Problem (2009G3): Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; z = foot i a b; y = foot i a c; g = on_line b y, on_line c z; r = parallelogram y c b r; s = parallelogram z b c s ? cong g r g s
HAGeo formalization: A B C = triangle; I = incenter A B C; AB = line A B; Z = foot I AB; AC = line A C; Y = foot I AC; BY = line B Y; CZ = line C Z; G = intersection BY CZ; R = parallelogram B C Y; S = parallelogram Z B C; Prove: cong G R G S
Difficulty score: 3.8