## $[$ Volume Ratios $]$ [Theorems of Ceva and Menelaus $]$

The base of the pyramid $P A B C D$ is a parallelogram $A B C D$. Point $K$ is the midpoint of edge $C P$, and point $M$ is located on edge $A B$, such that $A M: M B=1: 2$. Construct the section of the pyramid by a plane passing through points $K$ and $M$ and parallel to line $B D$. In what ratio does this plane divide the volume of the pyramid?
The base plane $ABCD$ passes through the line $BD$, which is parallel to the cutting plane, and intersects the cutting plane along a line $l$ passing through point $M$ parallel to $BD$. Let $L$ be the point of intersection of line $l$ with side $AD$ of parallelogram $ABCD$. Then, $\frac{AL}{LD} = \frac{AM}{MB} = \frac{1}{2}$. Let $F$ and $E$ be the points of intersection of line $I$ with the extensions of $BC$ and $CD$, respectively. Denote $BC = AD = a$. From the similarity of triangles $BMF$ and $AML$, we find that

$$
BF = AL \cdot \frac{BM}{AM} = \frac{1}{3} a \cdot 2 = \frac{2}{3} a, \quad CF = BC + BF = a + \frac{2}{3} a = \frac{5}{3} a
$$

The planes of the faces $BPC$ and $APD$ pass through parallel lines $BC$ and $AD$ and have a common point $P$. Therefore, they intersect along some line $m$ parallel to lines $BC$ and $AD$. Consider the plane of the face $BPC$. Let line $FK$ intersect edge $PB$ at point $N$ and line $m$ at point $T$. From the equality of triangles $PKT$ and $CKF$, we find that

$$
PT = CF = \frac{5}{3} a
$$

and from the similarity of triangles $PNT$ and $BNF$, it follows that

$$
\frac{PN}{NB} = \frac{PT}{BF} = \frac{\frac{5}{3} a}{\frac{2}{3} a} = \frac{5}{2}
$$

If $Q$ is the point of intersection of line $EK$ with edge $DP$, then similarly, we can prove that $\frac{PQ}{QD} = \frac{5}{2}$. Let $h$ be the height of the given pyramid, $S$ be the area of its base, and $V$ be the volume. Then, the height of the triangular pyramid $KCFE$ with vertex $M$ is $\frac{1}{2} h$, and the heights of the triangular pyramids $NBFM$ and $QDEL$ with vertices $N$ and $Q$ are $\frac{2}{7} h$. Further, we have:

$$
\begin{gathered}
S_{\triangle AML} = \frac{1}{9} S_{\triangle ABD} = \frac{1}{7} \cdot \frac{1}{2} S = \frac{1}{18} S, \\
S_{\triangle DEL} = S_{\triangle BFM} = 4 S_{\triangle AML} = \frac{2}{9} S \\
S_{\triangle CFE} = S_{ABCD} - S_{\triangle AML} + 2 S_{\triangle BFM} = S - \frac{1}{18} S + \frac{4}{9} S = \frac{25}{18} S .
\end{gathered}
$$

Thus,

$$
\begin{gathered}
V_{KCFE} = \frac{1}{3} S_{\triangle CFE} \cdot \frac{1}{2} h = \frac{1}{3} \cdot \frac{25}{18} S \cdot \frac{1}{2} h = \frac{25}{36} \cdot \frac{1}{3} S h = \frac{25}{36} V, \\
V_{NBFM} = \frac{1}{3} S_{\triangle BFM} \cdot \frac{2}{7} h = \frac{1}{3} \cdot \frac{2}{9} S \cdot \frac{2}{7} h = \frac{4}{63} \cdot \frac{1}{3} S h = \frac{4}{189} V \\
V_{CBMLDQKN} = V_{KCFE} - 2 V_{NBFM} = \frac{25}{36} V - \frac{8}{189} V = \frac{143}{252} V .
\end{gathered}
$$

Therefore, the cutting plane divides the volume of the given pyramid in the ratio 109:143.

## Answer

$109: 143$.