# Trisecting the Area of a Triangle

If we trisect the area of a triangle we will be dividing the triangle into three parts which are equal in area.  Let's begin with triangle ABC and look at two different ways to trisect it's area.

The first method of trisection consists of dividing triangle ABC into three triangles of equal area.  We begin by finding the centroid of triangle ABC.

Now we need to connect the centroid to each of the vertices to create the segments for the sides of each of our three triangles.

So we have that Area(ADB) = Area(BDC) = Area(CDA) and that Area(ADB) + Area(BDC) + Area(CDA) = Area(ABC)

We know this to be true because the centroid is formed by the intersection of the three medians of the triangle.  We know that the median of a triangle divides the triangle in half by connecting one vertex to the midpoint of the opposite side.  So, if we look at the median AM1, we know that Area(AM1B) = Area(AM1C)

Now the same is true for the area of triangles BM2C and BM2A as well as CM3A and CM3B.

Now if we look at the centroid, D, of triangle ABC, we know that each of our small triangle ADM3, M3DB, BDM1, M1DC, CDM2, and M2DA are all 1/6 the area of our triangle ABC.

Therefore, when we combine any two of these triangle, they are 1/6 + 1/6 = 1/3 the area of triangle ABC.  So we must have that each triangle ADB, BDC, and CDA are 1/3 the area of the triangle ABC.

Now we will look at the second method of trisecting the area of a triangle.  For this method we construct two segments parallel to the base of our triangle ABC in order to create three regions, one triangle and two trapezoids, that are each equal in area, and are each 1/3 the area of triangle ABC.

We begin by constructing a triangle that is 1/3 the area of the original triangle ABC.  We know that this triangle will be similar to our original triangle ABC by AAA.  So we must have that the ratio of the sides of our original triangle ABC to the new triangle is sqrt(3):1.  So we can begin our construction or this new triangle by first constructing a 30-60-90 triangle where one leg of the 30-60-90 triangle is equal in length to one segment of triangle ABC.  We will do this because we know that the ratio of the long leg to the short leg is sqrt(3):1.

In the above figures, we have that QR = AC, therefore the ratio of PR to AC is 1:sqrt(3).  So now we need to construct a circle around the point B with radius equal to the length or PR.  The point of intersection of the circle about B and the segment AC is one vertex of our new triangle.  We will call this point D.  So now we know that the ratio between AB and DB is 1:sqrt(3).  Finally we need to construct the point E so that DE is parallel to AC.

Now we have that triangle DBE is 1/3 the area of triangle ABC.  Next we need to construct the two trapezoids, each of which are also 1/3 the area of triangle ABC.  This means we want to divide the polygon ADEC in half.  So we want to construct a point G such that the ratio between BG and BD  be sqrt(2):1.  We know, from the properties of a 45-45-90 triangle, that if we construct a 45-45-90 triangle with leg length equal to BD we will be able to construct a circle about the point B so that its radius is equal to the hypotonuse of the 45-45-90 triangle giving us that the ratio between BG and BD is sqrt(2):1.  Then we can construct a segment through the point G parallel to the base, AC, of the original triangle ABC.  So now we have constructed the two trapezoids GDEH and AGHC, both of which are 1/3 the area of the original triangle ABC.

Finally we have that each trapezoid, AGHC and GDEH, as well as the triangle DBE are 1/3 the area of the original triangle ABC.

To view an interactive GSP sketch of the first method of trisection click here.
To view an interactive GSP sketch of the second method of trisection click here.