Posed Problem:

"Prove that the three medians of a triangle are concurrent and that the point of concurrence, the centroid, is two-thirds the distance from each vertex to the opposite side."

First, let us breakdown what the problem asks us to do.

1) Prove that the three medians of a triangle are concurrent.
- Median: Segment formed by connecting the midpoint of a side of a triangle and the vertex across from it.
- Concurrent: Existing or present at the same time

2) The point of concurrence, the centroid, is two-thirds the distance from each vertex to the opposite side.
- Point of concurrence: where the three medians meet; a shared point.
This is called a centroid.
- Centroid is two-thirds the distance from each vertex to the opposite side.

Next, let us construct a triangle in Geometer’s Sketch Pad. Use the previously built tools to construct an equilateral triangle, but any triangle shape or size will yield the same result. Construct the three midpoints and the three medians. Recall that the median is constructed by connecting the midpoint of a side with the vertex opposite it. Notice the centroid that is formed. Finally, measure the distance from one point to the centroid and then the distance from the centroid to the vertex point.

A result is presented below:

Please install Java (version 1.4 or later) to use JavaSketchpad applets.

We can move around the vertices of the triangle to make it larger or smaller. Finally, we have to show that the centroid is two-thirds the distance from each vertex to the other side. Observe the results obtained below. Because our trinagle is an equilateral triangle, CB, DE, and GF are the same length, but notice the relationship between the two segements connected by the centroid A. For instance, AB measures 3.537 and it is one-third of the entire length of CB. Similarly, AC is two-thirds of CB.

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