Let ABCD be a parallelogram with ∠ABC=120∘, AB=16 and BC=10. Extend CD through D to E so that DE=4. If BE intersects AD at F, then FD is closest to<spanclass=′latex−bold′>(A)</span>1<spanclass=′latex−bold′>(B)</span>2<spanclass=′latex−bold′>(C)</span>3<spanclass=′latex−bold′>(D)</span>4<spanclass=′latex−bold′>(E)</span>5[asy]
size(200);
defaultpen(linewidth(0.8));
pair A=origin,B=(16,0),C=(26,10*sqrt(3)),D=(10,10*sqrt(3)),E=(0,10*sqrt(3));
draw(A--B--C--E--B--A--D);
label("A",A,S);
label("B",B,S);
label("C",C,N);
label("D",D,N);
label("E",E,N);
label("F",extension(A,D,B,E),W);
label("4",(D+E)/2,N);
label("16",(8,0),S);
label("10",(B+C)/2,SE);
[/asy]