Problem (2011ARMOg10p6): Given is an acute triangle $ABC$. Its heights $BB_1$ and $CC_1$ are extended past points $B_1$ and $C_1$. On these extensions, points $P$ and $Q$ are chosen, such that angle $PAQ$ is right. Let $AF$ be a height of triangle $APQ$. Prove that angle $BFC$ is a right angle.
AlphaGeometry formalization: a b c = triangle; b1 = foot b a c; c1 = foot c a b; p = on_line b b1; q = on_line c c1, on_tline a a p; f = foot a p q ? perp b f f c
HAGeo formalization: A B C = acute_triangle; AB = line A B; AC = line A C; B1 = foot B AC; C1 = foot C AB; BB1 = line B B1; P = on_line BB1; AP = line A P; l = perpendicular_line A AP; CC1 = line C C1; Q = intersection CC1 l; PQ = line P Q; F = foot A PQ; BF = line B F; FC = line F C; Prove: perpendicular BF FC
Difficulty score: 1.5