Editing Angle trisection (section)

===With a marked ruler===
[[File:Trisecting angles three.svg|thumb|355px|Trisection of the angle using a marked ruler]]Another means to trisect an arbitrary angle by a "small" step outside the Greek framework is via a ruler with two marks a set distance apart. The next construction is originally due to [[Archimedes]], called a ''[[Neusis construction]]'', i.e., that uses tools other than an ''un-marked'' straightedge.  The diagrams we use show this construction for an acute angle, but it indeed works for any angle up to 180 degrees.

This requires three facts from geometry (at right):
# Any full set of angles on a straight line add to 180°,
# The sum of angles of any triangle is 180°, ''and'',
# Any two equal sides of an [[isosceles triangle]] will [[Pons asinorum|meet the third side at the same angle]].
{{Clear}}

Let {{mvar|l}} be the horizontal line in the adjacent diagram.  Angle {{mvar|a}} (left of point {{mvar|B}}) is the subject of trisection. First, a point {{mvar|A}} is drawn at an angle's [[ray (geometry)|ray]], one unit apart from {{mvar|B}}. A circle of [[radius]] {{mvar|AB}} is drawn.  Then, the markedness of the ruler comes into play: one mark of the ruler is placed at {{mvar|A}} and the other at {{mvar|B}}.  While keeping the ruler (but not the mark) touching {{mvar|A}}, the ruler is slid and rotated until one mark is on the circle and the other is on the line {{mvar|l}}.   The mark on the circle is labeled {{mvar|C}} and the mark on the line is labeled {{mvar|D}}.  This ensures that {{math|''CD'' {{=}} ''AB''}}.  A radius {{mvar|BC}} is drawn to make it obvious that line segments {{mvar|AB}}, {{mvar|BC}}, and {{mvar|CD}} all have equal length.  Now, triangles {{mvar|ABC}} and {{mvar|BCD}} are [[isosceles triangle|isosceles]], thus (by Fact 3 above) each has two equal angles.

[[Hypothesis]]: Given {{mvar|AD}} is a straight line, and {{mvar|AB}}, {{mvar|BC}}, and {{mvar|CD}} all have equal length,

[[logical consequence|Conclusion]]: angle {{math|''b'' {{=}} {{sfrac|''a''|3}}}}.

[[Mathematical proof|Proof]]:
# From Fact 1) above, <math> e + c = 180</math>°.
# Looking at triangle ''BCD'', from Fact 2) <math> e + 2b = 180</math>°.
# From the last two equations, <math> c = 2b</math>.
# Therefore, <math>a=c+b=2b+b=3b</math>.

and the [[theorem]] is proved.

Again, this construction stepped outside the [[Greek mathematics|framework]] of [[compass and straightedge constructions|allowed constructions]] by using a marked straightedge.