smc

First, let's draw a few auxiliary lines. Drop altitudes from $O$ to $AB$ and from $O$ to $BC$. We can label the points as $J$ and $K$, respectively. This forms square $OKBJ$. Connect $OB$. Without loss of generality, set the side length of the square $ABCD$ equal to $1$. Let $NK=x$, and since $N$ is the midpoint of $CB$, $KB$ would be $\frac{1}{2}-x$. With the same reasoning, $MJ=x$ and $JB=\frac{1}{2}-x$ We can also see that $△COK$ is similar to $△CMB$. That means $\frac{CK}{OK}=\frac{CB}{MB}$. Plugging in the values, we get: $\frac{\frac{1}{2}+x}{\frac{1}{2}-x}=\frac{1}{\frac{1}{2}}$. Solving, we find that $x=\frac{1}{6}$. Then, $OK=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$. The area of $△COB$ and $△AOB$ together would be $2(\frac{1}{2}\cdot 1\cdot \frac{1}{3})=\frac{1}{3}$. Subtract this area from the total area of $ABCD$ to get the area of $AOCD$. So, $AOCD=1-\frac{1}{3}=\frac{2}{3}$. The question asks for the ratio of the area of $AOCD$ to the area of $ABCD$, which is $\frac{\frac{2}{3}}{1}=\frac{2}{3}$. The answer is $\boxed{\frac{2}{3}}$.