Trying to Trisect a Triangle

By: Tim Lehman

Exporation 1: Dividing a triangle into 3 triangles of equal area

Method I: After trisecting one side of a triangle (side BC below), connect the opposite vertex (point A) to the two points that trisect the side (H and I). Because the side is divided into three equal segments, all three triangles have the same length of base and height. Thus, all three triangles are of equal area. Click here for an interactive GSP sketch of the example below.

Method II: In many ways, method II is simply a variation of method I. After trisecting one side of a triangle, construct a segment connecting the opposite vertex (point A) to only one of the points trisecting the side (I). We then have two triangles. One (ABI) is equal to 1/3 the area of the original triangle (ABC). The other triangle (AIC) is 2/3 the area of the original triangle (ABC).Click here for a GSP sketch of the below.

To trisect the original triangle, we need to divide the larger triangle (AIC) into two equal triangles. This can be accomplished by finding the midpoint of any side of the triangle and constucting the segment from them to the opposite vertex. The two possibilities can be seen below.

Method III:

Another way to divide a triangle into three triangles of equal area is to find the centroid (the point where the medians intersect). After determining the centroid (point G below), construct the segments connecting the vertices to the centroid. The three triangle created are of equal area.

Why does this work?

What are some other ways to divide the triangle into three equal triangles?

Exploration 2: Dividing a triangle into three equal pieces
The first exploration only considered cutting the original triangle into three equal triangles. Now we will examine two methods of dividing the triangle into three equal parts that are not all triangles.

Method I:

This method follows directly from method III above. The sides of the three equal sections are formed by connecting the midpoints of the three sides of the original triangle to the centroid. The reasoning of why it works is the same as in method III.