Problem

Source: 8-th Taiwanese Mathematical Olympiad 1999

Tags: geometry, circumcircle, geometry unsolved



Let $AD,BE,CF$ be the altitudes of an acute triangle $ABC$ with $AB>AC$. Line $EF$ meets $BC$ at $P$, and line through $D$ parallel to $EF$ meets $AC$ and $AB$ at $Q$ and $R$, respectively. Let $N$ be any poin on side $BC$ such that $\widehat{NQP}+\widehat{NRP}<180^{0}$. Prove that $BN>CN$.