(related to Proposition: Prop. 12.13: Volumes of Parts of Cylinder cut by Plane Parallel to Opposite Planes are as Parts of Axis)

- For let the cylinder $AD$ have been cut by the plane $GH$ which is parallel to the opposite planes (of the cylinder), $AB$ and $CD$.
- And let the plane $GH$ have met the axis at point $K$.
- I say that as cylinder $BG$ is to cylinder $GD$, so $EK$ (is) to $KF$.

- For let $EF$ have been produced in each direction to points $L$ and $M$.
- And let any number whatsoever (of lengths), $EN$ and $NL$, equal to $EK$, be set out (on the $EL$), and any number whatsoever (of lengths), $FO$ and $OM$, equal to () $FK$, (on the $KM$).
- And let the cylinder $PW$, whose bases (are) the circles $PQ$ and $VW$, have been conceived on $LM$.
- And let planes parallel to $AB$, $CD$, and the bases of cylinder $PW$, have been produced through points $N$ and $O$, and let them have made the circles $RS$ and $TU$ around the centers $N$ and $O$ (respectively).
- And since axes $LN$, $NE$, and $EK$ are equal to one another, the cylinders $QR$, $RB$, and $BG$ are to one another as their bases [Prop. 12.11].
- But the bases are equal.
- Thus, the cylinders $QR$, $RB$, and $BG$ (are) also equal to one another.
- Therefore, since the axes $LN$, $NE$, and $EK$ are equal to one another, and the cylinders $QR$, $RB$, and $BG$ are also equal to one another, and the number (of the former) is equal to the number (of the latter), thus as many multiples as $KL$ is of $EK$, so many multiples is cylinder $QG$ also of cylinder $GB$.
- And so, for the same (reasons), as many multiples as $MK$ is of $KF$, so many multiples is cylinder $WG$ also of cylinder $GD$.
- And if $KL$ is equal to $KM$ then cylinder $QG$ will also be equal to cylinder $GW$, and if the (is) greater than the then the cylinder (will also be) greater than the cylinder, and if (the is) less then (the cylinder will also be) less.
- So, there are four magnitudes - the axes $EK$ and $KF$, and the cylinders $BG$ and $GD$ - and equal multiples have been taken of $EK$ and cylinder $BG$ - (namely), $LK$ and cylinder $QG$ - and of $KF$ and cylinder $GD$ - (namely), $KM$ and cylinder $GW$.
- And it has been shown that if $KL$ exceeds $KM$ then cylinder $QG$ also exceeds cylinder $GW$, and if (the axes are) equal then (the cylinders are) equal, and if ($KL$ is) less then ($QG$ is) less.
- Thus, as $EK$ is to $KF$, so cylinder $BG$ (is) to cylinder $GD$ [Def. 5.5] .
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"