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

In the given figure, ABCD is a parallelogram CBSE 10th board IB

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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