23.9 In tetrahedron $ABCD$, $BD = DC = CB = \sqrt{2}$, $AC = \sqrt{3}$, $AB = AD = 1$. $M$ is the midpoint of $CD$. Then the cosine of the angle formed by the skew lines $BM$ and $AC$ is 
(A) $\frac{\sqrt{2}}{3}$.
(B) $\frac{\sqrt{2}}{2}$.
(C) $\frac{\sqrt{6}}{2}$.
(D) $\frac{\sqrt{3}}{9}$.
(China Sichuan Province High School Mathematics Competition, 1993)
[Solution] Take the midpoint $N$ of $AD$, and connect $BN, MN$. Then, by the midline theorem, we know $MN \parallel AC, \quad MN=\frac{AC}{2}=\frac{\sqrt{3}}{2}$.
$\angle BMN$ is the angle formed by the skew lines $BM$ and $AC$.

Let $\angle BMN=\alpha$.
From the side lengths of $\triangle ABD$, we know $\angle BAD=90^{\circ}$.
In the right $\triangle ABN$,
$$
BN=\sqrt{1^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{5}}{2},
$$
$BM$ is the altitude of the equilateral $\triangle BCD$ on side $DC$, $BM=\frac{\sqrt{3}}{2} \cdot \sqrt{2}=\frac{\sqrt{6}}{2}$. Applying the cosine rule in $\triangle BMN$ yields
$$
\begin{aligned}
\cos \alpha & =\cos \angle BMN=\frac{BM^{2}+MN^{2}-BN^{2}}{2 \cdot BM \cdot MN} \\
& =\frac{\left(\frac{\sqrt{6}}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}}{2 \cdot \frac{\sqrt{6}}{2} \cdot \frac{\sqrt{3}}{2}} \\
& =\frac{\sqrt{2}}{3} .
\end{aligned}
$$

Therefore, the answer is $(A)$.