The median AD of the $\Delta \mathrm{ABC}$  is bisector at E, BE meets AC in F, then  $\mathrm{AF}:FC$  is equal to 

Take A as origin 
Let $\overline{OB}=\overline{b}$  and  $\overline{OC}=\overline{c}$
Equation of lines BF and AC are 
$\overline{r}=\overline{b}+\lambda (\frac{\overline{b}+\overline{c}}{4}-\overline{b})$ and  $\overline{r}=\overline{O}+\mu \overline{C}$
For the point on intersection F. We have 
 $\overline{b}+\lambda \left(\frac{\overline{C}-3\overline{B}}{4}\right)=\mu \overline{C}$   $\lambda =\frac{4}{3},\mu =\frac{1}{3}$  $\therefore \overline{OF}\overline{r}=\frac{1}{3}\overline{C}$
Now  $\overline{AF}=\frac{\overline{C}}{3}=\frac{1}{3}\overline{AC}$
Hence  $AF:AC=\frac{1}{3}$