Teorema de Pitágoras.

$(A - B)\ \bot\ (A - C)\ \Rightarrow\ \|B - C\|^2 = \|A - B\|^2 + \|A - C\|^2$

Demonstração:

$\langle (A - B), (A - C) \rangle = 0\ \Rightarrow\ \langle A, A \rangle - \langle A, C \rangle - \langle B, A \rangle + \langle B, C \rangle = 0\ \Rightarrow$

$\Rightarrow \langle A, A \rangle + \langle B, C \rangle = \langle A, C \rangle + \langle B, A \rangle$ ${\Large (I)}$

$\|B - C\|^2 = \langle (B - C), (B - C) \rangle = \langle B, B \rangle + \langle C, C \rangle - 2\langle B, C \rangle$ ${\Large (II)}$

$(\|A - B\| + \|A - C\|)^2 = \langle (A - B), (A - B) \rangle + \langle (A - C), (A - C) \rangle + 2\|A - B\|\|A - C\| =$

$= \langle A, A \rangle + \langle B, B \rangle - 2\langle A, B \rangle + \langle A, A \rangle + \langle C, C \rangle - 2\langle A, C \rangle + 2\|A - B\|\|A - C\| \overset{\text{(II)}}{=}$

$\overset{\text{(II)}}{=} \|B - C\|^2 + 2\langle B, C\rangle + 2\langle A, A \rangle - 2\langle A, B \rangle - 2\langle A, C \rangle + 2\|A - B\|\|A - C\| \overset{\text{(I)}}{=}$

$\overset{\text{(I)}}{=} \|B - C\|^2 + 2\|A - B\|\|A - C\|$

Logo, $\fbox{$\|B - C\|^2 = \|A - B\|^2 + \|A - C\|^2$}$.

Quod Erat Demonstrandum.