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.

Fig.1.

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).

Fig.2.

The proof is over.