1990 AMC 12 Question 4

Let $ABCD$ be a parallelogram with $\angle ABC=120^\circ,$ $AB=16,$ and $BC=10.$ Extend $\overline{CD}$ through $D$ to $E$ so that $DE=4.$ If $\overline{BE}$ intersects $\overline{AD}$ at $F,$ then $FD$ is closest to