Thales's theorem/Linear algebra/Fact/Proof

Proof

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

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

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

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