難解的三角形比的煩人問題

Let area of ΔCDE be x, as AD : DC = 2 : 1, so
area of ΔAED : area of ΔCDE = 2 : 1
that is, area of ΔAED = 2x.
As E is the mid-pt of AB, ie. AE : EB = 1 : 1, so
area of ΔBEC
= area of ΔAEC
= area of ΔAED + area of ΔCDE
= 2x + x
= 3x
so, area of ΔBEC : area of ΔCDE : area of ΔAED
= 3x : x : 2x
= 3 : 1 : 2 (ans: B)