### Theorem

The three points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle.

### Proof

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 ∠CDE = 60° + β and F by making ∠BDF = 60° + γ. Then

∠EDF = 360° - (180° - β - γ) - (60° + β) - (60° + γ) = 60°.

Also

∠BFD = 180° - (60° + β + γ) = 60° + α.

Similarly, ∠CED = 60° + α. Since D is equidistant from BF and CE, DF = DE and ΔDEF is equilateral.

∠1 = (60° + α) - (β - γ) = 60° - β.

Similarly,

∠2 = 60° - γ.

Through F draw line r making ∠1' = ∠1. Through E draw a line s making angle ∠2' = ∠2.

∠3 = (60° + α) - (60° - β) = α + β

and

∠mr = (α + β) - β = α.

Similarly,

∠sn = (α + γ) - γ = α.

Further,

∠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. ### Morley's Miracle

#### Invalid proofs 