Solution :
Here, \(\triangle\) ABC is an isosceles with AB = AC.
\(\therefore\) \(\angle\) B = \(\angle\) C
In \(\triangle\)s ABD and ECF, we have
\(\angle\) ABD = \(\angle\) ECF [\(\because\) \(\angle\) B = \(\angle\) C]
\(\angle\) ADB = \(\angle\) EFC = 90
\(\therefore\) By AA similarity,
\(\triangle\) ABD ~ \(\triangle\) ECF