69
In the 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 $\triangle ABD \sim \triangle ECF$.
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In $\triangle ABC$, $AB = AC$ (Given)
$\therefore \angle ACB = \angle ABC$ ----- (1)
In $\triangle ABD$ and $\triangle ECF$
$\angle ADB = \angle EFC$ (each $90^\circ$)
$\angle ABD = \angle ACD$ (from (1))
$\therefore \triangle ABD \sim \triangle ECF$ (AA rule)
$\therefore \angle ACB = \angle ABC$ ----- (1)
In $\triangle ABD$ and $\triangle ECF$
$\angle ADB = \angle EFC$ (each $90^\circ$)
$\angle ABD = \angle ACD$ (from (1))
$\therefore \triangle ABD \sim \triangle ECF$ (AA rule)