The inscribed circle $\omega$ of the triangle $ABC$ touches its sides $BC, CA$ and $AB$ at points $A_1, B_1$ and $C_1$, respectively. Let $S$ be the intersection point of lines passing through points $B$ and $C$ and parallel to $A_1C_1$ and $A_1B_1$ respectively, $A_0$ be the foot of the perpendicular drawn from point $A_1$ on $B_1C_1$, $G_1$ be the centroid of triangle $A_1B_1C_1$, $P$ be the intersection point of the ray $G_1A_0$ with $\omega$. Prove that points $S, A_1$, and $P$ lie on a straight line.