Show that three point $G,H,G_1$ are collinear. Triangle $ABC$ has centroid $ G$ and orthcenter $H$. Line (through $A$) is perpendicular to $GA$, line (through $B$) is perpendicular to  $GB$, line (through $C$) is perpendicular to  $GC$ cut at three points which form a new triangle $A_1B_1C_1$. This new triangle has centroid $G_1$. Show that three point $G,H,G_1$ are collinear.
I have tried to so this problem with lots of theorems. But I can't find the way to solve. Or using any lemma? Help me to find and draw any auxiliary geometry element.
 A: 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.
