ΜΑΘΗΜΑΤΙΚΑ: MORLEY

Τετάρτη 22 Μαρτίου 2023

MORLEY - JHC Proof

I have the undisputedly simplest proof of Morley's Trisector Theorem. Here it is:

Let your triangle have angles 3a, 3b, 3c and let x* mean x + 60°, so that a + b + c = 0*. Then triangles with 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 180°. 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 (and the other two like it)	-	let me draw this one:

Let the angles at B, P, C be b, a**, c, and draw lines from P cutting BC at angle a* in the two senses, so forming an isosceles triangle PYZ. Choose the scale so that PY and PZ are both 1.

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

(in which the points X,Y should really be omitted). The points Y,Z are what I meant.

To make it a bit more clear, let me say that the angles of ΔBPR are b (at B), c* (at P), a* (at R).

Why do they all fit together? Well, at each internal vertex, the angles add up to 360°, as you'll easily check. And two coincident edges have either both been declared to have length 1, or are like the common edge BP of triangles BPR and BPC.

But ΔBPR is congruent to the subtriangle BPZ of ΔBPC, since PR = PZ = 1, ∠PBR = ∠PBZ = b, and ∠BRP = ∠BZP = a*.

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.

John Conway
Posted to geometry.puzzles group (1995)