ON THE SPHERE AND CYLINDER, BOOK I

the remaining cylindrical surface cut off by the straight lines $αγ$ and $βδ$ is greater than the parallelogram $αγβδ$.

If there are two straight lines on the surface of a right cylinder, and from the endpoints of these straight lines there are drawn tangent lines to the circle which forms the base of the cylinder, such that they lie in the same plane as the base and intersect, then the parallelograms contained by the tangent lines and the sides of the cylinder are greater than the surface of the cylinder between the straight lines lying on the surface of the cylinder. Let the circle $αβγ$ be the base of a right cylinder, and let there be two straight lines on its surface, whose endpoints are $α$ and $γ$. From $α$ and $γ$, let tangent lines be drawn to the circle, lying in the same plane, and let them intersect at $κ$. Let it also be understood that tangent lines have been drawn to the circle from the endpoints of the straight lines on the surface in the other base of the cylinder. It must be shown that the parallelograms contained by the tangent lines and the sides of the cylinder are greater than the cylindrical surface along the arc $αβγ$. For let the tangent $εζ$ be drawn, and from points $ε$ and $ζ$ let straight lines be drawn parallel to the axis of the cylinder, extending to the surface of the other base. Then the parallelograms contained by $ακ$, $κγ$ and the sides of the cylinder are greater than the parallelograms contained by $αε$, $εζ$, $ζγ$ and the sides of the cylinder. Since $εη$ and $ηζ$ are greater than $εζ$, let the common segments $αε$ and $ζγ$ be added; therefore, the whole lines $ακ$ and $κγ$ are greater than $αε$, $εζ$, and $ζγ$. Let the excess amount be the area $κ$. The area $κ$ is greater than half of the shapes contained by the straight lines $αε, εβ, βγ$ and the arcs $αδβ, βθγ$. The interior surface is greater than the surface composed of the parallelograms along $αε, εζ, ζγ$, and the trapezoid $αε, εβ, βγ$, and its opposite in the other base of the cylinder, the boundary of which is the entire perimeter of the parallelograms along $αγ$. The aforementioned surfaces happen to have the same boundary, which lies in a plane. Both are concave on the same side. One includes the other, and they have some parts in common. Therefore, the included one is smaller. Thus, removing the common segments $αβγ$ and their opposites, the surface of the cylinder along the arc $αβγ$ is less than the composite surface formed by the parallelograms along $αε, εζ, ζγ$ and the segments $αε, εβ, βζ, ζγ$ and their opposites. The segments of the aforementioned surfaces are smaller than the surface composed of the parallelograms along $ακ$ and $κγ$. For with $κ$ included, it is greater. But the segments were equal to them. It is therefore clear that the parallelograms contained by $ακ, κγ$ and the sides of the cylinder are greater than the cylindrical surface along the arc $αβγ$. If half of the area $κ$ is not greater than the aforementioned surfaces, tangent lines will be drawn to the surfaces so that the remaining segments become less than half of $κ$. The rest will be demonstrated in the same way as before. These things having been demonstrated, it is clear from the aforementioned that if a pyramid is inscribed in an isosceles cone, the surface of the pyramid excluding the base is less than the conical surface. For each of the triangles containing the pyramid is less than the conical surface between the sides of the triangle. Thus, the entire surface of the pyramid excluding the base is less than the surface of the cone excluding the base. And if a pyramid is circumscribed about an isosceles cone, the surface of the pyramid excluding the base is greater than the surface of the cone excluding the base, according to the continuity of that which was just shown. It is clear from what has just been demonstrated that if a prism is inscribed in a right cylinder, the surface of the prism, composed of the parallelograms, is less than the surface of the cylinder excluding the base. For each parallelogram of the prism is less than the portion of the cylindrical surface it corresponds to. And that if a prism is circumscribed about a right cylinder, the surface of the prism composed of the parallelograms...

A geometric diagram illustrates a proof involving the surface area of a cylinder and inscribed/circumscribed polygons. It shows an arc labeled α, β, γ representing a portion of the circular base of a cylinder. Tangents at points α and γ meet at point κ, forming a triangle ακγ. Within this triangle, additional tangent lines intersect at points ε, ζ, η, creating a secondary polygonal path between the arc and the outer tangents. The points are labeled with Greek letters α, β, γ, κ, ε, ζ, η.