Assignment 8

Altitudes and Circumcircles


Megan Dickerson

This assignment is looking at the relationships between altitudes, and circumcircles.

So, the first thing that I did was to find the altitudes of Triangle ABC and thus find the orthocenter (labeled H).

Then, I found the perpendicular bisectors of ABC in order to find the circumcenter and the circumcircle (It is purple in my picture).

Then, I looked at the circumcircles for the 3 new triangles, HAB (pink circle), HAC(orange circle), and HBC (light blue circle).

The orthocenter for Triangle HAB is C.

The orthocenter for Triangle HAC is B.

The orthocenter for Triangle HBC is A.

Fairly quickly I noticed that if I set A = H, the blue and purple circles are the same circle, the new circumcenter is the median of BC, and the orthocenter is now A.

Likewise, if I set H = B or C, the purple circle becomes the orange or pink circle, new circumcenter is median of AC or AB, and the orthocenter is the vertex that H is on.

If I exchange A and H, H is now located outside the triangle, Triangle ABC is now located inside the bigger Triangle HBC. The 2 circumcircles also exchange (the blue and purple). A similar picture occurs when B and H or C and H are exchanged. Click HERE to see a GSP animation of this happening.

Further Exploration

Next, I drew in some lines. I connected A and B to the circumcenter of HAB, A and C to the circumcenter of HAC, and B and C to the circumcenter of HBC.

Then, I reflected each segment from the vertex to H over the sides of the original Triangle ABC. My new picture looks like this:

I first looked at the Hexagon created by the green lines outside Triangle ABC(Hexagon BFAGCH). This hexagon was created out of the reflections of the lines from each vertex to H. All 6 of these points lay on the circumcircle. I claim that this Hexagon is equal to two times the original triangle ABC.


Triangle BHC is congruent to Triangle BJC by SSS

Triangle BHA is congruent to Triangle BFA by SSS

Triangle AHC is congruent to Triangle AGC by SSS

Hexagon BFAGCH can be decomposed in to 6 triangles (BHC, BJC, BHA, BFA, AHC, AGC). These 6 triangles are equal in content to the whole hexagon. But since BHC = BJC, BHA = BFA, and AHC = AGC, the area of the hexagon = 2BHC + 2BHA + 2AHC = 2(BHC + BHA + AHC) = 2 times the area of the original Triangle ABC.

Then, I looked at the radii of each circle created. To me they appeared to all be equal which would make the areas of all 4 circles equal. So, I claim that all 4 circles created have equal are.


HB is an arc on both the pink and blue circles (circumcircles for HAB and HBC).

When measured, these arcs are the same length.

Since equal arcs are subtended by the same segment then their areas are equal.

Similarly, HA is an arc on both pink and orange circles. Both arcs have equal length, thus their areas are equal.

Likewise, HC is an arc on both blue and orange circles. Both arcs have equal length, thus their areas of equal.

So, by transitive property all 3 circumcircles to triangles HAB, HBC, and HAC are equal in area.

If I then take arc BC as an arc to both the purple circle and blue circle, I can show that the arcs are equal, thus the areas are equal.

Therefore, all 4 circles have equal area and all radii are congruent.

The results of this proof are shown in this GSP Image: