The proof has been published (in Greek) in the bulletin of the department (Central Makedonia of Greece) of the Greek Mathematical Society "Diastasi": Nikolaos Dergiades, "A simple geometric proof of Morley's theorem", Diastasi 1991 is. 1-2 p. 37-38 Thessaloniki-Greece. Lemma 1

The external angle B of an isosceles (AB = AC) ΔABC is equal to 90° + A/2.

Lemma 2

The incenter I of ΔABC lies on the bisector of an angle, e.g A, and sees the opposite side BC with angle BIC = 90° + A/2 and conversely. Proof of Morley's Theorem

Given ΔABC, with angles A, B, C. If x = 60° + A/3, y = 60° + B/3, z = 60° + C/3, then x + y + z + 120° = 360°.

On the sides of an arbitrary equilateral triangle A1B1C1 we construct outwardly the triangles A'B1C1, A1B'C1, and A1B1C' such that the lines B1C' and B'C1 are symmetric with respect to the perpendicular bisector of B1C1 and also ∠B'C1A1 = ∠C'B1A1 = x and similarly the others as seen in the figure.

It is obvious that the angles of the three triangles A'B1C1, A1B'C1, A1B1C' at A', B', C' are A/3, B/3, C/3, respectively.

If A2 is the intersection of B'C1 and C'B1, then the triangle A2B1C1 is isosceles and A2A1 is the bisector of A2 = ∠B'A2C'.

By Lemma 1, we have that x + 60 = ∠B'C1B1 = 90° + A2/2.

Since ∠B'A1C' = x + 60° = 90° + A2/2, by Lemma 2 we conclude that A1 is the incenter of triangle A2B'C', and similarly B1, C1 are the incenters of triangles B2C'A', C2A'B' respectively. Hence ∠B'A'C1 = ∠C1A'B1 = ∠B1A'C' = A/3, or, finaly, A' = A, B' = B, C' = C which means that the triangles ABC, A'B'C' are similar and the trisectors of the angles of ΔABC form a triangle similar to A1B1C1 that is equilateral. Morley's Miracle

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