by Donny Thurston

In this write-up, we will be discussing the

mediansof a triangle, their point of concurrency, thecentroid, and a method by which we can use these constructions to create atrisected line segment. In particular, we're going to demonstrate how we can use these important geometric concepts to create a script in Geometer's Sketch Pad that will unfailingly provide a line trisected by three points, as well as to understandwhyit does so.

Proving the Properties of a CentroidSo, to begin, let us examine a triangle, its medians, and its centroid. As a reminder, a median is a segment to goes from one vertex of the triangle to the midpoint of the opposite side. The centroid is the point of concurrency of the medians, if one exists, and while we can assume that any two medians will certainly cross with a triangle (the reasoning for that can be found elsewhere), I suggest that

all three medians will intersect at the same point, the centroid, for every triangle.

Medians are Green Centroid is Orange

So, to prove this, let us consider the triangle ABC. Assume that each side has a midpoint, namely D, E, and F for AB, BC, and AC respectively, with the medians drawn. We will also construct the segment DE. Finally, the centroid, the intersection of medians BF and AE (and CD as proposed by this proof) will be G. The triangle is as follows.

Note that DE is a

midsegmentof the triangle, so it is parallel with AC, and is 1/2 the length of AC. So let us consider triangles AGC and EGD. Angle AGC is congruent to angle EGD due to their relationship as vertical angles. In addition angles GED and GDE are congruent to angles GAC and GCA, respectively, as they are alternate interior angles to parallel lines DE and AC. This means thattriangles ADC and EGD are similar triangles, due to the AAA postulate for similarity.In addition, we know that they have a similarity ratio of 2:1 (ADC to EGD), becasue AC has a ratio of 2:1 to DE. This suggests that the ratio of CG:GD and AG:GE are both 2:1.

In other words, point G (the centroid) is 2/3 of the distance from C to D, and 2/3 of the distance from point A to E.What about B and F?

Well, a similar proof can be proposed when constructing the segment EF (as opposed to DE). We use the midsegment EF to prove that AGB is similar to EGF, with the same similarity ratio of 2:1 to conclude that they intersect at a point that is 2/3 of the distance from B to F, and is 2/3 of the distance from A to E. But we have already identified that point (exactly 2/3 of the distance from A to E) as point G! Therefore,it must be that the intersection of segment BF and AE must be the same as between AE and CD.So, we have proven two things. First, that the medians of a triangle are concurrent at one point, the centroid, and second, that this point divides each median into two segments that have a ratio of 2:1. Or to say that another way,

that the centroid is a point that is exactly 1/3 the distance from one endpoint of the median, to the other.

This property of the centroid is going to prove essential to out method of trisecting any line segment.

Trisecting a Line Segment

This section will assume an adequate knowledge of GSP and how to create constructions using the program. Instructions will not be provided on details such as how to contruct a circle with a center and point.So, then, let us develop a method to create a tool in GSP that can provide for us a Trisected Line segment.

One key thing to remember about creating tools in GSP is to carefully understand that every "input" by the user is consider in the consruction of a tool. For example, if we were to create three arbitrary points in the construction of some figure, then that is exactly how many points someone who is using a tool based from that construction will need to provide to create the figure.

So, if our goal is two create a tool that will allow someone who create a trisected segment (without the need for any measuring or guessing), we need to consider how many points they will need to input, as well as what, exactly, those points represent. In this case, I believe that it is reasonable to assume that the user should be required to provide two points, namely, the two endpoints of the segment. The user should not be responsible for any other input.

So, let us begin with exactly that, a line segment we will call AB.

This is the segment that must be trisected, because it is the input (and only input) from the user. Therefore, it stands to reason that this line will represent a median in a triangle, as opposed to say, an side. This is because the centroid of a triangle has the potential to divide a distance between two points into thirds, and that no method of doing that for a side is quickly apparent.

Therefore, we need to create a triangle around our segment here, so that the segment represents a median. That means that one end point of the segment will be a vertex, and the other will be the midpoint of a side. Also remember that we cannot arbitrarily assign any more points, or else the user would be responsible for this when using the script.

Let us construct the side, to do this, we need two more vertices (we already have one). Since we cannot assign a point arbitrarily, let us develop a method to create a point that is dependent the information we already have. Fortunately, this will not be an obstacle to our intention of creating a trisected line, since, no matter where the vertices of the triangle end up (so long as the figure is still a triangle), the median we have constructed will always have a centroid in the proper place.

Therefore, I suggest constructing a circle with center A, and point B.

Then, place a point anywhere on that circle, so long as it is not on the line AB (as that would not result in a triangle). Then, construct a line that goes through that point and A. Place a point at the intersection of that line and the opposite end of the circle.

Because the radius of a circle is constant, point A is equidistant from the two points that we have just constructed on the circle. These two points would then make excellent vertices of our triangle. Construct segments from these two new points to point B.

Now we have made sure that our line, AB is a median of a triangle, thanks to the properties of the circle. Now, we need another median, so that we can identify a centroid. This is an easy task. Construct a midpoint on either of the new sides of our triangle (it does not matter which since the centroid is the point of concurrency for all three medians), and construct the intersection between that median and AB. This is the centroid.

We are almost done. Now that we have a centroid, we have split segment AB into two smaller segments, one twice as large as the other. In order to have three equally sized segments (to trisect the larger segment), we need to construct one more point. This can be done several different ways. I suggest constructing a circle with center at the centroid and point A. Because circles all have the same radius, the distance between the centroid and the intersection of our new circle and AB will be the same as the distance between A and the centroid. We will have "moved over" another 1/3. Create a point at that intersection.

Now, hide everything besides the segment AB and the four points on that segment (A, B, and the two points trisecting the line.

Here we have trisected line, and if we can use this contsruction to save a tool that will always allow us to trisect a line, using the essential properties of a centroid!

The script tool for the trisected line segment can be found here.