ALTITUDES AND ORTHOCENTERS
Presented by Godfried Lawson

In this assignment I will:

1. Construct a triangle ABC.

2. Construct the Orthocenter, H of triangle ABC.

3. Construct the Orthocenter of triangle HBC.

4. Construct the Orthocenter of triangle HAB.

5. Construct the Orthocenter of triangle HAC.

6. Construct the Circumcircles of triangles ABC, HBC, HAB, and HAC.

7. Conjectures? Proofs?

Below is the construction of steps 1 through 5.

with the circumcircles of triangles ABC, HBC, HAB, and HAC added to the above construction we obtain:

What would happen if any vertex of the triangle ABC was move to where the orthocenter H is located? Where
would H then be located?

When the orthocenter H of the triangle ABC is moved to any of the vertices of the triangle ABC, a right triangle is formed. This justify the definition of a right triangle where the base is perpendicular to the height.
When the orthocenter H of the triangle ABC is collinear to of the  circumcenter Co, It is also collinear to one of other circumcenter (C1, C2, or C3). An isosceles triangle is formed which resulted from the angle bisector of one of the vertices of the triangle ABC.
A reflection of the isosceles triangle from the base produces a similar triangle and the circumcenter C0 coincides with one of the circumcenter (C1, C2, C3). This is possible since all the circumcenter are congluent.
I can conclude that the circumcenters C1, C2, and C3 formed by the triangle HAB, HBC, and HAC are equidistance from the orthocenter of the triangle ABC.