smc

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 $△ADB,$ $\frac{AF}{FD}=\frac{AB}{DB}=\frac{6}{3}.$ Thus the answer is $\boxed{\text{(C)}2:1}$ Second way: Now, we use 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 $\frac{AF}{FD}=\frac{mD}{mA}=\boxed{2:1}$