In this exploration we will prove that:
1. The orthocenter, centroid, and circumcenter are collinear
2. The distance from the centroid and the orthocenter is two times the distance between the orthocenter and circumcenter
Definitions:
Orthocenter: (L) the point of intersection of the 3 altitudes of any triangle. An altitude is the perpendicular line constructed from the vertex to the line on the opposite side. The orthocenter may lie outside the triangle.
Centroid: (M) the point of intersection of 3 medians of any triangle. A median is the line constructed from the vertex to the midpoint of the opposite side of the triangle. We know that the centroid is located 2/3 of the way from the vertext to the opposite side.
Circumcenter: (N) the point of intersection of the 3 perpendicular bisectors of the sides of any triangle. The circumcenter may lie outside the triangle.
Now we will prove that the orthocenter, centroid, and circumcenter are collinear and the distance from LM is twice the distance from MN.
1. Let D be the midpoint of BC and construct AD. By definition of the centroid, M is on the line AD. Let X be the foot of the altitude from vertex A to side BC. Construct the altitude AX. By definition of an altitude, AX intersects BC at a right angle.
2. Construct the line MN. It appears to be going through L, but we must prove that. For now, call the point L'.
3. Construct the line ND. By definition, ND meets BC at a right angle. Since ND and AX meet BC at a right angle, this proves that ND and AX are parallel and AD is a transversal of ND and AX.
4. We now know that:
angle MAL' = angle MDN because they are alternate interior angles
angle AML' = angle DMN becaue they are vertical angles
5. M is the centroid, so we know that AM = (2/3)AD and MD = (1/3)AD. Therefore, AM = 2MD.
6. By Angle-Side-Angle, we know that triangle AML' is congruent to triangle DMN and that the distances L'M = 2MN.
7. Thus, we have proved that L' is collinear with line MN. We can prove that L' is located at the orthocenter because it is the intersection point of the three altitudes from vertices A, B, and C. By definition, L is the orthocenter of triangle ABC, so L = L'.
8. In conclusion: L, M, and N are collinear (they exist on the Euler Line) where LM = 2MN.
