In the following 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 \(\triangle\) ABD ~ \(\triangle\) ECF.

Given, ABC is an isosceles triangle.

AB = AC
\(\angle\) ABD = \(\angle\) ECF

In \(\triangle\) ABD and \(\triangle\) ECF,
\(\angle\) ADB = \(\angle\) EFC (Each 90°)
\(\angle\) BAD = \(\angle\) CEF (Already proved)
\(\triangle\) ABD ~ \(\triangle\) ECF (using AA similarity criterion)