Assignment 8

This exploration looks at the relationship between triangles, orthocenters and circumcirlces. If we begin with any triangle ABC and construct its orthocenter H, three more triangles are formed: ABH, ACH and BCH.

Now lets explore the construction of the orthocenters of each of these smaller triangles. I conjecture that the orthcenter of each triangle will coincide with the one point that does not make up that particular triangle. For example, the orthocenter of triangle ABH will coincide with point C, so lets prove that. To begin with, the defintiion of orthcenter is the intersection of the altitudes of each point in the triangle, basically the intersection of lines perpendicular from a vertex to its opposite side. Well we know that H is the orthocenter of triangle ABC, so H lies on three altitudes. These are the lines that make up each side of the triangle. So if we extend these segments into lines we construct:

This gives us the altitude for H in the triangle ABH because H is on the C altitude and shares the same opposite side as C. To construct the other two altitudes, we examine a vertex and the opposite altitude. We know that the line through B and H is an altitude, hence is perpendicular to the side opposite B and H, which is segment AC. Therefore the altitude for A in triangle ABH is the segment AC because when you extend its opposite side, the B altitude for triangle ABC, then segment AC becomes perpendicular. The same is true for the extended segment AH which is the A altitude for triangle ABC, where segment BC is the perpendicular, hence is the altitude for triangle ABH.

Therefore since the orthocenter for triangle ABH is where these three altitudes intersect then the orthocenter is also the point C, the only point not in the given triangle. The same can be true for triangles ACH and BCH where their orthocenters are point B and A respectively. Below are illustrations:

Now lets construct the triangle with the circumscribed circles constructed for each individual triangle ABC, ABH, ACH and BCH.

So if we contruct a ray extending from the center point of the circumscribed circle of triangle ABC through the midpoints of each side of ABC, then the ray also passes through the center point of each circumscribed circle for the smaller triangles as shown below.


Lets explore moving the vertex around to different places. If any vertex was moved to the orthocenter H we will have this sketch:

Here I have moved point B to where the orthocenter H was originally. As you can see, there 2 triangles formed, ABC and AHC, which share the segment AC as a side. The orthocenter H, which was orignally inside the triangle ABC is not outside because with the sengments extended into lines, the altitudes all intersect outside the triangle as shown. The same would be true if you moved A or C to the orthocenter, then H would be outside the box above the point that was moved like it is here.