2. As shown in Figure 1, in the right trapezoid $A B C D$, $A B=B C=4$, $M$ is a point on the leg $B C$, and $\triangle A D M$ is an equilateral triangle. Then $S_{\triangle C D M}$ : $S_{\triangle A B M}=$ $\qquad$ .
2.2.

As shown in Figure 6, draw $A E \perp$ $C D$ intersecting the extension of $C D$ at point $E$, then quadrilateral $A B C E$ is a square.
It is easy to prove
$\mathrm{Rt} \triangle A B M \cong \mathrm{Rt} \triangle A E D$.
Therefore, $B M=D E$.
Thus, $C M=C D$.
Let this value be $x$, then
$$
\begin{array}{l}
x^{2}+x^{2}=4^{2}+(4-x)^{2} \Rightarrow x^{2}=32-8 x . \\
\text { Hence, } \frac{S_{\triangle C D M}}{S_{\triangle B M}}=\frac{\frac{1}{2} x^{2}}{\frac{1}{2} \times 4(4-x)}=\frac{x^{2}}{16-4 x}=2 .
\end{array}
$$