Question

In given figure through the vertex D of a parallelogram ABCD, a line is drawn to intersect the sides BA and BC produced at E and F respectively. Prove that

$\frac{DA}{AE}=\frac{FB}{BE}=\frac{FC}{CD}$

Answer 1

In ${△}^{\prime }sEAD$ and $DCF$, we have

$\angle 1=\angle 2$ [$\because AB||DC\therefore$ Corresponding angles are equal]

$\angle 3=\angle 4$ [$\because AD||BC\therefore$ Corresponding angles are equal]

Therefore, by AA-criterion of similarity, we have

$△EAD\sim △DCF$

$\Rightarrow$ $\frac{EA}{DC}=\frac{AD}{CF}=\frac{DE}{FD}$

$\Rightarrow$ $\frac{EA}{DC}=\frac{AD}{CF}$

$\Rightarrow$ $\frac{AD}{AE}=\frac{CF}{CD}$...........(i)

Now, in ${△}^{\prime }sEAD$ and $EBF$, we have

$\angle 1=\angle 1$ [Common angle]

$\angle 3=\angle 4$

So, by AA-criterion of similarity, we have

$△EAD\sim △EBF$

$\Rightarrow$ $\frac{EA}{EB}=\frac{AD}{BF}=\frac{ED}{EF}$

$\Rightarrow$ $\frac{EA}{EB}=\frac{AD}{BF}$

$\Rightarrow$ $\frac{AD}{AE}=\frac{FB}{BE}$

From (i) and (ii), we have

$\frac{AD}{AE}=\frac{FB}{BE}=\frac{CF}{CD}$ $\left[Henceproved\right]$