**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, 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 Δ CLB (by AA similarity identity)

Hence, …………….. (i)

Let’s look at Δ CLM and Δ ALB:

∠ CLM = ∠ ALB ^{ }(vertically opposite angles)

∠ LAB = ∠ LCM (Interior angles)

Therefore, Δ CLM Δ ALB (by AA similarity identity)

Hence, ……… (ii)

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

…….. (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:

EL = 2 BL

**Hence Proved.**