To begin our exploration of orthocenters and altitudes, we will look at the orthocenter, H, of a triangle.

Click on this link to explore the orthocenter. The orthocenter is constructed by the intersection of the altitudes from the vertices.

Next, we will construct the orthocenters of the triangles, HBA, HAC, and HBC. If you'd like to participate, go to the orthocenter link and use the orthocenter tool to construct each one.

H1 is the orthocenter for triangle HBA, H2 is the orthocenter for triangle HAC, and H3 is the orthocenter for triangle HBC. Do you find it interesting that the orthocenters of the smaller triangles turn out to be the vertices of the larger triangle? Click here to explore if this is always true.

Next, let's look at the circumcircles of the triangles to determine if there is any relationship between the orthocenters and the circumcircles. Do you have any predictions?

Here are the circumcircles (in green) of all four of our triangles, triangles ABC, HBA, HAC, and HBC.

Notice that the circumcircles intersect at all four orthocenters of the four triangles. Also, notice that the circumcircle of the original triangle, ABC, intersects the three orthocenters of the three smaller triangles that form triangle ABC. So, if we extend this idea, any triangle formed by the orthocenter of a larger triangle and two points on the circumcircle of the larger triangle will have an orthocenter on the circumcircle of the large triangle.

Also, notice that when a vertex is moved to where the orthocenter is, the circumcircles for triangle ABC and triangle HBC are the same. Thus, there appears to only be three circumcircles. A similar situation occurs when moving the other two vertices to the orthocenter.

Another observation is that the radii of the circumcircles increase as triangle ABC gets closer to a degenerate triangle (a line). This is because the orthocenter is getting increasingly further from the triangle as triangle ABC becomes degenerate. The circumcircles reach a limit of all becoming a horizontal line. This limit is also established as the altitude of A increases to infinity.

Next, let's extend the altitudes of the vertices to the circumcircle of triangle ABC. Then, we will construct the orthotriangle and the triangle from the intersections of the extended altitudes and the circumcircle.

Above, you can see that these two triangles look similar. Let's explore further to see if they are always similar. Describe the cases where they are and are not similar. What are the characteristics of the original triangle?

Observations:

When the original triangle is a right triangle, the two triangles IJK and GHF become degenerate triangles and form a line.

When the original triangle is an obtuse triangle, there is only one point left from the orthotriangle. Triangle GHF still exists, but triangle IJK does not exist when the original triangle is an obtuse triangle.

When the original triangle is an acute triangle, the two triangles, IJK and GHF are similar.

Partial proof of similarity:

If we construct a parallel line to GH at I we get the following construction.

Notice that the parallel line contains the segment IJ (not sure how to prove).

Then, angle HGI is congruent to angle JIN by corresponding angles (using the altitude of N as the transversal).

Then, angle GHJ is congruent to angle IJM by corresponding angles (using the altitude of M as the transversal).

Now, we will construct a parallel line to segment GF at I.

Notice that the parallel line contains the segment IK (not sure how to prove).

Then, angle IGF is congruent to angle NIK by corresponding angles (using the altitude of N as the transversal).

Therefore, angle HGF is congruent to angle JIK.

Now, we will construct a parallel line to segment HF at J.

Notice that the parallel line contains the segment JK (not sure how to prove).

Then, angle FHJ is congruent to angle KJM by corresponding angles (using the altitude of M as the transversal).

Therefore, angle FHG is congruent to angle KJI.

Lastly, since two angles of the triangles are congruent, the third angle must be congruent.

So, triangle HGF is similar to triangle JIK.

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