'arbelos'
in greek means
'shoemaker's knife'
(gr: άρβηλος)

archimedes
(c.287-212BC)

the area between the three semicircles (arbelos)
is equal to the area of the circle which has as diameter the segment CD

areaADBECF=areaADB - areaAFC - areaCEB
\=π \frac{AB²}{8} - π \frac{AC²}{8} - π \frac{CB²}{8}
\=\frac{π}{8}[(AC+CB)² - AC² - CB²]
\=\frac{π}{8}(AC² + CB² + 2AC·CB - AC² - CB²)
\=π(\frac{CD}{2})²

proof

where is the circle?

proof

where is the circle?

\text{you can move the green points}

A_1 = (1, 6)
B_1 = (3.5, 6)
C_1 = (3.5, 9.33)
A = (9, 2)
B = (16, 2)
c = Semicircle(A, B)
a = Segment(A, B)
C = Point(a)
d = Semicircle(A, C)
e = Semicircle(C, B)
b = OrthogonalLine(C, a)
D = Intersect(c, b)
f = Segment(C, D)
E_1 = (1, 2.5)
F_1 = (3.5, 2.5)
G = (1, 5)
E = Point(e)
F = Point(d)
g = false
α = Angle(D, C, A)
H = Midpoint(f)
h = Circle(H, D)
i = false
j = Segment(A, D)
k = Segment(B, D)
β = Angle(A, D, B)