# Trisecting the Angle/Neusis Construction

## Theorem

Let $\alpha$ be an angle which is to be trisected.

This can be achieved by means of a neusis construction.

## Construction

Let there be a neusis ruler $S$ marked by two points $p$ and $q$.

Let the vertex of $\alpha$ be labelled $B$.

Let $A$ be the point of one of the branches of $\alpha$ such that the distance $AB$ equals $pq$.

Let the other branch of $\alpha$ be extended past $B$ through $D$.

Let a circle be drawn with center at $B$ with radius $AB$.

Let the straightedge $S$ be placed so that:

- $S$ passes over $A$
- $p$ lies on the circle at $C$
- $q$ lies on the straight line $BD$ at $D$.

Then the angle $3 \angle CBD = \alpha$

Thus $\alpha$ has been trisected.

## Proof

We have that $\angle BCD + \angle ACB$ make a straight angle.

As $CD = AB$ by construction, $CD = BC$ by definition of radius of circle.

Thus $\triangle BCD$ is isosceles.

By Isosceles Triangle has Two Equal Angles:

- $\angle CBD = \angle CDB$

From Sum of Angles of Triangle equals Two Right Angles:

- $\angle BCD + 2 \angle CBD$ equals two right angles.

Thus:

- $2 \angle CBD = \angle ACB$

Similarly, by Isosceles Triangle has Two Equal Angles:

- $\angle ACB = \angle CAB$

and again from Sum of Angles of Triangle equals Two Right Angles:

- $\angle ABC + 2 \angle ACB$ equals two right angles.

and so:

- $\angle ABC + 4 \angle CBD$ equals two right angles.

But $\alpha + \angle ABC + \angle CBD$ make a straight angle.

Thus:

- $\alpha + \angle ABC + \angle CBD = \angle ABC + 4 \angle CBD$

and so:

- $\alpha = 3 \angle CBD$

$\blacksquare$

## Also see

## Sources

- 2008: Ian Stewart:
*Taming the Infinite*... (previous) ... (next): Chapter $2$: The Logic of Shape: Problems for the Greeks