Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E. Prove that : EL = 2BL.

The parallelogram ABCD is shown in the figure below:

∠1 = ∠6 (Alternate angles)

∠2 = ∠3 (Vertically opposite angles)

DM = MC (M is the mid-point of CD)

△DEM ≅ △CBM (By AAS axiom)

∴ DE = BC (By C.P.C.T)

Also, AD = BC (As opposite sides of parallelogram are equal.)

From figure,

AE = AD + DE = BC + BC = 2BC.

Now,

∠1 = ∠6 (Proved above)

∠4 = ∠5 (Vertically opposite angles)

∴ △ELA ~ △BLC (By AA)

Since, corresponding sides of similar triangle are proportional to each other.

$\therefore \dfrac{EL}{BL} = \dfrac{EA}{BC} \\[1em] \Rightarrow \dfrac{EL}{BL} = \dfrac{2BC}{BC} \\[1em] \Rightarrow \dfrac{EL}{BL} = 2 \\[1em] \Rightarrow EL = 2BL.$

Hence, proved that EL = 2BL.