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This is an indirect proof. Consider an equilateral triangle A0B0C0 and 3 angles a, b, c such that a + b + c = π/3.
Let's build on the sides of this triangle 3 isosceles trapezoids such that their smaller base side is one side of the triangle, the non-parallel sides are congruent to the smaller base and the base angles are respective 2a, 2b and 2c (see fig.1) . So, for example A2A0 = A0B0 = B0B1, and ∡A0A2B1 = ∡B0B1A2 = 2c, a.s.o. Let us extend the line segments A1A2, B1B2 and C1C2 until they intersect in A, B and C.
Observe that A2B0 bisects angle ∡A0A2B1, because ∡A0A2B0 = ∡A0B0A2 = ∡B0A2B1.
Let us calculate ∡B1A2C. For this, ∡A1A0A2 = 2π – ∡A2A0B0 – ∡B0A0C0 – ∡C0A0A1 = 2π – (π – 2c) – π/3 – (π – 2b) = 2c + 2b – π/3 = 2(π/3 – a) – π/3 = π/3 – 2a.
Because the triangle A1A0A2 is isosceles (A2A0 = A0B0 = A0C0 = A0A1), ∡A0A2A1 = (π – ∡A1A0A2)/2 = π/3 + a, and from this ∡B1A2C = π – ∡A0A2A1 – 2c = 2π/3 – a – 2c.
Similarly ∡A2B1C = 2π/3 – b – 2c, so ∡B1A2C + ∡A2B1C = π – 3c, because a + b + c = π/3.
Two things follow from this. First, the angle in C is 3c. Second, ∡B0A2C + ∡B0B1C = π, so A2B0B1C is a cyclic quadrilateral. Obviously, A0 is also on the circle circumscribed.
CA0 and CB0 trisect the angle in C, because A2A0 = A0B0 = B0B1 (see fig.2).
The proof is over.