Taking @user10354138's comment, here's how we attack the problem: We will show that the midpoint $O$ of $GG_1$ is the circumcenter of $\triangle ABC$. In particular, we can actually show that $G$ is the midpoint of $HG_1$.
In the picture above, $A_2,B_2,C_2$ are midpoints of $GA_1,GB_1,GC_1$ respectively. Then $C_2$ is on the perpendicular bisector of $AB$. So it is sufficient to show that $C_2O\perp AB$, or $C_1G_1\perp AB$.
Now, if we consider a triangle $XYZ$ with sides (parallel to) the medians of $\triangle ABC$, then
- The sides of $\triangle XYZ$ and $\triangle A_1B_1C_1$ are pairwise perpendicular.
- The medians of $\triangle XYZ$ are parallel to the sides of $\triangle ABC$.
(The existence/construction of $\triangle XYZ$ and the proofs of the above statements are classical and left to you.)
From (1), it follows that $\triangle XYZ$ and $\triangle A_1B_1C_1$ are similar and their medians are pairwise perpendicular. This and (2) yield that $C_1G_1\perp AB$ and so on, which is what we are looking for.