18.87 As shown, $ABCD$ is a $2 \times 2$ square, $E$ is the midpoint of $AB$, $F$ is the midpoint of $BC$, $AF$ and $DE$ intersect at $I$, and $BD$ and $AF$ intersect at $H$. Then the area of quadrilateral $BEIH$ is 
(A) $\frac{1}{3}$.
(B) $\frac{2}{5}$.
(C) $\frac{7}{15}$.
(D) $\frac{8}{15}$.
(E) $\frac{3}{5}$.
(42nd American High School Mathematics Examination, 1991)
[Solution] It is easy to prove $\triangle A B F \cong \triangle D A E$, then $\angle A D E = \angle B A F$, and let it be $\alpha$. It can also be obtained that $\angle A I E = 90^{\circ}$.
Then $\sin \alpha = \frac{A E}{E D} = \frac{1}{\sqrt{1^{2} + 2^{2}}} = \frac{1}{\sqrt{5}}$, so $\cos \alpha = \frac{2}{\sqrt{5}}$. From the given, we have $S_{\triangle A B D} = 2, S_{\triangle D B E} = 1$.
By
$$
I D = A D \cos \alpha = \frac{4}{\sqrt{5}} \text{. }
$$

It is also easy to prove $D H = \frac{2}{3} B D = \frac{4 \sqrt{2}}{3}$.
(From the figure on the next page, it can be seen that $H$ is the midpoint of $B G$, so $H, G$ trisect $B D$).
Then
$$
\begin{array}{l}
S_{\triangle D H H} = \frac{1}{2} \cdot I D \cdot D H \sin \angle I D H \\
= \frac{1}{2} \cdot \frac{4}{\sqrt{5}} \cdot \frac{4 \sqrt{2}}{3} \sin \left(45^{\circ} - \alpha\right) \\
= \frac{1}{2} \cdot \frac{4}{\sqrt{5}} \cdot \frac{4 \sqrt{2}}{3} \cdot \frac{\sqrt{2}}{2} \left(\frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}}\right) \\
= \frac{8}{15} .
\end{array}
$$

Thus, $S_{B E I H} = 1 - \frac{8}{15} = \frac{7}{15}$. Therefore, the answer is $(C)$.