Talk:2023 IMO Problems/Problem 2

The proof is correct, but there is the necessity to show that points T, H, F are collinear ! In other words, It must be shown that corresponding angles AFH and ACB are congruent, in such a way that HF and BC are parallel, as TH and BC are builded parallel. It is true if AFB = 2 ACB , so ABF = ABC - ACB But ABF = 2 phi = 2 AES = 2 EAS' = 2 (BAS' - BAE) = 2 (1/2 BAC -(90° - ABC) ) = BAC - 180 ° + 2 ABC = BAC - (ABC + BCA + CAB) + 2 ABC = ABC - ACB , quod demonstrandum erat .