All heights of the pyramid $A B C D$, whose faces are acute-angled triangles, are equal to each other. It is known that $A B=9, B C=13$, and the angle $A D C$ is $60^{\circ}$. Find the edge $B D$.
Since the volume of a pyramid is one third of the product of the base and height, and all heights of the pyramid are equal, all faces of the pyramid are equal in area. We will prove that the opposite edges of such a pyramid are pairwise equal. Extend the tetrahedron $ABCD$ to a parallelepiped $AKBLNDMC$ ($AN\|KD\|BM\|LC$), by drawing through its opposite edges pairs of parallel planes (Fig.2). From the midpoint $G$ of edge $AB$, drop a perpendicular $GH$ to edge $CD$ (Fig.1). Consider the orthogonal projection $PA_1B_1$ of the tetrahedron $ABCD$ onto a plane perpendicular to $CD$, where $P$ are the projections of points $C, D$, and $H$; $A_1$ is the projection of vertex $A$, and $B_1$ is the projection of vertex $B$. From the equality of the areas of triangles $ADC$ and $BDC$, it follows that the heights from these triangles to the common side $CD$ are equal, and therefore, the orthogonal projections $A_1P$ and $B_1P$ of these heights onto the plane perpendicular to $CD$ are also equal. Since the projection $G_1$ of the midpoint of segment $AB$ is the midpoint of $A_1B_1$, the median $PG_1$ of the isosceles triangle $A_1B_1P$ is perpendicular to the base $A_1B_1$. Then, by the theorem of three perpendiculars, $GH \perp AB$. Thus, the common perpendicular to the skew lines $AB$ and $CD$ passes through the midpoint of $AB$. Similarly, the common perpendicular to the skew lines $AB$ and $CD$ passes through the midpoint of $CD$. Therefore, the segments connecting the midpoints of opposite edges of the tetrahedron $ABCD$ are perpendicular to these edges (Fig.2), and hence, to the faces of the parallelepiped $AKBLNDMC$. Therefore, the parallelepiped $AKBLNDMC$ is a rectangular parallelepiped. Consequently, the opposite edges of the tetrahedron $ABCD$ are pairwise equal as the diagonals of opposite faces of a rectangular parallelepiped. Thus,

$$
AD=BC=13, CD=AB=9
$$

Therefore,

$$
\begin{aligned}
BD=AC= & \sqrt{AD^2+BC^2-2AD \cdot BC \cos \angle ADC}= \\
& =\sqrt{169+81-13 \cdot 9}=\sqrt{133}
\end{aligned}
$$

## Answer

$\sqrt{133}$.