Editing Arbelos (section)

===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]]