Problem (2018SilkRoadp1): In an acute-angled triangle $ABC$ on the sides $AB$, $BC$, $AC$ the points $H$, $L$, $K$ so that $CH \perp AB$, $HL \parallel AC$, $HK \parallel BC$. Let $P$ and $Q$ feet of altitudes of a triangle $HBL$, drawn from the vertices $H$ and $B$ respectively. Prove that the feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and $H$ lie on the line $PQ$.
AlphaGeometry formalization: a b c = triangle; h = foot c a b; l = on_line b c, on_pline h a c; k = on_line a c, on_pline h b c; p = foot h b l; q = foot b h l; x = foot a h k; y = foot h a k ? coll p q x
HAGeo formalization: A B C = acute_triangle; AB = line A B; H = foot C AB; AC = line A C; l1 = parallel_line H AC; BC = line B C; L = intersection l1 BC; l2 = parallel_line H BC; K = intersection l2 AC; BL = line B L; P = foot H BL; HL = line H L; Q = foot B HL; HK = line H K; X = foot A HK; AK = line A K; Y = foot H AK; Prove: collinear P Q X Y
Difficulty score: 1.4