Morley's Trisector Theorem/Proof 2

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Theorem

Let $\triangle ABC$ be a triangle.

Let the internal angles of $\triangle ABC$ be trisected.

Let the points where these angle trisectors first intersect be $D$, $E$ and $F$.

Morleys-Theorem.png


Then $\triangle EDF$ is equilateral.


Proof


By comparing the given triangle $\triangle A'B'C' $ with the constructed triangle $\triangle ABC $, we shall prove that $ \triangle X'Y'Z' \sim \triangle XYZ $ where $\triangle XYZ $ is an equilateral triangle.


The Given Triangle $\triangle A'B'C'$

Morleys-Theorem-Fig1xxxx.png

The Constructed Triangle $\triangle ABC$

Morleys-Theorem-Fig2xx.png

We begin by constructing $\triangle XYZ$, an equilateral triangle such that:

$XY = YZ = XZ$

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

\(\ds \angle AXY\) \(=\) \(\ds 60 \degrees + \beta\)
\(\ds \angle XZB\) \(=\) \(\ds 60 \degrees + \gamma\)
\(\ds \therefore \angle XAY\) \(=\) \(\ds 180 \degrees - (60 \degrees + \beta + 60 \degrees + \gamma)\)
\(\ds \) \(=\) \(\ds 60 \degrees - \beta - 60 \degrees\)
\(\ds \) \(=\) \(\ds \alpha\)

Construct $\triangle BXZ$ such that:

\(\ds \angle ZXB\) \(=\) \(\ds 60 \degrees + \alpha\)
\(\ds \angle XZB\) \(=\) \(\ds 60 \degrees + \gamma\)
\(\ds \therefore \angle XBZ\) \(=\) \(\ds \beta\)


Construct $\triangle CYZ$ such that:

\(\ds \angle BXZ\) \(=\) \(\ds 60 \degrees + \beta\)
\(\ds \angle AYX\) \(=\) \(\ds 60 \degrees + \alpha\)
\(\ds \therefore \angle YCZ\) \(=\) \(\ds \gamma\)


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


$\angle AXB$ is calculated as follows:

\(\ds \angle AXB\) \(=\) \(\ds 360 \degrees - ( \angle AXY +\angle YXZ + \angle BXZ )\)
\(\ds \) \(=\) \(\ds 360 \degrees - ((60 \degrees + \beta) + 60 \degrees + (60 \degrees + \alpha))\)
\(\ds \) \(=\) \(\ds 360 \degrees - (180 \degrees + \beta + \alpha)\)
\(\ds \) \(=\) \(\ds 180 \degrees - \beta - \alpha\)

To proceed, it is necessary to prove that $ \triangle A'X'B' \sim \triangle AXB$.

We shall provide two alternate proofs for this preposition; a trigonometric proof and a geometric proof.


Trigonometric proof for $ \triangle A'X'B' \sim \triangle AXB$


Applying the Sine Rule for $\triangle XBZ$ and $\triangle XAY$, we have:

\(\text {(1)}: \quad\) \(\ds BX\) \(=\) \(\ds XZ \map \sin {\angle XZB} / \sin \beta = XZ \map \sin {60 \degrees + \gamma} / \sin \beta\)
\(\text {(2)}: \quad\) \(\ds AX\) \(=\) \(\ds XY \map \sin {\angle AYX } / \sin \alpha = XY \map \sin {60 \degrees + \gamma } / \sin \alpha\)

Dividing $(1)$ by $(2)$ and noting that by construction $XZ = XY$, we obtain:

\(\text {(3)}: \quad\) \(\ds \dfrac {BX } { AX}\) \(=\) \(\ds \dfrac {\sin \alpha} {\sin \beta}\)

Applying the Sine Rule to $\triangle A'X'B' $, we get:

\(\text {(4)}: \quad\) \(\ds \dfrac {B'X' } { A'X'}\) \(=\) \(\ds \dfrac {\sin \alpha} {\sin \beta}\)

Combining $(3)$ and $(4)$, yields:

\(\ds \dfrac {BX } { AX}\) \(=\) \(\ds \dfrac {B'X'} {A'X'}\)


For $\triangle A'X'B' $, we have:

\(\ds \angle A'X'B'\) \(=\) \(\ds 180 \degrees - \beta - \alpha\)

and we have already shown that:

\(\ds \angle AXB\) \(=\) \(\ds 180 \degrees - \beta - \alpha\)
\(\ds \leadsto \angle AXB\) \(=\) \(\ds \angle A'X'B'\)
\(\ds \therefore \triangle A'X'B'\) \(\sim\) \(\ds \triangle AXB\) Triangles with One Equal Angle and Two Sides Proportional are Similar


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


In a similar fashion, we can obtain the following triangle similarities:

\(\ds \triangle A'Y'C'\) \(\sim\) \(\ds \triangle CYC\)
\(\ds \triangle B'Z'C'\) \(\sim\) \(\ds \triangle BZC\)


--- End of the Trigonometric Proof for $ \triangle A'X'B' \sim \triangle AXB$ ---


Geometric proof for $ \triangle A'X'B' \sim \triangle AXB$

Morleys-Theorem-Auxiliary-Triangles.png

We consider 3 different cases for the geometric proof

[1] $\gamma < 30 \degrees $
[2] $\gamma > 30 \degrees $
[3] $\gamma = 30 \degrees $


The outline for proving case [1] is given as follows:

Construct $\triangle A''Y''X''$.
Construct $\triangle B''X''Z'' $.
Prove that $\triangle A''X''B'' \cong \triangle AXB$.
Prove that $\triangle A'X'B' \sim \triangle AXB$.


We construct triangle $\triangle A''Y''X''$ such that $\triangle A''Y''X'' \cong \triangle AYX $.

\(\ds \therefore \angle X''Y''A''\) \(=\) \(\ds \angle XYA\)
\(\ds \) \(=\) \(\ds 60 \degrees + \gamma\) by the construction of $\triangle XYA$


Next, we construct triangle $\triangle B''X''Z'' $ as follows:

\(\ds \angle Z''X''Y''\) \(=\) \(\ds 60 \degrees - 2 \gamma\) construct $\angle Z''X''Y''$
\(\ds \leadsto \angle X''Z''B'''\) \(=\) \(\ds 180 \degrees - \angle X''Y''A'' -\angle Z''X''Y'''\)
\(\ds \) \(=\) \(\ds 180 \degrees - (60 \degrees + \gamma) - (60 \degrees - 2 \gamma)\)
\(\ds \) \(=\) \(\ds 60 \degrees + \gamma\)
\(\ds \) \(=\) \(\ds \angle XZB\) by the construction of $\triangle XZB$
\(\ds \therefore X''Z''\) \(=\) \(\ds X''Y''\) $\triangle Y''X''Z''$ is an isosceles by Triangle with Two Equal Angles is Isosceles
\(\ds \) \(=\) \(\ds XY\) by $\triangle A''Y''X'' \cong \triangle AYX $
\(\ds \) \(=\) \(\ds XZ\) by equilateral $\triangle XYZ$ construction
\(\ds \angle B''X''Z''\) \(=\) \(\ds \angle BXZ\) construct $\angle B''X''Z''$
\(\ds \) \(=\) \(\ds 60 \degrees +\alpha\) by the construction of $\triangle XZB$
\(\ds \therefore \triangle B''X''Z''\) \(\cong\) \(\ds \triangle BXZ\) Angle-Side-Angle congruency


We shall now prove that that $\triangle A''X''B'' \cong \triangle AXB$.

We note that:

\(\ds \angle X''A''Y''\) \(=\) \(\ds \angle XAY\) by $\triangle A''Y''X'' \cong \triangle AYX $
\(\ds \) \(=\) \(\ds \alpha\)
\(\ds \angle X''B''Z''\) \(=\) \(\ds \angle XBZ\) by $\triangle B''X''Z'' \cong \triangle BXZ$
\(\ds \) \(=\) \(\ds \beta\)
\(\ds \angle AXB\) \(=\) \(\ds 180 \degrees - \angle XAY -\angle XBZ\)
\(\ds \) \(=\) \(\ds 180 \degrees - \alpha - \beta\)
\(\ds \angle A''X''B''\) \(=\) \(\ds 180 \degrees - \angle X''A''Y'' - \angle X''B''Z''\)
\(\ds \) \(=\) \(\ds 180 \degrees - \alpha - \beta\)
\(\ds \) \(=\) \(\ds \angle AXB\)
\(\ds B''X''\) \(=\) \(\ds BX\) by $\triangle B''X''Z'' \cong \triangle BXZ$
\(\ds A''X''\) \(=\) \(\ds AX\) by $\triangle A''Y''X'' \cong \triangle AYX $
\(\ds \therefore \triangle AXB\) \(\cong\) \(\ds \triangle A''X''B''\) Side-Angle-Side congruency


Consequently,

\(\ds \angle XBA\) \(=\) \(\ds \angle X''B''A''\)
\(\ds \) \(=\) \(\ds \beta\)
\(\ds \angle XAB\) \(=\) \(\ds \angle X''A''B''\)
\(\ds \) \(=\) \(\ds \alpha\)

and given that:

\(\ds \angle X'B'A'\) \(=\) \(\ds \beta\)
\(\ds \angle X'A'B'\) \(=\) \(\ds \alpha\)

yields the desired result:

\(\ds \triangle A'X'B'\) \(\sim\) \(\ds \triangle AXB\) Triangles with Two Equal Angles are Similar


For case [2], where $\gamma > 30 \degrees $, $\angle Z''X''Y'' = 2 \gamma - 60 \degrees $ and $\triangle X''Z''Y''$ is external to $\triangle A''Y''X''$ and $\triangle B''X''Z'' $.

In case [3], where $\gamma = 30 \degrees $, $\angle Z''X''Y'' = 0 \degrees $ and $\triangle X''Z''Y''$ degenerates into line segment $X''Z''$.

In either case, [2] or [3], the proof for $\triangle X'A'B' \sim \triangle XAB$ is very similar to the proof for case [1].

In a similar fashion, we can prove that:

$\triangle B'Z'C' \sim \triangle BZC$ and $\triangle A'Y'C' \sim \triangle AYC$

and that:

$\angle CAY = \alpha $, $\angle ACY = \gamma $, $\angle CBZ = \beta $ and $\angle BCZ = \gamma $.


--- End of the Geometric Proof for $\triangle A'X'B' \sim \triangle AXB$ ---


Because

\(\ds \angle ABC\) \(=\) \(\ds \angle ABX + \angle XBZ + \angle ZBC\)
\(\ds \) \(=\) \(\ds 3 \beta\)
and
\(\ds \angle BAC\) \(=\) \(\ds \angle BAX + \angle XAY + \angle CAY\)
\(\ds \) \(=\) \(\ds 3 \alpha\)

we have the following similarity:

\(\ds \triangle ABC\) \(\sim\) \(\ds \triangle A'B'C'\) Triangles with Two Equal Angles are Similar


Using $ \triangle ABC \sim \triangle A'B'C' $, $\triangle A'B'X' \sim \triangle ABX$ and $\triangle A'C'Y' \sim \triangle ACY$ triangle similarities, we observe that:

\(\ds \dfrac {AX} { A'X' }\) \(=\) \(\ds \dfrac {AB} { A'B' }\) by $\triangle A'B'X' \sim \triangle ABX$
\(\ds \dfrac {AY} { A'Y' }\) \(=\) \(\ds \dfrac {AC} { A'C' }\) by $\triangle A'C'Y' \sim \triangle ACY$
\(\ds \) \(=\) \(\ds \dfrac {AB} { A'B' }\) by $ \triangle ABC \sim \triangle A'B'C' $
\(\ds \leadsto \dfrac {AX} { A'X' }\) \(=\) \(\ds \dfrac {AY} { A'Y' }\) the 2 ratios are equal to $\dfrac {AB} { A'B'}$


Furthermore,

\(\ds \angle XAY\) \(=\) \(\ds \angle X'A'Y'\)
\(\ds \) \(=\) \(\ds \alpha\)
\(\ds \therefore \triangle XAY\) \(\sim\) \(\ds \triangle X'A'Y'\) Triangles with One Equal Angle and Two Sides Proportional are Similar
\(\ds \leadsto \dfrac { XY } { X'Y' }\) \(=\) \(\ds \dfrac {AX} { A'X' }\)
\(\ds \) \(=\) \(\ds \dfrac {AB} { A'B' }\) by $\triangle A'B'X' \sim \triangle ABX$


In a similar fashion, we can also prove the following triangle similarities:

\(\ds \triangle XBZ\) \(\sim\) \(\ds \triangle X'B'Z'\)
\(\ds \triangle YCZ\) \(\sim\) \(\ds \triangle Y'C'Z'\)

which yield the following:

\(\ds \dfrac {XZ} { X'Z' }\) \(=\) \(\ds \dfrac { BX } { B'X' }\) by $\triangle XBZ \sim \triangle X'B'Z'$
\(\ds \) \(=\) \(\ds \dfrac {AB} { A'B' }\) by $\triangle A'B'X' \sim \triangle ABX$
\(\ds \dfrac {YZ} { Y'Z' }\) \(=\) \(\ds \dfrac { CY } { C'Y' }\) by $\triangle YCZ \sim \triangle Y'C'Z'$
\(\ds \) \(=\) \(\ds \dfrac {AC} { A'C' }\) by $\triangle A'C'Y' \sim \triangle ACY$
\(\ds \) \(=\) \(\ds \dfrac {AB} { A'B'}\) by $\triangle A'B'C' \sim \triangle ABC$
\(\ds \therefore \dfrac {XY} { X'Y' }\) \(=\) \(\ds \dfrac {XZ} { X'Z' } = \dfrac {YZ} { Y'Z' }\) the 3 ratios are all equal to $\dfrac {AB} { A'B'}$

By construction:

\(\ds XY\) \(=\) \(\ds XZ = YZ\)
\(\ds \therefore \dfrac {XY} { X'Y' }\) \(=\) \(\ds \dfrac {XY} { X'Z' } = \dfrac {XY} { Y'Z' }\)
\(\ds \leadsto X'Y'\) \(=\) \(\ds X'Z' = Y'Z'\)

Hence, $\triangle X'Y'Z'$ is an equilateral triangle, which proves the theorem.

$\blacksquare$


Sources


  • The proof is provided by Joseph Hoshen in the spirit of and in the memory of his exceptional high school geometry teacher, Yehuda Klein (1914-1989).