Arbelos

The area of the arbelos is equal to the area of a circle with diameter ${\displaystyle HA}$.

Proof: For the proof, reflect the arbelos over the line through the points ${\displaystyle B}$ and ${\displaystyle C}$, and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters ${\displaystyle BA}$ ${\displaystyle AC}$) are subtracted from the area of the large circle (with diameter ${\displaystyle BC}$). 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 ${\displaystyle {\frac {\pi }{4}}}$), the problem reduces to showing that ${\displaystyle 2(AH)^{2}=(BC)^{2}-(AC)^{2}-(BA)^{2}}$. The length ${\displaystyle (BC)}$ equals the sum of the lengths ${\displaystyle (BA)}$ and ${\displaystyle (AC)}$, so this equation simplifies algebraically to the statement that ${\displaystyle (AH)^{2}=(BA)(AC)}$. Thus the claim is that the length of the segment ${\displaystyle AH}$ is the geometric mean of the lengths of the segments ${\displaystyle BA}$ and ${\displaystyle AC}$. Now (see Figure) the triangle ${\displaystyle BHC}$, being inscribed in the semicircle, has a right angle at the point ${\displaystyle H}$ (Euclid, Book III, Proposition 31), and consequently ${\displaystyle (HA)}$ is indeed a "mean proportional" between ${\displaystyle (BA)}$ and ${\displaystyle (AC)}$ (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 ${\displaystyle D}$ and ${\displaystyle E}$ be the points where the segments ${\displaystyle BH}$ and ${\displaystyle CH}$ intersect the semicircles ${\displaystyle AB}$ and ${\displaystyle AC}$, respectively. The quadrilateral ${\displaystyle ADHE}$ is actually a rectangle.

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

The line ${\displaystyle DE}$ is tangent to semicircle ${\displaystyle BA}$ at ${\displaystyle D}$ and semicircle ${\displaystyle AC}$ at ${\displaystyle E}$.

Proof: Since angle BDA is a right angle, angle DBA equals π/2 minus angle DAB. However, angle DAH also equals π/2 minus angle DAB (since angle HAB is a right angle). Therefore triangles DBA and DAH are similar. Therefore angle DIA equals angle DOH, where I is the midpoint of BA and O is the midpoint of AH. But AOH is a straight line, so angle DOH and DOA are supplementary angles. Therefore the sum of angles DIA and DOA is π. Angle IAO is a right angle. The sum of the angles in any quadrilateral is 2π, so in quadrilateral IDOA, angle 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 ${\displaystyle AH}$ 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.