3. As shown in Figure 3, in trapezoid $ABCD$, $DC \parallel AB$, $\frac{DC}{AB} = \frac{1}{3}$, $MN$ is the midline, $EF \parallel AB$ and passes through the intersection of $AC$ and $BD$, points $E$ and $F$ are on $AD$ and $BC$ respectively. Then the ratio of the areas of trapezoids $EFCD$, $MNEF$, and $ABNM$ is $\qquad$
3. 5: 7:20.

It is easy to prove that trapezoid $E F C D \backsim$ trapezoid $A B N M$, and trapezoid $M N C D \backsim$ trapezoid $A B F E$.
Let $D C=1$. Then $A B=3, M N=\frac{1}{2}(1+3)=2$.
Assume the area of trapezoid $E F C D$ is 1. Then the area of trapezoid $A B N M$ is 4. Let the area of trapezoid $M N E F$ be $x$. Note that, $\frac{M N}{A B}=\frac{2}{3}$. Therefore, $\frac{1+x}{4+x}=\left(\frac{2}{3}\right)^{2}=\frac{4}{9} \Rightarrow x=\frac{7}{5}$. Hence, the ratio of the areas of the three trapezoids is $1: \frac{7}{5}: 4=5: 7: 20$.