:700px Let there be cones and cylinders which are similar. Let the circles $c \left({ABCD}\right)$ and $c \left({EFGH}\right)$ be their bases. Let $BD$ and $FH$ be the diameters of the bases $c \left({ABCD}\right)$ and $c \left({EFGH}\right)$. Let $KL$ and $MN$ be the axes of the cones and cylinders. Let $L$ and $N$ be the apices of the cones. It is to be demonstrated that: :$c \left({ABCDL}\right) : c \left({EFGHN}\right) = BD^3 : FH^3$ where $c \left({ABCDL}\right)$ and $c \left({EFGHN}\right)$ are the cones on $c \left({ABCD}\right)$ and $c \left({EFGH}\right)$ respectively. Suppose it is not the case that: :$c \left({ABCDL}\right) : c \left({EFGHN}\right) = BD^3 : FH^3$ Then: :$c \left({ABCDL}\right) : O = BD^3 : FH^3$ where $O$ is either greater than $c \left({EFGHN}\right)$ or less than $c \left({EFGHN}\right)$. First suppose that $O < c \left({EFGHN}\right)$. Let the square $\Box EFGH$ be inscribed in $c \left({EFGH}\right)$. From Square Inscribed in Circle is greater than Half Area of Circle: :$\Box EFGH$ is greater than half $c \left({EFGH}\right)$. Let a pyramid be set up on the base $\Box EFGH$ of equal height as the cone. From: : {{EuclidPropLink|book = XII|prop = 6|title = Sizes of Pyramids of Same Height with Polygonal Bases are as Bases}} and: :Square Inscribed in Circle is greater than Half Area of Circle it follows that: :the pyramid so set up is greater than half the cone. Let the arcs $EF, FG, GH, HE$ be bisected at $P, Q, R, S$. Let $EP, PF, FQ, QG, GR, RH, HS, SE$ be joined. Using the reasoning of {{EuclidPropLink|book = XII|prop = 2|title = Areas of Circles are as Squares on Diameters}}: :each of $\triangle EPF, \triangle FQG, \triangle GRH, \triangle HSE$ is greater than half the segment of the circle $EFGH$ around it. On each of $\triangle EPF, \triangle FQG, \triangle GRH, \triangle HSE$, let pyramids be set up with the same apex as the cone. Again from the reasoning in {{EuclidPropLink|book = XII|prop = 2|title = Areas of Circles are as Squares on Diameters}}: :each of these pyramids is greater than half the segment of the cone around it. The operation: :bisecting the arcs remaining, joining the points of bisection with straight lines and setting up pyramids on the resulting triangles with the same apex as the cone can be repeated indefinitely. From {{EuclidPropLink|book = X|prop = 1|title = Existence of Fraction of Number Smaller than Given}}: :eventually we will leave some segments of the cone which will be less than the amount by which $c \left({EFGHN}\right)$ is greater than $O$. Let such segments be left, and let them be $EP, PF, FQ, QG, GR, RH, HS, SE$. Therefore the remainder, the pyramid $EPFQGRHSN$ whose base is the polygon $EPFQGRHS$ and whose apex is $N$, is greater than $O$. Let the polygon $ATBUCVDW$ similar to $EPFQGRHS$ be inscribed in $c \left({ABCD}\right)$. Let a pyramid $ATBUCVDWL$ be set up on the base $ATBUCVDW$ with the same apex as the cone $c \left({ABCDL}\right)$. Let $\triangle LBT$ be one of the triangular faces of $ATBUCVDWL$. Let $\triangle NFP$ be one of the triangular faces of $EPFQGRHSN$. Let $KT$ and $MP$ be joined. We have that $c \left({ABCDL}\right)$ is similar to $c \left({EFGHN}\right)$. Therefore from {{EuclidDefLink|XI|24|Similar Cones and Cylinders}}: :$BD : FH = KL : MN$ But: :$BD : FH = BK : FM$ Therefore: :$BK : FM = KL : MN$ So from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$BK : KL = FM : MN$ Then from {{EuclidPropLink|book = VI|prop = 6|title = Triangles with One Equal Angle and Two Sides Proportional are Similar}}: :$\triangle BKL$ is similar to $\triangle FMN$. Again, we have: :$BK : KT = FM : MP$ and again from {{EuclidPropLink|book = VI|prop = 6|title = Triangles with One Equal Angle and Two Sides Proportional are Similar}}: :$\triangle BKT$ is similar to $\triangle FMP$. We have that: :$BK : FM = KL : MN$ while: :$BK = KT$ and :$FM : PM$ We have that: :$\angle TKL = \angle PMN$ as both are right angles. So from {{EuclidPropLink|book = VI|prop = 6|title = Triangles with One Equal Angle and Two Sides Proportional are Similar}}: :$\triangle LKT$ is similar to $\triangle NMP$. We have that: :$\triangle LKB$ is similar to $\triangle NMF$ and so: :$LB : BK = NF : FM$ Also: :$\triangle BKT$ is similar to $\triangle FMP$. and so: :$KB : BT = MF : FP$ So from {{EuclidPropLink|book = V|prop = 22|title = Equality of Ratios Ex Aequali}}: :$LB : BT = NF : FP$ We also have that: :$\triangle LTK$ is similar to $\triangle NPM$. So: :$LT : TK = NP : PM$ Also: :$\triangle TKB$ is similar to $\triangle PMF$. So: :$KT : TB = MP : PF$ So from {{EuclidPropLink|book = V|prop = 22|title = Equality of Ratios Ex Aequali}}: :$LT : TB = NP : PF$ But it has already been proved that: :$TB : BL = PF : FN$ So from {{EuclidPropLink|book = V|prop = 22|title = Equality of Ratios Ex Aequali}}: :$TL : LB = PN : NF$ Therefore in $\triangle LTB$ and $\triangle NPF$ the sides are proportional. Therefore from {{EuclidPropLink|book = VI|prop = 5|title = Triangles with Proportional Sides are Similar}}: :$\triangle LTB$ and $\triangle NPF$ are equiangular. Therefore from {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}: :$\triangle LTB$ and $\triangle NPF$ are similar. Therefore from {{EuclidDefLink|XI|9|Similar Solid Figures}}: :the pyramid $BKTL$ is similar to the pyramid $FMPN$. But from {{EuclidPropLink|book = XII|prop = 8|title = Volumes of Similar Tetrahedra are in Triplicate Ratio of Corresponding Sides}}: :$BKTL : FMPN = BK^3 : FM^3$ In a similar manner, lines can be drawn: :from $A, W, D, V, C, U$ to $K$ and: :from $E, S, H, R, G, Q$ to $M$ and pyramids can be drawn with each of the triangles as bases which have the same apices as the cones. Hence it can be proved that the ratio of each of the similarly arranged pyramids to each of the similarly arranged pyramids is a ratio triplicate of that which $BD$ has to $FH$. From {{EuclidPropLink|book = V|prop = 12|title = Sum of Components of Equal Ratios}}: :$BKTL : FMPN = ATBUCVDWL : EPFQGRHSN$ Hence: :$ATBUCVDWL : EPFQGRHSN = BD^3 : FH^3$ But by hypothesis: :$c \left({ABCDL}\right) : O = BD^3 : FH^3$ Therefore: :$c \left({ABCDL}\right) : O = ATBUCVDWL : EPFQGRHSN$ So from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$c \left({ABCDL}\right) : ATBUCVDWL = O : EPFQGRHSN$ But: :$c \left({ABCDL}\right) > ATBUCVDWL$ Therefore: :$O > EPFQGRHSN$ But it has already been shown that: :$O < EPFQGRHSN$ Therefore it cannot be the case that: :$c \left({ABCDL}\right) : O = BD^3 : FH^3$ where $O < c \left({EFGHN}\right)$. In a similar way it can be proved that it cannot be the case that: :$c \left({EFGHN}\right) : O = BD^3 : FH^3$ where $O < c \left({ABCDL}\right)$. Next it is to be shown that it is not the case that: :$c \left({ABCDL}\right) : O = BD^3 : FH^3$ where $O > c \left({EFGHN}\right)$. Suppose it were possible that: :$c \left({ABCDL}\right) : O = BD^3 : FH^3$ where $O > c \left({EFGHN}\right)$. Therefore: :$O : c \left({ABCDL}\right) = FH^3 : BD^3$ But we have that: :$O : c \left({ABCDL}\right) = c \left({EFGHN}\right) : X$ where $X$ is some solid figure such that $X < c \left({ABCDL}\right)$. Therefore: :$c \left({EFGHN}\right) : X = FH^3 : BD^3$ where $X < c \left({ABCDL}\right)$. But this has been demonstrated to be impossible. Therefore it is not the case that: :$c \left({ABCDL}\right) : O = BD^3 : FH^3$ where $O > c \left({EFGHN}\right)$. But it has also been shown that: :$c \left({ABCDL}\right) : O = BD^3 : FH^3$ where $O < c \left({EFGHN}\right)$. Therefore: :$c \left({ABCDL}\right) : c \left({EFGHN}\right) = BD^3 : FH^3$ {{qed|lemma}} From {{EuclidPropLink|book = XII|prop = 10|title = Volume of Cone is Third of Cylinder on Same Base and of Same Height}}: :the cylinder on $c \left({ABCD}\right)$ is three times the cone on $c \left({ABCD}\right)$ and: :the cylinder on $c \left({EFGH}\right)$ is three times the cone on $c \left({EFGH}\right)$. Therefore the ratio of the cylinder on $c \left({ABCD}\right)$ to the cylinder on $c \left({EFGH}\right)$ equals the ratio triplicate of $BD$ to $FH$. {{qed}} {{Euclid Note|12|XII}}