Volumes of Parts of Cylinder cut by Plane Parallel to Opposite Planes are as Parts of Axis

Proof

 * Euclid-XII-13.png

Let the cylinder $AD$ be cut by the plane $GH$ which is parallel to the bases $AB$ and $CD$ of $AD$.

Let $GH$ meet the axis of $AD$ at the point $K$.

It is to be demonstrated that:
 * $BG : GD = EK : FK$

where:
 * $BG$ and $GD$ are the cylinder whose parts together make the cylinder $AD$
 * $EK$ and $FK$ are the axes of $BG$ and $GD$.

Let $EF$ be produced in both directions to $L$ and $M$.

Let there be set out:
 * any number of axes $EN, NL$ equal to $EK$

and
 * any number of axes $FO, OM$ equal to $FK$.

Let $PW$ be the cylinder on the axis $LM$.

Let $PQ$ and $VW$ be the bases of $PW$.

Let planes be defined parallel to $AB$ and $CD$ through the points $N$ and $O$.

Let the circles $RS$ and $TU$ be drawn in these planes with centers $N, O$ and with circumferences on $PV$ and $QW$.

We have that:
 * $LN = NE = EK$

Therefore from :
 * $QR : RB : BG = PQ : RS : AB$

But the bases $PQ, RS, AB$ are all equal.

Therefore the cylinders $QR, RB, BG$ are all equal.

So as:
 * the axes $LN, NE, EK$ are all equal

and:
 * the cylinders $QR, RB, BG$ are all equal

and:
 * the number of axes equals the number of cylinders

it follows that:
 * whatever multiple the axis $KL$ is of axis $EK$, the same multiple cylinder $QG$ will be of cylinder $GB$.

For the same reason:
 * whatever multiple the axis $MK$ is of axis $KF$, the same multiple cylinder $WG$ will be of cylinder $GD$.

If the axis is greater than the axis, the cylinder will also be greater than the cylinder.

If the axis is less than the axis, the cylinder will also be less than the cylinder.

The conditions of are seen to be fulfilled.

Hence the result.