In the given figure, ABCD is a parallelogram. BE bisects CD at M

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

Ans:

Given that:

BE bisects CD at M, $\therefore$ DM = MC

Let’s look at Δ ALE and Δ CLB:

∠ ALE = ∠ CLB (vertically opposite angles)

∠ EAC = ∠ ACB (Interior angles, given that AE ǁ BC)

Therefore, Δ ALE $\sim$ Δ CLB (by AA similarity identity)

Hence, $\frac{AL}{CL} = \frac{EL}{BL}$ …………….. (i)

Let’s look at Δ CLM and Δ ALB:

∠ CLM = ∠ ALB (vertically opposite angles)

∠ LAB = ∠ LCM (Interior angles)

Therefore, Δ CLM $\sim$ Δ ALB (by AA similarity identity)

Hence, $\frac{AL}{CL} = \frac{AB}{MC}$ ……… (ii)

From equation (i) and (ii), we get:

$\frac{EL}{BL} = \frac{AB}{MC}$ …….. (iii)

Since AB = DC (given that ABCD is parallelogram) and DM = MC (given),

Therefore AB = DC = 2MC

Putting this value in equation (iii), we get:

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

EL = 2 BL

Hence Proved.

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