Question

In given figure through the mid-point $M$ of the side $CD$ of a parallelogram $ABCD$, the line $BM$ is drawn intersecting $AC$ in $L$ and $AD$ produced in $E$. Prove that $EL=2BL$.

Answer 1

In $△BMC$ and $△EMD$, we have

$MC=MD$ [$\because$ M is the mid-point of CD]

$\angle CMB=\angle EMD$ [Vertically opposite angles]

and, $\angle MBC=\angle MED$ [Alternate angles]

So, by AAS-criterion of congruence, we have

$\therefore$ $△BMC\cong △EMD$

$\Rightarrow$ $BC=DE$......(i)

Also, $AD=BC$ [$\because$ABCD is a parallelogram].....(ii)

$AD+DE=BC+BC$

$\Rightarrow$ $AE=2BC$ ........(iii)

Now, in $△AEL$ and $△CBL$, we have

$\angle ALE=\angle CLB$ [Vertically opposite angles]

$\angle EAL=\angle BCL$ [Alternate angles]

So, by AA-criterion of similarity of triangles, we have

$△AEL\sim △CBL$

$\Rightarrow$ $\frac{EL}{BL}=\frac{AE}{CB}$

$\Rightarrow$ $\frac{EL}{BL}=\frac{2BC}{BC}$ [Using equation (iii)]

$\Rightarrow$ $\frac{EL}{BL}=2$

$\Rightarrow$ $EL=2BL$ $\left[Henceproved\right]$