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.
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: $\sin\theta=\sin(\pi-\theta)$.
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**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