Let $DM$ be the height of the equilateral triangle $ADC$. Then $M$ is the midpoint of $AC$ and

$$
DM = AD \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
$$

Since the plane of the face $ADC$ is perpendicular to the plane of the base $ABC$, the line $DM$ is perpendicular to the plane of the base. Therefore, $DM$ is the height of the pyramid $ABCD$. The median $BM$ of the isosceles triangle $ABC$ is its height. The line $BM$ is the orthogonal projection of the inclined line $DB$ on the plane of the base of the pyramid. Since $BM \perp AC$, by the theorem of three perpendiculars, $DB \perp AC$. From the right triangle $ACM$ we find that

$$
BM = \sqrt{BC^2 - CM^2} = \sqrt{7 - 1} = \sqrt{6}
$$

Therefore,

$$
BD = \sqrt{DM^2 + BM^2} = \sqrt{3 + 6} = 3
$$

Consider the section of the pyramid by a plane intersecting the edges $AB$, $CB$, $CD$, and $AD$ at points $P$, $Q$, $R$, and $T$ respectively. Suppose that $PQ \parallel RT$. Then through the parallel lines $PQ$ and $RT$ there are two planes $ABC$ and $ADC$, intersecting along the line $AC$. Thus, the lines $PQ$ and $RT$ are parallel to the line $AC$. Similarly, we can prove that if $PT \parallel QR$, then the lines $PT$ and $QR$ are parallel to the line $BD$. From the previously proven $AC \perp BD$. Therefore, if the plane intersecting the edges $AB$, $CB$, $CD$, and $AD$ gives a parallelogram in the section of the pyramid, then this parallelogram is a rectangle. Let $\frac{BP}{BA} = k$. Then

$$
PQ = kAC = 2k, \quad PT = (1-k)BD = 3(1-k).
$$

From the equation $2k = 3(1-k)$ we find that $k = \frac{3}{5}$. Thus, if the specified section is a square, then $PQ = PT = \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-22.jpg?height=81&width=2041&top_left_y=2735&top_left_x=0)

and $DB$, and in the section, a parallelogram is obtained. Similarly, we can prove that one pair of opposite sides of the parallelogram is parallel to the edge $AD$, and the other pair is parallel to the edge $BC$. Therefore, if a rectangle were obtained in the section, then the opposite edges $AD$ and $BC$ of the given pyramid would be perpendicular. We will prove that this is not the case. Suppose that $AD \perp BC$. Since the line $AC$ is the orthogonal projection of the inclined line $AD$ on the plane of the base of the pyramid, by the theorem of three perpendiculars, $AC \perp BC$, which is impossible ($\angle ACB < 90^\circ$). Similarly, we can prove that in the section of the pyramid $ABCD$ by a plane intersecting the edges $CA$, $CB$, $DA$, and $DB$, a square cannot be obtained.

## Answer

$3;$ 活
At the base of the pyramid lies an isosceles triangle $ABC$ with base $AC=2$ and lateral side $\sqrt{7}$. The face $ACD$ is perpendicular to the base plane and is an equilateral triangle. Find the edge $BD$, as well as the areas of all sections of the pyramid that are squares.