smc

Since $\angle AFB=\angle ADB=90^{\circ }$, quadrilateral $ABDF$ is cyclic. It follows that $\angle ADE=\angle ABF$, so $△ABF\sim △ADE$ are similar. In addition, $△ADE\sim △ACD$. We can easily find $AD=12$, $BD=5$, and $DC=9$ using Pythagorean triples. So, the ratio of the longer leg to the hypotenuse of all three similar triangles is $\frac{12}{15}=\frac{4}{5}$, and the ratio of the shorter leg to the hypotenuse is $\frac{9}{15}=\frac{3}{5}$. It follows that $AF=(\frac{4}{5}\cdot 13),BF=(\frac{3}{5}\cdot 13)$. Let $x=DF$. By Ptolemy's Theorem, we have $$13x+\left(5\cdot 13\cdot \frac{4}{5}\right)=12\cdot 13\cdot \frac{3}{5}\iff 13x+52=\mathrm{93.6.}$$ Dividing by $13$ we get $x+4=7.2⟹x=\frac{16}{5}$ so our answer is $\boxed{21}$.