ARCHIMEDES ON THE SPHERE

9

If a straight line falls into the circle which is the base of an isosceles cone, and from the endpoints of this line straight lines are drawn to the apex of the cone, the triangle enclosed by the falling line and the lines joined to the apex will be smaller than the surface of the cone intercepted between the lines joined to the apex. Let the base of the isosceles cone be the circle $ABΓ$, and the apex be $D$. And let a straight line $AΓ$ be drawn into it. And from the apex, let $AD$ and $DΓ$ be joined to $AΓ$. I say that triangle $ADΓ$ is smaller than the conic surface between $ADΓ$. Let the circumference $ABΓ$ be bisected at $B$, and let $AB, Γ B, DB$ be joined. The triangles $AB D$ and $BΓ D$ will be greater than triangle $ADΓ$. Let the stated triangles of triangle $ADΓ$ be $Θ$. Now, $Θ$ is either smaller than the segments $AB, BΓ$, or not. Let it not be smaller first. Since there are two surfaces—the conic one between $AD, B$ with the $AEB$ segments, and the $ABD$ triangle—having the same boundary as the perimeter of triangle $ABD$, the encompassing one will be greater than the encompassed. Therefore, the conic surface between $AD B$ with the $AEB$ segment is greater than the $ABD$ triangle. Similarly, the one between triangle $DBΓ$ with the $Γ ZB$ segment is greater than triangle $B DΓ$. Therefore, the whole conic surface with the segments is greater than the stated triangles. But the stated triangles are equal to triangle $ADΓ$ and the area $Θ$. Let area $Θ$ be removed as a common part. The remaining conic surface between $AD B$ is therefore greater than triangle $ADΓ$. Now let $Θ$ be smaller than the segments $AB, BΓ$. Bisecting the circumferences $AB, BΓ$ and their halves, we will take segments smaller than area $Θ$. Let those on the straight lines $AE, EB, BZ, ZΓ$ be left. And let $DE, DZ$ be joined. Again, therefore, by the same logic, the surface of the cone between $AD E$ with the segment under $AE$ is greater than triangle $ADE$. And the one between $ED B$ with the segment under $EB$ is greater than triangle $EDB$. Therefore, the surface between $AD B$ with the segments $AE, EB$ is greater than triangles $ADE, EBD$. And since triangles $ADE, DEB$ are greater than triangle $ABD$, as has been proven, the surface of the cone between $AD B$ with the segments under $AE, EB$ is much greater than triangle $ABD$. For the same reasons, the surface between $DBΓ$ with the segments under $ZB, ZΓ$ is greater than triangle $BDΓ$. Therefore, the whole conic surface between $ADΓ$ with the stated segments is greater than triangles $ABD, DBΓ$. But these are equal to triangle $ADΓ$ and area $Θ$, of which the stated segments are smaller than area $Θ$. The remaining conic surface between $ADΓ$ is therefore greater than triangle $ADΓ$.

A geometric diagram of a cone with apex δ. The base features points α, ε, β, ζ, γ connected to form a polygon. Lines extend from apex δ to these points. A point η outside the circle has tangents αη and γη, with a line connecting apex δ to η. Inside triangle αδβ, a small region is labeled θ.

10

If tangents are drawn to the circle which is the base of the cone in the same plane as the circle, and they meet each other, and from the points of contact and the intersection straight lines are drawn to the apex of the cone, the triangles enclosed by the tangents and the straight lines joined to the apex of the cone are greater than the surface of the cone intercepted by them. Let there be a cone whose base is the circle $ABΓ$ and whose apex is the point $D$. And let the tangents to circle $ABΓ$ be drawn in the same plane, $AH$ and $Γ H$, and from point $H$ and the apex of the cone, let $DA, DH, DΓ$ be joined to $A, H, Γ$. I say that triangles $ADH, H DΓ$ are greater...