By EGMO lemma 1.42 it suffices to prove that $X$ and $Y$ are midpoints of their respective arcs which can be proved by proving $C,I,X$ are collinear

Let's first prove that $X$ and $Y$ are the midpoint of their respective arcs. It suffices to prove that $AMIN$ is a rhombus. let $BC=a$ and define $b,c$ respectively. $AN=b*\frac {c}{a+b+c}=c*\frac {b}{a+b+c}=AM$. Thus, $IA$ is perpendicular to $XY$ and since $\angle XIY=\angle BAC=180-\angle XZY$(easy angle chasing), $I$ is the orthocenter of $\triangle XYZ$.

So first of all I will define some phantom points..... $Y'$ is the midpoint of the arc $AC$, and $X'$ is the midpoint of the arc $AB$ In this post I will show that $Y'\equiv Y$ By an easy angle-chase one can get that $CINY'$ is cyclic ( $\angle INC = \angle IY'C = \angle BAC$) So now we know that $\angle INY' = 90+ \frac{1}{2} \angle BAC$ Now from the problem we know that $ANIM$ is a rhombus (because of the angles $\angle IAN =\angle AIN$, thus $AN=NI$ same holds for $\triangle AMI$) Thus $\angle INM = 90 - \frac{1}{2}\angle BAC$ So now we see that $\angle INY' + \angle INM = 180$ Thus we see that $Y' \equiv Y$ So we do the same procedure for $X'$.Thus points $X,Y$ are midpoints of arcs. So now by the EGMO lemma 1.42, we have proven the problem.......

EulersTurban wrote: So first of all I will define some phantom points..... $Y'$ is the midpoint of the arc $AC$, and $X'$ is the midpoint of the arc $AB$ In this post I will show that $Y'\equiv Y$ By an easy angle-chase one can get that $CINY'$ is cyclic ( $\angle INC = \angle IY'C = \angle BAC$) So now we know that $\angle INY' = 90+ \frac{1}{2} \angle BAC$ Now from the problem we know that $ANIM$ is a rhombus (because of the angles $\angle IAN =\angle AIN$, thus $AN=NI$ same holds for $\triangle AMI$) Thus $\angle INM = 90 - \frac{1}{2}\angle BAC$ So now we see that $\angle INY' + \angle INM = 180$ Thus we see that $Y' \equiv Y$ So we do the same procedure for $X'$.Thus points $X,Y$ are midpoints of arcs. So now by the EGMO lemma 1.42, we have proven the problem....... What isEGMOlemma 1.42?

probably this Quote: Lemma 1.42 Let $ABC$ be an acute triangle inscribed in circle $\omega$. Let $X$ be the midpoint of the arc $BC$ not containing $A$ and define $Y,Z$ similarly. Show that the orthocenter of $\triangle{XYX}$ is the incenter $I$ of $\triangle{ABC}$ a lemma in the book Euclidean Geometry in Mathematical Olympiads by Evan Chen

parmenides51 wrote: probably this Quote: Lemma 1.42 Let $ABC$ be an acute triangle inscribed in circle $\omega$. Let $X$ be the midpoint of the arc $BC$ not containing $A$ and define $Y,Z$ similarly. Show that the orthocenter of $\triangle{XYX}$ is the incenter $I$ of $\triangle{ABC}$ a lemma in the book Euclidean Geometry in Mathematical Olympiads by Evan Chen Thanks

Or just note that $XI\perp YZ$ and the reflection of $I$ across $YZ$, which is $A$, lies on $(XYZ)$.

In fact, we can generalize so that $I$ does not have to be the incenter, it can be any point on the $A$-angle bisector.