 Investigating the CENTROID of a triangle

By: Erica Fletcher

The CENTROID (C) of a triangle is the common intersection of the three medians. A median of a triangle is the segment from a vertex to the midpoint of the opposite side. I have used Geometer’s Sketchpad (GSP) to construct the centroid and show that the medians divide the triangle into six small triangles. Thus showing that these triangles all have the same area. However, this is not enough to just investigate this so I will prove the following. From exploring with the triangle above I noticed that when I constructed 1 median, it divided the triangle into two triangles with equal area. Then I noticed when the second median was drawn, it divided the areas of the two triangles formed by the first median in the ratio 1:2. Lastly, I discovered that three medians of a triangle divide the triangle into six triangles that are all equal in area. I have shown this in my centroid gsp demonstration below. However, it is now time to prove this conjecture. Let us use triangle BDE.  Triangles ACE and ACB have the same altitude since they share the same vertex and are sitting on the same base EB. But we know that A is the midpoint of EB (AD is the median). Since A is the midpoint of EB then we know EA and AB congruent. It then follows that the area of triangle ACE is equal to the area of triangle ACB because they share a common altitude and base. Earlier we saw that the second median divided the two triangles formed by the first median in the ratio 2:1. Using that argument we know that the area of triangle ACE is 1/3 the area of triangle DEA. Similarly the triangles TEC and DTC are also equal in area. This means that the area of triangle TEC = area of triangle DTC= 1/3 area of the triangle DEA. Thus we can now conclude that the triangles DTC, TEC, ACE, and ACB are all equal in area. As a result the rest of the proof is trivial.

The GSP construction file can be found here. Exploring the Centroid.

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