All edges of the triangular pyramid ABCD touch a certain sphere. The three segments connecting the midpoints of the skew edges $A B$ and $C D, A C$ and $B D, A D$ and $B C$ are equal. The angle $D B C$ is $50^{\circ}$, and the angle $B C D$ is greater than the angle $B D C$. Find the ratio of the areas of the faces $A B D$ and $A B C$.
Let each of the segments connecting the midpoints of opposite edges of the tetrahedron $ABCD$ be equal to $a$. Extend the given tetrahedron to the parallelepiped $AKBLNDMC$ ($AN \| KD \| BM \| LC$) by drawing pairs of parallel planes through its opposite edges (Fig.1). The segments connecting the midpoints of opposite edges of the tetrahedron $ABCD$ are respectively equal and parallel to the edges of the parallelepiped $AKBLNDMC$, hence, all faces of this parallelepiped are rhombuses, all sides of which are equal to $a$. By the theorem on the sum of the squares of the diagonals of a parallelogram,

$$
AB^2 + CD^2 = 4a^2, \quad AC^2 + BD^2 = 4a^2, \quad AD^2 + BC^2 = 4a^2
$$

Therefore,

$$
AB^2 + CD^2 = AC^2 + BD^2 = AD^2 + BC^2
$$

Let a sphere touch the edges $AB, CD, AD$, and $BC$ of the tetrahedron $ABCD$ at points $E, F, G$, and $H$ respectively (Fig.2). Then

$$
AB + CD = (AE + BE) + (CF + DF) = (AG + BH) + (CH + DG) =
$$

$$
= (AG + DG) + (CH + BH) = AD + BC
$$

Similarly, $AC + BD = AB + CD$ and $AD + BC = AB + CD$. Consider the equalities

$$
AD^2 + BC^2 = AC^2 + BD^2, \quad AD + BC = AC + BD.
$$

Square both sides of the second equality and subtract the first from the result. We get the system

$$
\left\{
\begin{array}{l}
D^2 + BC^2 = AC^2 + BD^2 \\
2AD \cdot BC = 2BD \cdot AC
\end{array}
\right.
$$

from which it follows that $(AD - BC)^2 = (AC - BD)^2$. Therefore,

$$
AD - BC = AC - BD \quad \text{or} \quad AD - BC = BD - AC,
$$

and since $AD + BC = AC + BD$, in the first case $AD = AC$ and $BC = BD$, which is impossible because from the problem's condition it follows that $\angle BCD \neq \angle BCD$.

Thus, we have proved that

$$
AD = BD, \quad BC = AC, \quad AD = AB, \quad BC = CD
$$

therefore,

$$
AB = AD = BD, \quad BC = CD = AC
$$

i.e., $ABCD$ is a regular triangular pyramid with vertex $C$ (Fig.3). Let $T$ be the midpoint of $AB$. Then $DT$ and $CT$ are the heights of triangles $ABD$ and $ABC$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-19.jpg?height=83&width=654&top_left_y=1356&top_left_x=702)

## Answer

$\sqrt{3} \operatorname{tg} 40^{\circ}$.