Assume without loss of generality that $A$ is located on the $x$-axis and $C$ is located on the $y$-axis, let these points have coordinates of $A (a, 0)$ and $C (0, c)$. As the diagonals of a square bisect each other, we know that the intersection point $P$ of diagonal is also the mid-point of $AC$, i.e. $P(\frac{a}{2}, \frac{c}{2})$ and $\vec{PC} = (-\frac{a}{2}, \frac{c}{2})$.

As the diagonals of a square are perpendicular to each other and of the same length, the vectors $\vec{PB}$ and $\vec{PD}$ have the same length as the vector $\vec{PC}$ and are perpendicular to it. But for a given vector $\vec{u} = (s, t)$, there are exactly two vectors perpendicular to and having the same length as it: $\vec{v} = (-t, s)$ and $-\vec{v} = (t, -s)$. For the vector $\vec{u} = \vec{PC} = (-\frac{a}{2}, \frac{c}{2})$ we get $\vec{v} = (\frac{c}{2}, -\frac{a}{2})$ and w.l.o.g. we can assume that $\vec{PB} = \vec{v}$ and $\vec{PD} = -\vec{v}$. Now from here $B(\frac{a+c}{2}, \frac{a+c}{2})$ and $D(\frac{a-c}{2}, \frac{c-a}{2})$. Thus, we see that the point $B$ is located on the line $y = x$ and point $D$ is located on the line $y = -x$.

W.l.o.g., assume that the vertices of the square are labelled counter-clockwise with $A (a, 0)$, $C (0, c)$, where $a, c \ge 0$ (other cases are similar). Let $O$ be the origin, then $\angle AOC = 90^\circ$, i.e. the circumcircle (with diameter $AC$) of the square $ABCD$ passes through the origin $O$. Based on the assumptions made, $B$ definitely lies in the first quadrant and $D$ has to lie in the second quadrant (Fig. 4) or in the fourth quadrant (Fig. 5), otherwise the circle with the diameter $BD$ cannot pass the origin.

Fig. 4

Fig. 5

diameter *AC*) of the square *ABCD* passes through the origin *O*. Now note that the vertices of the square divide its circumcircle into four equal arcs of $90^\circ$, each having an inscribed angle of $45^\circ$ subtending on it. Thus, $\angle AOB = \angle BOC = 45^\circ$, i.e., $B$ lies on the line with equation $y = x$ (if $A = O$ or $C = O$, then one of those angles will lose its meaning, however, the other one is still $45^\circ$ and that is sufficient). Similarly, $\angle COD = 45^\circ$, if $D$ lies in the second quadrant, or $\angle AOD = 45^\circ$, if $D$ lies in the fourth quadrant. In both cases, $D$ lies on the line $y = -x$; this condition is also met in the special case $D = O$.