No squares... Complex bash: Let $(ABZ)$ be a unit circle. Note that $O$ lies on perpendicular bisectors to $CD$ and $XY$, so $O$ is center of $(CDXY)$ and so $OC=OX$. From squares we have $c=(a-b)i+b$ and $x=(z-a)i+a$. $OC^2=c\overline{c}=x\overline{x}=OX^2$, so after some easy manipulations we can get that $(b+iz)(a^2-bz)=0$. If $a^2=bz$ then $AB=AZ$ and so $AB=AX$. If $b+iz=0$ then $z=bi$. We want to prove that $AC\perp{}XY \parallel AZ$, so we want to prove that $C,A,-Z$ are collinear, there $-Z$ is antipode of $Z$. Note that we want to prove that $c-az\overline{c}=a-z$ or $\frac{c-a}{a\overline{c}-1}=z=bi$. But $\frac{c-a}{a\overline{c}-1}=\frac{(a-b)i-(a-b)}{a\frac{ai-bi+a}{ab}-1}=\frac{b(a-b)(i-1)}{(a-b)(i+1)}=bi$ and so we are done. $\blacksquare$