IVAN... Let $S$ and $T$ are the midpoint of $BC$ and arc $BAC$. Notice that $XI$ is the radical axis of $\odot (AIC)$ and $\odot (BI)$,so $XI^2=XA\cdot XC$. Similarly for $YI$,so $XY$ is the radical axis of $\odot (ABC)$ and point circle $\odot I$. Let $TA$ meets $XY$ at $M$,so $MI^2=MA\cdot MT$ and $AI\perp MT$,so $\angle AMI=\angle TIA$. On the other hand,it is well known that $\bigtriangleup IDS\backsim \bigtriangleup IAT$ and $IS\parallel AN$. So $\angle AIT=\angle DIS=180^\circ -\angle ANI$. So $M,A,N,I$ are concyclic and $MI$ is the diameter. So $M,V,A,N,I$ are concyclic.$\Box$.

It is well-known that $XY$ is the polar of the orthocenter of the intouch triangle, $H$ wrt $\omega$, hence $V$ is the image of $H$ under inversion in $\omega$. Inverting the problem in $\omega$, we want to show that $H$, the midpoint of the side of the intouch triangle not containing $D$, and $N$, the $D$-antipode in $\omega$ are collinear, which is trivial.

We know that the lines $IX$ and $IY$ are related to mixtilinear incircles (passing through the touching points of mixtilinear incircles to the sides). From this point of view, we recognise that the line $XY$ serves as the radical axis of the circumcircle and the point $I$. In other words, $XY$ is the line given with equation $ R^2 = PO^2 - PI^2$ (locus of $P$), where $O$ -- circumcircle of $ABC$ and $R$ -- its radius. Now, the problem is to show that the external bisector of $\angle BAC$ and the perpendicular line to $IN$ at $N$ meet on $XY$. Assume the external bisector of $\angle BAC$ and the perpendicular line to $IN$ at $N$ meet at $W$. Let $Q$ be the midpoint of the arc $BAC$ of the circumcircle, so that $AQ$ is the external bisector of $\angle BAC$. Now, \[ WO^2 - WI^2 = \mathrm{Pow}_{W}(O) - WI^2 + R^2 = WA\cdot WQ - WI^2 + R^2, \]and we need to show that $WI^2 = WA\cdot WQ$, in other words, we need to prove that $\angle WIA = \angle AQI$. Note that $W,A,N,I$ are concyclic. So $\angle WIA = \angle WNA$. If we assume $IQ$ intersects $(O)$ second time at $T$, which is the mixtilinear tangency point, then it remains to show that $\angle BAT = \angle CAN$. It is well known that $AT$ and $AN$ are isogonal conjugates of each other w.r.t. the angle $\angle BAC$. This completes our solution.

Let $IX \cap AB=X'$ and $IY \cap AC=Y'$, now from the angles notice that $\triangle AY'I \sim \triangle AIB$ and similarily $\triangle AX'I \sim \triangle AIC$, which means that if we consider an inversion $(A, AI)$ with reflection over $AI$ then $B \to Y'$ and $C \to X'$ which means that $(ABC) \to X'Y'$, now from ghost points its easy to tell that $X'Y'$ is the line parallel to $BC$ also tangent to incircle of $\triangle ABC$ which means that $X', N, Y'$ are colinear, now also notice line $XY$ is just the radical axis of $I, (ABC)$ from tangencies trivially, which means that if you let $U$ a point on $X'Y'$ such that $\angle UAI=90$, then $U$ is sent to midpoint of arc $BAC$ on the invert which means in fact from anti-parallels and tangency that $U$ also lies in the radical axis of $I, (ABC)$ therefore $X,Y,U$ are colinear, therefore $UANIV$ is cyclic with diameter $IU$ as desired thus we are done .

Let the tangent to \( \omega \) at \( N \) intersect \( AB \) and \( AC \) at points \( P \) and \( Q \), respectively. Let \( Z \) be the point of intersection of the external angle bisector of \( \angle PAQ \) and the tangent to \( \omega \) at \( N \). A simple angle chase reveals that \( IX \) and \( IY \) are the external angle bisectors of \( \triangle PAQ \), implying that points \( X \), \( Y \), and \( Z \) are collinear. Finally, we have \( \angle RAI = \angle RNI = \angle RVI = 90^\circ \), which implies \( IVRAN \) is cyclic.