2018 Oral Moscow Geometry Olympiad P3 wrote: On the extensions of sides $CA$ and $AB$ of triangle $ABC$ beyond points $A$ and $B$, respectively, the segments $AE = BC$ and $BF = AC$ are drawn. A circle is tangent to segment $BF$ at point $N$, side $BC$ and the extension of side $AC$ beyond point $C$. Point $M$ is the midpoint of segment $EF$. Prove that the line $MN$ is parallel to the bisector of angle $A$. First we will complete the parallelogram in this configuration. Extend $NM$ to a point $K$ such that $MK=MN$. Clearly $EKFN$ is a parallelogram. Now let $CA\cap NK=T$. Let $s=\frac{a+b+c}{2}$. So, $BN=(s-c)$. Hence, $FN=(b-s+c)$. Let $TE=x$. Hence by Menelaus Theorem we get that $$(\frac{a+x}{x})\cdot(\frac{b-s+c}{s})=1\implies \frac{a}{x}=\frac{s}{b-s+c}-1=\frac{a}{b-s+c}\implies x=TE=(b-s+c)=FN=EK$$Now as $EK\|AN$ we get that $\angle ATN=\angle TNA=\angle NAI_A$ where $I_A$ is the $A-\text{excenter}$ of $\triangle ABC$. Hence, $MN\|AI_A$. $\blacksquare$