4. As shown in Figure 1, in $\triangle A B C$, $M$ is the midpoint of $A C$, and $P, Q$ are the trisection points of side $B C$. If $B M$ intersects $A P$ and $A Q$ at points $D$ and $E$ respectively, then the length ratio of the three segments $B D, D E, E M$ is $(\quad$.
(A) $3: 2: 1$
(B) $4: 2: 1$
(C) $5: 3: 2$
(D) $5: 2: 1$
4.C.

As shown in Figure 5, draw $M H / / B C$ intersecting $A Q$ at $F$ and $A P$ at $H$. Therefore,
$$
\begin{array}{l}
\frac{M D}{B D}=\frac{M H}{B P}=\frac{1}{2} \cdot \frac{P C}{B P}=1, \\
\frac{E M}{B E}=\frac{M F}{B Q}=\frac{1}{2} \cdot \frac{C Q}{B Q}=\frac{1}{4} .
\end{array}
$$

Solving, we get $B D=\frac{5}{2} E M, D E=\frac{3}{2} E M$.
Thus, $B D: D E: E M=5: 3: 2$.