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 $AF$ : $FD$?

$\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2$
By the angle bisector theorem, $\frac{AB}{AE} = \frac{CB}{CE}$
$\frac{6}{AE} = \frac{7}{8 - AE}$ so $AE = \frac{48}{13}$
Similarly, $CD = 4$.
There are two ways to solve from here.
First way: 
Note that $DB = 7 - 4 = 3.$ By the angle bisector theorem on $\triangle ADB,$ $\frac{AF}{FD} = \frac{AB}{DB} = \frac{6}{3}.$ Thus the answer is $\boxed{\textbf{(C)}\; 2 : 1}$
Second way:
Now, we use [mass points](https://artofproblemsolving.com/wiki/index.php/Mass_points). Assign point $C$ a mass of $1$.
$mC \cdot CD = mB \cdot DB$ , so $mB = \frac{4}{3}$
Similarly, $A$ will have a mass of $\frac{7}{6}$
$mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}$
So $\frac{AF}{FD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}$