Problem (2019ARMOg10p6): Let $L$ be the foot of the internal bisector of $\angle B$ in an acute-angled triangle $ABC.$ The points $D$ and $E$ are the midpoints of the smaller arcs $AB$ and $BC$ respectively in the circumcircle $\omega$ of $\triangle ABC.$ Points $P$ and $Q$ are marked on the extensions of the segments $BD$ and $BE$ beyond $D$ and $E$ respectively so that $\measuredangle APB=\measuredangle CQB=90^{\circ}.$ Prove that the midpoint of $BL$ lies on the line $PQ.$
AlphaGeometry formalization: a b c = triangle; l = on_line a c, angle_bisector a b c; o = circumcenter a b c; d = angle_bisector a c b, on_circle o a; e = angle_bisector b a c, on_circle o b; p = on_line b d, on_dia a b; q = on_line b e, on_dia b c; m = midpoint b l ? coll p q m
HAGeo formalization: A B C = acute_triangle; l1 = angle_bisector A B C; AC = line A C; L = intersection l1 AC; O = circumcenter A B C; D = midarc_no B A C; E = midarc_no B C A; (AB) = circle_diameter A B; BD = line B D; P = intersection BD (AB); (BC) = circle_diameter B C; BE = line B E; Q = intersection BE (BC); M = midpoint B L; Prove: collinear M P Q
Difficulty score: 2.3