(by Polymath) This page is part of the Advanced High School Math Project.

If you don't know what Morley's theorem says, I urge you not to look it up just now. I'll tell you what it says at the end of this proof. This proof is not the shortest one (which you can find in many places, and is due to John Conway—I will post a link at the end to his proof), but it is the one I've found that's most accessible to high school students. Other proofs are either too advanced for most people (sometimes including myself), or confusing in their notations. This proof uses only basic, first-year high school geometry. It was found online by a student of mine, but I've tried to find it myself, and I can't. I am eager to give credit where it is due, and if anyone finds the link, I will happily cite it here. My proof is a reconstruction from some notes and from my memory, and the presentation will thus probably differ somewhat from that source anyway.

First, start with an equilateral triangle called XYZ. It is drawn in red below. Then replicate that triangle three times, and arrange them as shown in blue below.

Now choose 3 positive numbers called *a*, *b*, and *c*. But be sure those three numbers add up to 60. Draw a line from point X that makes an angle of *c* degrees from the blue triangle, a line from Z that makes an angle of *a* degrees, then one more from Z making an angle of *b* degrees, and one from Y making an angle of *c*
degrees again. Those lines are shown below in black, with the angles
labeled to show you what I mean. Call the intersection of those lines
B and A as indicated. Note that because the angles of the equilateral
triangles are each 60°, and no two of *a*, *b*, and *c* add up to more than 60°, the lines I drew can't be parallel, and thus the points B and A are well-defined.

So consider triangle BXZ. The two angles at X and Z add up to:

(60°+ *c*) + (60° + a) = 120° + (*a* + *c*) = 120° + (60° – *b*) = 180° – *b*.

Since the angles in that triangle must add to 180°, angle B must measure *b* degrees. Similarly, angle A must measure *a*
degrees. That is reflected in the next diagram below, as are two
extensions of the blue lines out to points S and T, which we will need
next.

Because triangles SXZ and TYZ both contain one angle of 60°, one angle of (60 + *c*)°
and a side of the original equilateral triangle, they are congruent by
ASA, and thus SZ = TZ. Furthermore, triangles SBZ and TZA each contain
an angle measuring *a* and *b*, and are thus similar. This
gives the first proportion below, and the second equals sign is from the
congruent segments we just proved.

Calculate next the measures of two angles. First, in triangle ZTA, angle T measures 180° – (*a *+ *b*). This angle is marked with a # below. Second, angle BZA is obtained by subtracting angles measuring *a* and *b* from the blue straight angle, making angle BZA also measure 180° – (*a* + *b*).
So if we draw in segment BA, the new triangle we formed (BZA) must be
similar to triangle ZTA, since they have one congruent angle, and the
sides creating that angle are proportional.

This makes the other two angles of triangle BZA also measure *b* and *a*.
And if we duplicate that same proof around points X and Y, we create
triangle ABC below. Note that this triangle has angles adding up to 3*a* + 3*b *+ 3*c*, which of course is 180°.

But imagine that you were given an arbitrary triangle to begin
with. If you trisect the angles of that triangle, the angles created
conform to the initial condition above that *a *+ *b* + *c *= 60°. Thus I could always create a diagram like the one above using the angles of any given triangle—divided by 3—as *a*, *b*, and *c*. If you scale that diagram appropriately, it will result in a triangle ABC that is the same as the arbitrarily given one.

Morley's theorem (1899): The three points of intersection of adjacent trisectors of the angles of any triangle form an equilateral triangle.

Morley proved this as a corollary to some other very complicated calculations of conic sections (I think) based on certain properties of triangles. But the simplicity of the fact (sometimes now known as Morley's Miracle) led to many different proofs, many understandably involving trigonometry. Conway's proof is the shortest, and doesn't use trigonometry or complex numbers or group theory (like some of the proofs), but it uses a notation that takes getting used to, and I doubt that high school students would be able to follow it as easily as this proof.

I, for one, find Morley's Miracle almost staggeringly unbelievable.