olymmath

ISL 2011 G6

    Let ABCA B C be a triangle with AB=ACA B=A C, and let DD be the midpoint of ACA C. The angle bisector of ∠BAC\angle B A C intersects the circle through D,BD, B, and CC in a point EE inside the triangle ABCA B C. The line BDB D intersects the circle through A,EA, E, and BB in two points BB and FF. The lines AFA F and BEB E meet at a point II, and the lines CIC I and BDB D meet at a point KK. Show that II is the incenter of triangle KABK A B.

    Dataset text NuminaMath olympiads. Quoted with source.

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