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.
Best Answer
This looks very close to true in diagrams, but it’s actually false.
For example, if
$$A = (4, -3), \quad B = (-4, 3), \quad C = (4, 3),$$
then
\begin{gather*} I = (2, 1), \quad D = (2, \sqrt{21}), \quad E = (2\sqrt6, 1), \quad F = \left(\frac{4 - 6\sqrt 6}5, \frac{-3 - 8\sqrt 6}{5}\right), \\ \quad O = \left(\frac{1 + 9\sqrt6 - 3\sqrt{14} + \sqrt{21}}{5}, 3\right), \quad L = \left(4, \frac{1 + \sqrt6 - 3\sqrt{14} + 4\sqrt{21}}{5}\right), \\ R = \left(\frac45, -\frac35\right), \quad N = (-22 + 10\sqrt 6 - 12\sqrt{14} + 10\sqrt{21}, 33 - 14\sqrt 6 + 6\sqrt{14} - 4\sqrt{21}), \end{gather*}
and
$$\mathrm{area}(\triangle NRF) = \frac{12}{5}(41 - 16\sqrt6 + 13\sqrt{14} - 11\sqrt{21}) ≈ 0.099306.$$