Napoleon's Triangle

by Beverly Hales

Objective: The purpose of this essay was to develop an understanding of Napolean's Triangle through exploration using the Geometer's Sketchpad and other math instructional software. This essay was written such that my geometry students at Pebblebrook High School could comprehend the material.

Given any triangle ABC, construct equilateral triangles on each side.

Find the center of each equilateral triangle. The center of a triangle is the incenter, which is the point of currency of the angle bisectors. The triangle formed by these three centers is Napoleon's Triangle.

Triangle HIJ is Napoleon's Triangle.

Since every triangle has an incenter and three non collinear points can be connected to form a triangle, the question is not of the existence of Napoleon's Triangle, but whether or not it is always equilateral. Using a GSP sketch, if randomly translating the vertices of the triangle ABC, the measurement of the sides of triangle HIJ are always equal.

The sketchpad gave visual confirmation which can be used to convince any high school geometry class, but it is not sufficient evidence for higher level mathematicians. I began examining other line segments associated with the triangles. I connected the vertex of my original triangle with the remote vertex of the equilateral triangles of the opposite side. This process was repeated at all three vertices of the given triangle. The line segments were concurrent. I found that this point is referred to as the Fermat Point.

Within the constructed equilateral triangles above, the circumscribed circles were constructed with the incenter as the center. It is important to note that in an equilateral triangles, the angle bisectors, altitudes, medians, and perpendicular bisectors are concurrent. Moreover, the incenter, the circumcenter and the centroid are also concurrent. These properties of the equilateral triangle are essential for determining the classification of Napoleon's Triangle as equilateral.

The goal was to determine if any congruent segments exist. By highlighting two triangles defined by the segments examined, it was easier to determine the triangles congruency.

Using the diagram above and the given statements concerning equilateral triangles, I was able to show the two shaded triangles congruent using the following statements.

1. Angle AEB=Angle FBC=60 degrees because it intercepts an arc determined by a chord which is one side of the equilateral triangle.
2. Using angle addition, Angle EBC is congruent to Angle ABF.
3. Using the congruent segments of the equilateral triangles, EB= BA and BC=BF.
4. The two triangles are congruent by SAS for congruent triangles.
5. EC = AF because they are corresponding parts of congruent triangles.

By using a similar proof with segment BD and either segment EC or segment AF, all three segments connecting the vertex of the given triangle to the remote vertex of the equilateral triangles on the opposite side are congruent.

Now that the segments that connect the vertex of the triangle to the remote vertices of the equilateral triangles are congruent, we can use them to further our exploration.

We continue our look at Napolean's Triangle by examining the different triangles formed by the segments connecting vertices of the original triangle and the attached equilateral triangles. By the SAS postulate for similar triangles, two triangles are similar if two pairs of corresponding sides are proportional and the included angles of the triangles are congruent.

Using a GSP sketch and calculate, I was able to show the angles congruent and corresponding sides proportional.

Let us look at the GSP sketch with a couple of the segments hidden and two specific triangles shaded.

In order to show the triangles similar it is necessary to referred to the special properties of equilateral triangles. As stated earlier, within an equilateral triangle, the centroid lies on the altitude. The centroid is 2/3 the distance to the opposite side. Therefore, AH is equal to 2/3(altitude) of the triangle.

The altitude of the equilateral triangle forms two right triangles having acute angle measures of 30 degrees and 60 degrees. By using the 30-60-90 triangle relationship, let x equal the length of the side opposite of the 30 degree angle. This segment is one-half the length of the side (AB) of the equilateral triangle.

x=1/2 AB

The side opposite the 60 degree angle is equal to the square root of 3 times x.

This segment relationship occurs with all equilateral triangles:

The ratio of the length of the segment connecting the centroid to the vertex on the altitude to a side of the equilateral triangle is square root of 3 over 1.

Segment AH and Segment AD lie on the angle bisector of the Triangle HAB and Triangle DAI. Since the angle measures of an equilateral triangle equal 60 degree, m<HAB =m< DAI = 30 degrees. Angle BAI is congruent to itself by the reflexive property, so by applying the angle addition postulate, m<HAI = m<BAD. Therefore, Triangle HAI is similar to Triangle BAD.

By stating proportional sides AB/AH = BD/HI = AI/AD = square root of 3/1.

Notice that BD corresponds to HI. This shows a relationship between a segment joining a vertex of the given (original) triangle and the remote vertex of the equilateral triangle on the opposite side and a side of Napolean's Triangle.

The same procedure would be used to show that EC/HJ equals the square root of 3/1 and BD/JI equals square root of 3/1.

For a detailed look at the proof of each pair of triangles, click here.

Based on the similarity ratios stated above:

Earlier we stated that the segments joining the vertex of the original triangle and the remote vertex of the equilateral triangle are congruent.

If BD= CE = AF, then HI = HJ = JI.

Therefore, Napolean's Triangle will always be equilateral.