prove:
quadrilateral's area = 1/2 the product of the lenghts of the diagonals and the sine of the angle between the diagonals ()thx
Recall that the area of the triangle built on vectors $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ that have the same beginning is $\displaystyle (1/2)|\vec{a}\times\vec{b}|$, i.e., half the length of the vector product. Let's introduce the signed vector product: it's the number whose absolute value equals the length of the vector product and the sign shows whether the two vectors plus their vector product form the right triple (i.e., if the shortest turn of the first vector towards the second one is counterclockwise). Let's denote it by $\displaystyle \langle\vec{a},\vec{b}\rangle$. This operation has properties similar to those of vector product:
$\displaystyle \langle\vec{a},\vec{a}\rangle=0$,
$\displaystyle \langle\vec{a},\vec{b}\rangle=-\langle\vec{b},\vec{a}\rangle$,
$\displaystyle \langle x\vec{a},\vec{b}\rangle= x\langle\vec{a},\vec{b}\rangle$,
and
$\displaystyle \langle\vec{a}+\vec{b},\vec{c}\rangle= \langle\vec{a},\vec{c}\rangle+ \langle\vec{b},\vec{c}\rangle$.
Now consider the following picture from MathWorld.
Find $\displaystyle \langle\vec{p},\vec{q}\rangle$ expressing $\displaystyle \vec{p}=\vec{b}+\vec{c}$ and $\displaystyle \vec{q}=\vec{a}+\vec{b}$ and using the fact that $\displaystyle \vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}$ as well as the properties above.