AND CYLINDER, BOOK I

point $D$, and let a pyramid be inscribed in the cone, having an equilateral triangle $ABΓ$ as its base. And let $D A, D Γ, D B$ be joined. I say that the triangles $AD B, AD Γ, B D Γ$ are equal to a triangle whose base is equal to the perimeter of triangle $ABΓ$, and the perpendicular from the apex to the base is equal to the perpendicular drawn from $A$ to $AB$. For let the perpendiculars $D K, D Λ, D M$ be drawn. These are therefore equal to one another. And let triangle $EZΘ$ be placed, having base $EZ$ equal to the perimeter of triangle $ABΓ$, and altitude $HΘ$ equal to $D Λ$. Since, therefore, the rectangle under $DK$ and $BΓ$ is double the triangle $DBΓ$, and the rectangle under $AB$ and $DK$ is double the triangle $AB D$, and the rectangle under $AΓ$ and $DM$ is double the triangle $AD Γ$, the rectangle under the perimeter of triangle $ABΓ$ and $DK$ (and $D Λ$, that is $HΘ$) is double the triangles $AD B, B D Γ, AD Γ$. And the rectangle under $EZ$ and $HΘ$ is also double the triangle $EZH$; therefore, triangle $EZH$ is equal to the triangles $AD B, B D Γ, AD Γ$.

A geometric diagram shows a cone with an inscribed triangular pyramid. The apex is labeled δ. The base circle contains triangle αβγ. Perpendiculars from the apex to the base sides are κ, λ, μ. A construction to the right shows a right-angled triangle with base εζ and altitude ηθ.

If a pyramid is circumscribed about an isosceles cone, the surface of the pyramid, excluding the base, is equal to a triangle having a base equal to the perimeter of the base, and height equal to the side of the cone. Let there be a cone whose base is the circle $ABΓ$, and let a pyramid be circumscribed so that its base—that is, the polygon $Δ EZ$—is circumscribed about the circle $ABΓ$. I say that the surface of the pyramid, excluding the base, is equal to the aforementioned triangle. For since the axis of the cone is at right angles to the base (that is, to the circle $ABΓ$), and the straight lines joined from the center of the circle to the points of contact are at right angles to the tangents, the straight lines joined from the apex of the cone to the points of contact will also be at right angles to $Δ E, EZ, Z Δ$—these are $KA, KB, KΓ$. Therefore, the stated perpendiculars are equal to one another, for they are the sides of the cone. Let triangle $Θ KΛ$ be placed, having $Θ K$ equal to the perimeter of triangle $Δ EZ$, and the perpendicular $Λ M$ equal to $AK$. Since, therefore, the rectangle under $Δ E$ and $AK$ is double the triangle $E Δ K$, and the rectangle under $EZ$ and $KB$ is double the triangle $EZK$, and the rectangle under $Z Δ$ and $KΓ$ is double the triangle $Δ KZ$, the rectangle under $Θ K$ and $AK$ (that is, $MΛ$) is double the triangles $E Δ K, Z Δ K, EZK$. And the rectangle under $Θ K$ and $Λ M$ is also double the triangle $Λ KΘ$. For these reasons, the surface of the pyramid, excluding the base, is equal to a triangle having a base equal to the perimeter of $Δ EZ$ and height equal to the side of the cone.

A geometric diagram shows a cone with a circumscribed triangular pyramid. The apex is κ. The circular base αβγ is inscribed within the triangular base δεζ. Below this, a separate triangle θκλ is depicted with height λμ.