yet another proof of parallelogram law

Define $g(ϵ)=⟨x+ϵy,x+ϵy⟩$, where $ϵ$ is real.Then $g(ϵ)=⟨x,x⟩+ϵ(⟨y,x⟩+⟨x,y⟩)+{ϵ}^2⟨y,y⟩.$Hence,

$${\parallel x+y\parallel }^2+{\parallel x-y\parallel }^2=g(1)+g(-1)=2⟨x,x⟩+2⟨y,y⟩=2{\parallel x\parallel }^2+2{\parallel y\parallel }^2.$$