alternate proof of parallelogram law

Proof of this is simple, given the cosine law:

c^{2}=a^{2}+b^{2}-2ab\\cos\\phi

where $a$ , $b$ , and $c$ are the lengths of the sides of the triangle, and angle $\\phi$ is the corner angle opposite the side of length $c$ .

Let us define the largest interior angles as angle $\\theta$ .Applying this to the parallelogram, we find that

	$\\displaystyle d_{1}^{2}$	$\\displaystyle=$	$\\displaystyle u^{2}+v^{2}-2uv\\cos\\theta$
	$\\displaystyle d_{2}^{2}$	$\\displaystyle=$	$\\displaystyle u^{2}+v^{2}-2uv\\cos\\left(\\pi-\\theta\\right)$

Knowing that

\\cos\\left(\\pi-\\theta\\right)=-\\cos\\theta

we can add the two expressions together, and find ourselves with

	$\\displaystyle d_{1}^{2}+d_{2}^{2}$	$\\displaystyle=$	$\\displaystyle 2u^{2}+2v^{2}-2uv\\cos\\theta+2uv\\cos\\theta$
	$\\displaystyle d_{1}^{2}+d_{2}^{2}$	$\\displaystyle=$	$\\displaystyle 2u^{2}+2v^{2}$

which is the theorem we set out to prove.