Trapezoid $ABCD$ has $AD||BC$, $BD = 1$, $\angle DBA = 23^{\circ}$, and $\angle BDC = 46^{\circ}$.  The ratio $BC: AD$ is $9: 5$.  What is $CD$?

$\mathrm{(A)}\ \frac 79\qquad \mathrm{(B)}\ \frac 45\qquad \mathrm{(C)}\ \frac {13}{15}\qquad \mathrm{(D)}\ \frac 89\qquad \mathrm{(E)}\ \frac {14}{15}$
Solution 1
Extend $\overline {AB}$ and $\overline {DC}$ to meet at $E$.  Then
\begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*}
Thus $\triangle BDE$ is isosceles with $DE = BD$.  Because $\overline {AD} \parallel \overline {BC}$, it follows that the triangles $BCE$ and $ADE$ are similar.  Therefore
\[\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,\]
so $CD = \boxed{\frac 45}.$

Solution 2
Let $E$ be the intersection of $\overline {BC}$ and the line through $D$ parallel to $\overline {AB}.$  By construction $BE = AD$ and $\angle BDE = 23^{\circ}$; it follows that $DE$ is the bisector of the angle $BDC$. So by the [Angle Bisector Theorem](https://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem) we get
\[CD = \frac {CD}{BD} = \frac {EC}{BE} = \frac {BC - BE}{BE} = \frac {BC}{AD} -1 = \frac 95 - 1 = \boxed{\frac 45}.\]
The answer is $\mathrm{(B)}$.