In the given figure, $ABCD$ is a parallelogram. $BE$ bisects $CD$ at $M$ and intersects $AC$ at $L$. Prove that $EL = 2BL$.

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Question

In the given figure, $ABCD$ is a parallelogram. $BE$ bisects $CD$ at $M$ and intersects $AC$ at $L$. Prove that $EL = 2BL$.

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Accepted Answer

Step 1: $	riangle ALE \sim 	riangle CLB$ (AA similarity)
$\Rightarrow \frac{AL}{CL} = \frac{EL}{BL}$ ... (i)
Step 2: Also $	riangle CLM \sim 	riangle ALB$ (AA similarity)
$\Rightarrow \frac{AL}{CL} = \frac{AB}{CM}$
Step 3: Since $AB = CD$ (parallelogram) and $CM = \frac{CD}{2}$ ($M$ is midpoint): $\frac{AL}{CL} = \frac{CD}{CD/2} = 2$ ... (ii)
Step 4: From (i) and (ii): $\frac{EL}{BL} = 2 \Rightarrow EL = 2BL$

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