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In $\Delta ABC$, $BE$ and $CF$ are the angular bisectors of $\angle B$ and $\angle C$ meeting at $I$. Prove that $AF/FI=AC/CI$

I have tried this question for hours but i am unable to hit the nut

I tried to using: (1)Similarity (2)Angle bisector theorem (3)Relation of lengths of angle bisector :- $BE^2+AE.EC=AB.BC$ $CF^2+AF.FB=AC.BC$

I tried the question with these many approaches but i unable to prove the above relation.please tell if any of these approach will help or please tell any other method to solve the question.

This is an olympiad book question (not homework problem)

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Hint: The ratio of the area of $\triangle AFI$ and the area of $\triangle ACI$ is given by $FI/CI$, and also by $AF/AC$.

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  • $\begingroup$ Thanks for your answer i got the question $\endgroup$ Jun 11 '19 at 21:08
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Since $AI$ is a bisector of $\Delta AFC$, we obtain: $$\frac{AF}{AC}=\frac{FI}{CI}$$ or $$\frac{AF}{FI}=\frac{AC}{CI}$$ and we are done!

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