Altitudes and Orthocenters
By: Kimberly Young
I began by doing the following:
1. Construct any triangle ABC.
2. Construct the Orthocenter H of triangle ABC.
3. Construct the Orthocenter (A) of triangle HBC.
4. Construct the Orthocenter (C) of triangle HAB.
5. Construct the Orthocenter (B) of triangle HAC.
The orthocenter of triangle HBC is A.
Proof:
By construction, AH is perpendicular to BC. From this, it
follows that the orthocenter of HBC must lie on the line AH.
Line AB is perpendicular to CH, by construction. The orthocenter
of HBC must line on the line AB.
Also, by construction, the line AC is perpendicular to BH.
The othocenter of HBC must line on the line AC.
It follows that the orthocenter of HBC is the intersection
of AH, AB, and AC. This is the point, A.
We could similarly prove that the orthocenter of HAB is C
and the orthocenter of HAC is B.
6. Construct the Circumcircles of triangles ABC, HBC, HAB,
and HAC.
7. Construct the nine point circles for ABC, HBC, HAB, and
HAC.
From this construction, my conjecture is that the Nine Point
Circle of triangles ABC, HBC, HAB, and HAC are the same.
Proof:
From the definition of the nine point circle, the nine point
circle bisects any line from the orthocenter to a point on the
circumcircle. (www.mathworld.com)
Also, the radius of the nine-point circle is one half of
the circumradius of the reference triangle. (www.mathworld.com)
By dilating (a 1/2 scale factor) each of the circumcircles
with its corresponding orthocenter, you will notice that the
circle that is created lies on top of the nine-point circle.
This occurrence may just seem coincidental.