# Assignment 8: Altitudes and Orthocenters

In this assignment, I will start out my explorations by constructing a triangle ABC, its orthocenter, H, and connecting H to each vertex. Next I will find the circumcircle of both the original triangle and the three smaller ones.

It appears that the circumcircles are all congruent! I wonder if H being the orthocenter is necessary for this to happen. I will now divide the triangle up using an arbitrary point in the interior of the triangle.

The circumcircles are not congruent! So now I will try to prove that if H is the orthocenter, the circumcircles of the four triangles will be congruent.

During my explorations for Assignment 4 involving the nine-point circle, I noticed that the nine-point circle always has half the radius of the circumcircle. Allow me to do a quick proof of this fact.

Because the nine-point circle includes the midpoints of the segments connecting the vertices to the orthocenter, a dilation of ½ about the orthocenter will form a triangle whose circumcircle is the original nine-point circle.  See below.

But the new circumcircle will have a radius one half as large as the radius of the original circumcircle. Hence one half the radius of the original circumcircle is congruent to the radius of the nine-point circle.

Now, returning to the original problem of proving that the circumcircles of the four triangles shown below are all the same, I can start by proving the nine-point circles are all the same. I will choose to compare the circumcircle of the smaller triangle outlined in red with the original triangle. However, the discussion would be identical if I had chosen either of the other two smaller triangles.

First of all, we can see that both triangles share M2, which is the midpoint of BC. Furthermore, M2 will line on the nine-point circle of each triangle. Secondly, we must note that O3 and O1 lie on the nine-point circle of triangle ABC because they are the midpoints of the segments connecting the orthocenter to the vertices B and C. But the orthocenter is one vertex of the red triangle. Hence O3 and O1 are midpoints for the red triangle, and therefore lie on its nine-point circle. Since any three non-collinear points determine a unique circle in Euclidean geometry, and the two nine-point circles share three points, M2, O3 and O1, the two nine-point circles must be one and the same.

To conclude, all four triangles will have the exact same nine-point circle. And the radius of the nine-point circle is half the radius of the circumcircle. Hence the circumcircles, although different, will have the same radius.  A sketch of the triangles, the circumcircles (in green) and the nine-point circle (in red) is given below.