91. The base of the right prism $A B C D A_{1} B_{1} C_{1} D_{1}$ is an isosceles trapezoid $A B C D$, in which $A D$ is parallel to $B C, |A D| / |B C| = n, n > 1$. Planes are drawn through the edges $A A_{1}$ and $B C$ parallel to the diagonal $B_{1} D$; planes are drawn through the edges $D D_{1}$ and $B_{1} C_{1}$ parallel to the diagonal $A_{1} C$. Determine the ratio of the volume of the triangular pyramid bounded by these four planes to the volume of the prism.
91. A plane passing through $A A_{1}$ parallel to $B_{1} D$ will be parallel to the plane $D D_{1} B_{1} B$. Similarly, a plane passing through $D D_{1}$ parallel to $A_{1} C$ will be parallel to the plane $A A_{1} C_{1} C$.

On the other hand, planes passing through the edges $B C$ and $B_{1} C_{1}$ will be parallel to the planes $A B_{1} C_{1} D$ and $A_{1} B C D_{1}$. Considering this fact, we will construct a section of our polyhedra by a plane parallel to the bases and passing through the midpoints of the lateral edges, and by a plane passing through the midpoints of the parallel sides of the prism base (see Fig. 10). In the figures, $L$ and $K$ are the midpoints of the opposite edges $E F$ and $H G$ of the triangular pyramid $E F G H$, and the edges $E F$ and $H G$ are perpendicular. Denoting $|B C|=x, |A D|=n x$, the height of the trapezoid $A B C D$ by $y$, and the height of the prism by $z$, we find

$$
|K S|=|S O|=\frac{y n}{n+1}, \quad|T L|=\frac{y}{2}
$$

$$
|K L|=y\left(\frac{3}{2}+\frac{n}{n+1}\right), \quad|E F|=\frac{5 n+3}{2} x, \quad|G H|=\frac{5 n+3}{n+1} z
$$

The volume of the prism is $\frac{(n+1) x y z}{2}$. The volume of the triangular pyramid is $\frac{1}{6}|E F| \cdot|G H| \cdot|K L|=\frac{(5 n+3)^{3}}{24(n+1)^{2}} x y z$.

Answer: $\frac{(5 n+3)^{3}}{12(n+1)^{3}}$.