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Midpoint of $YA^*$ lies on A-excircle

Source: 10.4 of XX Geometrical Olympiad in honour of I.F.Sharygin

August 7, 2024
geogeometry

Problem Statement

Let II be the incenter of a triangle ABCABC. The lines passing through AA and parallel to BI,CIBI, CI meet the perpendicular bisector to AIAI at points S,TS, T respectively. Let YY be the common point of BTBT and CSCS, and AA^* be a point such that BICABICA^* is a parallelogram. Prove that the midpoint of segment YAYA^* lies on the excircle of the triangle touching the side BCBC.