Problem

Source: 2024 Nepal TST P5

Tags: geometry, angles



Let $ABC$ be an acute triangle so that $2BC = AB + AC,$ with incenter $I{}.$ Let $AI{}$ meet $BC{}$ at point $A'.{}$ The perpendicular bisector of $AA'{}$ meets $BI{}$ and $CI{}$ at points $B'{}$ and $C'{}$ respectively. Let $AB'{}$ intersect $(ABC)$ at $X{}$ and let $XI{}$ intersect $AC'{}$ at $X'{}.$ Prove that $2\angle XX'A'=\angle ABC.{}$ (Proposed by Kang Taeyoung, South Korea)