In geometry, an **arbelos** is a plane region bounded by three semicircles with three apexes such that each corner of each semicircle is shared with one of the others (connected), all on the same side of a straight line (the *baseline*) that contains their diameters.^{[1]}

The earliest known reference to this figure is in Archimedes's *Book of Lemmas*, where some of its mathematical properties are stated as Propositions 4 through 8.^{[2]} The word *arbelos* is Greek for 'shoemaker's knife'. The figure is closely related to the Pappus chain.

Two of the semicircles are necessarily concave, with arbitrary diameters a and b; the third semicircle is convex, with diameter *a*+*b*.^{[1]}

The area of the arbelos is equal to the area of a circle with diameter HA.

**Proof**: For the proof, reflect the arbelos over the line through the points B and C, and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters BA, AC) are subtracted from the area of the large circle (with diameter BC). Since the area of a circle is proportional to the square of the diameter (Euclid's Elements, Book XII, Proposition 2; we do not need to know that the constant of proportionality is π/4), the problem reduces to showing that . The length |BC| equals the sum of the lengths |BA| and |AC|, so this equation simplifies algebraically to the statement that . Thus the claim is that the length of the segment AH is the geometric mean of the lengths of the segments BA and AC. Now (see Figure) the triangle BHC, being inscribed in the semicircle, has a right angle at the point H (Euclid, Book III, Proposition 31), and consequently |HA| is indeed a "mean proportional" between |BA| and |AC| (Euclid, Book VI, Proposition 8, Porism). This proof approximates the ancient Greek argument; Harold P. Boas cites a paper of Roger B. Nelsen^{[3]} who implemented the idea as the following proof without words.^{[4]}

Let D and E be the points where the segments BH and CH intersect the semicircles AB and AC, respectively. The quadrilateral ADHE is actually a rectangle.

*Proof*: ∠BDA, ∠BHC, and ∠AEC are right angles because they are inscribed in semicircles (by Thales's theorem). The quadrilateral ADHE therefore has three right angles, so it is a rectangle.*Q.E.D.*

The line DE is tangent to semicircle BA at D and semicircle AC at E.

*Proof*: Since ∠BDA is a right angle, ∠DBA equals π/2 minus ∠DAB. However, ∠DAH also equals π/2 minus ∠DAB (since ∠HAB is a right angle). Therefore triangles DBA and DAH are similar. Therefore ∠DIA equals ∠DOH, where I is the midpoint of BA and O is the midpoint of AH. But ∠AOH is a straight line, so ∠DOH and ∠DOA are supplementary angles. Therefore the sum of ∠DIA and ∠DOA is π. ∠IAO is a right angle. The sum of the angles in any quadrilateral is 2π, so in quadrilateral IDOA, ∠IDO must be a right angle. But ADHE is a rectangle, so the midpoint O of AH (the rectangle's diagonal) is also the midpoint of DE (the rectangle's other diagonal). As I (defined as the midpoint of BA) is the center of semicircle BA, and angle ∠IDE is a right angle, then DE is tangent to semicircle BA at D. By analogous reasoning DE is tangent to semicircle AC at E.*Q.E.D.*

The altitude 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 in each of these regions, known as the Archimedes' circles of the arbelos, have the same size.

The parbelos is a figure similar to the arbelos, that uses parabola segments instead of half circles. A generalisation comprising both arbelos and parbelos is the *f*-belos, which uses a certain type of similar differentiable functions.^{[5]}

In the Poincaré half-plane model of the hyperbolic plane, an arbelos models an ideal triangle.

The name *arbelos* comes from Greek ἡ ἄρβηλος *he árbēlos* or ἄρβυλος *árbylos*, meaning "shoemaker's knife", a knife used by cobblers from antiquity to the current day, whose blade is said to resemble the geometric figure.