Equilateral $\triangle ABC$ has side length $4$, and $M$ is the midpoint of $\overline{AC}$. Points $B,C,D$ are collinear, with $C$ between $B$ and $D$ and $CD=2BC$. What is the area of $\triangle CDM$?

1. Since $CD=2BC$ and the triangles $CDM$ and $CBM$ have the same altitude to line $BD$, $[CDM]=2[CBM]$.
2. Because $M$ is the midpoint of $AC$, $[CBM]=\frac12[ABC]$. Thus $[CDM]=[ABC]=\frac{\sqrt3}4\cdot4^2=\boxed{4\sqrt3}$.