Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

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.

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.

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]