Trisecting a Line Segment using Centroids

by Kassie Smith


First what is a centroid??

The centroid of a triangle is the point of concurrency of the three medians. Medians are those lines defined by one vertex of a triangle and the midpoint of the opposite side of the triangle.


We are going to try to trisect a line segment using the idea of centroids in place of alternate, more common approaches.

Steps to begin the construction.

1. Construct segment AB.

2. Construct 2 circles. One with center A and radius AB and the other with center B and radius BA. Label their intersections as C and D.

3. Construct a line through AC. This is a diameter. Label the intersection of this line with the circle as E.

4. Construct segments BC and AD. Label the intersection of AD with BE as F.

5. Construct line segment CF. Label the intersection of AB and CF as G.


G is the centroid of the triangle BCE.



So, G is of the way from A to B. WHY??


Let's look at what we have in this picture above. We have triangle BCE. A is the midpoint of CE by construction(it is the center of the circle and CE is the diameter). F is also the center of BE. Therefore, both CF and AB are medians of triangle BCE. Therefore, their intersection, G is the centroid of the triangle.

So what does that mean in terms of our goal? Why does G being the centroid guarantee that the point is of the way from A to B?

Let's take a moment away from our goal to prove that the medians intersect at the centroid, a point that splits up each median into a ratio of 2:1. There are many ways to prove this. I will show a proof that relies on the fact that the medians divide the large triangle into 6 smaller triangles, all with equal areas.

We will start with a new triangle, HIJ. I constructed the medians by finding the midpoint of each side and then constructing the line segment from the midpoint to the opposite vertex.

For the sake of this proof, I only need to look at half of this triangle that is split be a median. I chose to examine HLJ.

We will let our base be HL, one of the medians of the triangle. As you can see, we now have three smaller triangles inside HLJ. These are triangle LJG, JMG, and MHG. We are given that each of these triangles have the same area. Let triangle LJG have area X. Then triangle JHG has area 2X since JHG is composed of two of those two smaller triangles. Again, allowing HG to be our base, we see that these two triangles, LJG and JHG, have the same height, namely the length of a perpendicular line from J to HL.

Using the formula for the area of a triangle, we can finish our proof.

Area of triangle = * base * height

area of JHG = * HG * h = 2X

area of LJG = * HG * h = X

Now, substituting the second equation into the first, we get: * HG * h = 2 * * GL * h

Simplifying, we get: * HG = GL.

This tells us that HG is twice the length of GL or GL is half the length of HG. It also tells us that the centroid is of the way from H to L or G breaks up HL into a ratio of 2:1.

This same argument can be repeated and applied to the other two medians of the triangle, effectively proving that the medians meet at a point that is of the way from the midpoint to the vertex.


So now that we have proved that, to complete the trisection we need to find another point on AB. To do this, we can just repeat the construction we did to find our first point on the other side.


1. Construct the diameter of the right circle through C. Label the intersection of the diameter with the circle as H.

2. Construct line segments BD and AH. Label their intersection as I.

3. Construct CI. Label the intersection of CI with Ab as J.

This completes our trisection of a line segment.


Click here to open up the GSP file to view my work or to use a tool I created. I created one tool with AB just divided into the 2:1 ratio. I created another tool that trisects AB.