Solution:
Assume without loss of generality that the vertices $A, B, C, D$ are labeled in counterclockwise order. Let $A$ and $C$ have the coordinates $(a, 0)$ and $(0, c)$ respectively. The center $M$ of the square is the midpoint of $AC$ and therefore has the coordinates
$$\left(\frac{a}{2}, \frac{c}{2}\right).$$

To get from $M$ to $C$, we can move upward $c/2$ units and then leftward $a/2$ units (of course, these are signed distances, indicating movement in the opposite directions if they are negative). Since $B$ is the $90^{\circ}$ clockwise rotation of $C$ around $M$, we can get from $M$ to $B$ by moving rightward $c/2$ units and then upward $a/2$ units. This takes us from $M$ to the point
$$\left(\frac{a}{2} + \frac{c}{2}, \frac{c}{2} + \frac{a}{2}\right)$$
which clearly lies on the line $y = x$.