<s>[INST] Area of a quadrilateral proof Prove that the area of a quadrilateral is one half the product of the lengths of its diagonals and the sine of the angle between the diagonals.

 [/INST] The area of this quadrilateral can be calculated by summing up the area of its four triangles. For a triangle, its area can be calculated using the formula:
$A=\frac{1}{2}ab\sin\theta$
where $a$ and $b$ are the lengths of two of his sides and $\theta$ is the internal angle between them, so the total area of the quadrilateral is:
$A=\frac{1}{2}ac\sin\theta_1+\frac{1}{2}cb\sin\theta_2+\frac{1}{2}bd\sin\theta_3+\frac{1}{2}da\sin\theta_4$
where the four angles $\theta_1$, $\theta_2$, $\theta_3$ and $\theta_4$ are the internal angles between each couple of edges. By knowing that $\theta_1$ = $\theta_3$ and $\theta_2$ = $\theta_4$, and $\theta_1$ and $\theta_3$ are supplementary angles ($\theta_1 = \pi-\theta_3$), so all the angles have the same $sin$, hence:
$A=\frac{1}{2}ac\sin\theta+\frac{1}{2}cb\sin\theta+\frac{1}{2}bd\sin\theta+\frac{1}{2}da\sin\theta\\=\frac{1}{2}(ac+cb+bd+da)\sin\theta\\=\frac{1}{2}(a+b)(c+d)\sin\theta\\=\frac{1}{2}d1\cdot d2\sin\theta$
where $d1$ and $d2$ are the lengths of the two diagonlas and $\theta$ is any internal angle between them.
</s>