jmc

Extend segments $\overline{DA}$ and $\overline{CB}$ until they intersect at point $G$ as shown. Define $x=GB$ and $y=BF/3$ . Then $BF=3y$ and $FC=4y$. (We chose $y=BF/3$ so we could represent both $BF$ and $FC$ without fractions). Using the similarity of triangles $GBA$ and $GCD$, we have $\frac{x}{7}=\frac{x+7y}{10}$. Solving this equation for $x$, we find $x=49y/3$. Now using the similarity of triangles $GBA$ and $GFE$ and substituting $49y/3$ for $x$ we get $$\begin{array}{rl}\frac{x}{7}& =\frac{x+3y}{EF}\\ \frac{\frac{49}{3}y}{7}& =\frac{\frac{49}{3}y+3y}{EF}\\ \frac{49}{21}& =\frac{58}{3EF}\\ EF& =\boxed{\frac{58}{7}}\text{ units}.\end{array}$$

(Note: Intuitively, since $F$ is three-sevenths of the way from $B$ to $C$, $EF$ should be three-sevenths of the way from $7$ to $10$. This intuition is correct and may be proved using the same approach as above.)