Archimedis Syracusani Opera (1544 Basel Editio Princeps)

Cones that have the same height bear the same ratio to their bases. And those that have equal bases bear the same ratio to their heights. If a cylinder is cut by a plane parallel to the base, the cylinder is to the cylinder as the axis is to the axis. And in the same ratio as the cylinders are the cones, which have the same bases as the cylinders. And of equal cones, the bases are reciprocally proportional to their heights. And those whose bases are reciprocally proportional to their heights are equal. And those cones whose diameters of the bases bear the same ratio to the axes, that is to say, to the heights, are in the triplicate ratio of the diameters in the bases. All these were demonstrated by those before.

17

If there are two isosceles cones, and the surface of one cone is equal to the base of the other, and the perpendicular drawn from the center of the base to the side of the cone is equal to the height, the cones will be equal. Let there be two isosceles cones, α β γ and δ ε ζ. And let the base of α β γ be equal to the surface of δ ε ζ. And let the height α κ be equal to the perpendicular κ θ drawn from the center of the base of θ to one side of the cone, such as δ ε. I say that the cones are equal. Since the base of α β γ is equal to the surface of δ ε ζ, and equals bear the same ratio to the same thing, therefore as the base of α β γ is to the base of δ ε ζ, so is the surface of δ ε ζ to the base of δ ε ζ. But as the surface is to its own base, so is δ θ to κ θ. For it was shown that the surface of every isosceles cone bears the same ratio to the base as the side of the cone bears to the radius of the base, that is to say δ ε to θ ε. And as δ ε to θ ε, so is δ θ to θ κ. For the triangles are right-angled. And θ κ is equal to α κ. Therefore, as the base of α β γ is to the base of δ ε ζ, so is the height δ θ to the height α β γ. Therefore, the bases of α β γ and δ ε ζ are reciprocally proportional to their heights. Therefore, the cone α β γ is equal to the cone δ ε ζ.

Geometrical diagram showing two isosceles cones represented as triangles with their altitudes and radii. The top triangle is labeled with Greek letters alpha, beta, gamma, kappa. The second triangle is labeled delta, epsilon, zeta, theta, eta.

18

To every rhombus composed of isosceles cones, a cone is equal that has a base equal to the surface of one of the cones containing the rhombus, and a height equal to the perpendicular drawn from the vertex of the other cone to one side of the other cone. Let the rhombus composed of isosceles cones be α β γ δ. And the base is the circle about the diameter β γ. And the height is α δ. And let another cone κ θ κ be set out, having its base equal to the surface of the cone α β γ, and the height equal to the perpendicular drawn from point δ to α β, or to the line drawn straight with it. Let this be δ ζ. And let the height of the cone θ κ be θ λ. Let θ λ be equal to δ ζ. I say that the cone is equal to the rhombus. For let another cone μ ν ξ be set out, having its base equal to the base of the cone α β γ, and the height...

Geometrical diagram of a rhombus formed by two back-to-back isosceles cones (alpha, beta, gamma, delta) and a separate cone (kappa, theta, lambda) used for the proof of volume equivalence.