Bariz - Problem

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Tags: geometry, geometry unsolved

Moron

06.03.2015 12:51

Given acute triangle $ABC$ with circumcenter $O$ and the center of nine-point circle $N$. Point $N_1$ are given such that $\angle NAB = \angle N_1AC$ and $\angle NBC = \angle N_1BA$. Perpendicular bisector of segment $OA$ intersects the line $BC$ at $A_1$. Analogously define $B_1$ and $C_1$. Show that all three points $A_1,B_1,C_1$ are collinear at a line that is perpendicular to $ON_1$.

My solution: Let $ \ell_A, \ell_B, \ell_C $ be the perpendicular bisector of $ OA, OB, OC $, respectively . Let $ \triangle A_2B_2C_2 $ be a triangle with sides $ \ell_A, \ell_B, \ell_C $ ($ A_2=\ell_B \cap \ell_C, B_2=\ell_C \cap \ell_A, C_2=\ell_A \cap \ell_B $) . Since $ N_1 $ is the Kosnita point of $ \triangle ABC $ , so $ N_1 $ lie on $ AA_2, BB_2, CC_2 $ ( well-known) . From Desargue theorem ( for $ \triangle ABC $ and $ \triangle A_2B_2C_2 $ ) we get $ A_1, B_1, C_1 $ are collinear . From Sondat's theorem ( for $ \triangle ABC $ and $ \triangle A_2B_2C_2 $ ) we get $ ON_1 $ is perpendicular to their perspectrix $ \overline{A_1B_1C_1} $ . Q.E.D