Let $A(2x, 0)$ and $C(0, 2y)$. Note that $AC$ is the diagonal of the square $ABCD$. The midpoint of $AC$ is the center of the square, which we will label $E(x, y)$. We can translate this system by adding vector $(-x, -y)$ so that the center $E'$ is on the origin. This gives $A'(x, -y), C'(-x, y)$, and $E'(0, 0)$. We now want to rotate $A'$ by $90$ degrees CCW about the origin to get $B'$, which is $B'(y, x)$. Similarly, rotating $C$ gives $D'(-y, -x)$. Moving $B'(y, x)$ and $D'(-y, -x)$ back into our original coordinate system by subtracting the vector $(-x, -y)$ gives $B(x + y, x + y)$ and $D(x-y, y-x)$. Point $B$ is on $y=x$, while point $D$ is on $y=-x$.