Prove that for any triangle, H, G, and C are collinear, and

prove that HG = 2GC

Why are orthocenter(H), centroid(G) and circumcenter(C) collinear?

For any triangle let's construct an orthocenter(H) and centroid(G).

We can consider a line passing through H and G and expect that circumcenter is on the line.

If we define the intersect of the line and the perprndicular bisector of side BC as X,

What relation is there between triangle AHG and triangle DXG?

Since both line AH and line XD are perpendicular of side BC, they are parallel each other and by the theorem of parallel lines,

angle A of triangle AHG = angle D of triangle DXG

angle H of triangle AHG = angle X of triangle DXG

therefore triangle AHG is similar to triangle DXG

Also, since G is centroid, length of AG = 2 length of GD

therefore length of HG = 2 length of GX --- (*)

We have to show that X is circumcenter (i.e.the point Y which the perpendicular bisector of side AC intersect with the line HG is exactly to be X).

Likewise the discussion above,

triangle BHG is similar to triangle EYG

length of HG = 2 length of GY --- (**)

By (*) and (**), Y must be same with X. Therefore X is a circumcenter.

Our Result
For any triangle, H, G, and C are collinear

HG = 2GC

Do the various shapes of triangles satisfy the above relation?

Let's try to make sure by using measure in GSP for many cases.

Case 1. Acute triangle

Case 2. Right triangle

The point which is on right angle becomes a orthocenter and the midpoint of hypotenus becomes a cicumcenter.Therefore by the property of centroid HG=2GX

Case 3. Obtuse triangle