H. D. Grossman's Proof
The three points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle.
This proof was published in The American Mathematical Monthly, Vol. 50, No. 9 (Nov., 1943), p. 552.
Let the triangle have base BC and angles 3α, 3β,3γ. Let BDK, BF, CDH, CE be angle trisectors. E is determined by making
∠EDF = 360° - (180° - β - γ) - (60° + β) - (60° + γ) = 60°.
∠BFD = 180° - (60° + β + γ) = 60° + α.
Similarly, ∠CED = 60° + α.
Since D is equidistant from BF and CE,
∠1 = (60° + α) - (β - γ) = 60° - β.
∠2 = 60° - γ.
Through F draw line r making
∠3 = (60° + α) - (60° - β) = α + β
∠mr = (α + β) - β = α.
∠sn = (α + γ) - γ = α.
∠mn = (180° - 3β - 3γ) = 3α.
It remains only to prove that the lines m, n, r, and s converge to a point. The line KF joins the vertices of two isosceles triangles and therefore bisects ∠K. Then in triangle mBKs the bisector of ∠ms passes through F and being parallel to r, coincides with it. Similarly in triangle rHCn the bisector of ∠rn passes through E and being parallel to s, coincides with it.
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