Proof of Pythagorean Theorem in Euclidean Space

Write $A=(a_1,\dots,a_n)$, $B=(b_1,\dots,b_n)$, and $C=(c_1,\dots,c_n)$, and for $1\le i\le n$ set $u_i=b_i-a_i$ and $v_i=c_i-a_i$, so that, in the sense of the definition of the difference, $B-A=(u_1,\dots,u_n)$ and $C-A=(v_1,\dots,v_n)$.

By the definition of the Euclidean distance, for all points $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$ of $\mathbb{R}^n$ we have $d_E(x,y)=\sqrt{t}$ with $t=\sum_{i=1}^n (x_i-y_i)^2\ge 0$, where $\sqrt{t}$ is the nonnegative square root from Existence and Uniqueness of the Nonnegative Square Root; since that square root satisfies $(\sqrt{t})^2=t$, we obtain

Applying this identity three times, and using that $(x_i-y_i)^2=(y_i-x_i)^2$ for real numbers:

For each $i$ we have $b_i-c_i=(b_i-a_i)-(c_i-a_i)=u_i-v_i$, and expanding the square gives $(u_i-v_i)^2=u_i^2-2u_iv_i+v_i^2$. Summing over $i$ yields

By the definition of the dot product,

and this quantity is $0$ by the orthogonality hypothesis. Therefore

as claimed. $\blacksquare$