Proof of The Squared Norm of a Convex Combination of Two Points

Fix $x,y\in\mathbb{R}^{n}$ and $t\in\mathbb{R}$. Throughout, $z\cdot z'$ is the dot product; by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ one has $\lVert z\rVert^{2}=z\cdot z$ for every $z\in\mathbb{R}^{n}$, and the symmetry and bilinearity of the dot product used below are claims 1--5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$. Arithmetic with real numbers uses the axioms of the field of real numbers; we write $2$ for $1+1$, so that $s+s=2s$ for every real $s$.

Step 1 (expansion of the left-hand side). Put $z=t\,x+(1-t)\,y$. Additivity and homogeneity in the first argument (claims 2 and 4) give

and the corresponding statements in the second argument (claim 5) give $x\cdot z=t\,(x\cdot x)+(1-t)\,(x\cdot y)$ and $y\cdot z=t\,(y\cdot x)+(1-t)\,(y\cdot y)$. Since $y\cdot x=x\cdot y$ by claim 1, substituting and collecting terms yields

Step 2 (expansion of the squared distance). By claim 3 (differences in the first argument) and claim 5 (differences in the second argument), together with claim 1,

Step 3 (comparison). Substituting the identity of Step 2 into the right-hand side of the assertion and expanding,

Now $t-t(1-t)=t-t+t^{2}=t^{2}$ and $(1-t)-t(1-t)=(1-t)(1-t)=(1-t)^{2}$, so the right-hand side coincides with the expression obtained in Step 1, namely $\lVert t\,x+(1-t)\,y\rVert^{2}$. Since $x$, $y$ and $t$ were arbitrary, the identity holds in general.