18.84 As shown in the 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 quadrilateral $AOC D$ to the area of $ABCD$ is 
(A) $5: 6$.
(B) $3: 4$.
(C) $2: 3$.
(D) $\sqrt{3}: 2$.
(E) $(\sqrt{3}-1): 2$.
(26th American High School Mathematics Examination, 1975)
[Solution] Connect $A C, D B$. Since $A B C D$ is a square, $A C$ and $D B$ bisect each other perpendicularly at $P$. If $a$ represents the side length of the square, then
$$
A C=D B=a \sqrt{2} \text {. }
$$

Since $O$ is the intersection of the three medians of $\triangle A B C$, we have
$$
O P=\frac{1}{3} \cdot \frac{\sqrt{2}}{2} a \text {. Thus, } S_{\triangle A O C}=\frac{a^{2}}{6} .
$$

Therefore, $S_{A C X D}=\frac{a^{2}}{2}+\frac{a^{2}}{6}=\frac{2}{3} a^{2}$. Hence, the required ratio is
$$
2: 3 \text {. }
$$

Therefore, the answer is $(C)$.