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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 $∆ABD \sim ∆ECF$.
"
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.
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