by Donny Thurston

In this write-up, we will be exploring 4 different ways to trisect a line segment using Geometer's Sketchpad. In addition, we will be exploring the process in to "how" it was done, and why it works. It is assumed that all of GSP's abilities and tools will be available to the user, and that anyone could follow along with the explanation. Lets begin:

#1 - Construct A Median and find the Centroid

This method revolves around the fact that a centriod is 2/3 of the distance between a vertex of a triangle and the midpoint of the opposite side. This 2:1 ratio invites the ability to trisect a line.

This tool explains the construction and provides an example.

First, construct AB as the segment we want to trisect. Then we will create circles with centers A and B, and radius AB. This provies for us an "arbitrary" point that we can use to begin the construction of a triangle, without having to rely on simply designating a point somewhere (which can lead to problems when the software is in charge of finding a "random" point). This point, E, will be used to designate a line EF, through A, as seen in the figure. mEA must equal mAF since they are both radii of the same circle. Segments FB and EB complete the triangle (EB is optional, really).

This makes AB a median, as it connects a vertex, B, to the midpoint of EF, A. EG is constructed as another median, with G constructed as the midpoint between B and F. C is the point of concurrency between those medians, or the centroid. Therefore, the measure of AC is half of the measure of BC. A circle of radius AC can be constructed on C, and the intersection between that circle and CB will be D. AC must be equal to CD, because they are both radii of the same circle. Therefore BC must be twice CD, and so DB = CD = AC. The segment is trisected.

#2 - Parallel lines

In this instance, we are exploiting parallel lines to get the desired result.

Recall that when parallel lines intersect two or more transverals, they cut those transversals into proportional (but not always equal, of course) pieces.

In this instance (found here), we construct our segment AB, and then, similar to the last construction, draw our circles and find point E (above). Once again, this is done to get us a nice "random" point outside of the segment AB. In reality, for this construction, any point that is not on line AB will work.

Then, construct a circle of center E and radius AE. Extend AE, and the other intersection point will be F. Construct one more circle with center F and radius FE. Extend FE to find point G. Now we have 4 points on line AE, A, E, F, and G. Draw segment GB. Then construct lines parallel to GB at E and F.

This provides three parallel lines and two transverals, one of which is our segment AB. Since AG is cut into three equal pieces (thanks to the fact that they were all constructed to be the length of AB, each) then AB must also be cut into three equal pieces, since the proportions must be the same.

It is worth noting that this construction can be generalized to any number of equal sized segments on segment AB! By simply adding more circles (and so more parallel lines) past point G, and following the same procedure of linking the last point to point B, you can create any number of segments within AB. You will always require one less circle than the number of segments you would like. Therefore, to create a segment cut into n equal pieces, simply create (n-1) circles as described above, which will produce n parallel lines.

#3 - A Different Look at a Centroid

This construction uses the centroid again (there are probably countless constructions that use centroids that can trisect a line), but I like this one because it is short and elegant. The first image provides a lot of extra detail for the purposes of a proof, but the second provides only what is necessary to trisect a line.

In short, all it requires is to draw circles with center A and B and radius AB (like above). Connect A with either intersection of the two circles. It is worth noting that this intersection, if connected with a perpendicular line to AB, is halfway between A and B. This is a common property of two circles who share a radius. Find the midponit between

A and the intersection point, and connect it to the opposite intersection point. The intersection point between this midpoint and line segment AB trisects AB.

Why? Well, the proof is given above, but the quick answer is that C is the centroid of a triangle that is formed between A and the two intersection points of the circles. The segment between A and the midpoint of AB is a median, as well as the segment that includes the intersection point and C, projected onto the segment between A and the other intersection point. Since AC must be 2/3 of the distance between A and the halfway point between A and B, it must be 1/3 of the distance between A and B.

#4 - Similar Triangles

This construction takes advantage of similar triangles. Basically, we use our favorite "random point", D, as pictured above to construct a segment AE that is twice the length of AB. Then we use the other intersection of the triangle, point F, to construct a segment BF that is only one times the length of segment AB. The two triangles formed, AEC and BFC are similar, because they share vertical angles, and angle DAB and angle FBA are both 60 degrees. This is true because DAB and BFA are equilateral triangles since all of the sides of those triangles are equal (to the segment AB).

These two similar triangles have a ratio of 2:1. Therefore, AC is twice as long as BC, meaning that C trisects the line.

Interestingly, this can also be generalized! By manipulating the length of AE (such as by generating more circles moving E out by one measure of AB each), we can create similar triangles of any integer ratio! If we were to do 5 circles, we would have similar triangles of ratio 5:1, and we can 5-sect the line!