AHSME 1975 Problem 23

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

$\textbf{(A)}\ \frac{5}{6} \textbf{(B)}\ \frac{3}{4} \textbf{(C)}\ \frac{2}{3} \textbf{(D)}\ \frac{\sqrt{3}}{2} \textbf{(E)}\ \frac{(\sqrt{3}-1)}{2}$