已知矩形ABCD中,E是CD上一点,AD：AE＝1：2,CE：ED=1：3（1）F是AB中点,DF交AE于G,若CE =√3,求S△EFG.

CE:ED=1:3,CE=√3,DE=3√3
AD:AE=1:2,角D=90,角DAE=60,角FAE=30
AF=1/2 AB=1/2(CE+DE)=2√3
AG/sinF=FG/sin角FAE=FG/sin30
AG/sinD=GD/sin角DAE=GD/sin60
sinD/sinF=FGsin60/GDsin30=AF/AD
(√3/2)FG/(1/2)GD=2√3/3
FG/GD=2/3
FG=2/5FD
S△EFG=2/5S△FDE=(2/5)(1/2)AD*DE=(1/5)*3*3√3=9√3/5