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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.

enter image description here

enter image description here

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  • $\begingroup$ Yes, I just hided a few lines in geogebra . Lemme give a better diagram $\endgroup$ Jul 30, 2020 at 13:35
  • $\begingroup$ It is my own problem . Actually, It is inspired by another unanswered problem in MSE . $\endgroup$ Aug 3, 2020 at 1:59
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    $\begingroup$ Can you please provide the link to the mentioned unanswered problem in MSE? $\endgroup$
    – YNK
    Jan 2 at 14:42

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