THE(α) 1.1 Parabolick Space (i. e. in Fig. 133. that comprehended under the right line GH and the parabola GEH) is to a cir∣cumscribing Parallelogram GK, as 4 to 6 (or 2 to 3) but to an inscribed ▵ GEH as 4 to 3.
Suppose FH divided first into two then into four equal parts, and draw parallel to the ax EF the lines ef, ef, &c. dividing also EF into four parts, the first fg will be 3, the second 2, the third 1, by Prop. 34. lib. 1. but as ef is to ge so is ge to he, by Consect. 1. Prop. 4. Therefore he in the diameter EF is = o, in the first ef it is = ¼ (for as ef, 4, to ge, 1, so ge, 1, to he, ¼) in the second ef a portion of he is = 〈 math 〉〈 math 〉, in the third to 〈 math 〉〈 math 〉, and so the portions eh in the trilinear figure EhHK make a series in a duplicate arithmetical progression, viz. 1, 4, 9, 16: After the same manner, if the parts Ff, &c. are bisected, you'l find the portions eh in the external trilinear figure to make this series of numbers ⅛, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, and so onwards. Wherefore since the portions eh or the indivisibles of the trili∣near space circumscribed about the parabola are always in a duplicate arithmetical progression: the sum of them all will be to the sum of as many indivisibles of the parallelogram FK, ••qual to the line KH, i. e. the trilinear space it self to this pa∣ra••lelogram as 1 to 3, by Consect. 10. Prop. 21. lib. 1. Wherefore the semi-parabola ••EhH will be as 2, and the ▵ FEH as 1 ½; therefore the whole parabola as 4, and the whole ▵ GEH as 3, and the whole parallelogram GK as 6. Q E. D.