, $AE=AD$
$\Rightarrow \frac{AB}{AD}=\frac{AC}{AE}$
$\Rightarrow \frac{AB}{AD}=\frac{AD}{AE}$
Thus, in triangles ADE and ABC, we have
$\frac{AB}{AC}=\frac{AD}{AE}$
and $\angle BAC=\angle DAE$[common]
Hence, by SAS-criteria of similarity, we have
$△ADE\sim △ABC$