Let $I_a,I_b,I_c$ be the excenters of $\triangle ABC$ against $A,B,C.$ Circumcircle $(O)$ and incenter $I$ of $\triangle ABC$ become 9-point circle and orthocenter of $\triangle I_aI_bI_c.$ If $D,E$ denote the tangency points of the excircles $(I_a),(I_c)$ with $CB,BA,$ then $U \equiv DI_a \cap EI_c$ is the circumcenter of $\triangle I_aI_bI_c$ $\Longrightarrow$ $IOU$ is the Euler line of $\triangle I_aI_bI_c,$ thus $\overline{OU}=-\overline{OI}$ $\Longrightarrow$ $BUKI$ is a parallelogram with center $O,$ i.e. $KI \parallel BU \ (*).$ Since $\angle BDU=\angle BEU=90^{\circ},$ it follows that circumcenter $V$ of $\triangle BED$ is the midpoint of $\overline{UB}.$ But, it's clear that $L,N$ are the tangency points of $(I_a),(I_c)$ with $AB,BC$ $\Longrightarrow$ $NL,ED$ are antiparallel WRT $BA,BC,$ since $NLDE$ is an isosceles trapezoid with legs $NL=ED.$ Thus, $BV \perp NL$ and together with $(*),$ it follows that $KI \perp NL.$

Let D mipoint BI, (D) cuts BC, BA at E,F then we need OL^2-ON^2=DN^2-DL^2 so LB(LA-LF)=NB(NE-NB) so LB.AF=NB.EC obviously true by AF=NB, CE=LB so by 4 point lemma we have the conclusion

What Is the "4 point lemma",pls?

soryn wrote: What Is the "4 point lemma",pls? In the google search engine, the engine refers to Burnside's lemma.

Dear Mathlinkers, Sharygin I., Problema II 137, Problemas de geometria, Editions Mir (1986) 95 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=391586 Sincerely Jean-Louis