2016 AMC 12A Problem 12

Applying the Angle Bisector Theorem to $\triangle ABC$ gives $BD:DC=AB:AC=6:8,$ so $BD=\dfrac{6}{6+8}\cdot 7=3.$ Now $BF$ lies along the bisector of $\angle ABD$ in $\triangle ABD,$ so by the Angle Bisector Theorem again, $$\begin{gathered} AF:FD=AB:BD\\ =6:3=2:1. \end{gathered}$$ Thus, the correct answer is C.