Problem (2011CTSTp10): Let $AA',BB',CC'$ be three diameters of the circumcircle of an acute triangle $ABC$. Let $P$ be an arbitrary point in the interior of $\triangle ABC$, and let $D,E,F$ be the orthogonal projection of $P$ on $BC,CA,AB$, respectively. Let $X$ be the point such that $D$ is the midpoint of $A'X$, let $Y$ be the point such that $E$ is the midpoint of $B'Y$, and similarly let $Z$ be the point such that $F$ is the midpoint of $C'Z$. Prove that triangle $XYZ$ is similar to triangle $ABC$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; a1 = mirror a o; b1 = mirror b o; c1 = mirror c o; p = free; d = foot p b c; e = foot p a c; f = foot p a b; x = mirror a1 d; y = mirror b1 e; z = mirror c1 f ? simtri x y z a b c
HAGeo formalization: A B C = acute_triangle; O = circumcenter A B C; A1 = reflect_point_wrt_point A O; B1 = reflect_point_wrt_point B O; C1 = reflect_point_wrt_point C O; P = point; BC = line B C; D = foot P BC; AC = line A C; E = foot P AC; AB = line A B; F = foot P AB; X = reflect A1 D; Y = reflect B1 E; Z = reflect C1 F; Prove: similar X Y Z A B C
Difficulty score: 3.6