Ans:$\frac{80}{9}$ This year, APMO preliminary in Taiwan is at 11/22.

Sorry if I am being stupid rn, but isn't $C=F$??

Oh thanks, actually it's $D$ Fixed it.

Here is a bash sketch. Let $AF=x$. Then by PoP $FD=\sqrt{x(x+5)}$. Note that $\measuredangle FDC = \measuredangle DAC$ which gives $\triangle FDC \sim \triangle DAC$ and $\frac{DC}{x+5}=\frac{5}{DC}\Rightarrow DC=\sqrt{5(x+5)}$. Now use Stewart's on $\triangle FDC$ and cevian $AD$ to find $x$. If someone has anything nicer, please post. I tried for a long time and got nothing better than this, sadly. I though maybe Ptolemy but that didn't work.

Circle, radius $r=\frac{25}{2\sqrt{21}}$. Point $A(r\cos \alpha,r\sin \alpha)$ with $\alpha=\arcsin \frac{2}{5}$. Point $D(r\cos \delta,r\sin \delta)$ with $\delta=-\frac{\alpha}{3}$. Use $\triangle CDF \sim \triangle ABC$.

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