Let us work backwards. To prove the theorem, it suffices to show that if PQR is an equilateral triangle and we erect on its sides triangles RQA, PQB and QPC as in the diagram then we get a triangle ABC with angles 3α, 3β, 3γ and appropriate trisectors.

Let us reflect P in BR to get R₁, and Q in AR to get R₂; construct the points P₁, P₂, Q₁ and Q₂ similarly as in the diagram. All we need then is that the points R₁ and R₂ are on AB (and so P₁, P₂, Q₁ and Q₂ are also on the appropriate sides). That this so is easily seen by computing some angles. For example, ∠ARR₂ = β⁺, so ∠ARR₁ = β⁺ - (π - 2γ⁺), where "⁺" is a shorthand for "+ π/3".

Let us assume that α + β ≥ γ, i.e., that the points R₁ and R₂ are in the order shown in the diagram. Then

so ∠RR₂R₁ = (π - (π/3 - 2γ))/2 = γ⁺. This implies that point R₁ is on the segment AR₂. Similarly, R₂ is on the segment BR₁, and we are done.

If α + β < γ then R₁ and R₂ are interchanged on AB; to see that, we note that ∠R₁RR₂ = 2γ - π/3 and ∠RR₂R₁ = π - γ⁺, so R₁ is on BR₂. Similarly, R₂ is on AR₁, completing the solution.