3. As shown in Figure 16, 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 $S_{\triangle C D M}: S_{\triangle A B M}=$
(Hint: Complete the trapezoid into a square. Then it is easy to know that Rt $\triangle A B M \cong \mathrm{Rt} \triangle A E D$. Therefore, $E D=B M=x$, i.e., $C D=C M=4-x$. In $\triangle C D M$, $4^{2}+x^{2}=$ $2(4-x)^{2}$, then $x=8 \pm 4 \sqrt{3}(x=8+4 \sqrt{3}$ is discarded $)$. Hence $\left.S_{\triangle C D M}: S_{\triangle A B M}=1: 2.\right)$