Suppose otherwise. Let $M_{AB},...,M_{DA}$ be the four midpoints. We have the four cyclic quadrilaterals $AM_{AB}O_CM_{DA}$, and the other similar ones. Therefore $\measuredangle A=\pi-\measuredangle M_{DA}O_CM_{AB}=\measuredangle O_DO_CO_B=\measuredangle O_DO_AO_B=\measuredangle M_{BC}O_AM_{CD}=\pi-\measuredangle C$ However this implies that $ABCD$ is cyclic, which in turn implies that the four circumcenters coincide with each other and with the circumcenter of $ABCD$, which is a contradiction.

Solution during the test. Let $M_{IJ}$ be the midpoint of side $IJ$ in quadrilateral $ABCD$. Because $O_A$ and $O_C$ are in the perpendicular bisectors of $BD$, we have $M_{BD},O_A,O_C$ are collinear. Using the same argument, we get $O_D,O_A, M_{CD}$ collinear and for the other two sides. Then $O_AM_{BC} \perp BC$ and $O_AM_{CD} \perp CD$. Hence $O_B, M_{BC}, C, M_{CD}$ are concyclic and $O_A, M_{AD}, D, M_{CD}$ are also concyclic. Therefore, $\measuredangle O_B O_A D = \measuredangle BCD$ and $\measuredangle O_BO_CD = \measuredangle DAB$. It follows that $O_AO_BO_CO_D$ is cyclic if and only if $ABCD$ is cyclic. But if $ABCD$ is cyclic, then the four perpendicular bisector lie on one single point, and that contradicts the statement. $\blacksquare$

Notice that the perpendicular bisectors of $AB, BC, CD, AD$ are, respectively, the lines $O_CO_D, O_AO_D, O_AO_B, O_BO_C$. Hence, rotating $ABCD$ 90 degrees (direction doesn't matter) gives us $A'B'C'D'$ such that $A'B'\parallel O_CO_D, B'C'\parallel O_AO_D, C'D'\parallel O_AO_B, A'D'\parallel O_BO_C$. This, in turn, means that $A'B'C'D'$ shares the same angles, in order, as $O_CO_DO_AO_B$. As $A'B'C'D'$ also shares angles with $ABCD$, we get that, if $O_CO_DO_AO_B$ is cyclic, then $ABCD$ is cyclic, which would imply $O_A=O_B=O_C=O_D$, contradiction.