The Complete Works of Archimedes

...equal to $ad$, and let its height be $no$. Since $no$ is equal to $ad$, it is therefore as $no$ is to $θε$, so is $ad$ to $θε$. But as $ad$ is to $θε$, so is the rhombus $abcd$ to the cone $bcd$. And as $no$ is to $θε$, so is the cone $μνξ$ to the cone $θηλ$, because their bases are equal. Therefore, as the cone $μνξ$ is to the cone $θηλ$, so is the rhombus $abcd$ to the cone $bcd$. Therefore, the cone $μνξ$ is equal to the rhombus $abcd$. And since the surface of $abc$ is equal to the base of $kθ k$, therefore as the surface of $abc$ is to its own base, so is the base of $kθ k$ to the base of $μνξ$. For the base of $abc$ is equal to the base of $μνξ$. And as the surface of $abc$ is to its own base, so is $ab$ to $bε$, that is, $ad$ to $dζ$; for the triangles are similar. Therefore, as the base of $kθ k$ is to the base of $μνξ$, so is $ad$ to $dζ$. But $ad$ is equal to $no$, for it is so assumed; and $dζ$ to $θλ$. Therefore, as the base of $kθ k$ is to the base of $μνξ$, so is the height $no$ to $θλ$. In the cones $kθ k$ and $μνξ$, therefore, the bases are reciprocally proportional to the heights; thus the cones are equal. And it was shown that the cone $μνξ$ is equal to the rhombus $abcd$; therefore the cone $kθ k$ is also equal to the rhombus $abcd$.

If an isosceles cone is cut by a plane parallel to the base, and from the resulting circle a cone is described having the center of the base as its vertex: the resulting rhombus, being subtracted from the whole cone, will be equal to the assigned cone, which has a base equal to the surface of the cone 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 $abc$ be an isosceles cone. And let it be cut by a plane parallel to the base, and let it make the section $de$. Let $ζ$ be the center of the base. And from the circle about the diameter $de$, let a cone be described having its vertex at $ζ$. There will then be a rhombus $βδζε$, composed of isosceles cones. Let there be set out some cone $kθλ$, whose base is equal to the surface between $de$ and $ac$, and whose height is equal to $ζη$, the perpendicular drawn from point $ζ$ to $ab$. I say that if the rhombus $βδζε$ is subtracted from the cone $abc$, the remainder will be equal to the assigned cone $kθλ$. For let two cones $μνξ$ and $oπρ$ be set out, such that the base of $μνξ$ is equal to the surface of the cone $abc$, and the height is equal to $ζη$. By what was shown before, the cone $μνξ$ is equal to the cone $abc$. And as for the two isosceles cones, where the surface of the one 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 the base of the cone $oπρ$ be equal to the surface of the cone $abd$, and its height equal to $ζη$. For the same reasons, the cone $oπρ$ is equal to the rhombus $βδζ$; for it was previously demonstrated to be equal. Since the surface of the cone $abc$ is composed of the surface $βδε$ and the surface between $de$ and $ac$; but the surface of the cone $abc$ is equal to the base of the cone $μνξ$, and the surface $βδε$ is equal to the base of $oπρ$, and the surface between $de$ and $ac$ is equal to the base of $kθ k$; therefore the base of $μνξ$ is equal to the bases of $θκλ$ and $oπρ$, and the cones are under the same height. Therefore, the cone $μνξ$ is equal to the cones $θκλ$ and $oπρ$. And the cone $μνξ$ [is equal to...]