2004 AMC 10B Problem 20

Let $F$ be on $\overline{AC}$ with $DF \parallel BE,$ and write $ET = x,$ $BT = 4x.$ From $\triangle ATE \sim \triangle ADF,$ $\dfrac{DF}{x} = \dfrac{AD}{AT} = \dfrac{4}{3},$ so $DF = \dfrac{4x}{3}.$ From $\triangle BEC \sim \triangle DFC,$ $\dfrac{CD}{BC} = \dfrac{DF}{BE} = \dfrac{\frac{4x}{3}}{5x} = \dfrac{4}{15}.$ Therefore $$\begin{aligned} \dfrac{CD}{BD} &= \dfrac{\frac{CD}{BC}}{1 - \frac{CD}{BC}} \\ &= \dfrac{\frac{4}{15}}{\frac{11}{15}} = \dfrac{4}{11}. \end{aligned}$$ Thus, the correct answer is D.