Question
In the figure, $E$ is the point on side $CB$ produced on an isosceles triangle $ABC$ with $AB = AC.$ If $AD \bot BC$ and $EF \bot AC,$ prove that $\triangle ABD \sim \triangle ECF.$

Answer

$E$ is the point on side $CB$ produced on an isosceles triangle $ABC$ with $AB=AC.AD-BC \bot$ and$ EF \bot AC.$ with $AB=AC.$ Also, $AD \bot BC$ and $EF \bot AC.$
To prove:$ \triangle ABD \sim \triangle ECF$
Proof: In $\triangle ABD$ and $\triangle ECF,$
$\therefore AB = AC ........$Given
$\therefore \angle ACB = \angle ABC ......$Angle opposite to equal sides of a triangle are equal
$\Rightarrow \angle ABC = \angle ACB $
$ \Rightarrow \angle ABD = \angle ECF ..........(1)$
$ \angle ADB = \angle EFC .........(2) [$Each equal to $90^\circ$ In view of $(1)$ and $(2)]$
$\triangle ABD \sim \triangle ECF.............AA$ similarity criterion

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