# 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 AlternateProofOfParallelogramLaw 2013-03-22 12:43:52 2013-03-22 12:43:52 drini (3) drini (3) 5 drini (3) Proof msc 51-00 ProofOfParallelogramLaw2