Altitudes and Orthocenters
When an orthocenter is created for a triangle, it produces three smaller sub-triangles. The orthocenter of these smaller triangles will be explored in this write up to see how they are related to the original larger triangle.
The ORTHOCENTER (H) of a triangle is the common intersection of the three lines containing the altitudes. An altitude is a perpendicular segment from a vertex to the line of the opposite side.
Also, the circumcircles of triangle ABC, HBC, HAB and HAC will be constructed. The circumcircle for each triangle turns out to have the same radii. They are all the same size.
Exchanging vertices with orthocenter H
Triangle ABC is constructed with orthocenter H. Using this construction, each vertex of triangle ABC is switched with orthocenter H. This is like moving the vertex of the triangle to where the orthocenter H is located. The follow results are found when exploring this change:
Orthocenter of triangle HBC is vertex A
Orthocenter of triangle HAC is vertex B
Orthocenter of triangle HAB is vertex C
Why does this result happen? Looking at one specific example, ∆HBC, gives some insight.
Since HA is an altitude of ∆ABC, it is perpendicular to BC and passes through vertex A. This determines one of the altitudes of ∆HBC. The other two altitudes of ∆HBC are outside the triangle and can be found along sides AB and AC of ∆ABC. HB and HC pass through vertex B and C and are perpendicular to side AC and AB (from construction of orthocenter H). The common intersection of these three altitudes is vertex A. See diagram 1 and 2 for this case and diagram 3-6 for the other sub-triangles.
In a likewise manner the orthocenter of triangle HAB and triangle HAC will be vertex C and vertex B, respectively.
Diagram 1 Diagram 2
Diagram 3 Diagram 4
Diagram 5 Diagram 6
Circumcircle of triangle ABC, HBC, HAB and HAC
Construct the circumcircle of ∆ABC and all sub-triangles. The circles appear to have the same radius and all pass through the orthocenter H. To check out this conjecture, the circumcircle for a specific case will be reviewed. The specific case used is ∆HBC.
The circumcenters for ∆ABC and ∆HBC is constructed. The points E and F are the center, both lie on the perpendicular bisector of BC, because they both are equidistant from B and C. It can be seen that the two circles have the same radii. In fact, all circumcircles produced from the sub-triangles will be the same size.
Point E and F are both equidistant from BC and the angles at E and F both subtend the arc BC, therefore the angles are congruent.
When ∆ABC and ∆HBC are reflected about BC, they touch the opposite circumcircle.
Here is a sketchpad file with all the results pulled together (see several different tabs along bottom of file).