第12卷命题 11 · 等高锥柱体积比于底圆

Let there be cones and cylinders of the same height, let the circles ABCD, EFGH be their bases, KL, MN their axes and AC, EG the diameters of their bases; I say that, as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN. For, if not, then, as the circle ABCD is to the circle EFGH, so will the cone AL be either to some solid less than the cone EN or to a greater. First, let it be in that ratio to a less solid O, and let the solid X be equal to that by which the solid O is less than the cone EN; therefore the cone EN is equal to the solids O, X. Let the square EFGH be inscribed in the circle EFGH; therefore the square is greater than the half of the circle. Let there be set up from the square EFGH a pyramid of equal height with the cone; therefore the pyramid so set up is greater than the half of the cone, inasmuch as, if we circumscribe a square about the circle, and set up from it a pyramid of equal height with the cone, the inscribed pyramid is half of the circumscribed pyramid, for they are to one another as their bases, [XII. 6] while the cone is less than the circumscribed pyramid. Let the circumferences EF, FG, GH, HE be bisected at the points P, Q, R, S, and let HP, PE, EQ, QF, FR, RG, GS, SH be joined.

设两等高圆锥底圆为ABCD和EFGH，轴为KL和MN，直径AC和EG。假设圆ABCD与圆EFGH之比不等于圆锥AL与圆锥EN之比，而是等于一个小于圆锥EN的立体O与圆锥EN之比。

Therefore each of the triangles HPE, EQF, FRG, GSH is greater than the half of that segment of the circle which is about it. On each of the triangles HPE, EQF, FRG, GSH let there be set up a pyramid of equal height with the cone; therefore, also, each of the pyramids so set up is greater than the half of that segment of the cone which is about it. Thus, bisecting the circumferences which are left, joining straight lines, setting up on each of the triangles pyramids of equal height with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the solid X. [X. 1] Let such be left, and let them be the segments on HP, PE, EQ, QF, FR, RG, GS, SH; therefore the remainder, the pyramid of which the polygon HPEQFRGS is the base and the height the same with that of the cone, is greater than the solid O. Let there also be inscribed in the circle ABCD the polygon DTAUBVCW similar and similarly situated to the polygon HPEQFRGS, and on it let a pyramid be set up of equal height with the cone AL. Since then, as the square on AC is to the square on EG, so is the polygon DTAUBVCW to the polygon HPEQFRGS, [XII. 1] while, as the square on AC is to the square on EG, so is the circle ABCD to the circle EFGH, [XII. 2] therefore also, as the circle ABCD is to the circle EFGH, so is the polygon DTAUBVCW to the polygon HPEQFRGS.

在圆EFGH内作正方形EFGH，其面积大于半圆。以该正方形为底作与圆锥等高的棱锥，此棱锥大于圆锥的一半。继续平分弧，作三角形，并作等高的棱锥，反复操作，最终得到一些小于立体X的圆锥片段。

But, as the circle ABCD is to the circle EFGH, so is the cone AL to the solid O, and, as the polygon DTAUBVCW is to the polygon HPEQFRGS, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex. [XII. 6] Therefore also, as the cone AL is to the solid O, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex; [V. 11] therefore, alternately, as the cone AL is to the pyramid in it, so is the solid O to the pyramid in the cone EN. [V. 16] But the cone AL is greater than the pyramid in it; therefore the solid O is also greater than the pyramid in the cone EN. But it is also less: which is absurd. Therefore the cone AL is not to any solid less than the cone EN as the circle ABCD is to the circle EFGH. Similarly we can prove that neither is the cone EN to any solid less than the cone AL as the circle EFGH is to the circle ABCD.

剩余棱锥（底为多边形HPEQFRGS）大于立体O。在圆ABCD内作相似多边形DTAUBVCW，并作等高的棱锥。由比例关系，圆ABCD与圆EFGH之比等于多边形DTAUBVCW与多边形HPEQFRGS之比，也等于圆锥AL与立体O之比。

I say next that neither is the cone AL to any solid greater than the cone EN as the circle ABCD is to the circle EFGH. For, if possible, let it be in that ratio to a greater solid O; therefore, inversely, as the circle EFGH is to the circle ABCD, so is the solid O to the cone AL. But, as the solid O is to the cone AL, so is the cone EN to some solid less than the cone AL; therefore also, as the circle EFGH is to the circle ABCD, so is the cone EN to some solid less than the cone AL: which was proved impossible. Therefore the cone AL is not to any solid greater than the cone EN as the circle ABCD is to the circle EFGH. But it was proved that neither is it in this ratio to a less solid; therefore, as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN. But, as the cone is to the cone, so is the cylinder to the cylinder, for each is triple of each; [XII. 10] Therefore also, as the circle ABCD is to the circle EFGH, so are the cylinders on them which are of equal height.

通过比例转换，得出圆锥AL与内接棱锥之比等于立体O与圆锥EN内接棱锥之比。但圆锥AL大于其内接棱锥，故立体O也大于圆锥EN的内接棱锥，这与假设矛盾。类似可证圆锥AL不能与大于圆锥EN的立体成比例，因此原命题成立。圆柱同理。