# 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 $$\Delta JFK \sim \Delta BFA$$

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.

• Yes, I just hided a few lines in geogebra . Lemme give a better diagram – Sunaina Pati Jul 30 '20 at 13:35
• It is my own problem . Actually, It is inspired by another unanswered problem in MSE . – Sunaina Pati Aug 3 '20 at 1:59

This result is false! An IMOTCer said that when we use the relation tool, the geogebra shows that $$N \not\in RF$$ . 