$\begin{aligned} & {\left[\begin{array}{l}\text { Linear dependence of vectors } \\ \text { [ angles between lines and planes }\end{array}\right]}\end{aligned}$

The side of the base of a regular quadrilateral pyramid $\operatorname{ABCDP}$ ( $P$ - apex) is $4 \sqrt{2}$, and the angle between adjacent lateral faces is $120^{\circ}$. Find the area of the section of the pyramid by a plane passing through the diagonal $B D$ of the base parallel to the lateral edge $C P$.
Note that by the theorem of three perpendiculars $P C \perp B D$. Let $F$ be the projection of vertex $B$ onto the lateral edge $PC$. Since the line $PC$ is perpendicular to two intersecting lines ($BF$ and $BD$), this line is perpendicular to the plane $BDF$, hence $\angle B F D=120^{\circ}$. Let the line passing through the center $M$ of the square $ABCD$ parallel to the lateral edge $C$ intersect the lateral edge $AP$ at point $K$. Then the plane $DBK$ is parallel to $CP$, since it contains the line $MK$, parallel to $CP$. Therefore, triangle $BDK$ is the desired section. In this case, $MK$ is the midline of triangle $ACP$. From the right triangle $BMF$ we find that

$$
M F=M B \operatorname{ctg} \angle B F M=\frac{A B}{\sqrt{2}} \cdot \operatorname{ctg} 60^{\circ}=\frac{4}{\sqrt{3}}
$$

Then

$$
\sin \angle M C F=\frac{M F}{M C}=\frac{\frac{1}{\sqrt{3}}}{4}=\frac{1}{\sqrt{3}}, \cos \angle M C F=\sqrt{\frac{2}{3}}
$$

From the right triangle $PMC$ we find that

$$
C P=\frac{M C}{\cos \angle M C F}=\frac{4}{\sqrt{\frac{2}{3}}}=2 \sqrt{6}
$$

Therefore,

$$
S_{\triangle B D K}=\frac{1}{2} B D \cdot M K=\frac{1}{4} B D \cdot C P=\frac{1}{4} \cdot 8 \cdot 2 \sqrt{6}=4 \sqrt{6} .
$$

## Answer

$4 \sqrt{6}$