Show that XY is parallel to CD 
AB is parallel to CD. CD is not a diameter.
I want to show that $\triangle ZCD$ is similar to $\triangle ZXY$ but I don't know how to get there. The only thing I had in mind was using the arcs, but that didn't get me anywhere.
 A: 
Since $AB\parallel CD$, the small arc $AC$ is equal to the small arc $BD$. It follows that $$\angle PXQ=\angle PYQ.$$ 
Thus $\triangle ZXP$ is similar to $\triangle ZYQ$ (having two equal pairs of angles), and we get
$$ZX:ZY=ZP:ZQ.$$
It follows that $\triangle ZXY$ is similar to $\triangle ZPQ$, and so is similar to $\triangle ZCD$.
A: The statement above is an immediate consequence of Pascal's theorem whose statement is as follows

Let on a circle (on a conic section) be six distinct points given and arbitrarily named as $X,Y,U$ and $X',Y',U'$. Let's connect $X$ and $Y'$ and $Y$ and $X'$ with straights and name their intersection  by $xy$. Let's connect  $X$ and $U'$ and $U$ and $X'$ with straights and name their intersection  by $xu$, finally connect $Y$ and $U'$ and $U$ and $Y'$ with straights and name their intersectionby $yu$. Now, $xy$, $xu$, and $yu$ are collinear as illustratd in the figure below


If the point $X'$ moves upward then the $xy$ moves to the right, the point $xy$ disappears o the right side and the shows up on the left side. Between the two situations there one instant when $XY'$ and $YX'$ are parallel.
If the straights connecting $X$ and $Y'$ and $Y$ and $Y'$ are parallel (their intersection point, $xy$  is in the infinity) then the line connecting $xu$ and $yu$ has to meet $xy$ in the infinity. That is the straights $XY'$, $Y'X$, and the one connecting $xu$ and $yu$ are parallel. See, the figure below:

Apply Pascal's theorem for the original figure with $A=X$, $B=Y'$, $C=Y$, $D=Y'$, $P=U$, $Q=U'$, $yu=X$, and $xu=Y$. (Sorry for using the notations $X$ and $Y$ twice. I hope this will not cause any problem.)
Obviously, the triangles $Z\ Y\ X'$ and $yu \ xu \ Z$ are similar.
