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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.

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$AI$ 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.

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  • $\begingroup$ Can you clarify how the angle bisector theorem helps in deriving the above mentioned result? $\endgroup$ – Helix Feb 26 '18 at 14:49
  • $\begingroup$ Look at the statement of the angle bisector theorem given on wikipedia. This is exactly the result I have used. $\endgroup$ – B. Mehta Feb 26 '18 at 14:51
  • $\begingroup$ Oh so you're applying this on $\triangle AEC$. Thanks a lot. $\endgroup$ – Helix Feb 26 '18 at 14:55
  • $\begingroup$ I think you mean $AFC$, but yes. $\endgroup$ – B. Mehta Feb 26 '18 at 14:56
  • $\begingroup$ $AEC$ 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. $\endgroup$ – B. Mehta Feb 26 '18 at 15:00

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