The Complete Works of Archimedes

...are greater than the conical surface between the straight lines $AE, CE$ and the circumference $ABC$. For let $KBZ$ be drawn tangent to the circle and being parallel to $AC$, and let the circumference $ABC$ be bisected at $B$; and from $K, Z$ to $E$ let $KE, ZE$ be joined. And since $KD, DH$ are greater than $DB, BZ$, let $DA, ZC$ be added to both. Therefore the whole lines $AD, DC$ are [greater] than $AK, KZ, ZC$. And since the sides $AE, BE, CE$ are [sides] of the cone, they are equal, because the cone is isosceles. Similarly, they are also perpendiculars, as was shown in the lemma, [namely] those from the vertex to the bases of the triangles $AKE, KEZ, ZEC$. For $AK, KZ, ZC$ are less than $AD, DC$, while their heights are equal. It is clear, then, that the line joined from the vertex of a right cone to the point of contact on the base is perpendicular to the tangent; wherefore the triangles $AED, DEC$ are greater than the triangles $AEK, KEZ, ZEC$. Let there be an area $Θ$. Now, the area $Θ$ is either less than the remainders $AKB, BZC$, or not less. Let it first be less.

A geometric diagram showing a circle with a vertex $ε$ positioned above it. A triangle with vertex $ε$ and base $αγ$ is inscribed. Various other line segments and tangent lines are drawn, with points labeled $α, β, γ, δ, ε, ζ, η, θ, κ, λ, μ, ν, ξ, σ$. The diagram illustrates a proof regarding the surface area of a cone and the method of exhaustion using polygons.

Since, then, there are composite surfaces—namely, the surface of the pyramid of the trapezoid $AK, GC$ having $E$ as its vertex, and the conical surface between $AE, CE$ together with the segment $ABC$—and they have the same boundary, the perimeter of the triangle $AEC$, it is shown that the surface of the pyramid without the triangle $AEC$ is greater than the conical surface together with the segment $ABC$. Let the common segment $ABC$ be subtracted. Therefore the remaining triangles $AEK, KEZ, ZEC$, together with the remainders $AKB, BZC$, are greater than the conical surface between $AE, CE$. And the area $Θ$ is not less than the remainders $AKB, BZC$; therefore, by much more, the triangles $AEK, KEZ, ZEC$ together with $Θ$ will be greater than the conical surface between $AE, CE$. But the triangles $AEK, KEZ, ZEC$ together with $Θ$ are the triangles $AED, DEC$. Therefore the triangles $AED, DEC$ will be greater than the said conical surface. Now let $Θ$ be less than the remainders; then by always constructing polygons about the segments, by likewise bisecting the remaining circumferences and drawing tangents, we shall take certain remainders which will be less than the area $Θ$. Let them be left, and let $AMK, KNB, BXL, LSC$ be less than the area $Θ$, and let [lines] be joined to $E$. Again it is clear that the triangles $AEK, KEZ, ZEC$ are greater than the triangles $AEM, MEN, NEX, XES, SEC$. For the bases are greater than the bases, while the height is equal. Furthermore, the pyramid whose base is the polygon $AMKNBXLSG$ and whose vertex is $E$, without the triangle $AEC$, has a surface greater than the conical surface between $AE, CE$ together with the segment $ABC$. Let the common segment $ABC$ be subtracted; therefore the remaining triangles $AEM, MEN, NEX, XES, SEC$, together with the remainders $AMK, KNB, BXL, LSC$, will be greater than the conical surface between $AE, CE$. But the area $Θ$ is greater than the said remainders, and the triangles $AEK, KEZ, ZEC$ were shown to be greater than the triangles $AEM, MEN, NEX, XES, SEC$. Therefore, by much more, the triangles $AEK, KEZ, ZEC$ together with the area $Θ$—that is, the triangles $AED, DEC$—are greater than the conical surface between the straight lines $AE, CE$.

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