F3 Maths

Let BD meet EF at M :
Let the area of △ABD = 3 and △BCD = 7 ,
then △EBM = (1/4)△ABD .(Since EM :AD = 1 : 2 ,△EBM :△ABD= (1:2)^2)
i.e. △EBM = (1/4)3 = 3/4 ,
so the area of AEMD = 3 - 3/4 or 3*(3/4) = 9/4
;
△MDF = (1/4)△BCD = 7/4 ,
so the area of BMFC = 7 - 7/4 or 7*(3/4) = 21/4
the area of AEFD: the are of EBCF
= (AEMD + △MDF) : (△EBM + BMFC)
= (9/4 + 7/4) : (3/4 + 21/4)
= (9 + 7) : (3 + 21)
= 16 : 24
= 2 : 3