Continuing the series on triangle centers, let us consider ∆*ABC*, with circumcenter *O* and centroid *G*. Construct the line segment *OG*, and extend it out from *G* to the point *H* such that *GH*=2*OG*.

Next, construct the median from vertex *A* to the midpoint *M* of side *BC*. Then *G* lies on *AM*, with *AG*=2*GM*, as we showed here. Recalling that the circumcenter is the intersection of the perpendicular bisectors of the sides, we see that *OM*⊥*BC*.

Now, since *AG*=2*GM*, *GH*=2*OG*, and ∠*AGH*≅∠*MGO*, we see (by the SAS similarity condition) that ∆*AGH*~∆*MGO*. And since these triangles are similar, corresponding angles ∠*HAG* and ∠*OMG* are congruent. However, these are alternate interior angles for lines and cut by transversal , and therefore . And since *OM*⊥*BC*, we see , and is the triangle altitude from *A* to *BC*.

Analogous constructions show that *H* must also be on the triangle altitudes from *B* and *C*:

Thus, we see that *H* is the orthocenter of ∆*ABC*. So, we see that for any non-equilateral triangle, the circumcenter *O*, centroid *G* and orthocenter *H* are collinear, with *GH*=2*OG*; the line through these triangle centers is known as the Euler line of the triangle. (For an equilateral triangle, *O*, *G* and *H* are all the same point.)