Question

In the given figure, ABCD is parallelogram and E is the mid-point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L. Prove that
$\left(i\right)AF=2DC$
$\left(ii\right)DF=2DL$

Answer 1

Given, $E$ is mid point of $AD$

Also $EB\parallel DF$

$\Rightarrow$ $B$ is mid point of $AF$ [mid--point theorem]

so, $AF=2AB$ (1)

Since, $ABCD$ is a parallelogram,

$CD=AB$

$\Rightarrow AF=2CD$

$AD\parallel BC\Rightarrow LB\parallel AD$

In $\Delta FDA$

$\Rightarrow LB\parallel AD$

$\Rightarrow \frac{LF}{LD}=\frac{FB}{AB}=1$ from (1)

$\Rightarrow LF=LD$

so, $DF=2DL$