Now the space P = semicircle on AB − semicircle on AC: but the semicir. on AB = semicir. on AE + semicir. on BE, and the semicir. on AC = semicir. on AE − semicir. on CE, theref. semic. AB − semic. AC = semic. BE + semicir. CE, that is the space P is = semic. BE + semicir. CE; to each of these add the space Q, or the semicircle on BC, then P + Q = semic. BE + semic. CE + semic. BC, that is P + Q = double the semic. BE, or = the whole circle on BE.
5. In like manner, the two spaces PQ and RS together, or the whole space PQRS, is equal to the circle on the diameter BF. And therefore the space RS alone, is equal to the difference, or the circle on BF minus the circle on BE.
6. But, circles being as the squares of their diameters, BE2, BF2, and these again being as the parts or lines BC, BD, therefore the spaces PQ, PQRS, RS, TV, are respectively as the lines BC, BD, CD, AD, And if BC be equal to CD, then will PQ be equal to RS, as in the first or simplest case.
7. Hence, to find a circle equal to the space RS, where the points D and C are taken at random: From either end of the diameter, as A, take AG equal to DC, erect GH perpendicular to AB, and join AH; then the circle on AH will be equal to the space RS. For, the space PQ: the space RS ∷ BC ∶ CD or AG, that is as BE2: AH2 the squares of the diameters, or as the circle on BE to the cir∣cle on AH; but the circle on BE is equal to the space PQ, and therefore the circle on AH is equal to the space RS.
8. Hence, to divide a circle in this manner, into any number of parts, that shall be in any ratios to one another: Divide the diameter