The Complete Works of Archimedes

...cone is equal to the cone $abc$. And the [cone] $opr$ [is equal] to the rhombus $bdz$. Therefore, the remaining cone $θκλ$ is equal to the remainder.

20

If, in a rhombus composed of isosceles cones, one of the cones is cut by a plane parallel to the base, and from the resulting circle a cone is described having the same vertex as the other cone; and if from the whole rhombus the resulting rhombus is subtracted, the remaining solid will be equal to a cone having its base equal to the surface of the cone intercepted between the parallel planes, and its height equal to the perpendicular drawn from the vertex of the second cone to the side of the first cone. Let $abcd$ be a rhombus composed of isosceles cones. And let one cone be cut by a plane parallel to the base, and let it make the section $ez$. And from the circle about the diameter $ez$, let a cone be described having the point $d$ as its vertex. Thus the rhombus $ebzη$ will have been formed. And let it be conceived as having been subtracted from

A geometric diagram of a rhombus labeled with Greek letters α, β, γ, δ. Inside the rhombus, a horizontal line εζ divides the upper triangle (cone) αβγ. A smaller rhombus εβζη is formed. Dotted lines represent altitudes or heights from the vertices to the opposite sides and the central axis.

the whole rhombus. Let there be set out some cone $θκλ$, having its base equal to the surface between $ac$ and $ez$, and its height equal to the perpendicular drawn from point $d$ to $ba$, or to $ba$ produced. I say that the cone $θκλ$ is equal to the aforementioned remainder. For let two cones $mνξ$ and $opr$ be set out. And let the base of the cone $mνξ$ be equal to the surface of the cone $abc$, and its height be equal to $dη$. By what was shown before, the cone $mνξ$ is equal to the rhombus $abcd$. And of the cone $opr$, let the base be equal to the surface of the cone $ebz$, and its height be equal to $dη$. Likewise, the cone $opr$ is equal to the rhombus $ebzd$. Since the surface of the cone $abc$ is composed of the [surface] $ebz$ and the [surface] between $ez$ and $ac$; and the surface of the cone $abc$ is equal to the base of $mνξ$; and the surface of the cone $ebz$ is equal to the base of the cone $opr$; and the surface between $ez$ and $ac$ is equal to the base of $θκλ$; therefore, the base of $mνξ$ is equal to the bases of $opr$ and $θκλ$. And the cones are under the same height. Therefore, the cone $mνξ$ is equal to the cones $θκλ$ and $opr$. But the cone $mνξ$ is equal to the rhombus $abcd$, and the cone $opr$ is equal to the rhombus $ebzd$. Therefore, the remaining cone $θκλ$ is equal to the remaining remainder.

21

If an even-sided and equilateral polygon is inscribed in a circle, and straight lines are drawn joining the [vertices of the] sides of the polygon so that they are parallel to one of the lines subtending two sides of the polygon, all the joining lines together have to the diameter of the circle the same ratio as the chord subtending [an arc] less by one side than the semi-circle has to the side of the polygon. Let $abcd$ be a circle, and in it let the polygon $aezbηθγμνδλκ$ be inscribed, and let $eκ, zλ, bd, ην, θμ$ be joined. It is clear that they are parallel to the side subtending two sides of the polygon. I say then that all the said lines together have to the diameter $ac$ of the circle the same ratio that $γ e$ has to $ea$. For let $zη, λ b, ηδ, θν$ be joined. Then $zη$ is parallel to $ea$, and $bλ$ to $zη$. And since $eκ$ [is parallel to] $bλ$, and $bν$ to $δκ$, and $ημ$ to $θν$. And since $ea, η z$ are two parallels, and two lines $eη, aθ$ are drawn through them. It is therefore as...