Problem (2014IranGOAp3): Let $ABC$ be an acute triangle.A circle with diameter $BC$ meets $AB$ and $AC$ at $E$ and $F$,respectively. $M$ is midpoint of $BC$ and $P$ is point of intersection $AM$ with $EF$. $X$ is an arbitary point on arc $EF$ and $Y$ is the second intersection of $XP$ with a circle with diameter $BC$.Prove that $ \measuredangle XAY=\measuredangle XYM $.
AlphaGeometry formalization: a b c = triangle; m = midpoint b c; e = on_line a b, on_circle m b; f = on_line a c, on_circle m b; p = on_line a m, on_line e f; x = on_circle m b; y = on_line x p, on_circle m b ? eqangle x a a y x y y m
HAGeo formalization: A B C = acute_triangle; AB = line A B; M = midpoint B C; (M) = circle_center_point M B; E = intersection AB (M); AC = line A C; F = intersection AC (M); AM = line A M; EF = line E F; P = intersection AM EF; X = on_circle (M); PX = line P X; Y = intersection PX (M); Prove: equal_angle X A Y X Y M
Difficulty score: 3.0