# 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.