Archimedis Syracusani Opera (1544 Basel Editio Princeps)

in potentiality, that is to say, the ratio of γ to δ in length. The ratio that γ bears to δ in length, the polygon circumscribed about circle α bears to the surface of the pyramid circumscribed about the cone. For γ is equal to the perpendicular from the center to one side of the polygon. And the side of the cone. But the perimeters of the polygons are a common height to the halves of the surfaces. Therefore, the rectilinear figure about circle α bears the same ratio to the rectilinear figure about circle β as that figure itself bears to the surface of the pyramid circumscribed about the cone. So that the surface of the pyramid is equal to the rectilinear figure circumscribed about circle β. Since the rectilinear figure circumscribed about circle β bears a lesser ratio to the inscribed figure than the surface of the cone bears to circle β, the surface of the pyramid circumscribed about the cone will bear a lesser ratio to the rectilinear figure inscribed in circle β than the surface of the cone bears to circle β. This is impossible. For the surface of the pyramid is shown to be greater than the surface of the cone. And the inscribed rectilinear figure in circle β will be less than circle β. Therefore, circle β is not less than the surface of the cone. I say then that it is not greater either. If possible, let it be greater. Again, let a polygon be conceived as inscribed in circle β, and another circumscribed, such that the ratio of the circumscribed to the inscribed is less than the ratio that circle β bears to the surface of the cone. And let a polygon be conceived as inscribed in circle α, similar to the one inscribed in circle β, and let a pyramid be described from it having the same vertex as the cone. Since the figures inscribed in α and β are similar, they will bear the same ratio to one another as the squares of the radii original: αἱ ἐκ τῶν κέντρων δυνάμει; therefore, the polygon bears the same ratio to the polygon as γ bears to δ in length. And γ to δ bears a greater ratio than the polygon inscribed in circle α to the surface of the pyramid inscribed in the cone. For the radius of circle α to the side of the cone bears a greater ratio than the perpendicular drawn from the center to one side of the polygon to the perpendicular drawn from the vertex of the cone to the side of the polygons. Therefore, the polygon inscribed in circle α bears a greater ratio to the polygon inscribed in circle β than that same polygon bears to the surface of the pyramid. Therefore, the surface of the pyramid is greater than that inscribed in circle β. But the polygon circumscribed about circle β bears a lesser ratio to the inscribed one than circle β does to the surface of the cone. Therefore, by a wide margin, the polygon circumscribed about circle β bears a lesser ratio to the surface of the pyramid inscribed in the cone than circle β bears to the surface of the cone, which is impossible, for the circumscribed polygon is greater than circle β.
And the surface of the pyramid in the cone is less than the surface of the cone, for which reason the circle is not greater than the surface of the cone. And it was shown that it is not less, therefore it is equal.

15 The surface of every isosceles cone bears the same ratio to its base as the side of the cone bears to the radius of the base of the cone. Let there be an isosceles cone whose base is circle α. Let β be equal to the radius of α, and γ be the side of the cone. It must be shown that the surface of the cone bears the same ratio to circle α as γ bears to β. For let ε be the mean proportional between β and γ, and let circle δ be set out, having its radius equal to ε. Therefore, circle δ is equal to the surface of the cone. For this was shown in the preceding proposition. And it was shown that circle δ bears the same ratio to circle α as γ bears to β in length. For each is the same as the ratio of the square of ε to the square of β. Just as circles are to one another as the squares on their diameters are to one another. Similarly for the squares of the radii. For both the diameters and their halves, that is to say the radii, are such, and the radii are equal to β and ε, it is therefore clear that the surface of the cone bears the same ratio to circle α as γ bears to β in length.

A geometric diagram showing a circle with a center point labeled 'α'. To the right of the circle are three vertical line segments of different lengths, labeled from left to right as 'α', 'β', and 'γ'.