IMO 1995- Problem 1

Problem: Let $A,B,C,D$ be distinct points on a line, in that order. The circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$. $O$ is an arbitrary point on line $XY$ but not on $AD$. $CO$ intersects the circle with diameter $AC$ again at $M$, and $BO$ intersects the other circle again at $N$. Prove that $AM,DN$ and $XY$ are concurrent.
Solution: The result is trivial if $O$ coincides with $X$ or $Y$. Suppose it does not. From $ON.ON=OX.OY=OC.OM,$
$BCMN$ is cyclic. If $O$ is on the segment $XY$, then
$\angle{MAD}+\angle{MNB}+\angle{BND}=\angle{MAD}+\angle{MCA}+\angle{AMC}=180^0$.
If $O$ is not on the segment $XY$, then
$\angle{MAD}=180^0-\angle{AMC}-\angle{MAC}=180^0-\angle{BND}-\angle{ONM}=\angle{MND}$.
In either case, $ADNM$ is also cyclic. Let $AM$ and $DN$ intersect at $Z$. Let the line $ZX$ intersect the first circle at $Y_1$ and the second at $Y_2$. Then $ZX.ZY_1=ZA.ZM=ZD.ZN=ZX.ZY_2$.
Hence $Y_1=Y_2=Y$ and indeed $Z$ lies on $XY$.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf