MathDB
Another "triangle BIC" configuration

Source: Greece National Olympiad 2023, Problem 3

February 19, 2023
geometryperpendicularincentercircumcircle

Problem Statement

A triangle ABCABC with AB>ACAB>AC is given, ADAD is the A-angle bisector with point DD on BCBC and point II is the incenter of triangle ABCABC. Point M is the midpoint of segment ADAD and point FF is the second intersection of MBMB with the circumcirle of triangle BICBIC. Prove that AF⊥FCAF\bot FC.
Back to ProblemsView on AoPS