Question

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

Answer 1

In ΔBMC and ΔEMD, we have

MC = MC     [∵ M is the mid-point of CD]

∠CMB = ∠DME    [vertically opposite angles]

and ∠MBC = ∠MED    [Alternate angles]

So, by AAS criterion of congruence, we have

ΔBMC ≅ ΔEMD

= BC = AD   [CPCT]

Also, BC = AD   [∵ ABCD is a parallelogram]

Now, in ΔAEL and Δ CBL, we have

∠ALE = ∠CLB    [Vertically opposite angles]

∠EAL = ∠BCL    [Alternate angles]

So, by AA criterion of similarity of triangles,

we have 

ΔAEL ~ ΔCBL

= `(EL)/(BL) = (AE)/(CB)`

= `(EL)/(BL) = (2BC)/(BC)`   [∵ AE = AD + DE = BC + BC = 2BC]

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

= EL = 2BL