Editing Arbelos (section)

==Properties==
Two of the semicircles are necessarily concave, with arbitrary diameters {{mvar|a}} and {{mvar|b}}; the third semicircle is [[Convex curve|convex]], with diameter {{mvar|''a''+''b''.}}<ref name=wolfram/> Let the diameters of the smaller semicircles be {{mvar|{{overline|BA}}}} and {{mvar|{{overline|AC}}}}; then the diameter of the larger semircle is {{mvar|{{overline|BC}}}}.

[[File:Arbelos diagram with points marked.svg|right|thumb|320px|Some special points on the arbelos.]]
<!--In the following sections, the corners of the arbelos are labeled {{mvar|A}}, {{mvar|B}}, and {{mvar|C}}, such that the diameter of the outer semicircle is {{mvar|BC}}, assumed to have unit length; and the diameters of the inner semicircles are {{mvar|AB}} and {{mvar|AC}}, assumed to have lengths ''r'' and 1−''r'', respectively.  The letter {{mvar|H}} denotes the point where the outer semicircle intercepts the line that is [[perpendicular]] to the diameter {{mvar|BC}} through the point {{mvar|A}}.-->

===Area===
Let {{mvar|H}} be the intersection of the larger semicircle with the line perpendicular to {{mvar|BC}} at {{mvar|A}}. Then the [[area (geometry)|area]] of the arbelos is equal to the area of a circle with diameter {{mvar|{{overline|HA}}}}.

'''Proof''': For the proof, reflect the arbelos over the line through the points {{mvar|B}} and {{mvar|C}}, and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters {{mvar|{{overline|BA}}}}, {{mvar|{{overline|AC}}}}) are subtracted from the area of the large circle (with diameter {{mvar|{{overline|BC}}}}). Since the area of a circle is proportional to the square of the diameter ([[Euclid]]'s [[Euclid's Elements|Elements]], Book XII, Proposition 2; we do not need to know that the [[proportionality (mathematics)|constant of proportionality]] is {{math|{{sfrac|{{pi}}|4}}}}), the problem reduces to showing that <math>2|AH|^2 = |BC|^2 - |AC|^2 - |BA|^2</math>. The length {{mvar|{{abs|BC}}}} equals the sum of the lengths {{mvar|{{abs|BA}}}} and {{mvar|{{abs|AC}}}}, so this equation simplifies algebraically to the statement that <math>|AH|^2 = |BA||AC|</math>. Thus the claim is that the length of the segment {{mvar|{{overline|AH}}}} is the [[geometric mean]] of the lengths of the segments {{mvar|{{overline|BA}}}} and {{mvar|{{overline|AC}}}}. Now (see Figure) the triangle {{mvar|BHC}}, being inscribed in the semicircle, has a right angle at the point {{mvar|H}} (Euclid, Book III, Proposition 31), and consequently {{mvar|{{abs|HA}}}} is indeed a "mean proportional" between {{mvar|{{abs|BA}}}} and {{mvar|{{abs|AC}}}} (Euclid, Book VI, Proposition 8, Porism). This proof approximates the ancient Greek argument; [[Harold P. Boas]] cites a paper of [[Roger B. Nelsen]]<ref name="RBNelsen_2002">{{cite journal |last1=Nelsen |first1=R B |title=Proof without words: The area of an arbelos |journal=Math. Mag. |date=2002 |volume=75 |issue=2 |page=144|doi= 10.2307/3219152|jstor=3219152 }}</ref> who implemented the idea as the following [[proof without words]].<ref>{{cite journal| last=Boas | first=Harold P.| author-link1=Harold P. Boas | title=Reflections on the Arbelos | journal= [[The American Mathematical Monthly]]| year=2006| volume=113| issue=3| url=http://www.maa.org/programs/maa-awards/writing-awards/reflections-on-the-arbelos| pages=236–249 | doi=10.2307/27641891| jstor=27641891}}</ref>
[[File:Arbelos proof2.svg|center]]

===Rectangle===
Let {{mvar|D}} and {{mvar|E}} be the points where the segments {{mvar|{{overline|BH}}}} and {{mvar|{{overline|CH}}}} intersect the semicircles {{mvar|AB}} and {{mvar|AC}}, respectively.  The [[quadrilateral]] {{mvar|ADHE}} is actually a [[rectangle]].
:''Proof'': {{mvar|∠BDA}}, {{mvar|∠BHC}}, and {{mvar|∠AEC}} are right angles because they are inscribed in semicircles (by [[Thales's theorem]]). The quadrilateral {{mvar|ADHE}} therefore has three right angles, so it is a rectangle. ''Q.E.D.''

===Tangents===
The line {{mvar|DE}} is tangent to semicircle {{mvar|BA}} at {{mvar|D}} and semicircle {{mvar|AC}} at {{mvar|E}}.
:''Proof'': Since {{mvar|ADHE}} is a rectangle, the diagonals {{mvar|AH}} and {{mvar|DE}} have equal length and bisect each other at their intersection {{mvar|O}}. Therefore, <math>|OD| = |OA| = |OE|</math>. Also, since {{mvar|{{overline|OA}}}} is perpendicular to the diameters {{mvar|{{overline|BA}}}} and {{mvar|{{overline|AC}}}}, {{mvar|{{overline|OA}}}} is tangent to both semicircles at the point {{mvar|A}}. Finally, because the two tangents to a circle from any given exterior point have equal length, it follows that the other tangents from {{mvar|O}} to semicircles {{mvar|BA}} and {{mvar|AC}} are {{mvar|{{overline|OD}}}} and {{mvar|{{overline|OE}}}} respectively.

===Archimedes' circles===
The altitude {{mvar|AH}} divides the arbelos into two regions, each bounded by a semicircle, a straight line segment, and an arc of the outer semicircle.  The circles [[inscribed circle|inscribed]] in each of these regions, known as the [[Archimedes' circles]] of the arbelos, have the same size.