Let $AB, AC, BC$ be tangent to $\omega$ at $I, J, T$ respectively. Now polarity principle shows that $I,J,G$ are collinear and $A(I,J;M,G)=-1$, so $A(B,C;M,G)=-1$ which means $AG \parallel BC$, $M(A,G;H,C)=-1$. Let $MS$ be tangent to $\omega$ where $S$ is the point of tangency$(S\ne T)$. Polarity again shows $S,T,G$ collinear, and $M(P,G;S,T)=-1$. But $M(A,G;H,C)$ is also $-1$, so $H,M,S$ are collinear, and we are done.

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