In the given figure, E is a point on side CB produced of an isosceles triangle ABC with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, prove that $∆ABD \sim ∆ECF$.
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Given:

E is a point on side CB produced of an isosceles triangle ABC with $AB = AC$.

$AD \perp BC$ and $EF \perp AC$

To do:

We have to prove that $∆ABD \sim ∆ECF$.

Solution:

$\triangle ABC$ is an isosceles triangle.

$AB=AC$

This implies,

$\angle ABC=\angle ACB$          (Angles opposite to equal sides are equal)

In $\triangle ABD$ and $\triangle ECF$,

$\angle ABD=\angle ECF$             ($\angle BCA=\angle ECF$)

$\angle ADB=\angle EFC=90^o$

Therefore, by AA criterion,

$\triangle ABD \sim \triangle ECF$

Hence proved.