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Mansion point and incenter

Source: Korea National Olympiad 2019 P6

November 16, 2019
geometryKoreageometry solvedAngle Chasingsimilar trianglesincenterarc midpoint

Problem Statement

In acute triangle ABCABC, AB>ACAB>AC. Let II the incenter, Ω\Omega the circumcircle of triangle ABCABC, and DD the foot of perpendicular from AA to BCBC. AIAI intersects Ω\Omega at point M(≠A)M(\neq A), and the line which passes MM and perpendicular to AMAM intersects ADAD at point EE. Now let FF the foot of perpendicular from II to ADAD. Prove that ID⋅AM=IE⋅AFID\cdot AM=IE\cdot AF.
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