108. Old, short and nice proof of Euler's relation.

by Virgil Nicula, Sep 11, 2010, 4:44 PM

(C.D.P. - Gh. Titeica, 1935) Denote the diameter $ [NS]$ (north - south) of the circumcircle so that $ NS\perp BC$ and the sideline $ BC$ separates the points $ A$ and $ S$. The incircle touches the side $ [CA]$ in the point $ E$. Is well-known that $ SC = SI$ and the power of the incenter $ I$ w.r.t. the circumcircle $ w$ is $ p_{w}(I) =-IA\cdot IS = OI^{2}-R^{2}$, i.e. $ \boxed{OI^{2}= R^{2}-IA\cdot IS}$. Prove easily that $ \triangle AIE\sim\triangle NSC$, i.e. $ \frac{IA}{SN}=\frac{IE}{SC}$ $ \implies$ $ IA\cdot SC = 2Rr$ $ \implies$ $ IA\cdot IS = 2Rr$ $ \implies$ $ OI^{2}= R^{2}-2Rr$. In the my opinion, this proof is the shortest and very old !


Application (C.D.P. - Gh. Titeica, 1935). Let $\triangle ABC$ with the incircle $w=\mathbb C (I,r)$ and the circumcircle $\Omega=\mathbb C(O,R)\ .$ Denote $P\in (BC)$ so that

$IP\perp BC\ ,$ $\{A,I\}\cap \Omega=\{A,M\}\ ,$ $\{A,O\}\cap\Omega =\{A,D\}$ and $E\in MP\cap DI\ .$ Prove that $E\in \Omega$ and the ray $[EP$ is the $E$-bisector of $\triangle BEC\ .$

Proof. Let $R\in BC$ so that $AR\perp BC\ .$ Hence $IP\parallel AR\iff $ $\widehat{PIM}\equiv\widehat{RAM}\equiv\widehat{IAD}\iff $ $\boxed{\ \widehat{PIM}\equiv\widehat{IAD}}\ (1)\ .$ Thus, $\frac {PI}{IM}=\frac {IA}{AD}\iff$ $\frac r{IM}=\frac {IA}{2R}\iff$ $IA\cdot IM=2Rr\ ,$
what is true. Hence $\boxed{\ \frac {PI}{IM}=\frac {IA}{AD}\ }\ (2)\ .$ So from $(1)$ and $(2)$ get $\triangle PIM\sim\triangle IAD\iff$ $\widehat {PMI}\equiv\widehat{IDA}\iff$ $\widehat {EMA}\equiv\widehat{EDA}\ ,$ i.e. $E\in\Omega$ and the ray $[EP$ is the $E$-bisector of $\triangle BEC\ .$
This post has been edited 30 times. Last edited by Virgil Nicula, Dec 17, 2017, 9:22 AM

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