In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD=BC=5$, $AB=4$, and $DC=10$. The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in right triangle $DEF$. Then $CF=$

$\textrm{(A)}\ 3.25\qquad\textrm{(B)}\ 3.5\qquad\textrm{(C)}\ 3.75\qquad\textrm{(D)}\ 4.0\qquad\textrm{(E)}\ 4.25$
Drop perpendiculars from $A$ and $B$; then the triangle $DBY$ is similar to $DEF$ but with corresponding sides of half the length.

$XY=AB=4$ and $DX=YC=3$, hence $DY=7\implies DF=14\implies CF=\boxed{4.0}$.