137. Stewart's relation.

by Virgil Nicula, Oct 3, 2010, 3:15 PM

Consider a line $d$ and $\{A,B,C\}\subset d$ and a point $P$ . Then exists the Steiner's relation

$\boxed{PA^2\cdot\overline{BC}+PB^2\cdot\overline{CA}+PC^2\cdot\overline{AB}+\overline {BC}\cdot\overline{CA}\cdot\overline{AB}=0}$ .

Proof. Denote $T\in d$ , $PT\perp d$ and $PA=x$ , $PB=y$ , $PC=z$ . Using the generalized Pytagoras' relation obtain :

$\left\|\begin{array}{cc} x^2=z^2+AC^2-2\cdot \overline{AC}\cdot\overline{TC}=0\\\\ y^2=z^2+BC^2-2\cdot\overline{BC}\cdot\overline{TC}=0\end{array}\right|\left|\begin{array}{c} \odot\ \overline {BC}\\\\ \odot\ \overline{CA}\end{array}\right\|\ \bigoplus\implies$

$x^2\cdot\overline{BC}+y^2\cdot\overline{CA}=$ $z^2\cdot\left(\overline{BC}+\overline{CA}\right)+AC^2\cdot \overline{BC}+BC^2\cdot\overline{CA}-2\cdot\overline{AC}\cdot\overline{TC}\cdot\overline{BC}-2\cdot\overline{BC}\cdot\overline{TC}\cdot\overline{CA}$ $\iff$

$x^2\cdot\overline{BC}+y^2\cdot\overline{CA}=z^2\cdot \overline{BA}+\overline{CA}\cdot\overline{BC}\cdot\left(\overline{CA}+\overline{BC}\right)$ $\iff$ $x^2\cdot\overline{BC}+y^2\cdot\overline{CA}=z^2\cdot\overline{BA}+\overline{BC}\cdot\overline{CA}\cdot\overline{BA}$ $\iff$

$x^2\cdot\overline{BC}+y^2\cdot\overline{CA}+z^2\cdot\overline{AB}+\overline {BC}\cdot\overline{CA}\cdot\overline{AB}=0$ .
This post has been edited 16 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:11 AM

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