Altitudes and Orthocenters

by Michael Walliser

This write-up will deal with orthocenters and nine-point circles. If you wish to review orthocenters, click here. If you wish to review nine-point circles, go to http://mathworld.wolfram.com/Nine-PointCircle.html.

We will begin by constructing a triangle **ABC** and its orthocenter **H**. We will then subdivide **ABC** into three smaller triangles **HBC**, **HAC**, and **ABH**, as shown below.

Now we will construct the nine-point circle for triangle **ABC**, as shown below with the nine points and the center.

Now watch what happens when we construct the nine-point circles for triangles **HBC**, **HAC**, and **ABH** below.

The nine-point circle appears to be the same for all four triangles. Click here to open a GSP file with orthocenter and nine-point circle script tools. You will see that the circles are in fact the same, and that the concurrence holds for acute and obtuse triangles. For right triangles, since the orthocenter is located on the right angle vertex, triangles **HBC**, **HAC**, and **ABH** are degenerate.

So how are the nine points always the same for these different triangles? To answer that, we must remember where each of the nine points comes from. Three of the points are the midpoints of the sides, three of the points are the orthic points, and three of the points are the midpoints of the segments that connect the orthocenter to each vertex. Since each of these points can be constructed given only the three points of the original triangle ABC, we will examine each point's construction and how it fits on the nine-point circle. The figure below labels each of the nine points, and the table shows why each point is on the nine-point circle for each triangle.

Six of the points on the nine-point circle are the midpoints of segments AB, BC, AC, AH, HB, anc HC. Depending on which triangle we are looking at, each of these segments either runs from the orthocenter to a vertex or is a side of the triangle. Thus, the midpoint of each segment is on the circle by virtue of whichever of these two properties the segment satisfies.

The other three points are the orthic points of all the triangles. By the construction of the orthocenter, any side of a triangle is perpendicular to the line running through its opposite vertex and the orthocenter. Thus, AC is perpendicular to HB, AB is perpendicular to HC, and BC is perpendicular to AH. Each of these segments forms a side of two of the four given triangles, while its perpendicular counterpart contains the opposite vertex and the orthocenter of these two triangles (in one order for one triangle, and in opposite order for the other). For instance, AC forms a side of triangle ABC and a side of triangle HAC. Its perpendicular counterpart HB defines a line through the opposite vertex B and orthocenter H of triangle ABC, and therefore it defines a line through orthocenter B and opposite vertex H of triangle HAC.

Conversely, the perpendicular counterpart (HB in our example) forms a side of the *other* two of the four given triangles (HBC and ABH), while the original segment (AC) defines a line through the orthocenter (A for triangle HBC, C for triangle ABH) and the opposite vertex (A for triangle ABH, C for triangle HBC) of these two triangles. Therefore, the intersection of each pair of perpendiculars is an orthic point for all four triangles. This satisfies our conjecture that the nine points of each nine-point circle are concurrent.