If you're interested
in viewing constructions of triangle centers, see write
up #5. In this assignment, I'm proving a rather interesting phenomenon
about the orthocenter, the circumcenter and the centroid of a given triangle,
which is: the three points are collinear. The line defined by these
points is called the Euler line. The centroid lies along the segment
defined by the circumcenter and orthocenter, 1/3 of the way from the circumcenter
to the orthocenter.

In mathematical terms, let H be the orthocenter, C be the circumcenter and G be the centroid. Prove that for any triangle, H, G, and C are collinear, and prove that HC = 2GC.

If you want to manipulate this drawing to see for yourself, click here.

There's also an animated
version. These are GSP version 4, so they don't run on earlier
versions. The GSP sketch also has some measurements that illustrate
HC = 2GC.

Let BEF be a given triangle. Let H represent the orthocenter, G represent the centroid and C represent the circumcenter.

First of all, add some construction lines to aid in the proof. BI is the altitude from vertex B to side EF, BD is the median from B, and let CD be the perpendicular to EF from C. D is the midpoint of EF. It's worth noting here that CD and BI are parallel lines, since both were constructed as perpendiculars to EF. If we can prove that triangles GCD and GHB are similar with a similarity ratio of 1:2, we're home free!

Any two points determine
a line, so let CH determine a line and prove that G lies on that line.
CH is a transversal of parallel lines defined by CD and BI. As such,
angles DCG and GHB are congruent as alternate interior angles of transversal
CH. BD is also a transversal of the parallel lines defined by CD
and BI. That gives angle HBG congruent to angle CDG.

By Triangle Angle Sum, we know that the sum of angles in every triangle is 180 degrees, so angle CGD must be equal to angle HGB. If you want to see the arithmetic, click here. Since the two angles at G are equal, we have vertical angles at G, so G must lie on HC. We also have similar triangles GCD and GHB.

To figure out the similarity
ratio, remember that G is the centroid of the triangle, and lies on the
median BD 1/3 of the way from D to B. That gives DG:GB = 1:2, which
is the similarity ratio for triangles GCD and GHB. As corresponding
parts of similar triangles, CG:GH = 1:2, or HG = 2GC. QED.

The sum of angles for
any triangle is 180 degrees, so the measures of angles GDC + CDG + DCG
= 180 and GHB + HBG + HGB = 180. Setting the two equal gives CDG
+ CGD + DCG = GHB + HBG + HGB. We've proved that the measure of DCG = GHB,
and HBG = CDG (since that's what it means for angles to be congruent.)
Subtracting those angles from each side leaves CGD = HGB.