Question

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

Answer 1

In parallelogram ABCD,

AB || CD, BC || AD, AB = CD, BC = AD

Also, M is the midpoint of CD, so

DM = MC

Since E lies on AD produced, DE || BC.

Also, B, M, and E are collinear.

Now consider △DEM and △CBM:

• ∠DME = ∠CMB (vertically opposite angles)
• ∠DEM = ∠CBM (alternate interior angles, since DE || BC)
• DM = MC (M is the midpoint of CD)

Therefore, △DEM ≅ △CBM by AAS.

Hence, by CPCT, DE = BC

But in parallelogram ABCD,

AD = BC

So, AE = AD + DE = BC + BC = 2BC

Now consider △ELA and △BLC:

• ∠ELA = ∠BLC (vertically opposite angles)
• ∠EAL = ∠BCL (since AE || BC and AL, CL lie on AC)

Thus, △ELA ∼ △BLC by AA similarity.

Therefore, `(EL)/(BL) = (EA)/(BC)`

Substituting EA = 2BC,

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

So,

EL = 2BL

Hence proved.