Prove parallelogram has four triangles with same area using vectors

Question

I need to prove that a parallelogram has four triangles with same area using vectors only. Thought to prove it using area of triangle and parallelogram but with no success. May you help me please?

Answer 1

Let $a$ and $b$ be the vectors of two nonparallel sides of you parallelogram (as in your linked picture). Now the diagonals of the parallelogram are given by $a+b$ and $a-b$. The four triangles to think about are-up to translation-the triangles formed by the pairs $(a, \tfrac{1}{2}(a+b))$, $(b, \tfrac{1}{2}(a-b))$, $(a, \tfrac{1}{2}(a-b))$, and $(b, \tfrac{1}{2}(a+b))$. Noting the following basic properties of the cross product for all vectors $u, v, w$ and scalars $c$, one completes the proof by four simple computations (which I'll leave to you):

1) $u \times u = 0$

2) $u \times (cv) = c (u \times v)$

3) $u \times (v + w) = (u \times v) + (u \times w)$

Hint: The common area is $\tfrac{1}{4}|a \times b|$.