The Complete Works of Archimedes

...$ex$ to $xa$, $kx$ to $xo$; and as $kx$ to $xo$, [so] $zp$ to $po$. As $zp$ to $po$, [so] $lp$ to $pr$. As $lp$ to $pr$, [so] $thr$ to $rs$. [As $thr$ to $rs$], [so] $ms$ to $st$. As $ms$ to $st$, [so] $nu$ to $ut$. And as $ms$ to $st$, [so] $nu$ to $uph$. As $nu$ to $uph$, [so] $thch$ to $chph$. And as $thch$ to $chph$, [so] $mp$ to $chg$. And therefore all [the antecedents] to all [the consequents] are similarly as one ratio to one. Therefore, as $kx$ is to $xa$, so are $ek, zl, bc, n, thm$ to the diameter $ac$. As $kx$ is to $xa$, so is $hc, e$ to $ea$. Therefore it is also that as $ce$ is to $ea$, so are all the lines $ek, zl, bc, n, thm$ to the diameter $ac$.

A geometric diagram showing a circle with a horizontal diameter and an inscribed polygon. Vertical lines connect corresponding vertices across the diameter. Points are labeled with various Greek letters (α, β, γ, δ, ε, ζ, η, θ, κ, λ, μ, ν, ξ, ο, π, ρ, σ, τ, υ, φ, χ, ψ, ω).

22

If in a segment of a circle a polygon be inscribed having its sides (excluding the base) equal and even in number, and if straight lines be drawn parallel to the base of the segment joining the vertices of the polygon, then all the lines so drawn and half the base, to the height of the segment, have the same ratio as that which the straight line joined from the diameter of the circle to the vertex of the polygon has to the side of the polygon. For let there be a circle $abc$, let a certain straight line $ac$ be drawn through it, and on $ac$ let a polygon be inscribed in the segment $abc$, even-sided and having its sides equal, excluding the base $ac$. And let $zh, eth$ be joined, which are parallel to the base of the segment. I say that as $zh, eth, ax$ are to $bx$, so is $dz$ to $zb$. For again, let $kp, lr, ms, nt$ be joined similarly, which are [parallel] to $bz$. On account of these things, as $kp$ is to $kb$, both $kp$ is to $pl$, and $lr$ is to $ml$, and $mr$ is to $mn$, and $xa$ is to $xn$. And so as all are to all, in the ratio of one to one, as then $zh, eth, ax$ are to $bx$, so is $zk$ to $kb$. As $kz$ is to $kb$, let it be as $dz$ to $zb$. Therefore, as $dz$ is to $zb$, so are $zh, eth, ax$ to $bx$.

A second geometric diagram of a circle, similar in construction to the first, with horizontal segments and labeled points, used to demonstrate a three-dimensional geometric proof related to a sphere.

23

Let there be in a sphere a great circle $abcd$. And let there be inscribed in it an equilateral polygon, and let the number of its sides be measured by four. Let $ac, bd$ be the diameters. If indeed, while the diameter $ac$ remains fixed, the circle $abcd$ containing the polygon be revolved, it is clear that its circumference will describe the surface of the sphere, while the angles of the polygon (except for those at points $a, c$) will be described in the circumferences of circles drawn on the surface of the sphere perpendicular to the circle $abcd$. Their diameters will be the lines joining the angles of the polygon parallel to $bd$, and the sides of the polygon will then describe certain cones. The sides $az, an$ will describe the surface of a cone whose base is the circle about the diameter $zn$ and whose vertex is the point $a$. The sides $zh, mn$ will be carried along a certain conical surface [a frustum] whose base is the circle about the diameter $hm$, and the vertex...