Question

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.

Answer 1

∠1 = ∠6 (Alternate interior angles)

∠2 = ∠3 (Vertically opposite angles)

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

ΔDEM ≅ ΔCBM (AAS Congruence criterion)

So, DE = BC (Corresponding parts of congruent triangles)

Also, AD = BC (Opposite sides of a parallelogram)

AE = AD + DE = 2BC

Now, ∠1 = ∠6 and ∠4 = ∠5

ΔELA ∼ ΔBLC (AA similarity)

`=> (EL)/(BL)  = (EA)/(BC)`

`=> (EL)/(BL) = (2BC)/(BC) = 2`

`=>` EL = 2BL