by Mimii08, May 8, 2025, 10:36 PM
Hi! I found this interesting geometry problem and I would really appreciate help with the proof.
Let ABC be an acute triangle, and let T be the Fermat (Torricelli) point of triangle ABC. Let A1, B1, and C1 be the feet of the perpendiculars from T to the sides BC, AC, and AB, respectively. Let ω be the circle passing through points A1, B1, and C1. Let A2, B2, and C2 be the second points where ω intersects the sides BC, AC, and AB, respectively (different from A1, B1, C1).
Prove that triangle A2B2C2 is equilateral.