Proof of the Pythagorean theorem and its converse with the dot product

Proof

We expand a squared length, which proves both directions at once.

1. Let $x = A - B$ and $y = C - B$. With the Euclidean distance and the length, $d(A, B) = \norm{x}$, $d(B, C) = \norm{y}$ and $d(A, C) = \norm{x - y}$.
2. The dot product is bilinear and symmetric. So $\norm{x - y}^2 = \norm{x}^2 + \norm{y}^2 - 2 \, x \cdot y$.
3. By steps 1 and 2, $d(A, C)^2 = d(A, B)^2 + d(B, C)^2$ holds if and only if $x \cdot y = 0$.
4. The vertices of the triangle are not collinear, so they are distinct. Thus $x \neq 0$ and $y \neq 0$.
5. By definition, the angle $\angle ABC$ is the angle between $x$ and $y$. Since both are nonzero, $x \cdot y = 0$ holds if and only if this angle is $\pi / 2$.
6. Steps 3 and 5 give: $d(A, C)^2 = d(A, B)^2 + d(B, C)^2$ if and only if $\angle ABC = \pi / 2$.