It is easy to see that $ABCD$ is cyclic, whose circumcircle has diameter $BD$. Let the circumcircle be $x^2+y^2=1$, and $B(1,0),D(-1,0)$. Let $A(a,b),C(c,d)$, where $a,b,c,d$ are real numbers such that $a^2+b^2=c^2+d^2=1$. The equation of the circumcircle with diameter $AB$ is given by: $\left(x-\frac{a+1}{2}\right)^2+\left(y-\frac{b}{2}\right)^2=\frac{(a-1)^2}{4}+\frac{b^2}{4}$ This is equivalent to $x^2-(a+1)x+y^2-by+a=0$ The equation of the circumcircle with diameter $CD$ is given by: $\left(x-\frac{c-1}{2}\right)^2+\left(y-\frac{d}{2}\right)^2=\frac{(c+1)^2}{4}+\frac{d^2}{4}$ This is equivalent to $x^2-(c-1)x+y^2-dy-c=0$ Subtracting the two equations of circumcircles gives us line $XY$: $-(a+1)x+(c-1)x-by+dy+a+c=0$ It is easy to show that the midpoint of $AC$, which is $\left(\frac{a+c}{2},\frac{b+d}{2}\right)$, lies on this line.