Problem

Source: 38th Brazilian MO (2016) - First Day, Problem 1

Tags: geometry, incenter, Parallel Lines, Brazilian Math Olympiad 2016



Let ABC be a triangle. r and s are the angle bisectors of ABC and BCA, respectively. The points E in r and D in s are such that AD and AE \| CD. The lines BD and CE cut each other at F. I is the incenter of ABC. Show that if A,F,I are collinear, then AB=AC.