 ## Trisecting Areas of Triangles

#### by Robyn Bryant, Kaycie Maddox, and Kelli Nipper

1. Given a triangle ABC, find a point D such that line segments AD, BD, and CD trisect the area of the triangle into three regions with equal areas. Define D and prove that the triangle is divided into three regions of equal area. Show construction for finding D.  The above is an example of a triangle that has been divided into 3 equal areas by joining the three vertices to point D. To move the triangle around yourself, to make sure that these three areas are always the same, click here.

The next question is probably, how did you construct D? The above picture is to give you something to look at while we explain to you how we constructed point D. First we trisected each side of the triangle. Then we joined the trisected points of 2 sides that would form a segment parallel to and 1/3 of the distance away from the 3rd side. We did this to form all three dotted segments above. Point D is the point of concurrency of the three constructed segments. To form the three areas that are congruent, we joined the three vertices with D. Now that we have constructed D, we want to prove that the areas of triangles ADC, CDB, and BDA are congruent. Note, that we are not saying that the triangles themselves are congruent, just their areas. We have a theorem in geometry that is commonly refered to as the midsegment theorem. It states that the segment joining the midpoints of two sides of a triangle is parallel to and one half the length of the third side of the triangle. Well we are going to transfer that theorem to segments joining the points that are trisecting the sides of the triangles. For this activity we are only looking at the segments that are two thirds the length of the third side of the triangle.

The first thing we want to do is show that triangles QDR, WDX, and HDI are congruent. Since D is the midpoint of RX, QI, and HW, we know that we have some congruent segments. Since HI is one third of the base and RX is two thirds of the base, and D is the midpoint of RX, we can say that RD, XD and HI are all congruent. By using the same logic, we can say that XW, ID and QD are also congruent and so are RQ, WD, and HD. Therefore by SSS, we have three congruent triangles.

The next thing we want to do is to prove that the area of the following parallelograms are congruent: BWDQ, CXDI, and ARDH. We realize that these parallelograms are not congruent but their areas are. We are going to reprint the diagram here, simply because it will be easier to see. The area of a parallelogram is base times height. The heights of DXCI and DRAH are the same because RX is parallel to AC. The bases of these two parallelograms are also congruent because each base is one third of AC. Therefore we have two congruent areas. We can use the same process to prove that the area of the parallelograms BQDW and CXDI are also congruent.

The red segments are diagonals of the parallelograms, therefore cutting each into half.

Now look at the area of the three triangles that we are wanting to prove congruent; BDA, BCD, and ADC. Each triangle includes the smaller triangle that we first proved congruent; WDX, HID, and QRD. Also each triangle includes the total area of one of the congruent parallelograms. As a result, we have just proven that the segments joining the vertices to D form three triangles with equal areas.