Archimedis Syracusani Opera (1544 Basel Editio Princeps)

AND OF THE CYLINDER, BOOK I.

...equal to αδ. And let its height be θ. Since, therefore, θ is equal to αδ, it is therefore as θ is to αε, so is αδ to αε. But as αδ is to αε, so is the rhombus αβγδ to the cone βγδ. And as θ is to αε, so is the cone μνξ to the cone βγδ, because their bases are equal. Therefore, as the cone μνξ is to the cone βγδ, so is the rhombus αβγδ to the cone βγδ. Therefore, the cone μνξ is equal to the rhombus αβγδ. And since the surface of αβγ is equal to the base of πθκ, as the surface of αβγ is to its own base, so is the base of πθκ to the base of μνξ. For the base of αβγ is equal to the base of μνξ. And as the surface of αβγ is to its own base, so is αβ to βε, that is to say αδ to δζ. For the triangles are similar. Therefore, as the base of πθκ is to the base of μνξ, so is αδ to δζ. But αδ is equal to θ, for it was so posited; and δζ is equal to θλ. Therefore, as the base of πθκ is to the base of μνξ, so is θ to θλ. The bases of the cones πθκ and μνξ are therefore reciprocally proportional to their heights; therefore, the cones are equal. It was shown that μνξ is equal to the rhombus αβγδ, and therefore the cone πθκ is equal to the rhombus αβγδ.

If an isosceles cone is cut by a plane parallel to the base, and from the circle thus formed, a cone is inscribed having as its vertex the center of the base; and if the resulting rhombus is removed from the whole cone, the remainder will be equal to a cone that has a base equal to the surface of the cone lying between the parallel planes, and a height equal to the perpendicular drawn from the center of the base to one side of the cone. Let αβγ be an isosceles cone. And let it be cut by a plane parallel to the base, and let it make the section δε. Let ζ be the center of the base. And from the circle about the diameter δε, let a cone be inscribed having the vertex ζ. There will then be a rhombus βδζε, composed of isosceles cones. Let there be a cone κθλ, whose base is equal to the surface between δε and αγ. And let the height be equal to ζη, drawn as a perpendicular from point ζ to αβ. I say that if the rhombus βδζε is imagined as removed from the cone αβγ, the cone θκλ will be equal to the remainder. For let two cones be set out, μνξ and οπρ. Of these, let the base of μνξ be equal to the surface of the cone αβγ, and its height equal to ζη. For this reason, the cone μνξ is equal to the cone αβγ. For 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. And let the base of the cone οπρ be equal to the surface of the cone δβε, and its height equal to ζη. For this reason, the cone οπρ is equal to the rhombus βδζε. For this was previously demonstrated. Since the surface of the cone αβγ is composed of the surface βδε and the surface between δε and αγ. But the surface of the cone αβγ is equal to the base of the cone μνξ. And the surface of δεβ is equal to the base of οπρ. And the surface between δε and αγ is equal to the base of θκλ. Therefore, the bases of μνξ are equal to the bases of θκλ and οπρ, and the cones are under the same height. Therefore, the cone μνξ is also equal to the cones θκλ and οπρ. But the cone μνξ...

A geometric diagram showing a large triangle (representing a cone) with vertices labeled α, β, γ. A horizontal line segment δ ε cuts through the triangle parallel to the base α γ. From the center of the base ζ, lines are drawn to δ and ε, forming a rhombus-like structure (β δ ζ ε) within the cone. A perpendicular line segment ζ η is drawn from the center ζ to the side α β.