Problem (2021IranGOMp4): Let $ABC$ be a scalene acute-angled triangle with its incenter $I$ and circumcircle $\Gamma$. Line $AI$ intersects $\Gamma$ for the second time at $M$. Let $N$ be the midpoint of $BC$ and $T$ be the point on $\Gamma$ such that $IN \perp MT$. Finally, let $P $ and $Q$ be the intersection points of $TB $ and $TC$, respectively, with the line perpendicular to $AI$ at $I$. Show that $PB = CQ$.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; o = circumcenter a b c; m = on_line a i, on_circle o a; n = midpoint b c; t = on_circle o a, on_tline m i n; p = on_line t b, on_tline i a i; q = on_line t c, on_tline i a i ? cong p b c q
HAGeo formalization: A B C = acute_triangle; I = incenter A B C; O = circumcenter A B C; (O) = circle_center_point O A; AI = line A I; M = intersection AI (O); N = midpoint B C; IN = line I N; l1 = perpendicular_line M IN; T = intersection (O) l1; l2 = perpendicular_line I AI; BT = line B T; P = intersection BT l2; CT = line C T; Q = intersection CT l2; Prove: cong B P C Q
Difficulty score: 3.0