Editing Morley's trisector theorem (section)

==Proofs==
There are many [[Mathematical proof|proofs]] of Morley's theorem, some of which are very technical.<ref>{{citation|url=http://www.cut-the-knot.org/triangle/Morley/index.shtml|title=Morley's Miracle|publisher=[[Cut-the-knot]]|last=Bogomolny|first=Alexander|authorlink= Alexander Bogomolny |accessdate=2010-01-02}}</ref>  
Several early proofs were  based on delicate [[trigonometry|trigonometric]] calculations. Recent proofs include an [[algebra]]ic proof by {{harvs|first=Alain|last=Connes|authorlink=Alain Connes|txt|year=1998|year2=2004}} extending the theorem to general [[Field (mathematics)|fields]] other than characteristic three, and [[John Horton Conway|John Conway]]'s elementary geometry proof.<ref>{{Citation |last=Bogomolny |first=Alexander |title=J. Conway's proof |url=http://www.cut-the-knot.org/triangle/Morley/conway.shtml |publisher=[[Cut-the-knot]] |access-date=2021-12-03 |authorlink=Alexander Bogomolny}}</ref><ref>{{citation|chapter-url=http://thewe.net/math/conway.pdf|title=Power|editor1-last=Blackwell|editor1-first=Alan|editor2-last=Mackay|editor2-first=David|editor2-link=David J. C. MacKay|year=2006|chapter=The Power of Mathematics|last=Conway|first=John|author-link=John Horton Conway|publisher=Cambridge University Press|accessdate=2010-10-08|pages=36–50|isbn=978-0-521-82377-7}}</ref>  The latter starts with an equilateral triangle and shows that a triangle may be built around it which will be [[Similarity (geometry)|similar]] to any selected triangle.  Morley's theorem does not hold in [[spherical geometry|spherical]]<ref>[http://lienhard-wimmer.com/applets/dreieck/Morley.html Morley's Theorem in Spherical Geometry], [[Java applet]].</ref> and [[hyperbolic geometry]].

[[File:Morley Proof.svg|thumb|right|480px|Fig 1. &nbsp; Elementary proof of Morley's trisector theorem]]
One proof uses the trigonometric identity

{{NumBlk|::|<math>\sin(3\theta)=4\sin\theta\sin(60^\circ+\theta)\sin(120^\circ+\theta)</math>|{{EquationRef|1}}}}

which, by using of the sum of two angles identity, can be shown to be equal to

::<math>\sin(3\theta)=-4\sin^3\theta+3\sin\theta.</math>

The last equation can be verified by applying the sum of two angles identity to the left side twice and eliminating the cosine.

Points <math>D, E, F</math> are constructed on <math>\overline{BC}</math> as shown. We have <math>3\alpha+3\beta+3\gamma=180^\circ</math>, the sum of any triangle's angles, so <math>\alpha+\beta+\gamma=60^\circ.</math> Therefore, the angles of triangle <math>XEF</math> are <math>\alpha, (60^\circ+\beta),</math> and <math>(60^\circ+\gamma).</math>

From the figure

{{NumBlk|::|<math>\sin(60^\circ+\beta)=\frac{\overline{DX}}{\overline{XE}}</math>|{{EquationRef|2}}}}

and

{{NumBlk|::|<math>\sin(60^\circ+\gamma)=\frac{\overline{DX}}{\overline{XF}}.</math>|{{EquationRef|3}}}}

Also from the figure

::<math>\angle{AYC}=180^\circ-\alpha-\gamma=120^\circ+\beta</math>

and

{{NumBlk|::|<math>\angle{AZB}=120^\circ+\gamma.</math>|{{EquationRef|4}}}}

The law of sines applied to triangles <math>AYC</math> and <math>AZB</math> yields

{{NumBlk|::|<math>\sin(120^\circ+\beta)=\frac{\overline{AC}}{\overline{AY}}\sin\gamma</math>|{{EquationRef|5}}}}

and

{{NumBlk|::|<math>\sin(120^\circ+\gamma)=\frac{\overline{AB}}{\overline{AZ}}\sin\beta.</math>|{{EquationRef|6}}}}

Express the height of triangle <math>ABC</math> in two ways

::<math>h=\overline{AB} \sin(3\beta)=\overline{AB}\cdot 4\sin\beta\sin(60^\circ+\beta)\sin(120^\circ+\beta)</math>

and

::<math>h=\overline{AC} \sin(3\gamma)=\overline{AC}\cdot 4\sin\gamma\sin(60^\circ+\gamma)\sin(120^\circ+\gamma).</math>

where equation (1) was used to replace <math>\sin(3\beta)</math> and <math>\sin(3\gamma)</math> in these two equations. Substituting equations (2) and (5) in the <math>\beta</math> equation and equations (3) and (6) in the <math>\gamma</math> equation gives

::<math>h=4\overline{AB}\sin\beta\cdot\frac{\overline{DX}}{\overline{XE}}\cdot\frac{\overline{AC}}{\overline{AY}}\sin\gamma</math>

and

::<math>h=4\overline{AC}\sin\gamma\cdot\frac{\overline{DX}}{\overline{XF}}\cdot\frac{\overline{AB}}{\overline{AZ}}\sin\beta</math>

Since the numerators are equal

::<math>\overline{XE}\cdot\overline{AY}=\overline{XF}\cdot\overline{AZ}</math>

or

::<math>\frac{\overline{XE}}{\overline{XF}}=\frac{\overline{AZ}}{\overline{AY}}.</math>

Since angle <math>EXF</math> and angle <math>ZAY</math> are equal and the sides forming these angles are in the same ratio, triangles <math>XEF</math> and <math>AZY</math> are similar. 

Similar angles <math>AYZ</math> and <math>XFE</math> equal <math>(60^\circ+\gamma)</math>, and similar angles <math>AZY</math> and <math>XEF</math> equal <math>(60^\circ+\beta).</math> Similar arguments yield the base angles of triangles <math>BXZ</math> and <math>CYX.</math>

In particular angle <math>BZX</math> is found to be <math>(60^\circ+\alpha)</math> and from the figure we see that

::<math>\angle{AZY}+\angle{AZB}+\angle{BZX}+\angle{XZY}=360^\circ.</math>

Substituting yields

::<math>(60^\circ+\beta)+(120^\circ+\gamma)+(60^\circ+\alpha)+\angle{XZY}=360^\circ</math>

where equation (4) was used for angle <math>AZB</math> and therefore

::<math>\angle{XZY}=60^\circ.</math>

Similarly the other angles of triangle <math>XYZ</math> are found to be <math>60^\circ.</math>