Question

In triangle ABC, D and E are points on side
AB such that AD = DE = EB. Through D and
E, lines are drawn parallel to BC which meet
side AC at points F and G respectively. Through
F and G, lines are drawn parallel to AB which
meet side BC at points M and N respectively.
Prove that : BM = MN = NC.​

Answer 1

Given :  In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet side BC at points M and N respectively.

To find : Prove that : BM = MN = NC.​  

Solution:

Correction in figure  M & N are interchanged

AD = DE = EB

AD + DE + EB  = AB  

=> AD = DE  = EB  = AB/3

=> AE = 2AB/3

DF ||  GE ||  BC

=> ΔADF ≈ ΔAEG ≈ ΔABC

=> AF/AD  = AG/AE = AC/AB

=> 3AF/AB  = 3AG/2AB  = AC/AB

=>  3AF = 3AG/2  = AC

AF = AC/3

CF = AC - AF = 2AC/3

AG = 2AC/3

CG = AC - AG =  AC/3

GN || FM  ||  AB

=> ΔCGN ≈ ΔCFM ≈ ΔCAB

=> CN/CG  = CM/CF = CB/CA

=> 3CN/AC = 3CM/2AC = CB/AC

=> 3CN = 3CM/2 = BC

=> CN = BC/3

     CM = 2BC/3

     BM = BC - CM =  BC/3

  MN = CM - CN = BC/3

hence CN = MN  = BM   = BC/3

BM = MN = NC.​  

QED

hence proved

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