Cathetus theorem/Linear algebra/Fact/Proof2

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

${\displaystyle {}D=A+\lambda (v+w)\,,}$

with ${\displaystyle {}\lambda \in \mathbb {R} }$; the vector of the altitude has the form

${\displaystyle {}h=C-D=A+v-A-\lambda (v+w)=v-\lambda (v+w)\,.}$

The orthogonality condition for the altitude yields

${\displaystyle {}{\begin{aligned}0&=\left\langle v-\lambda (v+w),v+w\right\rangle \\&={\left(1-\lambda \right)}\left\langle v,v\right\rangle -\lambda \left\langle w,w\right\rangle \\&=\left\langle v,v\right\rangle -\lambda {\left(\left\langle v,v\right\rangle +\left\langle w,w\right\rangle \right)};\end{aligned}}}$

therefore,

${\displaystyle {}\lambda ={\frac {\left\langle v,v\right\rangle }{\left\langle v,v\right\rangle +\left\langle w,w\right\rangle }}\,.}$

Hence, we obtain

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