For every triangle, it can be shown that the orthocenter, geocenter, and circumcenter lie on a common line.  The simple fact seems to have escaped notice until Euler.  It is easy to show with a diagram.  In the figure O is the circumcenter, G is the geocenter.  We extend the segment OG twice its distance to the point H which we shall show is the orthocenter, and thus on the same line.  The proof comes easily by proving that triangle AHG is congruent to AOG.  The vertical angles at G, the well known fact that the median AA' is divided into a 2:1 ratio by G, and the given construction that GH=2 OG are all that is needed.  From this congruence we can see that AH is parallel to OA' and thus perpendicular to BC.  Since we could do the same construction from any other vertex without moving points OGH and get the same result, we are able to show that the three points lie on a single line.   
   Another well known point that lies on the Euler line is the center of the nine-point circle.  One way this can derived is from the fact that O and H are isogonal conjugates.  One of the known facts about such conjugates is that their pedal triangles have the same circumcircles, and that the center of that circle is at the midpoint of the two conjugates, so N, the center of the nine-point circle is the midpoint of OH.  The vertices of the pedal triangle of H is the "feet" of the three altitudes, the points where the altitudes meet the opposite sides.  The vertices of the Pedal triangle for O is the three midpoints, since O is on the perpendicular bisector of each side.
   Here are a few links to additional information about the Euler line.  The first is to "Interactive Mathematics", a site with lots of interactive Java links to see the relations involved in the Euler line and altitudes.  Here is a second from Dr. David Joyce at Clark Univ. with some good aplets to manipulate, but it does not show the nine point center.