Morley's Trisector Theorem

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.

Dijkstra's Proof

Choose angles $\alpha$, $\beta$ and $\gamma$ all greater than $0$ such that $\alpha + \beta + \gamma = 60 \degrees$.

Draw an equilateral triangle $\triangle XYZ$.

Construct $\triangle AXY$ and $\triangle BXZ$ with the angles as indicated:

We have that:

Because $\angle AXB = 180 \degrees - \paren {\alpha + \beta}$ , it follows that:

if $\angle BAX = \alpha + x$ then $\angle ABX = \beta - x$

Using the Sine Rule $3$ times, in $\angle AXB$, $\angle AXY$ and $\angle BXZ$, we have:

In the range in which these angles lie, the left hand side of the above is a strictly increasing function of $x$.

Thus we conclude that $x = 0$.

The result follows.

$\blacksquare$

Proof 2

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:

Construct $\triangle BXZ$ such that:

Construct $\triangle CYZ$ such that:

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

$\angle AXB$ is calculated as follows:

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:

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

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

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

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

and we have already shown that:

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

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

--- 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 $.

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

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

We note that:

Consequently,

and given that:

yields the desired result:

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

and

we have the following similarity:

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:

Furthermore,

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

which yield the following:

By construction:

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

$\blacksquare$

Also known as

is also known just as Morley's Theorem.

Source of Name

This entry was named for Frank Morley.

Historical Note

was proved by Frank Morley in $1899$.

Sources

• 1991: David Wells: Curious and Interesting Geometry ... (next): Morley's triangle
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Morley's theorem