ON THE SPHERE AND CYLINDER, BOOK I

A geometric diagram illustrates a mathematical proof. It shows a circular sector with center $ε$ and arc $α-β-γ$. Above the arc is a point $δ$, from which lines extend to the center $ε$ and points $α$, $β$, and $γ$. A polygonal chain $α-κ-β-ζ-γ$ is circumscribed about the arc, with $κ$ and $ζ$ being the vertices of the polygon outside the circle. Additional points $μ, ν, ξ, o$ are marked along the arc, with lines drawn from $ε$ to these points, creating smaller triangular segments within the main sector.

are greater than the conic surface between the straight lines $αε, γε$ and the arc $αβγ$. For let $κβ$ and $βζ$ be drawn as tangents to the circle, being parallel to $αγ$, and let the arcs $αβ$ and $βγ$ be bisected at $β$. And from $κ$ and $ζ$ let $κε$ and $ζε$ be joined to $ε$. And since $ακ, κβ$ are greater than $αβ, κβ$, let the common parts $κα, ζγ$ be added. The whole lines $αδ, δγ$ are therefore greater than $ακ, κβ, βζ, ζγ$. And since $αε, εβ, εγ$ are the sides of the cone, they are equal because the cone is isosceles. Similarly, they are also vertical perpendicular to the base, as was shown in the lemma, for the verticals from $ε$ to the bases of the triangles $ακε, κεβ, βεζ, ζεγ$. For the segments $ακ, κβ, βζ, ζγ$ are smaller than $γδ, δα$, but their heights are equal. It is clear that the perpendicular joined from the apex of the right cone to the surface of the base is perpendicular to the tangent. Thus, the triangles $αεδ, δγε$ are also greater than the triangles $αεκ, κεβ, βεζ, ζεγ$. Let there be an area $θ$. The area $θ$ is either smaller than the remainders of the arcs $ακβ$ and $βζγ$, or it is not smaller. Let it first be not smaller. Since there are composite surfaces—that of the pyramid with the trapezoid $κα, γζ$, having $ε$ as its apex, and the conic surface between $αεγ$ with the segment $αβγ$—and they have the same boundary, namely the perimeter of the triangles $αεγ$, it is clear that the surface of the pyramid, excluding the triangle $αεγ$, is greater than the conic surface with the segment $αβγ$. Let the segment $αβγ$ be removed as a common part. The remaining triangles $ακε, κεβ, βεζ, ζεγ$, with the remainders of the arcs $ακβ, βζγ$, are therefore greater than the conic surface between $αε, εγ$. But the area $θ$ is not smaller than the remainders of the arcs $ακβ, βζγ$; therefore, the triangles $αεκ, κεβ, βεζ, ζεγ$ with $θ$ will be much greater than the conic surface between $αε, εγ$. But the triangles $αεκ, κεβ, βεζ, ζεγ$ with $θ$ are sharper meaning the combined area is contained within than the triangles $αεδ, δεγ$. Therefore, the triangles $αεδ, δεγ$ will be greater than the stated conic surface. Let $θ$ be smaller than the remainders of the arcs, and because of this, by inscribing polygons around the segments while similarly bisecting the remaining arcs and drawing tangents, we will leave some remainders that will be smaller than the area $θ$. Let them be left behind, and let $αμκ, κνβ, βξζ, ζ oγ$ be smaller than the area $θ$, and let them be joined to $ε$. Again, it is clear that the triangles $ακε, κεβ, βεζ, ζεγ$ will be greater than the triangles $αεμ, μεν, νεξ, ξεζ, ζε o, oεγ$. For the bases are greater than the bases, but the height is equal. Furthermore, again in the same way, the pyramid—having the polygon $αμνξ oγ$ as its base and $ε$ as its apex—has a greater surface, excluding the triangle $αεγ$, than the conic surface between $αεγ$ with the segment $αβγ$. Let the segment $αβγ$ be removed as a common part; the remaining triangles $αεμ, μεν, νεξ, ξεζ, ζε o, oεγ$, with the remainders $αμκ, κνβ, βξζ, ζ oγ$, will be greater than the conic surface between $αε, εγ$. But the area $θ$ is greater than the stated remainders. And it was shown that the triangles $αεκ, κεβ, βεζ, ζεγ$ are greater than the triangles $αεμ, μεν, νεξ, ξεζ, ζε o, oεγ$. Therefore, the triangles $αεκ, κεβ, βεζ, ζεγ$ with the area $θ$, that is to say the triangles $αδε, δεγ$, are greater than the conic surface between the straight lines $αε, γε$.

11

If there are two straight lines on the surface of a right cylinder, the surface of the cylinder between the straight lines is greater than the parallelogram contained by the lines on the surface