"If AC = CD = DB = radius OA (see Fig. 3), the semi-circle ACE is ¼ of the semi-circle ACDB. We have now◐AB - 3◐AC = [trapezium] ACDB - 3 · meniscus ACE, [where the meniscus is the lunulae, i.e., lune] and each of these expressions is ¼◐AB or half the circle on ½AB as diameter. If then the meniscus AEC were quadrable so also would be the circle on ½AB as diameter. Hippocrates recognized the fact that the meniscus is not quadrable, and he made attempts to find other quadrable lunulae in order to make the quadrature of the circle depend on that of such quadrable lunulae."