Rotating the 4 sides:
Given ABCD, rotate AB by $\theta$ to get A'B', and rotate CD by $\theta$ to get C'D'.
(When we speak of rotating a segment by $\theta$, we mean rotating it by an angle of $\theta$, counter-clockwise around its midpoint.)
Then rotate the other two sides: rotate B'C' by $\theta$ to get B''C'', and rotate D'A' by $\theta$ to get D''A''.
To Prove:
The diagonals have rotated: (but not around their midpoints)
Theorem 1: Diagonal A''C'' is at an angle of $\theta$ from AC.
(And similarly for B''D'' and BD.)
They are rotating around a point on the line:
Theorem 2: The point A''C'' $\cap$ AC is independent of $\theta$.
Theorem 3: If the diagonals were perpendicular at P, then the new diagonals are also perpendicular at P.
Comments:
I have a proof but it is a mess of coordinates and trigonometry, completely unenlightening.
I would like a geometrical proof.
Rotation is such a simple operation to do, I am surprised I have never heard of any theorem like this.