In the figure, if \(\triangle\) ABE \(\cong\) \(\triangle\) ACD, show that \(\triangle\) ADE ~ \(\triangle\) ABC.

Solution :

It is given that \(\triangle\) ABE \(\cong\) \(\triangle\) ACD

\(\therefore\)   AB = AC

and  AE = AD

[ because corresponding parts of congruent triangles are equal ]

So,  \(AB\over AD\) = \(AC\over AE\)    or     \(AB\over AC\) = \(AD\over AE\)             ………(1)

\(\therefore\)  In triangles ADE and ABC, we have :

\(AB\over AC\) = \(AD\over AE\)

and  \(\angle\) BAC = \(\angle\) DAE

Thus, by SAS criterion of similarity, \(\triangle\) ADE ~ \(\triangle\) ABC.