Problem (2005G6): Let $ABC$ be a triangle, and $M$ the midpoint of its side $BC$. Let $\gamma$ be the incircle of triangle $ABC$. The median $AM$ of triangle $ABC$ intersects the incircle $\gamma$ at two points $K$ and $L$. Let the lines passing through $K$ and $L$, parallel to $BC$, intersect the incircle $\gamma$ again in two points $X$ and $Y$. Let the lines $AX$ and $AY$ intersect $BC$ again at the points $P$ and $Q$. Prove that $BP = CQ$.
AlphaGeometry formalization: a b c = triangle; m = midpoint b c; i = incenter a b c; d = foot i b c; k = on_line a m, on_circle i d; l = on_line a m, on_circle i d; x = on_pline k b c, on_circle i d; y = on_pline l b c, on_circle i d; p = on_line a x, on_line b c; q = on_line a y, on_line b c ? cong b p c q
HAGeo formalization: A B C = triangle; M = midpoint B C; I = incenter A B C; BC = line B C; D = foot I BC; (I) = circle_center_point I D; AM = line A M; K = intersection AM (I); L = intersection AM (I); BC = line B C; l1 = parallel_line K BC; X = intersection l1 (I); l2 = parallel_line L BC; Y = intersection l2 (I); AX = line A X; P = intersection AX BC; AY = line A Y; Q = intersection AY BC; Prove: cong B P C Q
Difficulty score: 3.0