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.

 A: Hint: 
Let $S$ be the area of a triangle, $a,b$ be the length of two edges and $\theta$ be the angle between them, you have the following formula
$$S=\frac{1}{2}ab\sin\theta$$
On more hint: $\sin\theta=\sin(\pi-\theta)$.
A: 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.
A: !**enter image description here
**
**The proof:
Area of a quadrilateral =
 area of the first triangle + area of the second triangle + area of the third triangle + area of the fourth triangle
= [0.5 .d . b . sin θ ] + [0.5 . b . c . sin (180 - θ) ] + [0.5 . a . c . sin θ ] + [0.5 . d . a . sin (180 - θ) ]
As, sin (180 - θ) = sin θ
Area = [0.5 . d . b . sin θ ] + [0.5 . b . c . sin θ ] + [0.5 . a . c . sin θ ] + [0.5 . d . a . sin θ ]
Area = 0.5 . sin θ . [ (d . b) + (b . c) + (a . c) + (d . a) ]
As, (d . b) + (b . c) = b . (c + d)
and  (a . c) + (d . a) = a . (c + d)
Area = 0.5 . sin θ . [ b . (c + d) + a . ( c + d) ]
Area = 0.5 . sin θ . (c + d) . [b + a]
As, (b + a) = p , (c + d) = q
Area = 0.5 . p . q . sin θ
