In a convex quadrilateral, the two diagonals are equal, their length is $e$, the sum of the two midlines is $s$, and the lengths of two adjacent sides are $a$ and $b$. What is the area of the quadrilateral?
Let the quadrilateral in question be $ABCD = N$, with the midpoints of its consecutive sides $AB, BC, CD, DA$ being $A_1, B_1, C_1, D_1$ respectively; thus $AC = BD = e$, $A_1C_1 + B_1D_1 = s$, $AB = a$, $BC = b$.

![](https://cdn.mathpix.com/cropped/2024_05_02_58935d6c153266941833g-1.jpg?height=393&width=729&top_left_y=275&top_left_x=687)

The quadrilateral $A_1B_1C_1D_1 = N_1$ is a parallelogram because the sides $A_1B_1$ and $C_1D_1$ are parallel to the diagonal $AC$ and half its length: $A_1B_1 \# C_1D_1 = e / 2$, since, for example, $A_1B_1$ is the midline of the triangle $ABC$ corresponding to the side $AC$. Similarly, $B_1C_1 = BD / 2 = e / 2$, so $N_1$ is a rhombus. It suffices to determine its area, because the area of $N$ is twice that of $N_1$. Indeed, connecting the vertices of $N_1$ to the intersection point $E$ of the diagonals of $N$, $N_1$ is divided into four triangles, and the areas of these triangles are equal to the areas of the triangles cut off by the corresponding sides of $N_1$ from $N$, for example, the areas of $E A_1 B_1$ and $B A_1 B_1$ are equal because they share the base $A_1B_1$ and the perpendicular heights are equal. Thus, the part of the area of $N$ inside the perimeter of $N_1$ is equal to the part outside. This is what we wanted to prove.

The diagonals of $N_1$ are perpendicular, so its area is equal to half the product of these diagonals, thus the area $t$ of $N$ is given by the product $A_1C_1 \cdot B_1D_1$. This can be expressed without determining the individual diagonals:

\[
\begin{gathered}
t = A_1C_1 \cdot B_1D_1 = \frac{1}{2} \left[ \left( A_1C_1 + B_1D_1 \right)^2 - \left( A_1C_1^2 + B_1D_1^2 \right) \right] = \\
= \frac{s^2}{2} - 2 \left[ \left( \frac{A_1C_1}{2} \right)^2 + \left( \frac{B_1D_1}{2} \right)^2 \right] = \frac{s^2}{2} - 2 \cdot A_1B_1^2 = \frac{1}{2} \left( s^2 - e^2 \right)
\end{gathered}
\]

(We used the fact that the center $O$ of $N_1$ forms right triangles with the endpoints of each side.)

$t$ must be positive, which is clearly true if $s > e$. We did not need to use the known side lengths $a$ and $b$. This does not mean that our result is valid for any set of data $s, e, a, b$, only if $N$ is formed and is convex.

The triangle $ABC$ is formed if and only if

\[
|a - b| < e < a + b
\]

Denoting the diagonals of $N_1$ by $g$ and $h$, from the system of equations $g + h = s$, $g^2 + h^2 = e^2$ we get

\[
g, h = \frac{1}{2} \left( s \pm \sqrt{2e^2 - s^2} \right)
\]

These are real if $e \geq s / \sqrt{2}$, or equivalently, with the above,

\[
s / \sqrt{2} \leq e < s
\]

If $g$ and $h$ are different, then the position of $O$ can be constructed in two ways, i.e., by swapping $g$ and $h$. Then $C_1$ is the reflection of $A_1$ over $O$, and finally, $D$ is the reflection of $C$ over $C_1$. $N$ is convex if and only if $D$ lies within the angle domain of $ABC$.

From the work of Zoltán Berkes (Budapest, Bolyai J. g. I. o. t.), with additional remarks.

Remark. By dissection, we obtain that the parallelogram $FGHJ$ in the figure is equal in area to $N$, and twice the area of $N_1$.