Given the triangle $ABC$, $M$ is the point inside that triangle. Construct $BMCD$ and $MAED$ parallelograms. Prove that the $ME$ line always goes through a fixed point when $M$ moves.
(In the drawing, I found the line $ ME $ always through the fixed point centroid $ G $ of the triangle ABC, but I have not proven it.)
$$\vec{EM}=\vec{EA}+\vec{AM}=\vec{DM}+\vec{AM}=\vec{DB}+\vec{DC}+\vec{AM}=$$ $$=\vec{CM}+\vec{BM}+\vec{AM}=\vec{CB}+\vec{BM}+\vec{BM}+\vec{AB}+\vec{BM}=3\vec{BM}+\vec{AB}+\vec{CB}.$$ In another hand, $$\vec{EG}=\vec{ED}+\vec{DI}+\vec{IG}=\vec{AM}+\frac{1}{2}(\vec{DB}+\vec{DC})+\frac{1}{3}\vec{IA}=$$ $$=\vec{AB}+\vec{BM}+\frac{1}{2}(\vec{CB}+2\vec{BM})+\frac{1}{3}(\frac{1}{2}\vec{CB}-\vec{AB})=$$ $$=2\vec{BM}+\frac{2}{3}\vec{AB}+\frac{2}{3}\vec{CB}=\frac{2}{3}\vec{EM}$$ and we are done!
Another way.
Let $EM\cap AI=\{G'\}$.
Thus, since $\Delta AG'E\sim\Delta IG'M,$ we obtain: $$\frac{AG'}{G'I}=\frac{AE}{MI}=2,$$ which says that $G'\equiv G$ and we are done!