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In the given figure, E is the point on side CB produced of an isosceles $\Delta ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, prove that $\frac{AC}{EC} = \frac{BD}{CF}$. Hence, prove that $\Delta ABD \sim \Delta ECF$.
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In $\Delta ABC$, $AB = AC$
n$\therefore \angle ABC = \angle ACB$ (1 Mark)
nAlso, $\angle ADB = \angle EFC = 90^\circ$ (Given) (1 Mark)
n$\therefore \Delta ABD \sim \Delta ECF$ (AA similarity) (1/2 Mark)
nThus, $\frac{AB}{EC} = \frac{BD}{CF} \implies \frac{AC}{EC} = \frac{BD}{CF}$ (1/2 Mark)
n$\therefore \angle ABC = \angle ACB$ (1 Mark)
nAlso, $\angle ADB = \angle EFC = 90^\circ$ (Given) (1 Mark)
n$\therefore \Delta ABD \sim \Delta ECF$ (AA similarity) (1/2 Mark)
nThus, $\frac{AB}{EC} = \frac{BD}{CF} \implies \frac{AC}{EC} = \frac{BD}{CF}$ (1/2 Mark)