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Let ABC be a triangle with incenter I, line AI meets the circumcircle of △ABC again at L. We let I_A be the reflection of I over L. Then, we have two facts, first fact is (a) the points I, B, C and I_A lie on a circle with diameter II_A and center L. In particular, LI = LB = LC = LI_A. The second fact is (b) lines BI_A and CI_A bisect the exterior angles of △ABC. In other words, they are the exterior angle bisectors of ∠B and ∠C. Can you prove this lemma? Well, this is a useful lemma in Euclidean geometry, especially if you are solving Math Olympiad geometry problems and I want to point out a great book in learning it which is the ”Euclidean Geometry in Mathematical Olympiad” by Evan Chen, there is a link below to this book. 🔥Subscribe to 1Psi3Colour: https://www.youtube.com/c/1Psi3Colour... Video Chapters: 0:00 The Lemma 1:09 First fact 3:31 Second fact 4:51 Outro + Subscribe! ”Euclidean Geometry in Mathematical Olympiad” by Evan Chen: https://web.evanchen.cc/geombook.html 💥Check out these videos: The integral everyone should know: • The MOST Influential Integral In ALL ... A deceivingly easy Putnam problem!: • Is This THE EASIEST Problem On THE HA... 👉Check out these playlists: Math Olympiad: • Math Olympiad Math puzzles: • Math puzzles College Math: • College Math This video is created using Manim: https://www.manim.community 👉Suggest a problem: https://forms.gle/hTibrUKz7QNyqoC48