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A decorative drop cap 'D'.Let there be a given portion ABC, whose diameter is BD. Upon the base AC, let a parallelogram AE be constructed, having two sides parallel and equal to the diameter BD, so that the remaining side touches the portion at its vertex. By continually dividing this parallelogram into two equal parts, there will finally be left a part that is smaller than a given space; let this be the parallelogram
A geometric diagram shows a parabolic segment ABC with inscribed and circumscribed parallelograms and triangles, labeled with points A, K, H, G, D, F, C at the base, and P, L, S, Q, M, R, N within the figure.
parallel to the diameter BD
BF, and let the base AC be divided into parts equal to the same DF, by points G, H, K, etc., and from there let lines GL, HM, KN, etc., be drawn to the section, and let the parallelograms DO, GP, HQ, KR, etc., be completed. I say that the figure composed of all these parallelograms (which hereafter shall be called "ordered circumscribed") exceeds the portion ABC by an amount less than any given space.
For let AN, NM, ML, LB, BS, etc., be joined; in this way, a certain rectilinear figure will also be inscribed in the portion. The excess of the circumscribed figure, which is composed of parallelograms, over the inscribed one will be greater than its excess over the portion ABC. However, the excess of the circumscribed figure over the inscribed one consists of triangles, whose parts on one side of the diameter, such as ARN, NQM,