Triangle $ABC$ has side lengths $AB=20, BC=21, CA=9$. The bisector of $\angle BAC$ meets $\overline{BC}$ at $D$ and the circumcircle at $E\ne A$. The circumcircle of $\triangle BED$ meets line $AB$ at $B$ and $F\ne B$. What is $CF$?

1. $\cos\angle BAC=\dfrac{9^2+20^2-21^2}{2\cdot9\cdot20}=\dfrac19$.
2. Using $\angle CEA=\angle CBA=\angle FEA$, triangle congruence gives $AF=AC=9$.
3. $CF^2=9^2+9^2-2(9)(9)\left(\frac19\right)=144$, so $CF=\boxed{12}$.