In the adjoining figure $AB$ and $BC$ are adjacent sides of square $ABCD$; $M$ is the midpoint of $AB$; 
$N$ is the midpoint of $BC$; and $AN$ and $CM$ intersect at $O$. The ratio of the area of $AOCD$ to the area of $ABCD$ is

$\textbf{(A)}\ \frac{5}{6}\qquad \textbf{(B)}\ \frac{3}{4}\qquad \textbf{(C)}\ \frac{2}{3}\qquad \textbf{(D)}\ \frac{\sqrt{3}}{2}\qquad \textbf{(E)}\ \frac{(\sqrt{3}-1)}{2}$
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 $\triangle COK$ is similar to $\triangle CMB$. 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 $\triangle COB$ and $\triangle AOB$ 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 $\boxed{\textbf{(C)}\ \frac{2}{3}}$. ~[jiang147369](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Jiang147369)