I guess...?
Invert with negative power with respect to the intersection of the diagonals of the quadrilateral in such a way that the circumcircle is preserved. We get that the four circles' radii are preserved by this map, so the four circles all have the same distance from the intersection of the diagonals. It follows that the centers of these four circles form a square (since they are concyclic and have diagonals which are perpendicular to each other at their center). But then, by symmetry arguments, the diagonals of \(ABCD\) are perpendicular, and \((ABCD)\) is tangent to all the small circles which can imply that it is symmetric over the angle bisectors of the diagonals, which gives us our conclusion that \(ABCD\) is a square.