5. Given that the angle bisector $A D$ of acute triangle $\triangle A B C$ intersects with the altitude $B E$ at point $M$, and $\triangle C D E$ is an equilateral triangle. Then $S_{\triangle D E M}: S_{\triangle A B M}=(\quad$.
(A) $\sqrt{2}: 2$
(B) $1: 2$
(C) $\sqrt{3}: 3$
(D) $1: 4$
5. D.

Since $B E \perp A C$ and $\triangle C D E$ is an equilateral triangle, we have
$$
\angle B E C=90^{\circ}, C D=D E.
$$

Therefore, $B D=D E=D C$.
Also, since $A D$ bisects $\angle B A C$, then $A B=A C$.
Thus, $\triangle A B C$ is an equilateral triangle.
Hence, $M$ is the centroid of the equilateral $\triangle A B C$.
Therefore, $S_{\triangle D E M}: S_{\triangle A B M}=(M D: A M)^{2}=1: 4$.