A regular hexagonal pyramid has its lateral faces forming a $45^{\circ}$ angle with the base plane. We place a plane $S$ on one edge of the base such that it intersects the lateral faces adjacent to this edge in parallel segments. What angle does the plane $S$ form with the base?
Let the vertices of the regular hexagon be $A, B, C, \ldots, F$, the apex of the pyramid be $G$, and choose the length of the base edge to be 2 units (Figure 1). The angle between the lateral face and the base can be read from the right triangle $G G^{\prime} O$: $\angle G G^{\prime} O = 45^{\circ}$, so $G^{\prime} O = G O = \sqrt{3}$ (from the equilateral triangle $F A O$). Place the plane $S$ on the edge $A B$, which intersects the lateral face $B G C$ at $B P$ and the lateral face $F G A$ at $A Q$, and by the given condition, $B P \| A Q$, so the projections of these segments onto the base plane, $B P^{\prime} \| A Q^{\prime}$, where $P^{\prime}$ is the projection of $P$ and $Q^{\prime}$ is the projection of $Q$. Since $P$ and $Q$ are points on the plane $F G C$ perpendicular to the base, their projections $P^{\prime}$ and $Q^{\prime}$ lie on the intersection line $F C$ of the two planes, thus $A B \| F C \| P^{\prime} Q^{\prime}$ and $B P^{\prime} \| A Q^{\prime}$, making the quadrilateral $A B P^{\prime} Q^{\prime}$ a parallelogram.

From the axial symmetry of the regular hexagon, it follows that $F Q^{\prime} = C P^{\prime}$, which is only possible if $B P^{\prime} \perp F C$ and $A Q^{\prime} \perp F C$ (Figure 2). The desired angle $\alpha$ can be determined from the right triangle $B P P^{\prime}$: $\operatorname{tg} \alpha = \frac{P P^{\prime}}{P^{\prime} B}$, $P^{\prime} B = \sqrt{3}$, and in the triangle $O G C$, $P P^{\prime} \| G O$ and $P^{\prime}$ bisects $O C$, so $P P^{\prime} = \frac{1}{2} G O = \frac{\sqrt{3}}{2}$. Substituting these values:

$$
\operatorname{tg} \alpha = \frac{1}{2} \quad \text{and} \quad \alpha = 26^{\circ} 34^{\prime}
$$

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