jmc

Use the angle bisector theorem to find $CD=\frac{21}{8}$, $BD=\frac{35}{8}$, and use Stewart's Theorem to find $AD=\frac{15}{8}$. Use Power of the Point to find $DE=\frac{49}{8}$, and so $AE=8$. Use law of cosines to find $\angle CAD=\frac{\pi }{3}$, hence $\angle BAD=\frac{\pi }{3}$ as well, and $△BCE$ is equilateral, so $BC=CE=BE=7$. I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines: $A{E}^2=A{F}^2+E{F}^2-2\cdot AF\cdot EF\cdot \cos \angle AFE.$ (1) $A{F}^2=A{E}^2+E{F}^2-2\cdot AE\cdot EF\cdot \cos \angle AEF.$ Adding these two and simplifying we get: $EF=AF\cdot \cos \angle AFE+AE\cdot \cos \angle AEF$ (2). Ah, but $\angle AFE=\angle ACE$ (since $F$ lies on $\omega$), and we can find $cos\angle ACE$ using the law of cosines: $A{E}^2=A{C}^2+C{E}^2-2\cdot AC\cdot CE\cdot \cos \angle ACE$, and plugging in $AE=8,AC=3,BE=BC=7,$ we get $\cos \angle ACE=-1/7=\cos \angle AFE$. Also, $\angle AEF=\angle DEF$, and $\angle DFE=\pi /2$ (since $F$ is on the circle $\gamma$ with diameter $DE$), so $\cos \angle AEF=EF/DE=8\cdot EF/49$. Plugging in all our values into equation (2), we get: $EF=-\frac{AF}{7}+8\cdot \frac{8EF}{49}$, or $EF=\frac{7}{15}\cdot AF$. Finally, we plug this into equation (1), yielding: $8^2=A{F}^2+\frac{49}{225}\cdot A{F}^2-2\cdot AF\cdot \frac{7AF}{15}\cdot \frac{-1}{7}$. Thus, $64=\frac{A{F}^2}{225}\cdot (225+49+30),$ or $A{F}^2=\frac{900}{19}.$ The answer is $\boxed{919}$.