Talk:Morley's Trisector Theorem

The proof given for Morley's theorem is very elegant. However, it is incomplete. It starts with an equilateral triangle and proves that the intersecting lines trisect the vertices of the triangle, which is the converse Morley's theorem. In the following draft I will complete the proof. I also modify the existing proof to clarify some issues and make it more rigorous. And if there is an agreement on my approach, I will modify the existing proof accordingly. (However, I need help with figures)


 * The given proof is a more-or-less accurate transcript (tidied up and presented neatly) of the one that appeared in the citation presented in the Sources section at the bottom of the page Morley's Trisector Theorem/Dijkstra's Proof.


 * I believe it is an important enough document to retain a record of, hence I would request that you do not replace the existing proof with your own, but instead please add your own proof as a new one on its own page. --prime mover (talk) 10:47, 11 February 2021 (UTC)


 * My issue is the proof given here does not prove Morley's Theorem. It proves the converse of the theorem but not the theorem itself. The fact that a reference is provided does not mean that the reference is correct. My argument is that the proof is incomplete. All what I did is completing the proof. I will not update the existing proof unless I am asked to do so. --Jhoshen1 (talk) 11:27, 11 February 2021‎ (UTC)


 * Whether it's incomplete or not is beside the point. It is of historical interest, so it merits inclusion here as is. Dijkstra is (was) a significant figure in the development of many aspects of technological advance in the 20th century, and hence his proof can be argued as at least having "celebrity" status.


 * Incidentally, when replying to a post in a talk page, it is usual to a) indent appropriately, and b) to sign your posts. Otherwise it can be difficult to work out who says what. Thank you for your cooperation. --prime mover (talk) 12:00, 11 February 2021 (UTC)


 * I understand your point, I will provide a second proof that is based on Dijkstra but is complete as I shown in my Draft --Jhoshen1 (talk) 19:01, 11 February 2021 (UTC)

=
======================DRAFT==================
 * Morleys-Theorem-2.png


 * Morleys-Theorem-Dijkstra-Proof.png

For proving Morley's Theorem, a triangle $\triangle ABC$ is constructed with an embedded $ \triangle XYZ$ equilateral triangle. We shall then prove that $ \triangle X'Y'Z' \cong \triangle XYZ $.

Construct $\triangle XYZ$, an equilateral triangle, with sides equal to $X'Y'$ such that:
 * $XY = YZ = XZ = X'Y'$

Noting that $\alpha + \beta + \gamma = 60 \degrees$, we construct $\triangle AXY$ such that


 * $\therefore \angle XAY = 180 \degrees - (60 \degrees + \beta + 60 \degrees + \gamma) = \alpha$

Construct $\triangle BXZ$ such that


 * $\therefore \angle XBZ = 180 \degrees - (60 \degrees + \alpha + 60 \degrees + \gamma ) = \beta$

Construct $\triangle CYZ$ such that


 * $\therefore \angle YCZ = 180 \degrees -( 60 \degrees + \beta + 60 \degrees + \alpha) = \gamma$

Construct $AB$, $BC$ and $AC$, the sides of $\triangle ABC$

Because

it follows that:
 * if $\angle BAX = \alpha + x$ then $\angle ABX = \beta - x$

Using the Sine Rule, we have:

Substituting $BX$ and $AX$ from $(2)$ and $(3)$ into $(1)$, respectively, and noting that $XZ=XY$, yields

We shall show that for $(4)$ to hold, we must have $x = 0$. In the range in which these angles lie, from $0 \degrees$ to $60 \degrees$, the $sine$ function is a strictly increasing function of its argument.

Under the assumption $x > 0 $ and $\beta - x > 0 $
 * $\map \sin {\alpha + x} > \sin \alpha$
 * $\sin \beta > \map \sin {\beta - x} $
 * $\leadsto \map \sin {\alpha + x} \sin \beta > \sin \alpha \map \sin {\beta - x}$
 * $\leadsto \dfrac {\map \sin {\alpha + x} } {\map \sin {\beta - x} }

> \dfrac {\sin \alpha} {\sin \beta}$ which contradicts $(4)$. If we now assume $x < 0 $, we obtain
 * $\dfrac {\map \sin {\alpha + x} } {\map \sin {\beta - x} }

< \dfrac {\sin \alpha} {\sin \beta}$ Again we have a contradiction with $(4)$. Thus we conclude that $x = 0$.

Consequently, $\angle BAX = \alpha $ and $\angle ABX = \beta $.

Given that $\angle B'A'X' = \alpha $ and $\angle A'B'X' = \beta $, we can establish the following triangle similarity.
 * $\triangle A'B'X' \sim \triangle ABX$


 * $\therefore (5) \;\; \dfrac { AB } { A'B' } = \dfrac { BX } { B'X' } = \dfrac {AX} { A'X' } $

In a similar fashion, we can be shown that $\angle CAY = \alpha $, $\angle ACY = \gamma $, $\angle CBZ = \beta $ and $\angle BCZ = \gamma $, which leads to the following triangle similarities:
 * $\triangle A'C'Y' \sim \triangle ACY$
 * $\therefore (6) \;\; \dfrac { AC } { A'C' } = \dfrac { CY } { C'Y' } = \dfrac {AY} { A'Y' } $


 * $\triangle B'C'Z' \sim \triangle BCZ$
 * $\therefore (7) \;\; \dfrac { BC } { B'C' } = \dfrac { CZ } { C'Z' } = \dfrac {BZ} { B'Z' } $

Because
 * $\angle ABC =\angle ABX + \angle XBZ + \angle ZBC = 3 \beta \;\;\;$ and
 * $\angle BAC =\angle BAX + \angle XAY + \angle CAY = 3 \alpha $

We have the following similarity
 * $ \triangle ABC \sim \triangle A'B'C' $
 * $\therefore (8) \;\; \dfrac { AB } { A'B' } = \dfrac { AC } { A'C' } = \dfrac { BC } { B'C' } $

Combining $(8)$ with $(5)$ and $(6)$ yields
 * $\dfrac { AX } { A'X' } = \dfrac { AY } { A'Y' } $
 * $\leadsto \triangle AXY \sim \triangle A'X'Y' \;\;\;$ Side-Angle-Side
 * $\therefore \dfrac { AY } { A'Y' } = \dfrac { AX } { A'X' } = \dfrac { XY } { X'Y' }  $

By construction $XY=X'Y'$
 * $\therefore \dfrac { AY } { A'Y' } = \dfrac { AX } { A'X' } = \dfrac { XY } { X'Y' } = 1  $

which leads to
 * $AY = A'Y' $
 * $ AX = A'X' $

Consequently $(5)$, $(6)$ and $(7)$ and $(8)$ can be written, respectively, as:
 * $\dfrac { AB } { A'B' } = \dfrac { BX } { B'X' } = \dfrac {AX} {A'X'} = 1 $
 * $\dfrac { AC } { A'C' } = \dfrac { CY } { C'Y' } = \dfrac {AY} { A'Y' } = 1 $
 * $\dfrac { BC } { B'C' } = \dfrac { CZ } { C'Z' } = \dfrac {BZ} { B'Z' } = 1 $
 * $\dfrac { AB } { A'B' } = \dfrac { AC } { A'C' } = \dfrac { BC } { B'C' } = 1 $

The preceding relationships imply that
 * $(9)\;\;\; B'Z' = BZ $
 * $(10)\;\; B'X' = BX $
 * $(11)\;\; C'Z' = CZ $
 * $(12)\;\; C'Y' = CY $

Equations $(9)$, $(10)$ and $\angle XBZ = \angle X'B'Z' = \beta $ establish the following triangle congruence:
 * $ \triangle X'B'Z' \cong \triangle XBZ\;\;\;$ Side-Angle-Side

Similarly $(11)$, $(12)$ and $\angle YCZ = \angle Y'C'Z' = \gamma$ yield:
 * $ \triangle Y'C'Z' \cong \triangle YCZ\;\;\;$ Side-Angle-Side

Thus
 * $Y'Z' = XY $

and
 * $Y'Z' = YZ$

Because by construction $XY=XZ=YZ=X'Y'$, we have
 * $Y'Z' = XY = X'Y' $
 * $Y'Z' = YZ = X'Y' $
 * $\therefore Y'Z' = X'Z' = X'Y' $

which proves that $\triangle X'Y'Z'$ is an equilateral triangle.