Solution:

To get a conclusion of the appropriate type (a rectangle OR a rhombus), we must divide up the problem into two cases. Here is one way of accomplishing this:

Case 1. The circumcenter $O$ of $ABC$ is the center of $ABCD$, the common midpoint of diagonals $AC$ and $BD$. Then since radii $OA$ and $OB$ are equal, we get $AC = 2OA = 2OB = BD$. Thus $ABCD$ is a parallelogram whose diagonals are congruent, i.e. a rectangle.

Case 2. The circumcenter $O$ of $ABC$ does not coincide with the midpoint $M$ of $AC$ and $BD$. Then since $O$ is on the perpendicular bisector of $AC$, we have $OM \perp AC$. But $O$ and $M$ are both on line $BD$, so $BD \perp AC$. Thus $ABCD$ is a parallelogram whose diagonals are perpendicular, i.e. a rhombus.