Here is a partly geometrical, partly mechanical derivation of the collinearity of the orthocenter, circumcenter and barycenter of a triangle ABC:

First, let's look at Ceva's theorem mechanically. In general, Ceva's theorem lets lines ("Cevians") through A, B, and C cut the (produced) opposite sides at points D, E, and F, and requires that the product of ratios AF/FB·BD/DC·CE/EA = 1, from which the conclusion is drawn that the three Cevians are concurrent. In mechanical proofs, masses are placed at A, B, and C, and these masses must be inversely proportional to their distances from D, E and F. The properties of the center of mass are then used to deduce the concurrence. In this context, the hypothesis of Ceva's theorem is really a compatibility condition that allows three mass values to exist which have the proper ratios needed by the mechanical proof. This compatibility condition is needed because since the ratio of mA/mB·mB/mC·mC/mA is clearly 1 for all choices of masses.

Now geometrically it is known that if D, E and F are chosen to be the midpoints of BC, CA, and AB, so that DEF is the medial triangle, then the construction of the orthocenter of DEF becomes identical to the construction of the circumcenter of ABC. I don't know a mechanical proof for this part, but the geometric proof is transparent. It is also clear that DEF is similar to ABC.

Since the construction of the orthocenter uses cevians, there exist masses mA, mB, mC and masses mD, mE, mF which have centers of mass at the orthocenter and circumcenter respectively of ABC. Since ABC and DEF are similar, the proportions between the three masses are identical in both triangles. Now the barycenter of ABC is constructed using the three cevians AD, BE, and CF, and it is known (and there is a simple mechanical proof) that the barycenter lies twice as far from A, B, and C as from D, E, and F respectively.

This all means that by choosing mD, mE, mF to be exactly 1/2mA, 1/2mB, 1/2mC, we have the center of mass of mA and mD at the barycenter, and similarly for mB and mE and mC and mF. Therefore the center of mass of all 6 masses is at the barycenter of ABC, while mA, mB, and mC have their center of mass at the orthocenter and mD, mE, and mF have theirs at the circumcenter. Therefore by properties of the center of mass, these three centers are collinear, and since the masses are in a 2:1 ratio, the barycenter is twice as far from the circumcenter as from the orthocenter.

I know that this has been proven many times by many methods, but I just felt like giving the mechanical method a try. It is not a "pure" mechanical proof, since it uses facts about the medial triangle that are derived by standard geometrical methods -- but then probably no mechanical proof is truly pure by that standard.

--Stuart Anderson

Not only that: they are also homothetic from the barycenter. So already at this point you can conclude that, since

1. the orthocenters of the two triangles correspond by the homothety and
   
2. the orthocenter of DEF coincides with the circumcenter of ABC,
   

the three points in ABC are collinear.

What you say below is correct but seems to me as disguising the above argument.

>Since the construction of the orthocenter uses cevians,
>there exist masses mA, mB, mC and masses mD, mE, mF which
>have centers of mass at the orthocenter and circumcenter
>respectively of ABC. Since ABC and DEF are similar, the
>proportions between the three masses are identical in both
>triangles. Now the barycenter of ABC is constructed using
>the three cevians AD, BE, and CF, and it is known (and there
>is a simple mechanical proof) that the barycenter lies twice
>as far from A, B, and C as from D, E, and F respectively.
>
>This all means that by choosing mD, mE, mF to be exactly
>1/2mA, 1/2mB, 1/2mC, we have the center of mass of mA and mD
>at the barycenter, and similarly for mB and mE and mC and
>mF. Therefore the center of mass of all 6 masses is at the
>barycenter of ABC, while mA, mB, and mC have their center of
>mass at the orthocenter and mD, mE, and mF have theirs at
>the circumcenter. Therefore by properties of the center of
>mass, these three centers are collinear, and since the
>masses are in a 2:1 ratio, the barycenter is twice as far
>from the circumcenter as from the orthocenter.
>
>I know that this has been proven many times by many methods,
>but I just felt like giving the mechanical method a try. It
>is not a "pure" mechanical proof, since it uses facts about
>the medial triangle that are derived by standard geometrical
>methods -- but then probably no mechanical proof is truly
>pure by that standard.
>
>--Stuart Anderson