Trisecting a Triangle

(or Morley's Theorem)

By Kelly Edenfield

Morley proved that given any triangle, the trisection of that triangle is always an equileral triangle. The first obstacle in proving this is the question of how to trisect the angles of a triangle. However, modern technology and Key Curriculum Press have armed us with Geometer's Sketchpad. This application enables the intrigued student to trisect angles with little difficulty.

Construction:

Construct an arbitrary triangle. Then measure each angle. In the calculator, divide each measure by 3. This gives the measure of one-third of the angle. Mark one of the points of the triangle as a center of rotation. Mark an adjacent edge and point at the other end of the side and highlight the one-third measure of that particular angle. Rotate by the highlighted angle. Repeat, rotating the rotation. You have now trisected one angle of a triangle. Repeat for the other two angles.

Connect the trisectors. That is, join the trisectors of angle A, the trisectors of angle B, and the trisectors of angle C at their points of intersection with trisectors of other angles. This will yield a triangle. Two triangles are possible. The desired triangle is composed of parallels to the sides of the original triangle. (In the following sketch, the blue triangle.) Measure the sides of this new triangle using the measure length function. All sides are equal; therefore, this new triangle is an equilateral triangle. Move the points of the arbitrary triangle around. The new triangle will always be an equilateral.

Below is a picture of the construction.

I constructed this triangle and the diagram because it seemed strange to me, yet interesting. When glancing at the statement, I did not believe that the proof would be too difficult. However, when I was unable to see a logical proof, I searched the Internet. It was there that I found a proof, claimed by its author to be the simplest proof of the theorem. Yet, it still baffles me. I will present it below.

Proof by John Conway of Princeton (15 February, 1995):

Let your triangle have angles 3a, 3b, 3c and let x* mean x + pi/3, so that a + b + c = 0. Then triangles with the angles

 0*, 0*, 0* a, b*, c* a*, b, c* a*, b*, c a**, b, c a, b**, c a, b, c**

exist abstractly, since in every case the angle-sum is pi.

Build them on a scale defined as follows:

0*, 0*, 0* - this is equilateral - make it have edge 1

a, b*, c* - make the edge joining the angles b* and c* have length 1 (similarly for a*, b, c and a*, b*, c)

a, b**, c - (similarly for a**, b, c and a, b, c**)

 X A Y, Z C

Let the angles at A, X, C be a, b**, c, and draw lines from X cutting AC at angle b* in the two senses, so forming an equilateral triangle XYZ. Choose the scale so that XY and XZ are both 1.

Now fit all these 7 triangles together! They'll form a figure:

 B P Q X A Y Z C

(in which points Y, Z should really be omitted.)

To make it a bit more clear, let me say that the angles of APX are a (at A), b* (at P), and c* (at X).

Why do they all fit together? Well, at each internal vertex, the angles add up to 2pi, as you'll easily check. And two coincident edges have either both been declared to have length 1, or are like the common edge AX of triangles APX and AXC.

But APX is congruent to the subtriangle AZX of AXC, since PX = ZX = 1, PAX = ZAX = a, and APX = AFX =b*.

So the figure formed by these 7 triangles is similar to the one you get by trisecting the angles of your given triangle, and therefore in that triangle the middle subtriangle must also be equilateral.

I'm not sure that I follow the proof completely, so I intend to continue searching for other proofs.