96
In the given figure, $E$ is a point on the side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, then prove that $\triangle ABD \sim \triangle ECF$.
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$ABC$ is an isosceles triangle
$\therefore AB = AC \Rightarrow \angle B = \angle C$
In $\triangle ABD$ and $\triangle ECF$,
$\angle ADB = \angle EFC$
$\angle ABD = \angle ECF$
$\therefore \triangle ABD \sim \triangle ECF$
$\therefore AB = AC \Rightarrow \angle B = \angle C$
In $\triangle ABD$ and $\triangle ECF$,
$\angle ADB = \angle EFC$
$\angle ABD = \angle ECF$
$\therefore \triangle ABD \sim \triangle ECF$