Given are the four sides of a quadrilateral and the angle enclosed by the diagonals. The area of the quadrilateral is to be calculated with these data!
Let the sides of the quadrilateral be $a, b, c, d$, the diagonals $e$ and $f$, and the angle between them $\omega$ (or $180^{\circ}-\omega$). Draw lines through the vertices of the quadrilateral parallel to the respective diagonals; the resulting parallelogram has sides $e$ and $f$, with the angle between them being $\omega$, and its area is $ef \sin \omega$. The area of the quadrilateral is half of this: $t=\frac{1}{2} e f \sin \omega$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5100f7e5ac4f1d72546fg-1.jpg?height=273&width=499&top_left_y=335&top_left_x=802)

Let the segments of the diagonals be $m$ and $n$, and $p$ and $q$ as shown in the diagram. Applying the cosine rule:

\[
\begin{aligned}
& a^{2}=m^{2}+p^{2}-2 m p \cos \omega \\
& b^{2}=n^{2}+p^{2}+2 n p \cos \omega \\
& c^{2}=n^{2}+q^{2}-2 n q \cos \omega \\
& d^{2}=m^{2}+q^{2}+2 m q \cos \omega
\end{aligned}
\]

From this:

\[
\begin{gathered}
b^{2}+d^{2}-a^{2}-c^{2}=2(m p+n p+n q+m q) \cos \omega= \\
=2(m+p)(p+q) \cos \omega=2 e f \cos \omega
\end{gathered}
\]

We express the value of $ef$ from the last equation and substitute it into the value of $t$. This results in:

\[
t=\frac{1}{4}\left(b^{2}+d^{2}-a^{2}-c^{2}\right) \operatorname{tg} \omega
\]

1 where $\omega$ is the angle between the diagonals that is opposite to the sides $a$ and $c$.

Donáth Géza (State Szent László g. VIII. o. Bp. X.)

Note: If $\omega=90^{\circ}$, then $b^{2}+d^{2}-\left(a^{2}+c^{2}\right)=0$, or $b^{2}+d^{2}=a^{2}+c^{2}$. In this case, the expression for $t$ does not yield a definite value. Imagine, for simplicity, a rhombus with a given side length and hinges at its vertices. Fix one side and rotate the others; during this process, the diagonals remain perpendicular to each other, the sum of the squares of the opposite sides remains equal, but the area varies. Let the side of the rhombus be $a$: its area varies between 0 and $a^{2}$.[^0]

[^0]:    ${ }^{1}$ If $\omega$ is the obtuse angle, then the sum of the squares of the opposite sides is: $a^{2}+c^{2}>b^{2}+d^{2}$.