imc

Since the heights of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$, $\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)$. $\frac{AB+EF}{2}=DC+EF$, so $AB+EF=2DC+2EF$. $EF$ is exactly halfway between $AB$ and $DC$, so $EF=\frac{AB+DC}{2}$. $AB+\frac{AB+DC}{2}=2DC+AB+DC$, so $\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB$, and $\frac{1}{2}AB=\frac{5}{2}DC$. $\frac{AB}{DC}=\boxed{5}$.