In $\triangle ABC$, $AB = 6$, $BC = 7$, and $CA = 8$. 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 $AF$ : $FD$?
$\textbf{(A)}\ 3:2\qquad$ $\textbf{(B)}\ 5:3\qquad$ $\textbf{(C)}\ 2:1\qquad$ $\textbf{(D)}\ 7:3\qquad$ $\textbf{(E)}\ 5:2$