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.


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 $\triangle ZCD$.
