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Answer :
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)