Problem (2017CTSTp14): Let $ABCD$ be a non-cyclic convex quadrilateral. The feet of perpendiculars from $A$ to $BC,BD,CD$ are $P,Q,R$ respectively, where $P,Q$ lie on segments $BC,BD$ and $R$ lies on $CD$ extended. The feet of perpendiculars from $D$ to $AC,BC,AB$ are $X,Y,Z$ respectively, where $X,Y$ lie on segments $AC,BC$ and $Z$ lies on $BA$ extended. Let the orthocenter of $\triangle ABD$ be $H$. Prove that the common chord of circumcircles of $\triangle PQR$ and $\triangle XYZ$ bisects $BH$.
AlphaGeometry formalization: a b c d = quadrangle; p = foot a b c; q = foot a b d; r = foot a c d; x = foot d a c; y = foot d b c; z = foot d a b; h = orthocenter a b d; o1 = circumcenter p q r; o2 = circumcenter x y z; k1 = on_circle o1 p, on_circle o2 x; k2 = on_circle o1 p, on_circle o2 x; m = midpoint b h ? coll m k1 k2
HAGeo formalization: A B C D = quadrilateral; BC = line B C; P = foot A BC; BD = line B D; Q = foot A BD; CD = line C D; R = foot A CD; AC = line A C; X = foot D AC; Y = foot D BC; AB = line A B; Z = foot D AB; H = orthocenter A B D; O1 = circumcenter P Q R; O2 = circumcenter X Y Z; (PQR) = circle_center_point O1 P; (XYZ) = circle_center_point O2 X; K1 K2 = intersection (PQR) (XYZ); M = midpoint B H; Prove: collinear M K1 K2
Difficulty score: 4.7