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