International Mathematical Olympiad

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 $Z^{\prime }$ on circle $ABC$ such that $A{Z}^{\prime }\parallel BC$. By the previous result, it suffices to prove that $CDX{Z}^{\prime }$ is cyclic. Notice that triangles $BAE$ and $C{Z}^{\prime }D$ are reflections in the perpendicular bisector of $BC$. Using this and that $A,O,E$ are collinear: $$\angle D{Z}^{\prime }C=\angle BAE=\angle BAO=90^{\circ }-\frac{1}{2}\angle AOB=90^{\circ }-\angle C=\angle DXC,$$ so $DX{Z}^{\prime }C$ is cyclic, giving $Z\equiv Z^{\prime }$ 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 $AXY$.

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Alternative solution.

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 $AXY$.