EMAT 6680, Professor Wilson
Exploration 8, Altitudes and Orthocenters by Ursula Kirk
a. Construct any triangle ABC
b. Construct the orthocenter of triangle ABC
c. Construct the orthocenter of triangle HBC
d. Construct the orthocenter of triangle HAB
e. Construct the orthocenter of triangle HAC
f. Construct the circumcircles of triangles ABC, HBC, HAB and HAC.
g. What would happen if any vertex of the triangle ABC was moved to where the orthocenter H is located?
When the orthocenter coincides with any vertex of triangle ABC, all the triangles HBC, HAC and HAC coincide with triangle ABC as it can be observed on the picture below. In addition, the circumcircles of triangle ABC and HBC coincide. Also the circumcircles for HAC and HAB become tangent at point A and H. Therefore A = H and orthocenterABC=orthocenterHAB
The nine point circle
It is a circle that passes through nine critical points of a triangle. These points are
1. The 3 midpoints of the side of the triangle
2. The 3 feet of the altitudes of the triangle
3. The midpoint of the line segment that joints each vertex to the orthocenter.
For triangle ABC, the nine point circle goes trough points L, M and N which are the midpoints of the triangle. Also goes through X, Y and Z which are the midpoints of the segments AH, BH, CH. Finally it also goes through the points D, E and F which are the feet of the altitudes of triangle ABC. Using O as the center of my nine point circle, I can construct a circle that passes through all of these critical points.
Another interesting construction in the picture above is the Euler segment. By constructing the orthocenter, centroid, and circumcenter of triangle ABC, I find a segment where the orthocenter, the centroid and the circumcenter are collinear. The incenter is not on the Euler line except when the triangle is isosceles.