# Proving ratios using the angle bisector theorem

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.
• Oh so you're applying this on $\triangle AEC$. Thanks a lot. – Helix Feb 26 '18 at 14:55
• I think you mean $AFC$, but yes. – B. Mehta Feb 26 '18 at 14:56
• $AEC$ is a line, since $BE$ is the bisector of $\angle B$, so $E$ lies on $AC$. Hence $AEC$ is a line and $AFC$ is a triangle. – B. Mehta Feb 26 '18 at 15:00