Write Up #4

Six Smaller Triangles from the Centroid

Before reading this write up, it might be helpful to open up this GSP file to be able to follow along!!

When you first open the GSP file, you will see that our whole triangle is split into 6 smaller triangles labeled 1 through 6.  These have been created by the three medians, which as defined above, create the centriod.  The calculated areas, which were found simply using our measurement tool in GSP, are shown for each of the 6 triangles.  Notice that these values are all equal to each other no matter where or how you drag the vertices of the whole triangle to change its shape/appearance.  However, this isn't enough to prove how or why these 6 smaller triangles are equal in area, so this is what we will discuss below. 

All 6 triangles share a common vertex: the centroid, which in our GSP file is the point G.  The three medians create midpoints D, E, and F on each of the 3 sides of the big triangle.  Now, let's start breaking the 6 smaller triangles into pairs and discuss their areas. 

Let's first look at triangles 2 and 3.  They share altitude FG, which we can also call their height.  Their bases, AF and CF are equal in length because we know F is the midpoint of segment AC.  Using the area formula for a triangle, A = 1/2bh, we can say their areas are equal because their bases and heights are equal to each other, therefore, the area formula holds true to be equal for both triangles 2 and 3.  The same concept holds true for the other two pairs of triangles.  Triangles 4 and 5 share height EG and midpoint E means their bases, CE and BE, are of equal length, therefore, the areas of triangles 4 and 5 are equal to each other.  Triangles 1 and 6 share height DG and midpoint D means their bases, AD and BD, are of equal length, therefore, the areas of triangles 1 and 6 are equal to each other.  So, we now have established 3 pairs of triangles with equal areas, but how do we know that all 6 triangles have areas equal to each other? 

Going back to triangles 2 and 3 and also considering parts of the whole picture, it's fair to say:

     (Δ2 + Δ3) = (Δ2 + Δ3 + Δ4) - Δ4 where(Δ2 + Δ3 + Δ4) is the right half of the big triangle

Now consider the left half of the big triangle (Δ1 + Δ6 + Δ5) and another pair of small triangles that we know have equal areas (triangles 1 and 6).  We can say:

     (Δ1 + Δ6) = (Δ1 + Δ6 + Δ5) - Δ5

Now that we have established equations about the right half and left half of our whole triangle, which are two other triangle shapes within our figure, we can discuss these two triangles in a way similar to the way we were discussing the three pairs of the six small triangles.  These two triangles are equal in area as well.  The right and left triangle share the same altitude/height of AE, one of our medians.  Then, the same bases for our triangle 4 and 5 pair, CE and BE, are the bases for our right and left half triangles as well, therefore we know the bases are of equal length.  So finally, we can say that their areas are equal and write it as:

     (Δ2 + Δ3 + Δ4) = (Δ1 + Δ6 + Δ5)

So this shows that our six small triangles are of equal area to each other!   


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