(diagram 1) Let there be a segment BQG enclosed by a straight-line and a section of a right-angled cone. First, let, in fact, BG be at right angles to the diameter, and let BD be drawn from point B parallel to the diameter, and let GD from G be touching the section of the cone at G. In fact, triangle BGD will be right-angled. (diagram 2) Let, in fact, BG be divided into equal segments, however many, BE, EZ, ZH, HI, IG, and let ES, ZT, HU, IX be drawn from the section parallel to the diameter, but let them be joined from the points at which these cut the section of the cone to G and extended. (diagram 3) I say, in fact, that triangle BDG is (diagram 4) less than three-times trapezoids KE, LZ, MH, NI and triangle XIG, (diagram 5) but is larger than three-times trapezoids ZF, HQ, IP and triangle IJG.

(diagram 6 = general diagram)

(diagram 6 = general diagram) For let a straight-line ABG be drawn and let AB equal to BG be demarcated, and let a balance AG be conceived, and B will be its middle. And let it be suspended from B, but let BDG also be suspended from the balance at B, G, and let areas R, C, Y, W, D, J be suspended from the other side of the balance at A, (diagram 7) and let area R incline equally to trapezoid DE holding in this way, while C inclines equally to trapezoid ZS, Y to TH, W to YI, DJ to triangle XIG. In fact, the whole will incline equally to the whole. Thus, triangle BDG would be three-times area RCYWDJ. (Prop. 6) (diagram 8) Since BGQ is a segment enclosed by a straight-line and segment of a right-angled cone, and BD has been drawn from B parallel to the diameter, while GD has been drawn from G as touching the section of the cone at G, and some other line too, SE, has been drawn parallel to diameter, BG has the same ratio to BE that SE has to EF. (Prop. 5) (diagram 9) Thus BA also has the same ratio to BE that trapezoid DE has to KE. (For, by similar triangles, BD : KB = SE : EF, or BD+SE : KB+EF = BD : BE. However, the trapezoids with the same height have the ratio as BD+SE : KB+EF = SE : EF = BG : BE (see above) = AB : BE. Hence, trapezoid DE : KE = AB : BE. (diagram 10) AB will be similarly proved to have the same ratio to BZ that trapezoid SZ has to LZ, and to BH that TH has to MH, to BI that UI has to NI.

(diagram 11) And so since DE is a trapezoid having right angles at point B, E, with sides converging to G, some area, R, suspended from the balance at A, will incline equally to it, with the trapezoid holding as it is now positioned, and further as BA is to BE, so is trapezoid DE to KE, therefore, area KE is larger than area R. For this has been proved. (Prop. 10) (diagram 12) Again, ZS is a trapezoid having right angles at Z, E and ST converging to G, and area C when suspended from the balance at A inclines equally to it, with the trapezoid holding as it is now positioned. Also, AB is to BE, so is trapezoid ZS to ZF. (for it was proved that SE : EF = AB : BE, while SE : EF = TZ : ZZ', so that trapezoid SZ : trapezoid FZ = SG : FE = AB : BE) And as AB is to BZ so is trapezoid ZS to LZ. And so area C would be smaller than trapezoid LZ and larger than ZF. For this too has been proved. (Prop. 12)

(diagram 6 = general diagram)

(diagram 13) For the same reasons, in fact, area Y is also smaller than trapezoid MH but larger than QH, and area W is smaller than trapezoid NOIH but larger than PI, and similarly area D' is also smaller than triangle XIG but larger than GIO.(Prop. 8) (diagram 14: large trapezoids) And so, since trapezoid KE is larger than area R, LZ than area C, MH than Y, NI than W, and triangle XIG than DJ, it is obvious that all the mentioned areas are also larger than area RCYWDJ. (diagram 15: triangle BDG) But RCYWDJ is a third part of triangle BGD. Therefore, it is clear that triangle BGD is smaller than three-times trapezoids KE, LZ, MH, NI and triangle XIG. (diagram 16: small trapezoids) Again since trapezoid ZF is smaller than area C, QH than Y, IP than W, and triangle IOG than DJ, it is obvious that all these mentioned figures are also smaller than area DJWYC. And so it is obvious that triangle BDG is also larger than three-times trapezoids FZ, QH, IP and triangle IFO, but less than three-times those drawn earlier.

The basic idea will be to compress the inner and outer trapezoids and triangle into the parabola (requires html5 browser support):

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