Notice that since $\angle DBM= \angle DCM = 90^\circ$ $MB,MC$ are tangent to $\odot (BIC)$. Also it is easy to show that $FI$ is tangent to $\odot (BIC)$. Let $FN'$ be the other tangent from $F$ to $\odot (BIC)$. From la hire's theorem see that since $F$ lies on polar of $M$ WRT $\odot (BIC)$, $M\in IN'$. So $N=N'$. $\blacksquare$.

Since: $MB$, $MC$ tangent ($BIC$) at $B$, $C$, we have: $BICN$ is harmonic quadrilateral So: tangents at $I$, $N$ of ($BIC$) and $BC$ concurrent or $FN$ tangents ($BIC$) at $N$

Please note that point $A'$ isn't needed at all. So let's first of define an inversion ,$\psi(\Gamma_{\triangle BIC})$. Now we have the following $B \xrightarrow{\psi} B$, $C \xrightarrow{\psi} C$, $I \xrightarrow{\psi} I $ $\Gamma_{\triangle ABC} \xrightarrow{\psi} \overline{BC}$ and $M \xrightarrow{\psi} M'$, $F \xrightarrow{\psi} F'$ The center of $\Gamma_{\triangle BIC} $ is obviously $D$ So let's start with the angle-chase: We have the following $\angle DF'M = \angle DM'F = 90$ and $\angle DF'I = \angle DIF = 90$, thus the points $M,I,F',N$ are all collinear,thus $ 90 = \angle DF'N = \angle DNF $ Thus the line $FN$ is tangent to $\Gamma_{\triangle BIC} $.....

Notice that $M$ is the polar point of $BC$ and $FI$ is tangent to $\odot BIC$

Since $M, I, N$ are collinear and $MB$ and $MC$ are tangent to $(BIC), BICN$ is harmonic which implies that $FN$ is tangent to $(BIC$)

By incenter-excenter lemma $D$ is center of $(BIC)$ thus $FI$ its tangent to $(BIC)$ and we take polars w.r.t.$(BIC)$ Clearly $\mathcal P_M=BC$ and $\mathcal P_I=FI$ thus by la'hire since $M,I,N$ are colinear we have that $BC, FI, \mathcal P_N$ are concurrent thus $\mathcal P_N=NF$ as desired...

Oh, how to use tikz in AoPS?

Let the $A$-mixtilinear incircle touch $(ABC)$ at $T$. It's well-known that $T \in MI$ and $TI = TN$. By the Incenter-Excenter Lemma, $D$ is the center of $(BIC)$. Because $\angle FID = 90^{\circ}$, it follows that $FI$ is tangent to $(BIC)$. Since $\angle DTI = \angle DTM = 90^{\circ}$, we know $T$ lies on the circle with diameter $DI$. Hence, the Radical Axis Theorem on $(DTI), (BIC), (ABC)$ implies $II \equiv IF, DT, BC$ are concurrent at $IF \cap BC = F$. Thus, $F$ lies on $DT$, which is obviously the perpendicular bisector of $IN$. But this implies $FI = FN$, which clearly suffices. $\blacksquare$ Remark: The article linked below contains all of the well-known facts I cited in this solution. https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2020-02/mr_2_2020_mixtilinear.pdf GeoGebra: https://www.geogebra.org/geometry/nwmgedzz.

We can easily find that $MB,MC$ is tangent to the circle $\varGamma$. So quadrilateral $BICN$ is a harmonic quadrilateral. So $B,C;S,F$ are harmonic points. So $MN$ is the polar of $F$. $\boxed{}$

We have that $MB,MC$ are tangent to$(BIC)$. Now, $F$ lies on the polar of $M$,so by La Hire's theorem,$M$ lies on the polar of $F$. Now,by construction $FI$ is tangent to $(BIC)$. So $IM$ is the polar of $F$ and hence $N$ is the other tangency point from $F$.