Question

In given figure, E is a point on side CB produced of an isosceles triangle ABC with AB=AC. If $AD\perp BCandEF\perp AC$, prove that

(i) $△ABD\sim △ECF$

(ii) $AB\times EF=AD\times EC$

Answer 1

It is given that $△ABC$ is isosceles with

$AB=AC$

$\therefore$ $\angle B=\angle C$

Now, in ${△}^{\prime }sABD$ and $ECF$, we have

$\angle ABD=\angle ECF$ [$\because \angle B=\angle C$]

$\angle ADB=\angle EFC={90}^{\circ }$ [$\because AD\perp BC$ and $EF\perp AC$]

So, by AA-criterion of similarity, we have

$△ABD\sim △ECF$ $\left[Henceproved\right]$

$\Rightarrow$ $\frac{AB}{EC}=\frac{AD}{EF}$

$\Rightarrow$ $AB\times EF=AD\times EC$ $\left[Henceproved\right]$