Solution:

First, we can deduce that $BD =5, AD = 12, CD = 9$ by inspecting Pythagorean triples.

This yields the following diagram:

Then, we get $\angle ADE = \angle ACD$ by the similarity of \triangleriangle ADC and \triangleriangle AED . Since \triangleriangle ABF and \triangleriangle ADF are both right triangles, they both have circumcircles with diameter $AB,$ making $ABDF$ cyclic. Thus, $$\angle ABF = \angle ADF = \angle ACD,$$ making $$\cos (\angle ABF) = \dfrac 35,$$$$\sin (\angle ABF) = \dfrac 45 .$$ As such, $$AF = \dfrac 35 \cdot 13,$$$$BF = \dfrac 45 \cdot 13.$$

By Ptolemy's Theorem, we get $$AB \cdot DF + DB \cdot AF = BF\cdot AD .$$ Therefore, $$13\cdot DF + 5\cdot 13\dfrac 45 =12 \cdot 13\dfrac 35 .$$

This makes $$DF + 5\cdot \dfrac 45 = 12 \cdot \dfrac 35 ,$$ so $DF + 4 = \dfrac {36}5.$ As such, $DF=\dfrac {16}5,$ making $m+n=21.$

Thus, the correct answer is B .