If a cylinder is cut by a plane parallel to its opposite planes, then the cylinder is to the cylinder as the axis is to the axis.

Let the cylinder *AD* be cut by the plane *GH* parallel to the opposite planes *AB* and *CD.* Let the plane *GH* meet the axis at the point *K.*

I say that the cylinder *BG* is to the cylinder *GD* as the axis *EK* is to the axis *KF.*

Produce the axis *EF* in both directions to the points *L* and *M.* Set out any number whatever of axes *EN* and *NL* equal to the axis *EK,* and any number whatever *FO* and *OM* equal to *FK.* Construct the cylinder *PW* on the axis *LM* with the circles *PQ* and *VW* as bases.

Carry the planes through the points *N* and *O* parallel to *AB* and *CD* and to the bases of the cylinder *PW,* and let them produce the circles *RS* and *TU* about the centers *N, O.*

Then, since the axes *LN, NE,* and *EK* equal one another, therefore the cylinders *QR, RB,* and *BG* are to one another as their bases.

But the bases are equal, therefore the cylinders *QR, RB,* and *BG* also equal one another.

Since then the axes *LN, NE,* and *EK* equal one another, and the cylinders *QR, RB,* and *BG* also equal one another, and the multitude of the former equals the multitude of the latter, therefore, the multiple the axis *KL* is of the axis *EK* is the same multiple the cylinder *QG* is of the cylinder *GB.*

For the same reason, the multiple the axis *MK* is of the axis *KF* is the same multiple the cylinder *WG* is of the cylinder *GD.*

And, if the axis *KL* equals the axis *KM,* then the cylinder *QG* also equals the cylinder *GW*; if the axis is greater than the axis, then the cylinder is also greater than the cylinder; and if less, less. Thus, there being four magnitudes, the axes *EK* and *KF* and the cylinders *BG* and *GD,* there have been taken equimultiples of the axis *EK* and of the cylinder *BG,* namely the axis *LK* and the cylinder *QG,* and equimultiples of the axis *KF* and of the cylinder *GD,* namely the axis *KM* and the cylinder *GW,* and it has been proved that, if the axis *KL* is in excess of the axis *KM,* the cylinder *QG* is also in excess of the cylinder *GW*; if equal, equal; and if less, less. Therefore the axis *EK* is to the axis *KF* as the cylinder *BG* is to the cylinder *GD.*

Therefore, *if a cylinder is cut by a plane parallel to its opposite planes, then the cylinder is to the cylinder as the axis is to the axis.*

Q.E.D.