:600px 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 $KL$ and $MN$ be the axes of the cones and cylinders. Let $L$ and $N$ be the apices of the cones. Thus: :let $KL$ be the height of the cone $c \left({ABCDL}\right)$ and the cylinder $AO$ :let $MN$ be the height of the cone $c \left({EFGHM}\right)$ and the cylinder $EP$. It is to be shown that: :$c \left({ABCD}\right) : c \left({EFGH}\right) = MN : KL$ That is, the bases of the cones and cylinders are reciprocally proportional to their heights. Either $LK = MN$ or $LK \ne MN$. First suppose $LK = MN$. We have that $AO = EP$. From {{EuclidPropLink|book = XII|prop = 11|title = Volume of Cones or Cylinders of Same Height are in Same Ratio as Bases}}: :$c \left({ABCD}\right) = c \left({EFGH}\right)$ Thus: :$c \left({ABCD}\right) : c \left({EFGH}\right) = MN : KL$ Suppose WLOG that $LK \ne MN$. Suppose $MN > LK$. Let $QN$ be cut off from $MN$ equal to $KL$. Let the cylinder $EP$ be cut by the plane $TUS$ through $Q$ parallel to the planes holding the circles $c \left({EFGH}\right)$ and $c \left({RP}\right)$. Let the cylinder $ES$ be described with the circle $c \left({EFGH}\right)$ as its base and with height $NQ$. We have that the cylinder $AO$ equals the cylinder $EP$. Therefore from {{EuclidPropLink|book = V|prop = 7|title = Ratios of Equal Magnitudes}}: :$AO : ES = EP : ES$ But from {{EuclidPropLink|book = XII|prop = 11|title = Volume of Cones or Cylinders of Same Height are in Same Ratio as Bases}}: :$AO : ES = c \left({ABCD}\right) : c \left({EFGH}\right)$ and from {{EuclidPropLink|book = XII|prop = 13|title = Volumes of Parts of Cylinder cut by Plane Parallel to Opposite Planes are as Parts of Axis}}: :$EP : ES = MN : QN$ Therefore from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$c \left({ABCD}\right) : c \left({EFGH}\right) = MN : QN$ But: :$QN = KL$ Therefore: :$c \left({ABCD}\right) : c \left({EFGH}\right) = MN : KL$ Therefore in the cylinders $AO$ and $EP$, the bases are reciprocally proportional to their heights. {{qed|lemma}} Let the bases of the cylinders $AO$ and $EP$ be reciprocally proportional to their heights: :$c \left({ABCD}\right) : c \left({EFGH}\right) = MN : KL$ It is to be proved that cylinders $AO$ and $EP$ are equal. Let $QN$ be cut off from $MN$ equal to $KL$. Let the cylinder $EP$ be cut by the plane $TUS$ through $Q$ parallel to the planes holding the circles $c \left({EFGH}\right)$ and $c \left({RP}\right)$. Let the cylinder $ES$ be described with the circle $c \left({EFGH}\right)$ as its base and with height $NQ$. We have that $KL = QN$. Thus: :$c \left({ABCD}\right) : c \left({EFGH}\right) = MN : QN$ But from {{EuclidPropLink|book = XII|prop = 11|title = Volume of Cones or Cylinders of Same Height are in Same Ratio as Bases}}: :$c \left({ABCD}\right) : c \left({EFGH}\right) = AO : ES$ and from {{EuclidPropLink|book = XII|prop = 13|title = Volumes of Parts of Cylinder cut by Plane Parallel to Opposite Planes are as Parts of Axis}}: :$MN : QN = EP : ES$ Therefore from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AO : ES = EP : ES$ Therefore from {{EuclidPropLink|book = V|prop = 9|title = Magnitudes with Same Ratios are Equal}}: :$AO = EP$ {{qed}} {{Euclid Note|15|XII}}