Prove that $N,R,F$ are collinear In a triangle $ABC$, let I be the incentre. Let $D$, $E$, $F$ be the intersections of $(ABC)$. with the lines through $I$ perpendicular to $BC$, $CA$, $AB$, respectively.
Define $O= BC \cap DE$ and $L= AC \cap DE$ . Define $IF\cap AB= R$ . Let $N=(BOF) \cap (LAF)$ .Prove that $N$,$R$,$F$ are collinear.
My progress: Since $F\in (ABC) $ , I thought of using simson points . So I took points $J$,$R$,$K$ as the simson points in $BC$,$BA$,$AC$ wrt point $F$ respectively. ( as shown in the diagram )
Then since $NBFO$ and $AFLN$ is cyclic, we get that $180- \angle ONF=\angle OBF=\angle CBF=180- \angle FAC=180 -\angle FAL = \angle FNL $ .
Hence points $O$,$N$,$L$ are collinear .
Now, I am stuck . I tried using phantom points but couldn't proceed . I am thinking of using Radical axis but still confused.
Here are some more observations which might be trivial but still, we have $BJFR$, $RFKA$,$CJFK$ concyclic . We also have $\Delta JFK \sim \Delta BFA $
Please post hints if possible.
Thanks in advance.
Ps: This is my own observation, so there is a very high chance that I might be wrong.
Below are a few diagrams for the problem.
This result is false! An IMOTCer said that when we use the relation tool, the geogebra shows that $N \not\in RF$ . 
Thank you show much to everyone for trying this problem .
This result is false! An IMOTCer said that when we use the relation tool, the geogebra shows that $N \not\in RF$ . 
Thank you show much to everyone for trying this problem .