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)

∴ ∆DEM ≅ ∆CBM   ...(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

∴ ∆ELA ~ ∆BLC   ...(AA similarity)

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

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

`=>` EL = 2BL