ALTITUDES AND ORTHOCENTERS

1. Construct any triangle ABC.

2. Construct the Orthocenter H of triangle ABC.

3. Construct the Orthocenter of triangle HBC, H1.

4. Construct the Orthocenter of triangle HAB, H2.

5. Construct the Orthocenter of triangle HAC, H3.

6. Construct the Circumcircles of triangles ABC, HBC, HAB, and HAC.

7. Conjectures? Proof?

8. Construct the nine point circles for triangles ABC, HBC, HAC, and
HAB.

The picture is messy, but we see that the nine point circles are the same for all these triangles!

9. Construct triangle ABC, its incircle, its three excircles, and its nine-point
circle. Conjecture?

The green circles are the excircles, the ninepoint circle is the red circle, the incircle is the black circle. It appears that the excircles are all tangent to the nine point circle and the incircle is tangent always to one of the excircles and the nine point circle simulaneously! Click here to examine this yourself. Move any of the triangle vertices around to check that the nine point circle is always tangent to the three excircles and the incircle is always tangent to one of the excircles and the nine point. Why is this so?

Well, the nine point circle always contains the midpoints of the sides of the triangle. If you don't remember that this is true by construction, it can be examined here. Use the script to make a nine point circle. Use a GSP sketch to verify that the circle does indeed contain the midpoints of the sides...hmmm... what I say is true, but I see that it doesn't matter in this discussion. Anyway...

Constructing a segment from the center of the excircle to the center of the nine point circle, we have

As you can see, the segment intersects the circles at their point of tangency, as expected.

Let's see... I want to show that the nine point and the excircle must always be tangent. I must show why both circles must be at that point (hopeully by construction). Look here for further investigation.

I have learned that indeed my conjecture is true. However, the proof I desire is rather difficult and makes use of inversive geometry. It would be adviseable to accept this conjecture and move on. So let's look at number 13:

13. The internal angle bisectors of triangle ABC are extended to meet the
circumcircle at points L, M, and N, respectively. Find the angles of triangle
LMN in terms of the angles A, B, and C.

Above is the desired construction. The red triangle LMN is the triangle we want to measure the interior angles of (in terms of angles A, B, C).

Now point M is contained by the angle bisector of angle B (by construction), thus all points on that angle bisector (blue) are equidistant from side AB and side BC. Therefor the green segments a and b are congruent.

Well, back up and start again. I am not getting where I want to be. Let's look at this: Here is triangle ABC, its circumcircle, and the requested construction of triangle LMN.

Consider the angle at vertex M. Note that one side (not one half) of the angle at vertex M (defined by the angle bisector of angle ABC) subtends the same chord (or arc length) as one half the angle at vertex A (defined by the angle bisector of angle BAC).

So we conclude that one part of angle M is congruent to 1/2 of angle A.

But what about the remaining portion of angle M? Well, the other side of angle M subtends the same arclength and chord as one half of the angle at vertex C.

So we conclude that the remaining portion of the angle at M is congruent to one half of angle C.

Thus the measure of angle M can be expressed as the sum of one half of angle A plus one half of angle C.

Generalizing, we say that the measure of any of the interior angles of the constructed red triangle can be expressed as one half the sum of the original triangle's adjacent angles. That is;

angle L is one half angle C plus one half angle B

angle N is one half angle B plus one half angle A

angle M is one half angle A plus one half angle C.

We have expressed the angles of triangle LMN in terms of the angles of triangle ABC, as desired.