jmc

Label the angles as shown in the diagram. Since $\angle DEC$ forms a linear pair with $\angle DEA$, $\angle DEA$ is a right angle. Let $\angle DAE=\alpha$ and $\angle ADE=\beta$. Since $\alpha +\beta =90^{\circ }$, and $\alpha +\angle BAF=90^{\circ }$, then $\beta =\angle BAF$. By the same logic, $\angle ABF=\alpha$. As a result, $△AED\sim △BFA$. By the same logic, $△CFB\sim △DEC$. Then, $\frac{BF}{AF}=\frac{3}{5}$, and $\frac{CF}{BF}=\frac{5}{7}$. Then, $7CF=5BF$, and $5BF=3AF$. By the transitive property, $7CF=3AF$. $AC=AF+CF=10$, and plugging in, we get $CF=3$. Finally, plugging in to $\frac{CF}{BF}=\frac{5}{7}$, we get $BF=\boxed{4.2}$