Prove that these two lines are perpendicular. Consider a parallelogram $WXYZ$, with points $A$ and $B$ on sides $WX$ and $XY$ respectively, so that $\angle WAZ = \angle YBZ$. Let the midpoint of $WY$ be $M$. Prove that $OM$, where $O$ is the centre of the circle $AXB$, is perpendicular to $WY$.

EDIT: In response to Mick's solution. I think you need to explain why the equal angles means the two lines are parallel. I think your solution breaks down when you say that KMN is a straight line without proof. Here is a picture where your first two paragraphs are correct, but doesn't solve the problem because the original angles are not the same. 

 A: Let there be a red circle passing through $\triangle WXY$ with its circum-center at K. This circle will cut the blue circle (the circum-circle that passes through $\triangle AXB$ with center O) at G.

Since M is the midpoint of the chord WY, $KM \bot WY$.
Since KO (extended) is the line of centers and GX is the common chord of the red and blue circles, we get $KO \bot GX$ at N and N is the midpoint of GX.
From the above - (1) KMN is a straight line joining the centers to the midpoints of two parallel chords; and (2) KON is the line of centers and therefore is also a straight line, we can say that KMON is straight line. Result follows because $\angle KMY$ is right-angled.
Remarks:-
1) The green dotted line is the angle bisector of $\angle WZY$ in order to have $\angle WAZ = \angle YBZ$.
2) The dotted circle can be disregarded.
3) $\angle WAZ$ [edit: $= \angle YBZ$] restricts how the blue circle can be drawn but has no other bearings on the conclusion. [edit: The same color shaded angles will of course vary accordingly but they also have no bearings on the conclusion.]
