yet another proof of parallelogram law

Define $g(\epsilon)=\langle x+\epsilon y,x+\epsilon y\rangle$ , where $\epsilon$ is real. Then $g(\epsilon)=\langle x,x\rangle+\epsilon(\langle y,x\rangle+\langle x,y\rangle)% +\epsilon^{2}\langle y,y\rangle.$ Hence,

\|x+y\|^{2}+\|x-y\|^{2}=g(1)+g(-1)=2\langle x,x\rangle+2\langle y,y\rangle=2\|% x\|^{2}+2\|y\|^{2}.

Title	yet another proof of parallelogram law
Canonical name	YetAnotherProofOfParallelogramLaw
Date of creation	2013-03-22 16:08:23
Last modified on	2013-03-22 16:08:23
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	5
Author	Mathprof (13753)
Entry type	Proof
Classification	msc 46C05