smc

Draw $\overline{DG}\parallel \overline{FC}$ with $G$ on $\overline{AB}$. We know that $GF=BF=5$, since $△BFE\sim △BGD$. Likewise, since $△ADG\sim △ACF$, we know that $AG=5$. Thus, $AF=AG+GF+FB=5+5+5=15$, which is answer $\boxed{15}$.