by: Doris Santarone

Proof of Napolean's Triangle using Trigonometry

 

Let's start with our original sketch (with the equilateral triangles constructed outward).

 

First, I will construct segments AM and AN, to create triangle AMN. I will call the sides of this triangle h, j, and k as seen below. (I have also named the sides of the original triangle as a, b, and c.)

Because points M and N were constructed as centers, I know that j bisects Angle YAC, and h bisects angle BAZ. So, Angle MAC = Angle NAB = 30 degrees, or Angle MAC + Angle NAB = 60 degrees.

 

We can apply the Law of Cosines to get:

(1)

 

Now, I will extend h to intersect BZ at point E, which is also the perpendicular bisector of BZ (based on construction of "center"). See below:

Using the fact that h = 2/3 (AE), or AE = (3/2)h and the Pythagorean Theorem, we can say:

 

Similarly,

 

Now, going back to apply this into equation (1), we get

 

 

Using the Cosine of the Sum Rule and the fact that cos60 degrees = 1/2 and sin 60 degrees = (sqrt3)/2, we get:

(2)

 

Now, I will apply the Law of Cosines to Triangle ABC (the original triangle).

I will use this, along with the fact that (found in the derivation of the Law of Sines) to substitute back into the previous equation (2).

Since this equation is symmetrical in a, b, and c, then we can say that the triangle connecting the 3 centroids is equilateral.

---

Back

---