Proving the orthocenter, circumcenter and centroid of a triangle are collinear. I just had a problem on a test related to Euler’s Line theorem,
(Euler Line Theorem: The orthocenter H, the circumcenter O, and the centroid G of any triangle are collinear.  Furthermore, G is between H and O (unless the triangle is equilateral, in which case the three points coincide) and HG = 2GO.)
On the test problem, the professor gave us diagram of a triangle with three points in the interior of the triangle.  The question said,
“Use the following diagram to prove synthetically that the circumcenter O, the centroid G, and the orthocenter H of a triangle are collinear.”
Nobody in the class got full credit on it.  He said it should be done this way:
Draw a line segment from O to G, and extend it such that OG=1/2 GH.  Then prove that H is the orthocenter.
This didn’t sit well with me, and I don’t know what to think about his method.

*

*Instead of a proof by contradiction, it seems like a proof by, I don’t know, acceptance?  Accept that what you are trying to prove is true, then use half of what the theorem says to work backwards to prove something in the given.  Is this a valid way to prove something?  I have never heard of this, and it seems like it wouldn’t always be robust.


*It rubs me the wrong way to be trying to prove that 3 points are collinear, and then your first step is to draw them so that they are collinear.  It seems like a circular argument, although in the end he proved that H was the orthocenter.  I like ending with what I am trying to prove, but maybe I am not right here.


*Given the theorem, it seems like you are accepting that half of it is true, constructing them in this manner.  If you are using half of the proof, why wouldn’t  it be valid to accept the whole thing is true and say, “By the Euler Line theorem, these three points are collinear.”
Any help would be greatly appreciated.  This is really nagging at me.
 A: The proof you described is perfectly valid: it follows from the similitude of triangles $AGH$ and $MGO$ (see diagram, where $M$ is the midpoint of $BC$), which in turn implies $OM\parallel AH$, so that $AH$ lies on the altitude from $A$ (and the same can reasoning can be repeated for $B$ and $C$). Triangles $AGH$ and $MGO$ are similar because $AG/MG=HG/OG=2$ and $\angle AGH=\angle OGM$ (SAS criterion).
Constructing point $H$ on line $OG$ and then showing that it is the orthocenter is indeed a very simple and clever way to avoid a direct collinearity proof. Compare it, for instance, with the proof given at cut-the-knot, which is more cumbersome.

A: Here is sketch of a different way to prove it.
Find the midlines of ABC.

The circumcenter of ABC is the othocenter of PQR
The centroid of ABC is the centroid of PQR
PQR is similar to ABC
Construct Euler line between the two orthocenter / Circumcenter of PQR / ABC and the Centroid, creating more similar triangles.  And there are corresponding points between the othocenter of PQR and the orhtocenter of ABC along that line.   
