Let $ABCD$ be a convex quadrilateral which is not a parallelogram. Draw the line joining the intersection points of the perpendicular bisectors of the two pairs of the opposite sides of $ABCD$. Is it true that this line is parallel to the radical axis of the circles whose diameters are the diagonals of the quadrilateral?
I think somehow this is the case, leading to the fact that the line joining the two intersection points arisen by the four perpendicular bisectors is therefore perpendicular to the common chord of the two above-mentioned circles.
Thanks for any comment, hint, or solution!