Let $A O$ intersect $B C$ at $E$. As $E D W$ is a right-angled triangle and $O$ is on $W E$, the condition $O W=O D$ means $O$ is the circumcentre of this triangle. So $O D=O E$ which establishes that $D, E$ are reflections in the perpendicular bisector of $B C$. Now observe:
$$180^{\circ}-\angle D X Z=\angle Z X Y=\angle Z A Y=\angle Z C D$$
which shows $C D X Z$ is cyclic. !
We next show that $A Z \| B C$. To do this, introduce point $Z^{\prime}$ on circle $A B C$ such that $A Z^{\prime} \| B C$. By the previous result, it suffices to prove that $C D X Z^{\prime}$ is cyclic. Notice that triangles $B A E$ and $C Z^{\prime} D$ are reflections in the perpendicular bisector of $B C$. Using this and that $A, O, E$ are collinear:
$$\angle D Z^{\prime} C=\angle B A E=\angle B A O=90^{\circ}-\frac{1}{2} \angle A O B=90^{\circ}-\angle C=\angle D X C,$$
so $D X Z^{\prime} C$ is cyclic, giving $Z \equiv Z^{\prime}$ as desired. Using $A Z \| B C$ and $C D X Z$ cyclic we get:
$$\angle A Z D=\angle C D Z=\angle C X Z=\angle A Y Z,$$
which by the converse of alternate segment theorem shows $D Z$ is tangent to circle $A X Y$.