Problem (2018G2): Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
AlphaGeometry formalization: a b c = iso_triangle; m = midpoint b c; p = on_pline a b c; x = on_line p b; y = on_line p c, eqangle3 y p m x m p ? cyclic a p x y
HAGeo formalization: A B C = isos_triangle; M = midpoint B C; BC = line B C; l1 = parallel_line A BC; P = on_line l1; PB = line P B; X = on_line PB; ω1 = angle_equal2 M P P X M; PC = line P C; Y = intersection PC ω1; Prove: concyclic A P X Y
Difficulty score: 2.7