# In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB $$\perp$$ AC, prove that $$\triangle$$ ABD ~ $$\triangle$$ ECF.

## Solution :

Here, $$\triangle$$ ABC is an isosceles with AB = AC.

$$\therefore$$  $$\angle$$ B = $$\angle$$ C

In $$\triangle$$s ABD and ECF, we have

$$\angle$$ ABD = $$\angle$$ ECF        [$$\because$$ $$\angle$$ B = $$\angle$$ C]

$$\angle$$ ADB = $$\angle$$ EFC = 90

$$\therefore$$  By AA similarity,

$$\triangle$$ ABD ~ $$\triangle$$ ECF