How can MN be parallel to BC when N lies on the side BC?

I believe there are some typos (which I edited below). MathPassionForever wrote: Let points $M$ and $N$ lie on sides $AB$ and $BC$ of triangle $ABC$ in such a way that $MN||AC$. Points $M'$ and $N'$ are the reflections of $M$ and $N$ about $BC$ and $AB$ respectively. Let $M'A$ meet $BC$ at $X$, and let $N'C$ meet $AB$ at $Y$. Prove that $A,C,X,Y$ are concyclic. We need to prove that $M,N,Y,X$ are concyclic. We fix $B,N,M$ and animate $A$ on $BM$. Now, $A\mapsto AM'\cap BC = X$ is a projective map and $A\mapsto C\mapsto CN'\cap BA = Y\mapsto X'$ is a projective map where $X'Y$ is antiparallel to $MN$ and $X'$ lies on $BC$. Now we need to work for three cases of $A$. If $A = B$, we get $X = Y = B$ which is a degenerate case. If $A = M$, $X$ and $Y$ are the feet from $M$ and $N$ on $BC$ and $BA$, so $M,N,X,Y$ are clearly concyclic. Finally we pick $A$ as reflection of $M$ in $B$ which gives $X$ and $Y$ are points of infinity on $BC$ and $BA$ respectively and we are done. $~\square$ P.S. This is my first solution using "moving points", so I'm not sure if this works.

MathPassionForever wrote: Let points $M$ and $N$ lie on sides $AB$ and $BC$ of triangle $ABC$ in such a way that $MN||AC$. Points $M'$ and $N'$ are the reflections of $M$ and $N$ about $BC$ and $AB$ respectively. Let $M'A$ meet $BC$ at $X$, and let $N'C$ meet $AB$ at $Y$. Prove that $A,C,X,Y$ are concyclic. \(\Delta ABM' \sim \Delta CBN' \Rightarrow \angle BAM' = \angle BCN'\) done.

From the definition of the points $M'$, $N'$, $X$ and $Y$ follows that $XB$ is the external angle bisector of triangle $MXA$ and $YB$ is the external angle bisector of triangle $NYC$. Let $XP$ and $YQ$ be respectively internal angle bisectors in triangles $MXA$ and $NYC$. Using the property of external and internal angle bisectors of triangle and the fact, that $MN \parallel AC$, we get that $\dfrac{MP}{PA} = \dfrac{BM}{AB} = \dfrac{BN}{BC} = \dfrac{NQ}{QC}$, so, by Thales, $PQ \parallel AC$. Also angles $PXQ$ and $PYQ$ are both equal to $90^\circ$, as angles between respectively external and internal angle bisectors of angles $MXA$ and $NYC$. So the quadraliteral $PYXQ$ is cyclic. And since $PQ \parallel AC$, we have that the quadraliteral $AYXC$ is cyclic too, Q. E. D.