# 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

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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

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- 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).