The Pedal Triangle

and the Sim(p)son Line

Homer Simpson creates the Simson line - Look mom! No ruler!


Okay, so we are not going to talk about Homer's contributions to mathematics. But in a topic almost as risky as when Homer became the "Beer Baron", we will be discussing explorations with the pedal triangle and the true Simson line.

Lets begin with the construction of the pedal triangle. For any triangle ABC , the pedal triangle is formed by first choosing the pedal point, an arbitrary point P in the same plane as ABC. Then, construct the perpendiculars from P to each of the sides of triangle ABC. (You may need to extend the segments of the triangle into lines.) The pedal triangle is formed by connecting the three points where the perpendicular lines intersect each side of the triangle, in the case below, R,S,T.

One of the amazingly cool things about pedal triangles, is that as you move the pedal point, your triangle changes as well. Click here to open a GSP file where you can manipulate the location of the pedal point above.

One of the things you may have noticed as you moved the pedal point of the triangle is at times, the pedal triangle seems to compress, or degenerate, into a line segment. This segment is a part of the famous "Simson Line".

There are various locations in which the pedal point can be to form the Simson line for a given triangle. The question is, what is the set of all of these points? Some places that are usually discovered for the degenerate triangle are at each of the vertices of the triangle. Hmmmm, what set of points do we know that contain all of the vertices of the triangle? How about the circumcircle of the triangle? Lets construct it and see.

As you can see, when P is on the circumcircle of triangle ABC, the pedal triangle becomes the Simson line. Click here for a GSP file to see the animation of the pedal point.

As you watched the animation, you might have noticed that there are times when the Simson line coincided with one of the sides of triangle ABC.

What is it about these points on the circle that allow this to happen? Let's explore a bit and find out...


It appears that there is only one point for each side of the triangle, that can produce the simson line on that segment. That point is located somewhere on the arc of the circumcircle cut off by that side of the triangle. In the case above, for the simson line to be segment BC, P must be in a particular place on arc BC. It appears that when P is in the above location, that it lies on the ray containing A and the circumcenter of the triangle. Lets construct that ray and check our hypothesis.

It does indeed appear to be on that ray. Let's check the pedal point on the other two rays as well.

Okay, so we now know that when the pedal point coincides with the point of intersection of the ray containing the opposite vertex and the circumcenter, and the circumcircle of the triangle, that the simson line will be the side of the triangle opposite of the vertex. There must be a simpler way to describe these points, but to my knowledge, these rays do not have a special name such as median, or angle bisector. They do contain "cevians", which are segments drawn from a vertex of a triangle to any point on the opposite side.

The term cevian, was coined by an Italian mathematician, Giovanni Ceva, who proved a very cool theorem known today as Ceva's theorem. The theorem states the following:

Since we drew each of these rays through the same point, the circumcenter, we know that the three rays are concurrent. This fact would allow us to state that for our given triangle, the above ratios would hold. Of course, this would imply we could prove Ceva's theorem. Click here for a link to a proof of Ceva's theorem.




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