# Morley's trisector theorem Morley's trisector theorem If each vertex angle of the outer triangle is trisected, Morley's trisector theorem states that the purple triangle will be equilateral.

In plane geometry, Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, called the first Morley triangle or simply the Morley triangle. The theorem was discovered in 1899 by Anglo-American mathematician Frank Morley. It has various generalizations; em particular, if all of the trisectors are intersected, one obtains four other equilateral triangles.

Conteúdo 1 Provas 2 Side and area 3 Morley's triangles 4 Related triangle centers 5 Veja também 6 Notas 7 Referências 8 External links Proofs 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 Alain Connes (1998, 2004) 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.

Fig 1. Elementary proof of Morley's trisector theorem One proof uses the trigonometric identity {pecado de estilo de exibição(3teta )=4sin theta sin(60^{círculo }+teta )pecado(120^{círculo }+teta )} (1) que, by using of the sum of two angles identity, can be shown to be equal to {pecado de estilo de exibição(3teta )=-4sin ^{3}theta +3sin 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 {estilo de exibição D,E,F} are constructed on {estilo de exibição {overline {BC}}} as shown. Nós temos {displaystyle 3alpha +3beta +3gamma =180^{círculo }} , the sum of any triangle's angles, assim {displaystyle alpha +beta +gamma =60^{círculo }.} Portanto, the angles of triangle {displaystyle XEF} são {alfa de estilo de exibição ,(60^{círculo }+beta ),} e {estilo de exibição (60^{círculo }+gama ).} From the figure {pecado de estilo de exibição(60^{círculo }+beta )={fratura {overline {DX}}{overline {XE}}}} (2) e {pecado de estilo de exibição(60^{círculo }+gama )={fratura {overline {DX}}{overline {XF}}}.} (3) Also from the figure {displaystyle angle {AYC}=180^{círculo }-alpha -gamma =120^{círculo }+beta } e {displaystyle angle {AZB}=120^{círculo }+gama .} (4) The law of sines applied to triangles {displaystyle AYC} e {displaystyle AZB} yields {pecado de estilo de exibição(120^{círculo }+beta )={fratura {overline {CA}}{overline {AY}}}sin gamma } (5) e {pecado de estilo de exibição(120^{círculo }+gama )={fratura {overline {AB}}{overline {AZ}}}sin beta .} (6) Express the height of triangle {estilo de exibição ABC} in two ways {displaystyle h={overline {AB}}pecado(3beta )={overline {AB}}cdot 4sin beta sin(60^{círculo }+beta )pecado(120^{círculo }+beta )} e {displaystyle h={overline {CA}}pecado(3gama )={overline {CA}}cdot 4sin gamma sin(60^{círculo }+gama )pecado(120^{círculo }+gama ).} where equation (1) was used to replace {pecado de estilo de exibição(3beta )} e {pecado de estilo de exibição(3gama )} in these two equations. Substituting equations (2) e (5) no {beta de estilo de exibição } equation and equations (3) e (6) no {gama de estilo de exibição } equation gives {displaystyle h=4{overline {AB}}sin beta cdot {fratura {overline {DX}}{overline {XE}}}cdot {fratura {overline {CA}}{overline {AY}}}sin gamma } e {displaystyle h=4{overline {CA}}sin gamma cdot {fratura {overline {DX}}{overline {XF}}}cdot {fratura {overline {AB}}{overline {AZ}}}sin beta } Since the numerators are equal {estilo de exibição {overline {XE}}cdot {overline {AY}}={overline {XF}}cdot {overline {AZ}}} ou {estilo de exibição {fratura {overline {XE}}{overline {XF}}}={fratura {overline {AZ}}{overline {AY}}}.} Since angle {displaystyle EXF} and angle {displaystyle ZAY} are equal and the sides forming these angles are in the same ratio, triangles {displaystyle XEF} e {displaystyle AZY} are similar.