The Complete Works of Archimedes

9If, in an isosceles cone, straight lines [chords] fall into the circle which is the base of the cone, and from their extremities straight lines are drawn to the vertex of the cone, the triangle comprised by the chord and the lines joined to the vertex will be less than the surface of the cone intercepted between the lines joined to the vertex. Let the base of an isosceles cone be the circle ABC, and the vertex D. And let some straight line AC be joined in it, and from the vertex to A and C let AD and DC be joined. I say that triangle ADC is less than the conical surface between AD and CD. Let the arc ABC be bisected at B, and let AB, CB, and DB be joined. Then triangles ABD and
A geometric diagram representing a cone construction. Point δ is the apex at the top. The base is shown as a circular arc containing points α, ε, β, ζ, and γ. Lines radiate from the apex δ to each of these points on the base. Below the arc, tangent lines from α and γ meet at an external point η, and a line δ η connects the apex to this intersection. To the left of the main diagram, a small detached circular segment is labeled with the letter θ, representing a magnitude or area discussed in the proof.
BCD are [together] greater than triangle ADC; let the amount by which the said triangles exceed triangle ADC be Th. Now, Th is either less than the segments AB and BC, or it is not. Let it first be not less. Since, then, there are two surfaces—namely, the conical surface between AD and DB together with the segment AEB, and the triangle ADB—having the same boundary as the perimeter of triangle ADB, the containing surface will be greater than the contained; therefore the conical surface between AD and DB together with the segment AEB is greater than triangle ADB. Likewise, the conical surface between DB and DC together with the segment BZC is greater than triangle BDC. Therefore the whole conical surface together with the segments is greater than the aforementioned triangles. But the aforementioned triangles are equal to triangle ADC together with the area Th; and the segments are not less than Th. Therefore the conical surface is greater than triangle ADC. Now let Th be less than the segments AB and BC. Bisecting then the arcs AB and BC, and bisecting their halves, let us take segments which are [together] less than the area Th. Let these be the segments [subtended] by the straight lines AE, EB, BZ, and ZC; and let DE and DZ be joined. Again, therefore, by the same reasoning, the surface of the cone between AD and DE, together with the segment upon AE, is greater than triangle ADE. And the surface between ED and DB, together with the segment upon EB, is greater than triangle EDB. Therefore the surface between AD and DB, together with the segments AE and EB, is greater than triangles ADE and EDB. But since triangles ADE and DEB are [together] greater than triangle ADB, as has been proved, much more then is the surface of the cone between AD and DB, together with the segments of AE and EB, greater than triangle ADB. For the same reasons, the surface between DB and DC, together with the segments upon BZ and ZC, is greater than triangle BDC. Therefore the whole conical surface between AD and DC, together with the said segments, is greater than triangles ABD and DBC. And these are equal to triangle ADC and the area Th, whereas the said segments are less than the area Th. Therefore the remaining conical surface between AD and DC is greater than triangle ADC.

If tangents be drawn to the circle which is the base of the cone, being in the same plane as the circle and meeting one another, and from the points of contact and the intersection straight lines be drawn to the vertex of the cone, the triangles contained by the [tangent] surfaces and the straight lines joined to the vertex of the cone are greater than the conical surface intercepted by them. Let there be a cone whose base is the circle ABC, and the vertex the point D. And let AH and CH, being in the same plane, be drawn touching the circle ABC, and from the point H and the vertex of the cone to A and C, let AD, HD, and CD be joined. I say that triangles ADH and DCH are [together] greater