Orthocenter is one of the nice centers of a triangle among others. In this essay, we try to explore some relations between the orthocenter, altitudes, and some circles of a triangle such as circumcircle, nine-point circle and excircles by the help of the GSP software.

Let's take a triangle ABC. And construct its orthocenter H (HABC). And then let's construct the orthecenters HHBC, HHAB, HHAC for the triangles HBC, HAB and HAC respectively. then let's construct the circumcircles with circumcenters CABC ,CHBC ,CHAB, CHAC for these triangles. Figure 1a shows all of these constructions.

As we may observe from the Figure 1a, the orthecenters of the triangles HBC, HAB and HAC are the vertices A, C and B of the original triangle ABC. This is not a coincide. (Do not believe me? Click Here to explore it yourself).

Conjecture 1: Let ABC be a triangle and let H be the orthecenter of it. Then the orthecenters of the triangles HBC, HAB and HAC are the vertices A, C and B of the original triangle ABC.

Proof: Let ABC be a triangle and H be the ortocenter of it. For the triangle HBC, since the lines passing through A&H, B&H, and C&H are the perpendiculars for the sides BC, CA and AB respectively, then the lines passing through A&B, C&A are the perpendiculars for the segments CH, HB. Since the lines passing through A&B, C&A intersects at A, it is the orthecenter for the triangle HBC. In other words, A = HHBC. (See Figure 2 below).

Applying similar arguments for the triangles HAB and HAC, we may prove that C = HHAB, B = HHAC.

At first looking to Figure 1b, we may observe that, when we connect the each orthecenter of the four triangles (i.e. ABC, HBC, HAB and HAC) to their circumcenters, we get an prism whose sides are congruent rhombuses. Figure 3 shows the picture of the prism with some calculation showing that all sides of the prism are equal.

Click Here for a demonstration of this observation with GSP.

Conjecture 2: Let ABC be a triangle and let H be the orthecenter of it. If we connect the each orthecenter of the four triangles (i.e. ABC, HBC, HAB and HAC) to their circumcenters, we get an prism whose sides are congruent rhombuses.

Proof: Just aplly Conjecture 1, and the fact that the distances from circumcenter to the vertices of a triangle equal.

Now if we construct the nine-point circle for each triangle ABC, HBC, HAB and HAC, we see that they have a common nine-point circle. Figure 4 illustrates this fact.

Conjecture 3: Let ABC be a triangle and let H be the orthecenter of it. The nine-point circles of the trianles ABC, HBC, HAB and HAC are the same.

Proof: Let C be the nine point circle of the triangle ABC.

For the triangle HBC: Since the orthocenter of the triangle HBC is the point A;

the midpoints of the segments AB, AC,are the midpoints of the feet of the altitudes. Moreover, the midpoints of the segments HB and HC and HA are the midpoints of the segments connecting the vertices B, C and A to H in the triangle ABC. Since the orthecenter of HBC is A (from Conjecture 1), the the feet of the altitudes of the triangle ABC are also the feet of the altitudes of the triangle ABC.

So, the 9-point are in the same location but in different properties in triangles ABC and HBC. So, the the nine-point circle is exactly the same for both triangles.

Exercise: Prove in the same way that the nine-point circle is same for HAB, HAC and ABC.

This page created October 10, 1999

This page last modified October 16, 1999