jmc

Let $F$ be a point on $\overline{AC}$ such that $\overline{DF}$ is parallel to $\overline{BE}$. Let $BT=4x$ and $ET=x$.



Because $△ATE$ and $△ADF$ are similar, we have $$\frac{DF}{x}=\frac{AD}{AT}=\frac{4}{3},$$and $$DF=\frac{4x}{3}.$$Also, $△BEC$ and $△DFC$ are similar, so $$\frac{CD}{BC}=\frac{DF}{BE}=\frac{4x/3}{5x}=\frac{4}{15}.$$Thus $$\frac{CD}{BD}=\frac{CD/BC}{1-(CD/BC)}=\frac{4/15}{1-4/15}=\boxed{\frac{4}{11}}.$$