jmc

First we draw diagonal $\overline{BD}$, and let the diagonals intersect at $T$, as shown:



Since $\overline{CT}$ and $\overline{BM}$ are medians of $△BCD$, point $O$ is the centroid of $△BCD$, so $OC=(2/3)CT$. Since $T$ is the midpoint of $\overline{AC}$, we have $CT=AC/2$, so $OC=(2/3)CT=(2/3)(AC/2)=AC/3$. Since $\overline{OC}$ is $\frac{1}{3}$ of $\overline{AC}$, we know that $\overline{OA}$ is the other $\frac{2}{3}$ of $\overline{AC}$, which means $OC/OA=\boxed{\frac{1}{2}}$.