PlanetMath: alternate proof of parallelogram law

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alternate proof of parallelogram law

(Proof)

Proof of this is simple, given the cosine law:

$\displaystyle c^2 = a^2 + b^2 - 2ab\cos\phi$

where

, and

are the lengths of the sides of the triangle, and angle $\phi$ is the corner angle opposite the side of length

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

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

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

$\displaystyle d_1^2+d_2^2$	$\displaystyle =$	$\displaystyle 2u^2 + 2v^2 - 2uv\cos\theta + 2uv\cos\theta$
$\displaystyle d_1^2+d_2^2$	$\displaystyle =$	$\displaystyle 2u^2 + 2v^2$