EMAT 6690- Essay # 1
Trisecting the Area of a Triangle
Given any triangle ABC, we can find a point D such that line segments AD, BD and CD trisect the area of the triangle into three regions with equal areas.
Since point D is the centroid of triangle ABC, we can construct point D by constructing the medians of triangle ABC. The intersection of the medians, or the point of concurrency, is the centroid. Click here to see the construction of the centroid in GSP.
We can prove that the centroid trisects the area of triangle ABC into three regions of equal area. First we need to trisect the area of one of the smaller triangle such as triangle ADB. If we trisect the base of this triangle, AB, then we can trisect the area of the triangle. We trisect the segment AB by creating the midpoint of segment AD, point F.
Then we construct a line parallel to AC through point B. We construct points F’ and D’ on the parallel line through B on the opposite side of AB from F and D such that AF = BF’ and AD = BD’. When we construct segments FF’ and DD’ we obtain the intersection of these segments and segment AB, points H and G, respectively. These points trisect segment AB into equal segments. Click here to see this construction in GSP. If we connect points H and G to vertex D we obtain the trisection of the area of triangle ABD.
We can repeat this construction for the other two triangles ACD and BCD. When we trisect the area of all three triangles we see that the line HD intersects segment AC at a point of trisection for triangle ACD. Similarly we can conclude the same for all the lines of trisection of the triangles and we obtain the following construction.
We can now apply a modified version of the mid-segment theorem. Since CI and CM are 2/3 the length of CA and CB, respectively, then segment IM is 2/3 the length of AB and parallel to AB. We also know that ID = DM. Therefore ID and DM is 1/3 the length of AB and is equal to GH. Applying the same theorem to the other segments, GK and JH, we obtain that KD = DG = JI and DJ = DH = KM. Now we can say by SSS that triangles DIJ, DKM and DGH are congruent and therefore their areas are equal.
We still need to prove that all the adjacent non-black triangles are congruent. We will start by showing that triangles BDH and BDM are congruent. Since DM is parallel to AB and DH is parallel to CB then DMBH is a parallelogram. Then DB is the diagonal which divides the parallelogram into congruent triangles BDH and BDM. We can apply this same argument to obtain the following congruencies and therefore equal area triangles.
BDH ≈ BDM
ADG ≈ ADI
CDJ ≈ CDK
We need to make one final claim. We know that BH and GA are both 1/3 of AB and the altitude for both triangles is the same. Now we can claim that the area of triangle ADG and BHD are the same. Similarly, we can claim that the areas of all the following triangles are the same:
ADG and BHD
BDM and ADK
CDJ and ADI
By transitivity then all the colored triangles have the same area. Therefore triangle ABC is trisected into three triangles all of equal area, where each triangle is made up of two congruent colored triangles and one black triangle. Click here to see this construction in GSP.