Assignment #8:

Altitudes and Orthocenters

By Amber Krug

Below is the construction of:

1.  Any triangle ABC;

2.  Orthocenter H of triangle ABC;

3.  Orthocenter A of triangle HBC;

4.  Orthocenter C of triangle HAB; and

5.  Orthocenter B of triangle HAC.

My conjecture is that A is the Orthocenter of triangle HBC, B is the Orthocenter of triangle HAC, and C is the Orthocenter of triangle HAB.

Proof:

By construction: BC is perpendicular to AH, AB is perpendicular to CH, and AC is perpendicular to BH; therefore, the Orthocenter of triangle HBC lies on the lines AH, AB, and AC.  This must occur at the intersection of these three lines which is A.  Thus, A is the orthocenter of triangle HBC.

Similarly, B is the Orthocenter of HAC, and C is the Orthocenter of triangle HAB.

We can then construct the circumcircles of ABC, HAB, HAC, and HBC:

If we construct the Nine Point Circles of triangles ABC, HAB, HAC, HBC, we find that the Nine Point Circles are the same circle for each triangle.

However, this is not the case.  This only occurs when the Orthocenter H is in the center of the circumcircle of ABC.