4. 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 $CDEF$, $FEMN$, and $NMAB$ is equal to
4.5:7:20.

It is easy to prove that trapezoid $CDEF$ is similar to trapezoid $NMAB$,
and trapezoid $CDMN$ is similar to trapezoid $FEAB$.
Let $DC=1$. Then $AB=3, MN=2, EF=\frac{3}{2}$.
Let the area of trapezoid $CDEF$ be 1. Then the area of trapezoid $NMAB$ is 4. Let the area of trapezoid $FEMN$ be $x$.
Notice that, $\frac{MN}{AB}=\frac{2}{3}$.
By conclusion (1) we have
$$
\frac{1+x}{x+4}=\left(\frac{2}{3}\right)^{2}=\frac{4}{9} \Rightarrow x=\frac{7}{5} \text {. }
$$

Therefore, the required ratio is $1: \frac{7}{5}: 4=5: 7: 20$.
Figure 7

In the right triangle $\triangle PCO$, by the projection theorem we know
$$
PC^{2}=PQ \cdot PO \text {. }
$$

Also, by the secant-tangent theorem we know
$$
PC^{2}=PE \cdot PF \text {. }
$$

Thus, $PQ \cdot PO=PE \cdot PF$.
Therefore, points $E, F, O, Q$ are concyclic.
Hence $\angle EQF=\angle EOF=2 \angle BAD$.
Also, $\angle PQE=\angle OFE=\angle OEF=\angle OQF$, and $CQ \perp PD$, then
$$
\angle EQC=\angle FQC \text {. }
$$

Since $\angle AEC=\angle PQC=90^{\circ}$, points $B, E, C, Q$ are concyclic.
$$
\begin{array}{l}
\text { Then } \angle EBC=\angle EQC=\frac{1}{2} \angle EQF \\
=\frac{1}{2} \angle EOF=\angle BAD .
\end{array}
$$

Therefore, $CB \parallel AD$.
Thus, $BO=DO$, meaning quadrilateral $ABCD$ is a parallelogram.
Hence, $AB=DC, BC=AD$.