As ΔABC is similar to ΔDEC, BEC is a straight line and AB//DE.
Since AB:DE = 2:1,
area of ΔABC : area of ΔDEC = 4:1
In ΔDEC, DF : FC = 3:4, so
area of ΔDFE : area ofΔFCE= 3:4
Since ΔDEC = ΔDFE + ΔFCE
If area of ΔDEC = 1, area of ΔDFE = 3/7
Therefore, area of ΔABC : area of ΔDFE = 4 : 3/7 = 4X7 : 3 = 28 :3