4. A regular square pyramid $S-A B C D$. Extend one side of the base $C D$ to $E$, such that $D E=2 C D$. A plane is drawn through $B, E$ and the midpoint $F$ of edge $S C$, dividing the pyramid into two parts. The ratio of the volumes of these two parts is $\qquad$
4. $31: 29$.

Let the side length of the base of a regular quadrilateral pyramid be $a$, and the height $h$. Then $S_{\triangle B C E}=\frac{1}{2} a \cdot 3 a=\frac{3}{2} a^{2}$.
Since $I D: a=2: 3$,
Therefore, $S_{\triangle I D E}=\frac{1}{2} \cdot \frac{2}{3} a \cdot 2 a=\frac{2}{3} a^{2}$.
Let the volume of the tetrahedron $F-B C E$ be $V_{1}$, then
$$
\begin{aligned}
V_{1}=\frac{1}{3} & S_{\triangle B C E} \cdot h_{1} \\
& =\frac{1}{3} \cdot \frac{3}{2} a^{2} \cdot \frac{h}{2}, \\
& =\frac{1}{4} a^{2} h .
\end{aligned}
$$

Take the midpoint $M$ of $D C$, and connect
$M F$. Then
$$
\begin{array}{l}
\frac{K D}{F M}=\frac{E D}{E M}=\frac{4}{5}, \\
K D=\frac{4}{5} F M=\frac{2}{5} S D .
\end{array}
$$

The volume of the tetrahedron $K-I D E$
$$
V_{2}=\frac{1}{3} \cdot \frac{2}{3} a^{2} \cdot \frac{2}{5} h=\frac{4}{45} a^{2} h,
$$

The volume of the polyhedron $B F C-I K D$ $V=V_{1}-V_{2}=\frac{29}{180} a^{2} h$.
Since $V_{s-A B C D}=\frac{1}{3} a^{2} h$. The volume of the remaining part
$$
V=\frac{31}{180} a^{2} h
$$

Therefore, the ratio of the volumes of the two parts is $31: 29$.