# Morley's Trisector Theorem/Proof 2

## 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 $F$.

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 $\triangle A'B'C'$

The Constructed Triangle $\triangle ABC$

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 $\triangle A'X'B' \sim \triangle AXB$

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).