Okay, I reconstructed that proof of Morley's theorem. 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 most presentable to people who know high school geometry. I find other proofs 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 (thanks to commenter for catching that)] 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 ratio 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 appropiately, 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 equliateral 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.

Doesn't this imply that you can trisect an arbitrary angle? At lease one that is less than 60 degrees?

Very handsomely done, though.

Posted by: Foxy | January 02, 2007 at 09:26 AM

To clarify, doesn't this imply you can construct a trisection of an arbitrary angle? Every step seems to be possible using just a compass and straightedge.

Posted by: Foxy | January 02, 2007 at 09:30 AM

i see why you might think that, but you can't trisect an arbitrary angle. the reason that this proof doesn't contradict that is that it starts with the equilateral triangle in the middle and builds the triangle from there. if, however, you start with an arbitrary triangle, no compass/straightedge construction can locate that equilateral triangle. the classic counter-example to trisectability is the 60° angle, which can't be trisected. if you try to use this proof to do it, you'd need the 20° angle FIRST, which is exactly the object you were trying to construct.

Posted by: Polymath | January 02, 2007 at 01:56 PM

Oh, true. I guess I didn't think it through.

Anyway, though. Very nice proof for a very excellent theorem.

Posted by: Foxy | January 02, 2007 at 03:43 PM

Hey Polymathematics, I was just wondering if anything huge and lifechanging happened to you lately? Or not so much?

Posted by: Your Wife | January 10, 2007 at 08:19 AM

That previous comment really cracks me up.

Posted by: Michael | January 24, 2007 at 12:58 PM

I've one comment on the Morley proof: You have to add that a & b & c are > 0 !

Otherwise one can choose, for example, c = 30 , a = 30 and b = 0 (all in degrees). Now point B is not defined properly!

The rest of the proof is ok!

Posted by: student | January 29, 2007 at 03:35 AM

how do you trisect a 56 degree angle

Posted by: andrew | February 01, 2007 at 03:07 PM

It is proven impossible to trisect an arbitrary angle.

Posted by: Mathlete #1 | February 08, 2007 at 03:04 PM

i'm not sure if mathlete is claiming there's a flaw in the proof or responding to andrew, but i should make it clear that mathlete is correct.

in general, it isn't possible to trisect an arbitrary angle. that means that the equilateral triangle in this proof can't be constructed (given the original triangle) with a compass and straightedge. but that doesn't invalidate the theorem.

Posted by: Polymath | February 08, 2007 at 04:08 PM

Can I have the initial of your first name and your last name so that I could cite your work in my research prospectus? It would be great if you could.

Posted by: Mathlete #1 | February 10, 2007 at 04:44 PM

To clarify, Initial of first name and full last name.

Posted by: Mathlete #1 | February 10, 2007 at 05:02 PM

Thank you Polymath.

Posted by: Mathlete #1 | February 14, 2007 at 05:14 PM

In the third diagram, shouldn't the letter next to point B (giving the size of the angle) be "b"? The text says this angle is b, but the diagram says c.

Posted by: Charlie Huttar | October 16, 2007 at 09:41 AM

By the way, this proof was very helpful. Thanks!

Posted by: Charlie Huttar | October 16, 2007 at 09:42 AM

I'm glad it was helpful, Charlie. The number next to the B in the third diagram should indeed be a "b", as the text indicates, but I think it is. I do see, though, that the "b" overlaps the diagram a little, and it might look like a "c". Correcting the diagram is tricky, and I'm not going to try it, but thanks for bringing it up. Sorry for any confusion.

Posted by: Polymath | October 16, 2007 at 10:55 AM

wow my discrete math teacher just gave us this one today and i figured out a proof similar to this in that you start with the equilateral triangle in 30 minutes. mine may not have been exactly correct though, somehow i did it only with geometry and no math or variables...

Posted by: student | November 21, 2007 at 12:48 AM

I've a new owen proof, how I show it?

Posted by: Magdy Essafty | December 03, 2007 at 05:09 PM

Referring to an earlier comment, and merely dealing with integral numbers of degrees, one can of course trisect any angle for which the trisection of the angle is 3 degrees or a multiple thereof. This follows from the fact that one can construct 15 degrees from an equilateral triangle, and 18 degrees from a regular pentagon.

Posted by: John Trainin | May 26, 2008 at 04:42 AM

I do not quite understand your last statement:

If you scale that diagram appropiately, it will result in a triangle ABC that is the same as the arbitrarily given one.

Of course, if the given triangle satisfies Morley's Theorem, then by construction, the triangle constructed is unique (up to scaling), so it will indeed result in a triangle ABC that is the same as the arbitrarily given one (up to scaling).

However, isn't it still possible that some arbitrary given triangle does not satisfy Morley's Theorem, so if you use its angles a, b, c to construct as you did above, you end up with a different triangle with the one you started with?

To summarize, I think there might be something wrong with the logic you are employing.

Posted by: Ben | December 17, 2008 at 05:55 PM

I retract my previous comment. The proof does indeed make sense.

Posted by: Ben | December 17, 2008 at 05:59 PM

I don’t know if this is true, i argued with my friend about this, and he thinks it’s true, maybe it is or maybe it ain’t but i’m not absolutely convinced

Posted by: A disappointed One | January 19, 2011 at 10:09 AM