# 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.