4. As shown in Figure 2, in $\square A B C D$, if points $E$ and $F$ on side $A B$ satisfy $A E=E F=F B, C E$ intersects $D F$ and $D B$ at points $M$ and $N$ respectively, then $E M: M N: N C$ equals ( ).
(A) $2: 1: 4$
(B) $4: 3: 5$
(C) $5: 3: 12$
(D) $5: 4: 12$
4.C.

Let $A E=E F=F B=1$, set $E M=x, M N=y, N C=$ $z$. From $\triangle C D M \subset \triangle E F M$, we get $\frac{x}{y+z}=\frac{1}{3}$, i.e., $3 x-y-z=0$.

From $\triangle C D N \backsim \triangle E B N$, we get $\frac{x+y}{z}=\frac{2}{3}$, i.e., $3 x+3 y-2 z=0$.
Then we have $x=\frac{5}{12} z, y=\frac{1}{4} z$.