Thales's theorem/Linear algebra/Fact/Proof

Without loss of generality, we may assume that ${\displaystyle {}E=\mathbb {R} ^{2}}$ and ${\displaystyle {}P=0}$; we denote the radius by ${\displaystyle {}r}$. We write ${\displaystyle {}w=C}$, ${\displaystyle {}v=A}$, ${\displaystyle {}B=-v}$. The vector from ${\displaystyle {}A}$ to ${\displaystyle {}C}$ is ${\displaystyle {}-v+w}$, and the vector from ${\displaystyle {}B}$ to ${\displaystyle {}C}$ is ${\displaystyle {}v+w}$. Therefore,

${\displaystyle {}{\begin{aligned}\left\langle -v+w,v+w\right\rangle &=\left\langle -v,v\right\rangle +\left\langle -v,w\right\rangle +\left\langle w,v\right\rangle +\left\langle w,w\right\rangle \\&=-\left\langle v,v\right\rangle -\left\langle v,w\right\rangle +\left\langle v,w\right\rangle +\left\langle w,w\right\rangle \\&=-\left\langle v,v\right\rangle +\left\langle w,w\right\rangle \\&=-r^{2}+r^{2}\\&=0.\end{aligned}}}$

This means that the two sides of the triangle meeting at ${\displaystyle {}C}$ are perpendicular to each other.