In acute triangle ABC, AB>AC. Let I the incenter, Ω the circumcircle of triangle ABC, and D the foot of perpendicular from A to BC. AI intersects Ω at point M(=A), and the line which passes M and perpendicular to AM intersects AD at point E. Now let F the foot of perpendicular from I to AD.
Prove that ID⋅AM=IE⋅AF. geometryKoreageometry solvedAngle Chasingsimilar trianglesincenterarc midpoint