This is the write-up of Assignment #8
Brian R. Lawler
EMAT 6680

Altitudes and Orthocenters.


1.-6. Construct any triangle ABC. Construct the Orthocenter H of triangle ABC, triangle HBC, triangle HAB, and triangle HAC. Construct the Circumcircles of triangles ABC, HBC, HAB, and HAC.

Demonstrating a moment of incredible brilliance, I used Geometer's Sketchpad to perform these constructions. The results looked something like the sketch below. (Click HERE for another drawing with the altitudes showing.)

The Problem

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. Does your result hold only for acute triangles?

Analysis and Proof

Again, to begin my investigation, I performed this construction in Geometer's Sketchpad.

I asked the program to measure the three interior angles of triangle ABC as well as the three interior angles of LMN.

I'm not sure why I suspected (probably from work earlier in assignment 8), but I guessed that the interior angles of triangle LMN would be equal to the sums of pairs of half-angles of triangle ABC. I asked GSP to compute these calculations and found this to be true as I dragged the various points of ABC around the sketchpad. (Click on the image at right to investigate this for yourself.)

So restating my claim:

To prove the above, I printed a full-page version of the sketch above, and identified the angles I knew, in terms of a, b, and c. With some work in labeling what I knew, I was able to establish the above statements. To simplify labeling, I named the measure of an angle to equal the middle letter. For example, I labeled the measure of angle CAB as a. The following is a proof that . The other 2 statements in the claim are proven in the same manner.

(Use the sketch at right to identify additional labels.)

First, name the measure of each angle of triangle ABC as a, b, and c. The bisections create two angles, each equal to half the original angle.
Next, see that first by the angles of a triangle sum to 180 and then that a, b, and c sum to 180.
Since angle CEL is supplemental to angle CEP, and that a, b, and c sum to 180, .
Angle CZL is a right angle. (I will not attempt to prove this in this problem, but would make an excellent follow-up or continuation of my work. The property holds based upon the construction of points L and M, along with the fact Z lies on the perpendicular bisector of angle ABC.)
and by vertical angles, .
Also by vertical angles, .
By an argument similar to steps 4 and 5, .
And by arguments similar to step 6, .
Thus , or equivalently .

Click here to see a picture of these measures labeled.

Comments? Questions? e-mail me at

Last revised: December 28, 2000

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