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.


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.
