Problem 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.

## △ABC should look similar to the triangle depicted below.

## First, I let D be the point of concurrency of the angle bisectors. The triangle was trisected into three smaller triangles and it looked like the triangle we were trying to mimic.

## However, the areas of the three triangles were not the same.

Next, I let D be the point of concurrency of the medians of the triangle. We see that the areas of the three smaller triangles are equal as desired.

The Centroid of a triangle is the point of concurrency of the three medians.

## Recall, a median of a triangle is the segment from a vertex to the midpoint of the opposite side. The medians divide the triangle into six smaller triangles.

## We want to show that △ADB, △ADC, and △BDC have equal areas.

## Lets start by showing that the six triangles formed by the medians have the same area.

## For this proof, Click Here.

## We found that

The area of ΔAGF is ⅙ of the area of ΔABC.## The areas of the remaining 5 remaining triangles are each equal to ⅙ the area of ΔABC, proving that the medians divide the triangle into 6 smaller triangles of equal area.

## △ADB, △ADC, and △BDC are each made up of two of these smaller triangles.

## Since these 6 small triangles have equal area, we can conclude that △ADB, △ADC, and △BDC have equal areas as well.

Problem 2## Given a triangle ABC, and given a point E, find points B' and C' such that line segments AE, B'E, and C'E trisect the area of the triangle into three regions with equal areas.

## △ABC should look similar to the triangle depicted below.

Base Case:## First, let E be located on vertex A. If EB' and EC' trisect the area of the △ABC, then B' and C' must be the trisection points of BC.

## From Problem 1 and our explorations of the base case, we can conclude that point E must lie on a median of △ABC between one of the triangle's vertexs and its centroid, G. So point E cannot lie closer than (1/3) the length of median AI (point I is identified in the picture below) to side BC. If point E was closer than this, the triangle would not be trisected into three region with equal areas.

## Let AE be (1/3) the distance from a vertex, to the midpoint of the opposite side.

## First, I found the centroid, G, of the triangle and then found the midpoint of AG. E is the midpoint of AG.

## Problem 3

## Given a triangle ABC. Construct two line segments parallel to the base BC to divide the triangle into three regions with equal areas.

## △ABC should look similar to the triangle depicted below.

## Now, prove that the construction divides the triangle into three regions of equal area.

## So, we have shown that the Area △ADE = (1/3) (Area △ABC) and the Area △AFG = (2/3) (Area △ABC).

## Therefore, △ADE and quadrilaterals DEGF and BFGC each have an area eqaul to (1/3) the Area of △ABC as desired.

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