2. 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) $\frac{\sqrt{3}}{9}$.
2. (A).

Take the midpoint $N$ of $AD$, and connect $BN, MN$. Then, by the midline theorem, we know $MN \parallel AC, MN = \frac{AC}{2} = \frac{\sqrt{3}}{2}$. 
$\angle BMN$ 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 height from $D$ to $DC$ in the equilateral $\triangle BCD$, $BM = \frac{\sqrt{3}}{2} \times \sqrt{2} = \frac{\sqrt{6}}{2}$. Applying the cosine rule in $\triangle BMN$, we get
$$
\begin{array}{l}
\cos \alpha = \cos \angle BMN \\
= \frac{BM^2 + MN^2 - BN^2}{2 \times BM \times MN} \\
= \frac{\left(\frac{\sqrt{6}}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{5}}{2}\right)^2}{2 \times \frac{\sqrt{6}}{2} \times \frac{\sqrt{3}}{2}} \\
= \frac{\sqrt{2}}{3}.
\end{array}
$$