1 In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of $A B$, and $\triangle A C M$ is folded along $C M$ such that the distance between $A$ and $B$ is $2 \sqrt{2}$. At this moment, the volume of the tetrahedron $A-B C M$ is $\qquad$ (1998, National High School League)
Solution: The tetrahedron $A-BCM$ after folding is shown in Figure 4. Take the midpoint $D$ of $CM$, and connect $AD$. It is easy to see that $AD \perp CM$. In $\triangle BCM$, draw $DE \perp CM$ intersecting $BC$ at point $E$, and connect $AE$. We know that $AE \perp CM$. Then $AD=\sqrt{3}$, $DE=CD \tan 30^{\circ}=\frac{\sqrt{3}}{3}, CE=2 DE=\frac{2 \sqrt{3}}{3}$.
In $\triangle ABC$,
$$
\begin{array}{l}
\because AC=2, AB=2 \sqrt{2}, BC=2 \sqrt{3}, \\
\therefore AC^{2}+AB^{2}=BC^{2} .
\end{array}
$$

Thus $\angle BAC=90^{\circ}$.
Also, $AC^{2}=4=\frac{2 \sqrt{3}}{3} \times 2 \sqrt{3}=CE \cdot BC$, by the projection theorem we know
$AE \perp BC$, and $AE=\frac{AC \cdot AB}{BC}=\frac{2 \sqrt{6}}{3}$.
Since $AE \perp BC, AE \perp CM$,
therefore, $AE \perp$ plane $BCM$.
So, $V_{A-BCM}=\frac{1}{3} AE \cdot S_{\triangle BCM}=\frac{2 \sqrt{2}}{3}$.