, 2023

Let $AO$ meet $BC$ at the point $E$. Since $EDW$ is a right triangle and $O$ lies on $WE$, the condition $OW = OD$ shows that $O$ is the circumcenter of triangle $EDW$. Hence $OD = OE$, so $D$ and $E$ are symmetric with respect to the perpendicular bisector of side $BC$.
Observe that:
$$180^\circ - \angle DXZ = \angle ZXY = \angle ZAY = \angle ZCD,$$
so $CDXZ$ are concyclic.

Next we prove that $AZ \parallel BC$. To do this, introduce an auxiliary point $Z'$ on the circle $ABC$ satisfying $AZ' \parallel BC$. By the proof of the previous paragraph, we know it suffices to prove that $CDXZ'$ are concyclic. Note that triangle $BAE$ and triangle $CZ'D$ are symmetric with respect to the perpendicular bisector of side $BC$. Using this fact together with the collinearity of $A, O, E$, we obtain
$$\angle DZ'C = \angle BAE = \angle BAO = 90^\circ - \frac{1}{2}\angle AOB = 90^\circ - \angle C = \angle DXC.$$

Hence $DXZ'C$ are concyclic, which implies that $Z$ and $Z'$ are the same point, as desired.
Finally, since $AZ \parallel BC$ and $CDXZ$ are concyclic, we have
$$\angle AZD = \angle CDZ = \angle CXZ = \angle AYZ,$$
and by the tangent-chord angle, $DZ$ is tangent to the circle $AXY$. □

Note that the point $Z$ is the Miquel point of the lines $AC, BC, BA$ and $DY$. Hence $BDZY$ are concyclic and $CDXZ$ are concyclic. Moreover, there is a spiral similarity centered at $Z$ that maps $BC$ to $YX$.
Since $BC \perp YX$, the rotation angle of the above spiral similarity is $90^\circ$, so the circle $ABCZ$ and the circle $AXYZ$ are orthogonal, which means the radius $OZ$ of the circle $ABCZ$ is perpendicular to the circle $AXYZ$.
Since $OW = OD$, triangle $OWD$ is isosceles, and
$$\angle ZOA = 2\angle ZBA = 2\angle ZBY = 2\angle ZDY = \angle ODW + \angle DWO,$$
so $D$ lies on the line $ZO$, which is tangent to the circle $AXY$. □