Proposition XIII

Let the square αβγδ [Fig. 12] be the base of a perpendicular prism with square bases and let a cylinder be inscribed in the prism whose base is the circle ζηθ which touches the sides of the parallelogram αβγδ at, ζ, η, and θ. Pass a plane through its center and the side in the square opposite the square αβγδ corresponding to the side γδ; this will cut off from the whole prism a second prism which is 1/4 the size of the whole prism and which will be bounded by three parallelograms and two opposite triangles. In the semicircle ζη describe a parabola whose origin is η and whose axis is ζκ, and in the parallelogram δη draw μν∥κζ; this will cut the circumference of the semicircle at ξ, the parabola at λ, and μν × νλ = νζ² (for this is evident [Apollonios, Con. I, 11]). Therefore μν: νλ = κη²: λσ². Upon μν construct a plane parallel to η; this will intersect the prism cut off from the whole prism in a right-angled triangle one side of which is μν and the other a straight line in the plane upon γδ perpendicular to γδ at ν and equal to the axis of the cylinder, but whose hypotenuse is in the intersecting plane. It will intersect the portion which is cut off from the cylinder by the Fig. 12. plane passed through η and the side of the square opposite the side γδ in a right-angled triangle one side of which is μξ and the other a straight line drawn in the surface of the cylinder perpendicular to the plane κν, and the hypotenuse. . . . . . . . . . . . . . . . and all the triangles in the prism: all the triangles in the cylinder-section = all the straight lines in the parallelogram δη: all the straight lines between the parabola and the straight line η. And the prism consists of the triangles in the prism, the cylinder-section of those in the cylinder-section, the parallelogram δη of the straight lines in the parallelogram δη∥κζ and the segment of the parabola of the straight lines cut off by the parabola and the straight line η; hence prism: cylinder-section = parallelogram ηδ: segment ζη that is bounded by the parabola and the straight line η. But the parallelogram δη = 3/2 the segment bounded by the parabola and the straight line η as indeed has been shown in the previously published work, hence also the prism is equal to one and one half times the cylinder-section. Therefore when the cylinder-section = 2, the prism = 3 and the whole prism containing the cylinder equals 12, because it is four times the size of the other prism; hence the cylinder-section is equal to 1/6 of the prism, Q. E. D.