Morley's trisector theorem

There are many proofs of Morley's theorem, some of which are very technical. Several early proofs were based on delicate trigonometric calculations. Recent proofs include an algebraic proof by extending the theorem to general fields other than characteristic three, and John Conway's elementary geometry proof. The latter starts with an equilateral triangle and shows that a triangle may be built around it which will be similar to any selected triangle. Morley's theorem does not hold in spherical and hyperbolic geometry. One proof uses the trigonometric identity which, by using of the sum of two angles identity, can be shown to be equal to ::\sin(3\theta)=-4\sin^3\theta+3\sin\theta. The last equation can be verified by applying the sum of two angles identity to the left side twice and eliminating the cosine. Points D, E, F are constructed on \overline{BC} as shown. We have 3\alpha+3\beta+3\gamma=180^\circ, the sum of any triangle's angles, so \alpha+\beta+\gamma=60^\circ. Therefore, the angles of triangle XEF are \alpha, (60^\circ+\beta), and (60^\circ+\gamma). From the figure {{NumBlk|::|\sin(60^\circ+\beta)=\frac{\overline{DX}}{\overline{XE}}|}} and {{NumBlk|::|\sin(60^\circ+\gamma)=\frac{\overline{DX}}{\overline{XF}}.|}} Also from the figure ::\angle{AYC}=180^\circ-\alpha-\gamma=120^\circ+\beta and {{NumBlk|::|\angle{AZB}=120^\circ+\gamma.|}} The law of sines applied to triangles AYC and AZB yields {{NumBlk|::|\sin(120^\circ+\beta)=\frac{\overline{AC}}{\overline{AY}}\sin\gamma|}} and {{NumBlk|::|\sin(120^\circ+\gamma)=\frac{\overline{AB}}{\overline{AZ}}\sin\beta.|}} Express the height of triangle ABC in two ways ::h=\overline{AB} \sin(3\beta)=\overline{AB}\cdot 4\sin\beta\sin(60^\circ+\beta)\sin(120^\circ+\beta) and ::h=\overline{AC} \sin(3\gamma)=\overline{AC}\cdot 4\sin\gamma\sin(60^\circ+\gamma)\sin(120^\circ+\gamma). where equation (1) was used to replace \sin(3\beta) and \sin(3\gamma) in these two equations. Substituting equations (2) and (5) in the \beta equation and equations (3) and (6) in the \gamma equation gives ::h=4\overline{AB}\sin\beta\cdot\frac{\overline{DX}}{\overline{XE}}\cdot\frac{\overline{AC}}{\overline{AY}}\sin\gamma and ::h=4\overline{AC}\sin\gamma\cdot\frac{\overline{DX}}{\overline{XF}}\cdot\frac{\overline{AB}}{\overline{AZ}}\sin\beta Since the numerators are equal ::\overline{XE}\cdot\overline{AY}=\overline{XF}\cdot\overline{AZ} or ::\frac{\overline{XE}}{\overline{XF}}=\frac{\overline{AZ}}{\overline{AY}}. Since angle EXF and angle ZAY are equal and the sides forming these angles are in the same ratio, triangles XEF and AZY are similar. Similar angles AYZ and XFE equal (60^\circ+\gamma), and similar angles AZY and XEF equal (60^\circ+\beta). Similar arguments yield the base angles of triangles BXZ and CYX. In particular angle BZX is found to be (60^\circ+\alpha) and from the figure we see that ::\angle{AZY}+\angle{AZB}+\angle{BZX}+\angle{XZY}=360^\circ. Substituting yields ::(60^\circ+\beta)+(120^\circ+\gamma)+(60^\circ+\alpha)+\angle{XZY}=360^\circ where equation (4) was used for angle AZB and therefore ::\angle{XZY}=60^\circ. Similarly the other angles of triangle XYZ are found to be 60^\circ. ==Side and area==

The first Morley triangle has side lengths a^\prime=b^\prime=c^\prime=8R\,\sin\tfrac13A\,\sin\tfrac13B\,\sin\tfrac13C, where R is the circumradius of the original triangle and A, B, and C are the angles of the original triangle. Since the area of an equilateral triangle is \tfrac{\sqrt{3}}{4}a'^2, the area of Morley's triangle can be expressed as \text{Area} = 16 \sqrt{3}R^2\, \sin^2\!\tfrac13A\, \sin^2\!\tfrac13B\, \sin^2\!\tfrac13C. ==Morley's triangles==

Morley's theorem entails 18 equilateral triangles. The triangle described in the trisector theorem above, called the first Morley triangle, has vertices given in trilinear coordinates relative to a triangle ABC as follows: \begin{array}{ccccccc} A \text{-vertex} &=& 1 &:& 2 \cos\tfrac13 C &:& 2 \cos\tfrac13 B \\[5mu] B \text{-vertex} &=& 2 \cos\tfrac13 C &:& 1 &:& 2 \cos\tfrac13 A \\[5mu] C \text{-vertex} &=& 2 \cos\tfrac13 B &:& 2 \cos\tfrac13 A &:& 1 \end{array} Another of Morley's equilateral triangles that is also a central triangle is called the second Morley triangle and is given by these vertices: \begin{array}{ccccccc} A \text{-vertex} &=& 1 &:& 2 \cos\tfrac13(C - 2\pi) &:& 2 \cos\tfrac13(B - 2\pi) \\[5mu] B \text{-vertex} &=& 2 \cos\tfrac13(C - 2\pi) &:& 1 &:& 2 \cos\tfrac13(A - 2\pi) \\[5mu] C \text{-vertex} &=& 2 \cos\tfrac13(B - 2\pi) &:& 2 \cos\tfrac13(A - 2\pi) &:& 1 \end{array} The third of Morley's 18 equilateral triangles that is also a central triangle is called the third Morley triangle and is given by these vertices: \begin{array}{ccccccc} A \text{-vertex} &=& 1 &:& 2 \cos\tfrac13(C + 2\pi) &:& 2 \cos\tfrac13(B + 2\pi) \\[5mu] B \text{-vertex} &=& 2 \cos\tfrac13(C + 2\pi) &:& 1 &:& 2 \cos\tfrac13(A + 2\pi) \\[5mu] C \text{-vertex} &=& 2 \cos\tfrac13(B + 2\pi) &:& 2 \cos\tfrac13(A + 2\pi) &:& 1 \end{array} The first, second, and third Morley triangles are pairwise homothetic. Another homothetic triangle is formed by the three points X on the circumcircle of triangle ABC at which the line XX −1 is tangent to the circumcircle, where X −1 denotes the isogonal conjugate of X. This equilateral triangle, called the circumtangential triangle, has these vertices: \begin{array}{lllllll} A \text{-vertex} &=& \phantom{-}\csc\tfrac13(C - B) &:& \phantom{-}\csc\tfrac13(2C + B) &:& -\csc\tfrac13(C + 2B) \\[5mu] B \text{-vertex} &=& -\csc\tfrac13(A + 2C) &:& \phantom{-}\csc\tfrac13(A - C) &:& \phantom{-}\csc\tfrac13(2A + C) \\[5mu] C \text{-vertex} &=& \phantom{-}\csc\tfrac13(2B + A) &:& -\csc\tfrac13(B + 2A) &:& \phantom{-}\csc\tfrac13(B - A) \end{array} A fifth equilateral triangle, also homothetic to the others, is obtained by rotating the circumtangential triangle /6 about its center. Called the circumnormal triangle, its vertices are as follows: \begin{array}{lllllll} A \text{-vertex} &=& \phantom{-}\sec\tfrac13(C - B) &:& -\sec\tfrac13(2C + B) &:& -\sec\tfrac13(C + 2B) \\[5mu] B \text{-vertex} &=& -\sec\tfrac13(A + 2C) &:& \phantom{-}\sec\tfrac13(A - C) &:& -\sec\tfrac13(2A + C) \\[5mu] C \text{-vertex} &=& -\sec\tfrac13(2B + A) &:& -\sec\tfrac13(B + 2A) &:& \phantom{-}\sec\tfrac13(B - A) \end{array} An operation called "extraversion" can be used to obtain one of the 18 Morley triangles from another. Each triangle can be extraverted in three different ways; the 18 Morley triangles and 27 extravert pairs of triangles form the 18 vertices and 27 edges of the Pappus graph. ==Related triangle centers==

The Morley center, X(356), centroid of the first Morley triangle, is given in trilinear coordinates by \cos\tfrac13A + 2\cos\tfrac13B\,\cos\tfrac13C \,:\, \cos\tfrac13B + 2\cos\tfrac13C\,\cos\tfrac13A \,:\, \cos\tfrac13C + 2\cos\tfrac13A\,\cos\tfrac13B 1st Morley–Taylor–Marr center, X(357): The first Morley triangle is perspective to triangle \sec\tfrac13A \,:\, \sec\tfrac13B \,:\, \sec\tfrac13C ==See also==