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.

Title	alternate proof of parallelogram law
Canonical name	AlternateProofOfParallelogramLaw
Date of creation	2013-03-22 12:43:52
Last modified on	2013-03-22 12:43:52
Owner	drini (3)
Last modified by	drini (3)
Numerical id	5
Author	drini (3)
Entry type	Proof
Classification	msc 51-00
Related topic	ProofOfParallelogramLaw2