Cathetus theorem/Linear algebra/Fact/Proof

We set ${\displaystyle {}v=C-A}$, and ${\displaystyle {}w=B-C}$. The vector from ${\displaystyle {}A}$ to ${\displaystyle {}B}$ equals ${\displaystyle {}v+w}$. Due to fact, the foot of the altitude is

${\displaystyle {}{\begin{aligned}D&=A+p_{(v+w)\mathbb {R} }(v)\\&=A+\left\langle v,{\frac {v+w}{\Vert {v+w}\Vert }}\right\rangle {\frac {v+w}{\Vert {v+w}\Vert }}\\&=A+{\frac {\left\langle v,v\right\rangle }{\Vert {v+w}\Vert ^{2}}}(v+w)\\&=A+{\frac {\left\langle v,v\right\rangle }{\left\langle v,v\right\rangle +\left\langle w,w\right\rangle }}(v+w).\end{aligned}}}$

Hence, we obtain

${\displaystyle {}{\begin{aligned}d(A,D)\cdot d(A,B)&={\frac {\left\langle v,v\right\rangle }{\left\langle v,v\right\rangle +\left\langle w,w\right\rangle }}\Vert {v+w}\Vert \cdot \Vert {v+w}\Vert \\&={\frac {\left\langle v,v\right\rangle }{\left\langle v,v\right\rangle +\left\langle w,w\right\rangle }}\left\langle v+w,v+w\right\rangle \\&=\left\langle v,v\right\rangle \\&=\Vert {v}\Vert ^{2}\\&=d(A,C)^{2}.\end{aligned}}}$