In $\triangle ABC$, $BE$ and $CF$ are the angular bisectors of $\angle B$ and $\angle C$ meeting at $I$. Prove that $\frac{AF}{FI}=\frac{AC}{CI}$.
By the angle bisector theorem we have $\frac{AC}{AE} = \frac{CB}{BE}$ and $\frac{AB}{AF} = \frac{BC}{FC}$. How do I proceed after this? Hints would be appreciated.
$AI$ is an angle bisector of $\angle A$, so by the angle bisector theorem, $\frac{AF}{AC} = \frac{FI}{IC}$. This easily rearranges to the required result.