2009 AMC 12B Problem 16

Draw the line through $D$ parallel to $AB,$ meeting $BC$ at $E,$ so $ABED$ is a parallelogram with $BE = AD.$ Then $\angle BDE = \angle DBA = 23^\circ,$ and since $\angle BDC = 46^\circ,$ segment $DE$ bisects $\angle BDC.$ By the angle bisector theorem in $\triangle BDC,$ $\dfrac{EC}{BE} = \dfrac{DC}{DB},$ so $$\begin{aligned} CD &= DB \cdot \dfrac{BC - AD}{AD} \\ &= 1 \cdot \left(\dfrac{9}{5} - 1\right) \\ &= \dfrac{4}{5}. \end{aligned}$$ Thus, the correct answer is B.