Ratio of Segments on the Euler Line

by

Mike Rosonet

This page is devoted to proving that, for any triangle,

- the centroid, orthocenter, and circumcenter are collinear, and

- the distance between the centroid and the orthocenter is twice the distance between the centroid and the circumcenter.

There are several special points in the center of a triangle, but focus on four of them: the centroid, the orthocenter, the circumcenter, and the incenter. Defining them first is necessary in order to see their relationship with each other.

The centroid (G) of a triangle is the point of intersection of the three medians of the triangle. A median is a segment constructed from a vertex to the midpoint of the subtending side of the triangle.

The centroid is located 2/3 of the way from the vertex to the midpoint of the opposite side.

The orthocenter (H) of a triangle is the point of intersection of the three altitudes of the triangle. An altitude is a line constructed from a vertex to the subtending side of the triangle and is perpendicular to that side. It should be noted that the orthocenter, in different cases, may lie outside the triangle; in these cases, the altitudes extend beyond the sides of the triangle.

The circumcenter (C) of a triangle is the point of intersection of the three perpendicular bisectors of the triangle. A perpendicular bisector is a line constructed at the midpoint of a side of a triangle at a right angle to that side. It should be noted that the circumcenter, in different cases, may lie outside the triangle; in these cases, the perpendicular bisectors extend beyond the sides of the triangle.

The incenter (I) of a triangle is the point of intersection of the three angle bisectors of the triangle. An angle bisector is a line whose points are all equidistant from the two sides of the angle. Thus, the incenter (I) is equidistant from all three sides of the triangle.

Now, we will prove that the centroid (G), the orthocenter (H) and the circumcenter (C) are collinear and that HG is congruent to 2GC.

1. Let

Xbe the midpoint ofEF. Construct the medianDX. SinceGis the centroid,Gis onDXby the definition of centroid. Also, construct the altitudeDM. SinceHis the orthocenter,His onDMby the definition of orthocenter. Therefore,DMmeetsEFat a right angle.

2. Construct a line through points

CandGso that it intersectsDM. Note, it appears that lineCGintersectsDMat the pointH. However, this has not been proven yet. So, label the point of intersectionH'.

3. Construct

CX. SinceCis the circumcenter,CXmeetsEFat a right angle by the definition of circumcenter.

4. Since

DMandCXboth meetEFat right angles,DMis parallel toCX. Then, the medianDXis a transversal that cutsDMandCX.5.

6. Since

Gis the centroid,DG= (2/3)DXandGX= (1/3)DX. Therefore,DG= 2GX.7.

8. Thus,

H'is located at the intersection ofDMandGCand is in fact on the lineGCsuch thatGH'= 2GC. Thus,G,H', andCare collinear. Similarly,H'can be proved to be located on the altitudes constructed from verticesEandFso thatGH'= 2GC. Therefore,H'lies on all three altitudes.9. Thus,

H'is the orthocenter because it is lies on all three altitudes. Yet, by the given hypothesis,His the orthocenter. Thus,H'=H.10. Therefore,

G,H, andCare collinear, andGH= 2GC.

The line that connects the centroid (G), the orthocenter (H), and the circumcenter (C) is called the Euler Line.