Denote the lengths $A B, B C, C D$, and $D A$ by $a, b, c$, and $d$ respectively. Because $A B C D$ is cyclic, $\triangle A B X \sim \triangle D C X$ and $\triangle A D X \sim \triangle B C X$. It follows that $\frac{A X}{D X}=\frac{B X}{C X}=\frac{a}{c}$ and $\frac{A X}{B X}=\frac{D X}{C X}=\frac{d}{b}$. Therefore we may write $A X=a d k, B X=a b k, C X=b c k$, and $D X=c d k$ for some $k$. Now, $\angle X D C=\angle B A X=\angle Y X B$ and $\angle D C X=\angle X B Y$, so $\triangle B X Y \sim \triangle C D X$. Thus, $X Y=$ $D X \cdot \frac{B X}{C D}=a b d k^{2}$. Analogously, $X Y=a c d k^{2}$. Note that $X Y / X Z=C B / C D$. Since $\angle Y X Z=\pi-\angle Z A \stackrel{c}{Y}=\angle B C D$, we have that $\triangle X Y Z \sim \triangle C B D$. Thus, $Y Z / B D=X Y / C B=a d k^{2}$. Finally, Ptolemy's theorem applied to $A B C D$ gives $$(a d+b c) k \cdot(a b+c d) k=a c+b d$$ It follows that the answer is $$\frac{a d(a c+b d)}{(a b+c d)(a d+b c)}=\frac{20 \cdot 23}{22 \cdot 26}=\frac{115}{143}$$