168. Some properties of the Lemoine's point.

by Virgil Nicula, Oct 27, 2010, 1:57 PM

Quote:
PP1. Let $\triangle ABC$ with centroid $G$ , circumcenter $O$ and symmedian center $S$ (Lemoine's point). Prove that $AS$ is the bisector of $\angle BSC\ \iff\ OG\perp AG$ .
Proof. Denote $D\in AS\cap BC$ , $E\in BS\cap CA$ , $F\in CS\cap AB$ . Using the well-known relations $\boxed{\frac {DB}{DC}=\frac {c^2}{b^2}}$ a.s.o. prove easily that $AD=\frac{2bc}{b^2+c^2}\cdot m_a$

and $SA=\frac{b^2+c^2}{a^2+b^2+c^2}\cdot AD$ $\implies$ $\boxed{SA=\frac {2bc\cdot m_a}{a^2+b^2+c^2}}$ a.s.o. Thus, $AS$ is the bisector of $\angle BSC\iff \frac {SB}{SC}=\frac {DB}{DC}\iff$ $\frac {\frac {2ca}{a^2+b^2+c^2}\cdot m_b}{\frac {2ab}{a^2+b^2+c^2}\cdot m_c}=\frac {c^2}{b^2}$ $\iff$

$b\cdot m_b=c\cdot m_c$ $\iff$ $b^2\cdot\left[2\left(a^2+c^2\right)-b^2\right]=c^2\cdot\left[2\left(b^2+a^2\right)-c^2\right]$ $\iff$ $\boxed {b^2+c^2=2a^2}\ (*)$ . But $OG\perp AG$ $\iff$ $GO^2+GA^2=OA^2$ $\iff$

$R^2-OG^2=GA^2$ $\iff$ $\frac 19\cdot\left(a^2+b^2+c^2\right)=\frac 49\cdot m_a^2$ $\iff$ $a^2+b^2+c^2=2\left(b^2+c^2\right)-a^2$ $\iff$ $b^2+c^2=2a^2$ , i.e. the relation $(*)$ .

Generalization.
This post has been edited 17 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:24 PM

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