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The proof below is another backwards proof based
on the work of Dergiades dating from 1991.

According to Alexander Bogomolny, this proof is first
published in Greek in the bulletin of Diastasi,
a Greek Mathematical Society.<ref 2>

The original reference is from "A simple geometric
proof of Morley's Theorem", Diastasi 1991 issue
1-2 pp 37-38. Thessaloniki-Greece. <ref 2>

The proof uses the following lemma:

LEMMA Let BAC = a, point D is the incentre of ABC if and only if (a) the line DA bisects a and (b) CDB = 90 + a/2.

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PROOF Suppose that angles , and are given with ++ = 180. Dergiades' approach is to construct a triangle with these angles for which Morley's Theorem is true.

According to Dergiades, suppose we let
Then 0<x,y,z<60 and x+y+z=120.

Consider an equilateral triangle. Construct 3 isoceles triangles with base angles x, y, z respectively ( with the constraints x+y+z =120 and 0<x,y,z<60) on the sides on the equilateral triangle.

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Extend the sides of the isosceles triangles so that they intersect as shown:

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Since D is the vertex of the equilateral triangle, and I is the vertex of the isosceles triangle with base angle x, it follows that the line ID is the angle bisector.

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Join B and C, note that BDC = 60+z+y (vertically opposite angles)

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BDC = 60+z+y = 60+x+y+z-x=180-x. Since BIC = 180-2x. and D is bisects BIC, the Lemma shows that D is the incentre of CIB.

It follows that the lines drawn at points A B and C are angle trisectors.

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The last step is to prove ABC has angles , and . In the following figure:

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we can see that
a+s+t+x+x= 180
s+x+z+60=180, i.e. s+x+z=120
t+x+y+60=180, i.e. t+x+y=120

by substitution, we have

It follows that:
a=60-x= /3.

Hence ABC has angles , and .