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 $ riangle ABD \sim riangle ECF$.

Question

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 $	riangle ABD \sim 	riangle ECF$.

Accepted Answer

Step 1: $ABC$ is an isosceles triangle. $\because AB = AC \Rightarrow \angle B = \angle C$.
Step 2: In $	riangle ABD$ and $	riangle ECF$: $\angle ADB = \angle EFC = 90°$ (given $AD \perp BC$ and $EF \perp AC$).
Step 3: $\angle ABD = \angle ECF$ (since $\angle B = \angle C$). $	herefore 	riangle ABD \sim 	riangle ECF$ (AA similarity).

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