# 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 YetAnotherProofOfParallelogramLaw 2013-03-22 16:08:23 2013-03-22 16:08:23 Mathprof (13753) Mathprof (13753) 5 Mathprof (13753) Proof msc 46C05
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