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
[asy]
draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0));
label("A", (0,0), S);
label("B", (2,0), S);
label("C", (2,2), N);
label("D", (0,2), N);
label("M", (1,0), S);
label("N", (2,1), E);
label("O", (1.2, .8));
[/asy]
$ \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} $
1. **Define the square and midpoints:**
   - Let the side length of square \(ABCD\) be \(s\).
   - Coordinates of vertices: \(A(0,0)\), \(B(s,0)\), \(C(s,s)\), \(D(0,s)\).
   - Midpoint \(M\) of \(AB\) is \(\left(\frac{s}{2}, 0\right)\).
   - Midpoint \(N\) of \(BC\) is \(\left(s, \frac{s}{2}\right)\).

2. **Find the equations of lines \(AN\) and \(CM\):**
   - Slope of \(AN\) is \(\frac{\frac{s}{2} - 0}{s - 0} = \frac{1}{2}\).
     - Equation of \(AN\): \(y = \frac{1}{2}x\).
   - Slope of \(CM\) is \(\frac{s - 0}{s - \frac{s}{2}} = 2\).
     - Equation of \(CM\): \(y - s = 2(x - s)\) or \(y = 2x - s\).

3. **Find the intersection point \(O\) of \(AN\) and \(CM\):**
   - Set the equations equal: \(\frac{1}{2}x = 2x - s\).
   - Solve for \(x\):
     \[
     \frac{1}{2}x = 2x - s \implies s = \frac{3}{2}x \implies x = \frac{2s}{3}
     \]
   - Substitute \(x = \frac{2s}{3}\) into \(y = \frac{1}{2}x\):
     \[
     y = \frac{1}{2} \left(\frac{2s}{3}\right) = \frac{s}{3}
     \]
   - Coordinates of \(O\) are \(\left(\frac{2s}{3}, \frac{s}{3}\right)\).

4. **Calculate the area of quadrilateral \(AOCD\):**
   - Use the Shoelace Theorem or divide into triangles.
   - Vertices of \(AOCD\): \(A(0,0)\), \(O\left(\frac{2s}{3}, \frac{s}{3}\right)\), \(C(s,s)\), \(D(0,s)\).
   - Area using Shoelace Theorem:
     \[
     \text{Area} = \frac{1}{2} \left| 0 \cdot \frac{s}{3} + \frac{2s}{3} \cdot s + s \cdot s + 0 \cdot 0 - \left(0 \cdot \frac{2s}{3} + \frac{s}{3} \cdot s + s \cdot 0 + s \cdot 0\right) \right|
     \]
     \[
     = \frac{1}{2} \left| 0 + \frac{2s^2}{3} + s^2 + 0 - \left(0 + \frac{s^2}{3} + 0 + 0\right) \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{2s^2}{3} + s^2 - \frac{s^2}{3} \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{2s^2}{3} + \frac{3s^2}{3} - \frac{s^2}{3} \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{4s^2}{3} \right|
     \]
     \[
     = \frac{2s^2}{3}
     \]

5. **Calculate the area of square \(ABCD\):**
   - Area of \(ABCD\) is \(s^2\).

6. **Find the ratio of the areas:**
   \[
   \text{Ratio} = \frac{\text{Area of } AOCD}{\text{Area of } ABCD} = \frac{\frac{2s^2}{3}}{s^2} = \frac{2}{3}
   \]

The final answer is \(\boxed{\frac{2}{3}}\)