Problem Statement: The CENTROID (G) 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.
First I describe how to create the centroid:
The First step is to create any triangle in GSP:
Then construct the midpoints of each side:
Third, Construct the medians (the line from the midpoint to the opposite vertex, i.e. the midpoint of CA to Point B) and select the intersection of the medians. This is your Centroid:
Well that was easy! But we really want to go further in this exploration. Thus I have chosen to (attempt to) Prove the three medians of a triangle are concurrent and the centroid is
the distance from a vertex to the midpoint of the opposite side
First, let's set up a triangle as so:
We can assume, without loss of generality, that A has coordinates (0,0), B is located at (b,0) and C is at (c,d). I am going to snapshot the proof, which I did in GSP, since the formulas look better there:
Is it true that the centroid is
This is very easily seen. For example, Consider median AD. D has coordinates of ([c+b]/2, d/2) whereas the centroid has coordinates ([b+c]/3,d/3). Since A is at (0,0) it is easy to see that the centroid is 2/3 of the way there. By similar triangles, this works for all of the other medians as well.
One More exploration: Let us prove that the medians of a triangle partition the original triangle into 6 triangles with the same area.
Let G be the centroid. We know that CE=EA because E is the midpoint. We also know that GE = GE by reflexive property. Finally CG = GA by the property of the Centroid. Thus Triangle CGE = AGE by Side-Side-Side congruency.
(Going Forward equality means areas are equal)
By the same token, Triangle CGF = Triangle BGF and Triangle BGD = Triangle DGA. We also know that Triangle CAG = CDA - DGA. and Triangle CGB = Triangle CDB - Triangle BGD. Doing some algebra, with the facts that Triangle CAG = Triangle CGB and Triangle CGE = Triangle AGE, we get Triangle CGE = 1/2 (Triangle AGC).
Also, since Triangle CGF = Triangle BGF, Triangle CGE = 1/2(Triangle CGB) = Triangle CGE.
Therefore, Triangle CGF= Triangle BGF = Triangle CGE = Triangle AGE, and thus it follows that all the triangles have the same area.
