smc

Extend $\overline{AB}$ and $\overline{DC}$ to meet at $E$. Then $$\begin{array}{rl}\angle BED& =180^{\circ }-\angle EDB-\angle DBE\\ & =180^{\circ }-134^{\circ }-23^{\circ }=23^{\circ }.\end{array}$$ Thus $△BDE$ is isosceles with $DE=BD$. Because $\overline{AD}\parallel \overline{BC}$, it follows that the triangles $BCE$ and $ADE$ are similar. Therefore $$\frac{9}{5}=\frac{BC}{AD}=\frac{CD+DE}{DE}=\frac{CD}{BD}+1=CD+1,$$ so $CD=\boxed{\frac{4}{5}}.$