Question

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

Answer 1

In ${△}^{\prime }sBMC$ and $EMD$, we have

$\angle BMC=\angle EMD$..........(opposite angles)

$MC=MD$........ (mid point)

$\angle MCB=\angle MDE$..........(alternate angles)

So, by AAS-congruence criterion, we have

$△BMC\cong △EMD$

$\Rightarrow$ $BC=ED$ [$\because$ Corresponding parts of congruent triangle a equal]

In ${△}^{\prime }sAEL$ and $CBL$, we have

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

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

So, by AA-criterion of similarity, we have

$△AEL\sim △CBL$

$\Rightarrow$ $\frac{AE}{BC}=\frac{EL}{BL}=\frac{AL}{CL}$

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

$\Rightarrow$ $\frac{EL}{BL}=\frac{AD+DE}{BC}=\frac{BC+DE}{BC}=\frac{2BC}{BC}=2$ [$\because$ AD=BC and DE=BC]

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