Let $AO$ intersect $BC$ at $E$. As $EDW$ is a right-angled triangle and $O$ is on $WE$, the condition $OW = OD$ means $O$ is the circumcentre of this triangle. So $OD = OE$ which establishes that $D, E$ are reflections in the perpendicular bisector of $BC$.

Now observe:
$$180^{\circ} - \angle DXZ = \angle ZXY = \angle ZAY = \angle ZCD,$$
which shows $CDXZ$ is cyclic.

We next show that $AZ \parallel BC$. To do this, introduce point $Z'$ on circle $ABC$ such that $AZ' \parallel BC$. By the previous result, it suffices to prove that $CDXZ'$ is cyclic. Notice that triangles $BAE$ and $CZ'D$ are reflections in the perpendicular bisector of $BC$. Using this and that $A, O, E$ are collinear:
$$\angle DZ'C = \angle BAE = \angle BAO = 90^{\circ} - \frac{1}{2} \angle AOB = 90^{\circ} - \angle C = \angle DXC,$$
so $DXZ'C$ is cyclic, giving $Z \equiv Z'$ as desired.

Using $AZ \parallel BC$ and $CDXZ$ cyclic we get:
$$\angle AZD = \angle CDZ = \angle CXZ = \angle AYZ,$$
which by the converse of alternate segment theorem shows $DZ$ is tangent to circle $AXY$.

Notice that point $Z$ is the Miquel-point of lines $AC$, $BC$, $BA$ and $DY$; then $B, D, Z, Y$ and $C, D, X, Y$ are concyclic. Moreover, $Z$ is the centre of the spiral similarity that maps $BC$ to $YX$.

By $BC \perp YX$, the angle of that similarity is $90^{\circ}$; hence the circles $ABCZ$ and $AXYZ$ are perpendicular, therefore the radius $OZ$ in circle $ABCZ$ is tangent to circle $AXYZ$.

By $OW = OD$, the triangle $OWD$ is isosceles, and
$$\angle ZOA = 2 \angle ZBA = 2 \angle ZBY = 2 \angle ZDY = \angle ODW + \angle DWO,$$
so $D$ lies on line $ZO$ that is tangent to circle $AXY$.