Given a square $A B C D$ in the plane, a line is to be constructed such that the orthogonal projections of the square's vertices onto this line are arranged in such a way that the three consecutive distances determined by the projection points are equal.
Denoting the projections of points $\mathrm{A} B C D$ as $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, our task can be satisfied in four ways: $A^{\prime} B^{\prime}=$ $B^{\prime} D^{\prime}=D^{\prime} C^{\prime}, A^{\prime} D^{\prime}=D^{\prime} B^{\prime}=B^{\prime} C^{\prime}, B^{\prime} A^{\prime}=A^{\prime} C^{\prime}=C^{\prime} D^{\prime}$, and finally $B^{\prime} C^{\prime}=C^{\prime} A^{\prime}=A^{\prime} D^{\prime}$. In the first case, $A^{\prime} D^{\prime}=2 A^{\prime} B^{\prime}=2 B^{\prime} D^{\prime}$, which means the projection $D_{1}^{\prime}$ of the midpoint $D_{1}$ of side $A D$ coincides with $B^{\prime}$. This implies that the sought line is perpendicular to the projection ray $D_{1} B$. Similarly, we obtain 3 more directions. Therefore, the complete solution is: infinitely many lines perpendicular to 4 directions.

Remark: The problem can be easily generalized to any parallelogram.