8. (10 points) The volume of tetrahedron $ABCD$ is $4(m+n)$. Point $M$ divides edge $AB$ in the ratio $m: n$. A section is made through point $M$ and the midpoints of edges $BC$ and $AD$. Find the area of this section if the distance from point $D$ to it is $h$.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_e38b8a325740b6549434g-08.jpg?height=820&width=1188&top_left_y=435&top_left_x=480)

Let $N$ and $L$ be the midpoints of sides $AD$ and $BC$ respectively. $K$ is the point of intersection of edge $DC$ with the plane of the section. The plane of the section intersects the planes $ABD$ and $CBD$ at point $E$. We apply Menelaus' theorem twice:

$$
\frac{EB}{ED} \cdot \frac{DN}{NA} \cdot \frac{AM}{MB}=1 \quad \text{and} \quad \frac{EB}{ED} \cdot \frac{DK}{KC} \cdot \frac{CL}{LB}=1
$$

From which we obtain

$$
\frac{DK}{KC}=\frac{ED}{EB}=\frac{AM}{MB}=\frac{m}{n}
$$

Thus, point $K$ divides side $DC$ in the same ratio as point $M$ divides side $AB$.

Notice,

$$
\frac{V_{DALB}}{V_{DALC}}=\frac{S_{ALB}}{S_{ALC}}=\frac{LB}{LC}=1 ; \quad \frac{V_{LDM A}}{V_{LDM B}}=\frac{S_{DMA}}{S_{DM B}}=\frac{MA}{MB}=\frac{m}{n} ; \quad \frac{V_{LMDN}}{V_{LMAN}}=\frac{S_{MDN}}{S_{MAN}}=\frac{DN}{AN}=1
$$

From these ratios, we find that

$$
V_{LMDN}=\frac{1}{2} \cdot V_{LDM A}=\frac{m}{2(m+n)} \cdot V_{DALB}=\frac{m}{4(m+n)} \cdot V_{DABC}=m
$$

Similarly, using the ratio in which point $K$ divides side $DC$, we can prove that $V_{LKDN}=m$. Therefore,

$$
2m=V_{LMDN}+V_{LKDN}=V_{DKLMN}=\frac{1}{3} \cdot h \cdot S_{KLMN} \Longrightarrow S_{KLMN}=\frac{6m}{h}
$$

## Answers:

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