AMC 12 2016a Problem 12

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 $AF$ : $FD$?

pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C);
draw(A--B--C--A--D^^B--E);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,NE);
label("$E$",E,NW);
label("$F$",F,1.5*N);

\textbf{(A)}\ 3:2 \textbf{(B)}\ 5:3 \textbf{(C)}\ 2:1 \textbf{(D)}\ 7:3 \textbf{(E)}\ 5:2