In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$, $E$ as the midpoint of $\overline{BC}$, and $F$ as the midpoint of $\overline{DA}$. The area of $ABEF$ is $1.5$ times the area of $FECD$. What is $AB/DC$?

1. Segment $EF$ is the midline of the trapezoid, with length $EF=\frac{AB+DC}2$, and it splits the trapezoid's height in half.
2. $[ABEF]=\frac{AB+EF}2\cdot\frac h2$, $[FECD]=\frac{EF+DC}2\cdot\frac h2$.
3. Setting $[ABEF]=k\cdot[FECD]$ and simplifying: $AB+EF=k(EF+DC)$, i.e. $3AB+DC=k(AB+3DC)$, giving $\dfrac{AB}{DC}=\dfrac{3k-1}{3-k}$.
4. For $k=\frac32$: $\dfrac{AB}{DC}=\dfrac{\frac92-1}{3-\frac32}=\dfrac{\frac72}{\frac32}=\frac73$.