jmc

First note that $FE=(AB+DC)/2$. Because trapezoids $ABEF$ and $FECD$ have the same height, the ratio of their areas is equal to the ratio of the averages of their parallel sides. Since $$AB+\frac{AB+DC}{2}=\frac{3AB+DC}{2}$$and $$\frac{AB+DC}{2}+DC=\frac{AB+3DC}{2},$$we have $$3AB+DC=2(AB+3DC)=2AB+6DC,\text{and}\frac{AB}{DC}=\boxed{5}.$$