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.

Click **HERE** to see the
GSP sketch of the above construction.