Assignment 8: Altitudes and Circumcircles

By Krista Floer

Examine the triangle formed by the points where the extended altitudes meet the circumcircle. How is it related to the Orthic triangle? Proof? Will the relationship still hold if the original triangle is obtuse?

First, here is the GSP sketch that I made. I have measured the perimeter of both triangles and calculated the ratio between them. It seems like the orthic triangle is always half of the extended altitude triangle.

Now let's look at a proof:

We want to show that .5(BC + CD + BD) = EF + FG + EG. First consider this property of the orthocenter: if the orthocenter is reflected over a side of its triangle, then it will lie on the circumcircle. We see that point B is the reflection of the orthocenter. Since it is a reflection then that means the distance from the orthocenter to E is equal to the distance from B to E. So we have BE = EOrtho. This can be applied to show that FOrtho = FC and GOrtho = GD in a similar fashion.

Now let's consider triangle BOrthoC, as seen here:

Because BE = EOrtho and FC = FOrtho, by definition we have that EF is a midsegment of the triangle. This means that EF is 1/2 of the length of BC . Using similar logic, we have that FG = 1/2CD and EG = 1/2BD.

By summing like sides of each equation, we obtain EF + FG + EG = 1/2BC + 1/2CD + 1/2BD. Simplifying this we have EF + FG + EG = 1/2(BC + CD + BD). Thus, we have shown the perimeter of the orthic triangle is 1/2 that of the extended altitude triangle.

This is for when the triangle is acute. What happens when it is obtuse? I found that the ratio stays at 1/2 until a certain point when the angle becomes too obtuse, and then the ratio gets smaller and smaller. I have included a picture and a GSP sketch of my findings.

Go Back to HOME