In $\triangle ABC$, $AB=9$, $BC=8$, and $CA=10$. Point $D$ lies on $\overline{BC}$, and $\overline{AD}$ bisects $\angle BAC$. Point $E$ lies on $\overline{AC}$, and $\overline{BE}$ bisects $\angle ABC$. The bisectors intersect at $F$. What is the ratio $AF$ : $FD$?

By the angle bisector theorem, $BD=\dfrac{AB}{AB+AC}\cdot BC=\dfrac9{19}(8)=\dfrac{72}{19}$.

Since $\overline{BE}$ bisects $\angle ABD$ (as $D$ lies on segment $BC$), by the angle bisector theorem in $\triangle ABD$: $AF:FD=AB:BD=9:\dfrac{72}{19}=\dfrac{9\cdot19}{72}=\dfrac{171}{72}=\dfrac{19}8$.

In general $AF:FD=(AB+AC):BC$; here $(9+10):8=\boxed{19:8}$.