Trisecting the Angle/Neusis Construction
This can be achieved by means of a neusis construction.
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 the straightedge $S$ be placed so that:
Then the angle $3 \angle CBD = \alpha$
Thus $\alpha$ has been trisected.
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.
- $\angle CBD = \angle CDB$
- $\angle BCD + 2 \angle CBD$ equals two right angles.
- $2 \angle CBD = \angle ACB$
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.
- $\angle ABC + 4 \angle CBD$ equals two right angles.
But $\alpha + \angle ABC + \angle CBD$ make a straight angle.
- $\alpha + \angle ABC + \angle CBD = \angle ABC + 4 \angle CBD$
- $\alpha = 3 \angle CBD$