Archimedis Syracusani Opera (1544 Basel Editio Princeps)

...is equal to the cone αβγ. And the cone πορ is equal to the rhombus βδεζ. The remaining cone θκλ is therefore equal to the remainder.

20

If, in a rhombus composed of isosceles cones, the other cone is cut by a plane parallel to the base, and from the circle thus formed a cone is inscribed, having the same vertex as the other cone; and if the resulting rhombus is removed from the whole rhombus, the remainder will be equal to a cone that has a base equal to the surface of the cone between the parallel planes, and a height equal to the perpendicular drawn from the vertex of the other cone to the side of the other cone. Let there be a rhombus composed of isosceles cones, αβγδ. And let the other cone be cut by a plane parallel to the base, and let it make the section εζ. And from the circle about the diameter εζ, let a cone be inscribed having the vertex at point δ. There will then be a formed rhombus εβζδ. And let it be imagined as removed from the whole rhombus. Let there be a cone θκλ, having its base equal to the surface between αγ and εζ. And let the height be equal to the perpendicular drawn from point δ to βα, or to the line on the same straight line as it. I say that the cone θκλ is equal to the aforementioned remainder. For let two cones be set out, μνξ and οπρ. And let the base of the cone μνξ be equal to the surface of αβγ. And let the height be equal to δη. For the reasons mentioned above, the cone μνξ is equal to the rhombus αβγδ. And of the cone οπρ, let the base be equal to the surface of the cone εβζ. And let the height be equal to δη. Similarly, the cone οπρ is equal to the rhombus εβζδ. And since the surface of the cone αβγ is composed of the surface εβζ and the surface between εζ and αγ. But the surface of the cone αβγ is equal to the base of μνξ. And the surface of the cone εβζ is equal to the base of the cone οπρ. And the surface between εζ and αγ is equal to the base of θκλ. Therefore, the base of μνξ is equal to the bases of οπρ and θκλ. And the cones are under the same height. And the cone μνξ is therefore equal to the cones θκλ and οπρ. But the cone μνξ is equal to the rhombus αβγδ. And the cone οπρ is equal to the rhombus εβζδ. The remaining cone θκλ is therefore equal to the remaining remainder.

21

If an equilateral and equiangular polygon with an even number of sides is inscribed in a circle, and straight lines are drawn joining the sides of the polygon, so that they are parallel to any one of the lines subtending two sides of the polygon, all such joining lines have the same ratio to the diameter of the circle as the line subtending one less than half the sides has to the side of the polygon. Let αβγδ be a circle, and in it let a polygon be inscribed, αεζβηθγμνδλκ. And let εκ, ζλ, βδ, ην, θμ be joined. It is clear that they are parallel to the line subtending two sides of the polygon. I say that all the aforementioned have the same ratio to the diameter αγ as γε has to εα. For let ζκ, ηλ, βδ, θν be joined. Therefore, ζκ is parallel to εα, and βλ is parallel to ζη. And further, δη is parallel to βλ, and θν is parallel to δη. And γμ is parallel to θν. And since εα and κζ are two parallels, and εκ and αζ are two lines drawn through, it is therefore as...

A geometric diagram showing a rhombus-like figure (alpha, beta, gamma, delta) composed of two cones sharing a base. Horizontal lines (epsilon-zeta, eta-theta) intersect the figure. A central vertical axis connects points beta and delta. Various points are labeled with Greek letters: alpha, beta, gamma, delta, epsilon, zeta, eta, theta, iota, kappa, lambda, mu, nu, xi, omicron, pi, rho.