Problem (2016BalkanMOp2): Let $ABCD$ be a cyclic quadrilateral with $AB<CD$. The diagonals intersect at the point $F$ and lines $AD$ and $BC$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $AD$ and $BC$ respectively, and let $M$, $S$ and $T$ be the midpoints of $EF$, $CF$ and $DF$ respectively. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the segment $CD$.
AlphaGeometry formalization: a b c = triangle; d = on_circum a b c; f = on_line a c, on_line b d; e = on_line a d, on_line b c; k = foot f a d; l = foot f b c; m = midpoint e f; s = midpoint c f; t = midpoint d f; o1 = circumcenter m k t; o2 = circumcenter m l s; p = on_circle o1 m, on_circle o2 m ? coll c d p
HAGeo formalization: A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); AC = line A C; BD = line B D; F = intersection AC BD; AD = line A D; BC = line B C; E = intersection AD BC; K = foot F AD; L = foot F BC; M = midpoint E F; S = midpoint C F; T = midpoint D F; O1 = circumcenter K M T; O2 = circumcenter L M S; (O1) = circle_center_point O1 M; (O2) = circle_center_point O2 M; X = intersection (O1) (O2); CD = line C D; Prove: collinear X C D
Difficulty score: 2.3