Let \(ABC\) be an acute-angled triangle with \(AC > AB\), let \(O\) be its circumcentre, and let \(D\) be a point on the segment \(BC\). The line through \(D\) perpendicular to \(BC\) intersects the lines \(AO, AC,\) and \(AB\) at \(W, X,\) and \(Y,\) respectively. The circumcircles of triangles \(AXY\) and \(ABC\) intersect again at \(Z \ne A\).
Prove that if \(W \ne D\) and \(OW = OD,\) then \(DZ\) is tangent to the circle \(AXY.\)
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 \(A Z | B C\). To do this, introduce point \(Z^{\prime}\) on circle \(ABC\) such that \(AZ^{\prime} | BC\). By the previous result, it suffices to prove that \(CDXZ^{\prime}\) is cyclic. Notice that triangles \(BAE\) and \(CZ^{\prime} D\) are reflections in the perpendicular bisector of \(BC\). Using this and that \(A,O,E\) are collinear:\[\angle DZ^{\prime}C = \angle BAE = \angle BAO = 90^{\circ} - \frac{1}{2} \angle AOB = 90^{\circ}-\angle C = \angle DXC\]
so \(DXZ^{\prime}C\) is cyclic, giving \(Z \equiv Z^{\prime}\) as desired. Using \(AZ | BC\) 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 \(AXY\).