Let $O_1$ be the $O$-antipode in the circumcircle of $\triangle OAD$, and define $O_2$ similarly. Since $\angle O_1AD=\angle OBC$, $\angle O_1DA=\angle OCB$, we have $\triangle O_1AD\sim \triangle OBC$. Similarly, $\triangle DAO\sim \triangle CBO_2$, so $AO_1DO$ and $BOCO_2$ are similar cyclic quadrilaterals. By "Mean Geometry," the quadrilateral formed by the midpoints of corresponding points in the two similar cyclic quads is similar to both of them, so the relevant quadrilateral is cyclic. We're done.