Answer: 919

Small Hint:

Let $E'$ be diametrically opposite $E$ on $\omega;$ angles in semicircles give $\angle DFE = \angle E'FE = 90^\circ,$ so $F$ lies on line $E'D$

Big Hint:

Place $B = (0, 0)$ and $C = (7, 0);$ then $D,$ $E,$ $E'$ have clean coordinates, and $F$ comes from intersecting line $E'D$ with $\omega$

Solution:

Let $E'$ be the point of $\omega$ diametrically opposite $E.$ Since $\overline{DE}$ is a diameter of $\gamma,$ the angle $\angle DFE = 90^\circ,$ and since $\overline{EE'}$ is a diameter of $\omega,$ also $\angle E'FE = 90^\circ.$ Both $FD$ and $FE'$ are perpendicular to $FE,$ so $D$ lies on line $E'F:$ the point $F$ is the second intersection of line $E'D$ with $\omega.$

Set $B = (0, 0)$ and $C = (7, 0);$ then $A = \left(\frac{65}{14}, \frac{15\sqrt{3}}{14}\right).$ The bisector gives $\frac{BD}{DC} = \frac{AB}{AC} = \frac{5}{3},$ so $D = \left(\frac{35}{8}, 0\right).$ Since $E$ is the midpoint of arc $BC$ not containing $A,$ both $E$ and $E'$ lie on the vertical line $x = \frac{7}{2}$ through the center $O = \left(\frac{7}{2}, -\frac{7\sqrt{3}}{6}\right),$ which satisfies $|OB| = |OA|$ with $R^2 = \frac{49}{3}.$ Thus $E = \left(\frac{7}{2}, -\frac{7\sqrt{3}}{2}\right)$ and $E' = \left(\frac{7}{2}, \frac{7\sqrt{3}}{6}\right).$

The direction from $E'$ to $D$ is proportional to $(3, -4\sqrt{3}),$ and the point $E' + t\,(3, -4\sqrt{3})$ lies on $\omega$ when $57t^2 - 56t = 0,$ so $t = \frac{56}{57}$ gives $F = \left(\frac{245}{38}, -\frac{105\sqrt{3}}{38}\right).$ Then $$\begin{aligned} AF^2 &= \left(\frac{240}{133}\right)^2 + 3\left(\frac{510}{133}\right)^2 \\ &= \frac{837900}{17689} = \frac{900}{19}, \end{aligned}$$ so $m + n = 900 + 19 = 919.$