In the figure,$E$ is a point on the side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$,prove that $\Delta ABD \sim \Delta ECF$.

(N/A) Given: $ABC$ is an isosceles triangle with $AB = AC$.
Since $AB = AC$,the angles opposite to these sides are equal,so $\angle ABC = \angle ACB$.
Since $E$ is a point on $CB$ produced,$\angle ABC + \angle ABE = 180^{\circ}$ (linear pair).
Also,$\angle ACB + \angle ECF = 180^{\circ}$ is not directly applicable,but we know $\angle ABC = \angle ACB$.
In $\triangle ABD$ and $\triangle ECF$:
$1$. $\angle ADB = \angle EFC = 90^{\circ}$ (Given).
$2$. $\angle ABD = \angle ECF$ is incorrect; rather,$\angle ABC = \angle ACB$. Since $\angle ABC + \angle ABE = 180^{\circ}$ and $\angle ACB + \angle ECF$ is not the relation,we observe that $\angle ABC = \angle ACB$. Since $\angle ACB = \angle ECF$ (vertically opposite is not applicable here),we use $\angle ABC = \angle ACB$. In $\triangle ABC$,$\angle B = \angle C$. Thus,$\angle ABD = 180^{\circ} - \angle ABC$ and $\angle ECF = \angle ACB$. Actually,$\angle B = \angle C$,so $\angle ABD = 180^{\circ} - \angle B$ and $\angle ECF$ is just $\angle C$. Wait,$\angle ABC = \angle ACB$. Therefore,$\angle ABD = 180^{\circ} - \angle ABC = 180^{\circ} - \angle ACB$. This is not equal to $\angle ECF$. Let us re-evaluate: $\angle ABC = \angle ACB$. Since $E$ is on $CB$ produced,$\angle ECF$ is the same as $\angle ACB$. Thus $\angle B = \angle C$. In $\triangle ABD$ and $\triangle ECF$,$\angle ADB = \angle EFC = 90^{\circ}$ and $\angle B = \angle ECF$ is not correct. The correct relation is $\angle ABC = \angle ACB$. Since $\angle ABC = \angle ACB$,and $\angle ECF$ is the exterior angle,$\angle ECF = \angle ABC$. Thus $\angle B = \angle ECF$. By $AA$ similarity,$\Delta ABD \sim \Delta ECF$.