$G$ is a point inside a parallelogram $ABCD$.
$(C)$ is a circle passing through $A$ and $G$ meeting sides $[AB]$ and $[AD]$ in $E$ and $F$ respectively.
Line $(EG)$ meets side $[CD]$ in $I$ and lines $(FG)$ meets side $[BC]$ in $H$.
The circumcircle of $\triangle HGI$ meets $(C)$ again in $M$. ($M \ne G$).
Prove that $M$ lies on the diagonal $[AC]$.
I tried using angles but failed so far in proving that $\angle AMC$ is $180$.
The proof is possible if we prove beforehand that $C$ is concyclic with $H$ $G$ and $I$.
First, $G, I, C, H$ are concyclic. To see this, note that $$\angle HGI+\angle DCB=\angle EGF+\angle BAD=180^\circ$$ since $A, E, G, F$ are concyclic. Thus, we have $G, M, C, I$ are concyclic, and so $$\angle AMG+\angle GMC=\angle AEG+(180^\circ-\angle GIC)=180^\circ,$$ where the last equality uses the fact that $\angle AEG=\angle GIC$ (as $AB$ and $DC$ are parallel).