Editing Angle trisection (section)

===With a right triangular ruler ===
[[File:01-Dreiteilung-des-Winkels-Bieberbach.svg|thumb|upright=1.35|Bieberbach's trisection of an angle (in blue) by means of a right triangular ruler (in red)]]

In 1932, [[Ludwig Bieberbach]] published in ''[[Crelle's Journal|Journal für die reine und angewandte Mathematik]]'' his work ''Zur Lehre von den kubischen Konstruktionen''.<ref name="Ludwig Bieberbach">Ludwig Bieberbach (1932) "Zur Lehre von den kubischen Konstruktionen", ''Journal für die reine und angewandte Mathematik'', H. Hasse und L. Schlesinger, Band 167 Berlin, p. 142–146 [http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN243919689_0167&DMDID=DMDLOG_0020 online-copie (GDZ)]. Retrieved on June 2, 2017.</ref> He states therein (free translation):

:"''As is known ... every cubic construction can be traced back to the trisection of the angle and to the multiplication of the cube, that is, the extraction of the third root. I need only to show how these two classical tasks can be solved by means of the right angle hook.''"

The construction begins with drawing a [[circle]] passing through the vertex {{mvar|P}} of the angle to be trisected, centered at {{mvar|A}} on an edge of this angle, and having {{mvar|B}} as its second intersection with the edge. A circle centered at {{mvar|P}} and of the same radius intersects the line supporting the edge in {{mvar|A}} and {{mvar|O}}.

Now the ''[[set square|right triangular ruler]]'' is placed on the drawing in the following manner: one [[cathetus|leg]] of its right angle passes through {{mvar|O}}; the vertex of its right angle is placed at a point {{mvar|S}} on the line {{mvar|PC}} in such a way that the second leg of the ruler is tangent at {{mvar|E}} to the circle centered at {{mvar|A}}. It follows that the original angle is trisected by the line {{mvar|PE}}, and the line {{mvar|PD}} perpendicular to {{mvar|SE}} and passing through {{mvar|P}}. This line can be drawn either by using again the right triangular ruler, or by using a traditional [[straightedge and compass construction]]. With a similar construction, one can improve the location of {{mvar|E}}, by using that it is the intersection of the line {{mvar|SE}} and its perpendicular passing through {{mvar|A}}.

''Proof:'' One has to prove the angle equalities <math>\widehat{EPD}= \widehat{DPS}</math> and <math>\widehat{BPE} = \widehat{EPD}.</math> The three lines {{mvar|OS}}, {{mvar|PD}}, and {{mvar|AE}} are parallel. As the [[line segment]]s {{mvar|OP}} and {{mvar|PA}} are equal, these three parallel lines delimit two equal segments on every other secant line, and in particular on their common perpendicular {{mvar|SE}}. Thus {{math|1=''SD{{'}}'' = ''D{{'}}E''}}, where {{mvar|D'}} is the intersection of the lines {{mvar|PD}} and {{mvar|SE}}. It follows that the [[right triangle]]s {{mvar|PD{{'}}S}} and {{mvar|PD{{'}}E}} are congruent, and thus that <math>\widehat{EPD}= \widehat{DPS},</math> the first desired equality. On the other hand, the triangle {{mvar|PAE}} is [[isosceles triangle|isosceles]], since all [[radius]]es of a circle are equal; this implies that <math>\widehat{APE}=\widehat{AEP}.</math> One has also <math>\widehat{AEP}=\widehat{EPD},</math> since these two angles are [[alternate angles]] of a transversal to two parallel lines. This proves the second desired equality, and thus the correctness of the construction.