Prove that O,G,H are colinear and OG/GH = 1/2 pending

Prove that Circumcenter(O),Centroid(G) and Orthocenter(H) in a triangle are colinear - also known as Euler's line - and OG/GH = 1/2.

Proof:
Simplest way to see it in action is in a right triangle.

In a right triangle:

|\
| \
|   \  O
| G  \
|       \
-------
H

It's clear that OGH is a straight line.
Why?
Orthocenter is simply the intersection of base and height.
Circumcenter is the midpoint of hypotenuse and HO is the median hence G lies on HO.
Centroid divides the median in 2:1 hence GH = 2OG.

Now, a formal proof.
In a triangle ABC, we know that AH = 2OM where M is midpoint of BC.
Also: AH || OM why?
Since AH is perpendicular to BC by definition.
BC is a chord in the circumcircle and O is the center so OM will be perpendicular bisector of BC.
Hence AH || OM

And of course A,G,M are colinear.
Now we claim that triangles AGH and MGO are similar.
Why?
AH = 2MO
AG = 2MG

Now the traversal AGM intersects the || lines AH and OM => angles HAG and OMG are equal.
=> AGH ~ MGO
=> GH = 2OG by similarity ratio.

Now, the collinearity of O,G,H.
Since angles AGH = MGO and AGM is a straight line it's only possible when OGH is a straight line. H.P.

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