jmc

We first draw a diagram: Since $F$ is the midpoint of $\overline{AB}$, the area of $△AFC$ is half of the area of $△ABC,$ or 12 square units. Following the same reasoning, we see that $E$ is the midpoint of $\overline{AC},$ so the area of $△CEF$ is half that of $△AFC,$ or $\boxed{6}$ square units.