In the pyramid $A B C D$, the area of the face $A B C$ is four times the area of the face $A B D$. A point $M$ is taken on the edge $C D$, such that $C M: M D=2$. Planes are drawn through the point $M$, parallel to the faces $A B C$ and $A B D$. Find the ratio of the areas of the resulting sections.
Let $S_{\triangle A B D}=S$. Then $S_{\triangle A B C}=4 S$. Suppose a plane passing through point $M$ parallel to plane $A B C$ intersects edges $A D$ and $B D$ at points $L$ and $K$, respectively, and a plane passing through point $M$ parallel to plane $A B D$ intersects edges $B C$ and $A C$ at points $P$ and $Q$, respectively. Then triangle $L K M$ is similar to triangle $A B C$ with a ratio of $\frac{D M}{D C}=\frac{1}{3}$, and triangle $Q P M$ is similar to triangle $A B D$ with a ratio of $\frac{C M}{C D}=\frac{2}{3}$. Therefore,

$$
\left.S_{\triangle L K M}=\left(\frac{1}{3}\right)^2 S_{\triangle A B C}=\frac{1}{9} \cdot 4 S=\frac{4}{9} S, S_{\triangle Q P M}=\left(\frac{2}{3}\right)^2 \cdot S_{\triangle A B D}=\frac{4}{9} S .
$$

## Answer

1.