Morley's Miracle
J. Arioni's Proof

John Arioni
May 27, 2017

Let's take an arbitrary $\Delta ABC\,$ and trisect the angles $\alpha,\,$ $\beta,\,$ $\gamma\,$ of $\Delta ABC.\,$ (fig. 1)

angles in triangle trisected

Let's construct three triangles, each around one copy of the same equilateral triangle as shown in fig. 2. The red, blue and green angles are the same angles $\alpha/3,\,$ $\beta/3,\,$ $\gamma/3\,$ of fig. 1.

angles in triangle trisected. New triangles constructed

Rearranging the triangles of fig.2 by superimposing the equilateral triangles yields the six-pointed star of fig.3 here below:

six-pointed star

Inside the six-pointed star there are represented six angles, highlighted with different shades of grey: these angles will play a key role in the problem. Equal shades of grey are used here to point out the pairs of angles that are congruent (as we'll see further on).

The triangles of fig. 2 will help us to evaluate such angles simply applying the rule of the sum of angles in some triangle.

Fig.4 below provides an example applied to one of the triangles of fig.2. In red, blue and green are respectively angles $\alpha/3,\,$ $\beta/3,\,$ $\gamma/3\,$ of fig. 1.

Sum of angles of a triangle equals $\pi.$

six-pointed star, specifying angles

Repeating the same steps as in fig. 4 with the other two triangles of fig. 2, what we eventually get is represented, in fig. 5. The three-pointed star (fig.5 bottom right) is what is left of the six-pointed after removing the parts we don't need any longer.

six-pointed star, putting together

Inside the three-pointed star of fig. 5 let's construct the $\Delta X'Y'Z'\,$ as shown in fig. 6.1 (the triangle with blue sides).Then let's complete $\Delta X'Y'Z'\,$ with the vertex angles, simply applying the sum of angles rule to the three grey triangles of fig. 6.2 - it's plain that $\Delta X'Y'Z'\,$ and $\Delta ABC\,$ of fig. 1 are similar. Now let's construct (fig. 6.3) the three-pointed star $X'Y'Z';\,$ the three-pointed stars $X'Y'Z'\,$ and $XYZ\,$ have parallel sides (*) as well as congruent vertex angles, so they are similar.

Finally let's construct the blue triangle as in fig 6.4: it's plain that the blue and and yellow triangles are similar, then the blue triangle is equilateral, which proves Morley's theorem.

six-pointed star, almost finished

$\Delta ABC\,$ compared with $\Delta X'Y'Z':$

Morley, final

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