In the given figure, \(E\) lies on the extension of side
\(CB\) of an isosceles triangle \(ABC\) where \(AB=AC\). If
\(AD⊥BC\) and \(
EF⊥AC\). Derive that
\(
△ADB∼△ECF\).
In isosceles triangle \(ACB\),
where \(AB = AC\)
Angles opposite to equal sides are equal.
\(\angle C = \angle\) - - - - (1)
In \(\Delta ABD\) and \(\Delta ECF\),
\(\angle ABD = \angle\) (from (1))
\(\angle ADB = \angle\) (Both \(90^\circ\))
Thus, \(\Delta ABD \sim \Delta ECF\) (by similarity)
Hence proved.