Let's construct a quadrilateral if the lengths of the two diagonals, the lengths of two opposite sides, and the distance between the midpoints of the other two sides are given.
Imagine the problem is solved, let the sought quadrilateral be $A B C D$, where $A C$ and $B D$ are diagonals, the lengths of sides $A B$ and $C D$ are respectively equal to the given segments $e, f, a$, and $c$, and the distance between the midpoints $E$ and $F$ of sides $B C$ and $D A$ is equal to the given segment $g$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4ab2886a8c5e8ca6a357g-1.jpg?height=383&width=809&top_left_y=299&top_left_x=650)

Reflect the quadrilateral over point $E$. Thus, $B$ and $C$ coincide, and the reflections of $A, D$, and $F$ are $A^{\prime}, D^{\prime}, F^{\prime}$, respectively. The quadrilateral $A D A^{\prime} D^{\prime}$ is a parallelogram because its diagonals bisect each other. Therefore, the quadrilateral $A F F^{\prime} D^{\prime}$ has opposite sides $F^{\prime} D^{\prime}$ and $F A$ parallel and equal, as they are equal to $F D$ due to the reflection and the midpoint. Thus, $A F F^{\prime} D^{\prime}$ is a parallelogram. Consequently, $A D^{\prime}=F F^{\prime}=2 F E=2 g$. Since, moreover, by reflection $B D^{\prime}=C D=c$ and $D^{\prime} C=D B=f$, the triangles $A D^{\prime} B$ and $A D^{\prime} C$ have all three sides known, and after choosing the position of their common side $A D^{\prime}$, they can be constructed. The missing vertex $D$ can be obtained either by reflecting $D^{\prime}$ over the midpoint of $B C$, or by completing the triangle $B D^{\prime} C$ to form the parallelogram $B D^{\prime} C D$.

The condition for constructibility is that the triplets of segments $2 g, a, c$ and $2 g, e, f$ satisfy the triangle inequality. If both conditions are met, then $D$ can be constructed.

For the construction of the triangles $A D^{\prime} B$ and $A D^{\prime} C$, it is sufficient to consider one solution for the first, for example, $A D^{\prime} B$. However, for the second triangle, for example, $C$, we must consider both sides of $A D^{\prime}$. Thus, the number of solutions is 2, as the construction of $D$ is unique.

Pál Greguss (Budapest, Fazekas M. gyak. g., II. o. t.)

Note. Some found 4 or even 8 solutions by constructing $D$ on the base $B C$ from the sides $f$ and $c$ - on the side of $B C$ opposite to $D^{\prime}$ - in two ways. However, this swaps the roles of the segments $f$ and $c$ as diagonals and sides. Considering the two possibilities for $B$ on both sides of $A D^{\prime}$, we only get solutions that differ in traversal direction (i.e., mirror-symmetric solutions).