Commentary on Proposition XI.34¶
Back to Proposition XI.34.
In this proposition Euclid makes two assumptions which require notice,
that, if two parallelepipeds are equal, and have equal bases, their heights are equal, and (2) that, if the bases of two equal parallelepipeds are unequal, that which has the lesser base has the greater height. In justification of the former statement Euclid says, according to what Heiberg holds to be the genuine reading, “for parallelepipedal solids of the same height are to one another as their bases” [XI. 32]. This apparently struck some very early editor as not being sufficient, and he added the explanation appearing in Simson’s text, “For if, the bases EH, NQ being equal, the heights AG, CM”
were not equal, neither would the solid AB be equal to CD. But it is by hypothesis equal. Therefore the height CM is not unequal to the height AG; therefore it is equal.” Then, it being perceived that there ought not to be two explanations, the genuine one was erased from the inferior MSS. While the interpolated explanation does not take us very far, the truth of the statement may be deduced with perhaps greater ease from XI. 31 than from XI. 32 quoted by Euclid. For, assuming one height greater than the other, while the bases are equal, we have only to cut from the higher solid so much as will make its height equal to that of the other. Then this part of the higher solid is equal to the whole of the other solid which is by hypothesis equal to the higher solid itself. That is, the whole is equal to its part: which is impossible.
The genuine text contains no explanation of the second assumption that, if the base EH be greater than the base NQ, while the solids are equal, the height CM is greater than the height AG; for the added words “for, if not, neither again will the solids AB, CD be equal; but they are equal by hypothesis” are no doubt interpolated. In this case the truth of the assumption is easily deduced from XI. 32 by reductio ad absurdum. If the height CM were equal to the height AG, the solid AB would be to the solid CD as the base EH is to the base NQ, i.e. as a greater to a less, so that the solids would not be equal, as they are by hypothesis. Again, if the height CM were less than the height AG, we could increase the height of CD till it was equal to that of AB, and it would then appear that AB is greater than the heightened solid and a fortiori greater than CD: which contradicts the hypothesis.
Clavius rather ingeniously puts the first assumption the other way, saying that, if the heights are equal in the equal parallelepipeds, the bases must be equal. This follows directly from XI. 32, which proves that the parallelepipeds are to one another as their bases; though Clavius deduces it indirectly from
The advantage of Clavius’ alternative is that it makes the second assumption unnecessary. He merely says, if the heights be not equal, let CM be the greater, and then proceeds with Euclid’s construction.
It is also to be observed that, when Euclid comes to the corresponding proposition for cones and cylinders [XII. 15], he begins by supposing the heights equal, inferring by XII. 11 (corresponding to XI. 32) that, the solids being equal, the bases are also equal, and then proceeds to the case where the heights are unequal without making any preliminary inference about the bases. The analogy then of XII. 15, and the fact that he quotes XI. 32 here (which directly proves that, if the solids are equal, and also their heights, their bases are also equal), make Clavius’ form the more convenient to adopt.
The two assumptions being proved as above, the proposition can be put shortly as follows.
Suppose the edges terminating at the corners of the base to be perpendicular to it.
Then (a), if the base EH be equal to the base NQ, the parallelepipeds being also equal, the heights must be equal (converse of XI. 31), so that the bases are reciprocally proportional to the heights, the ratio of the bases and the ratio of the heights being both ratios of equality.
(b) If the base EH be greater than the base NQ, and consequently (by deduction from XI. 32) the height CM greater than the height AG, cut off CT from CM equal to AG, and draw the plane TV through T parallel to the base NQ, making the parallelepiped CV, with CT ($=$ AG) for its height.
Then, since the solids AB, CD are equal,
$$(\text{solid } AB) : (\text{solid } CV) = (\text{solid } CD) : (\text{solid } CV)\text{.} \qquad \text{[V. 7]}$$
But
$$\begin{aligned} (\text{solid } AB) : (\text{solid } CV) &= \text{▱ } HE : \text{▱ } NQ, & \text{[XI. 32]} \\ \text{and}\qquad (\text{solid } CD) : (\text{solid } CV) &= \text{▱ } MQ : \text{▱ } TQ & \text{[XI. 25]} \\ &= CM : CT. & \text{[VI. 1]} \end{aligned}$$
Therefore
$$\begin{aligned} \text{▱ } HE : \text{▱ } NQ &= CM : CT \\ &= CM : AG. \end{aligned}$$
Conversely (a), if the bases EH, NQ be equal and reciprocally proportional to the heights, the heights must be equal.
Consequently
$$\text{(solid } AB\text{)} = \text{(solid } CD\text{)}. \qquad \text{[XI. 31]}$$
(b) If the bases EH, NQ be unequal, if, e.g. ▱ EH > ▱ NQ, then, since
$$\text{▱ } EH : \text{▱ } NQ = CM : AG,$$
$$CM > AG.$$
Make the same construction as before.
Then
$$\begin{aligned} \text{▱ } EH : \text{▱ } NQ &= (\text{solid } AB) : (\text{solid } CV), & \text{[XI. 32]} \\ \text{and}\qquad CM : AG &= CM : CT \\ &= \text{▱ } MQ : \text{▱ } TQ & \text{[VI. 1]} \\ &= (\text{solid } CD) : (\text{solid } CV). & \text{[XI. 25]} \end{aligned}$$
Therefore
$$(\text{solid } AB) : (\text{solid } CV) = (\text{solid } CD) : (\text{solid } CV),$$
whence
$$\text{(solid } AB\text{)} = \text{solid } CD. \qquad \text{[V. 9]}$$
Suppose that the edges terminating at the corners of the bases are not perpendicular to it.
Drop perpendiculars on the bases from the corners of the faces opposite to the bases.
We thus have two parallelepipeds equal to AB, CD respectively, since they are on the same bases FK, RO and of the same height respectively. [XI. 29, 30]
If then (1) the solid AB is equal to the solid CD,
$$\text{(solid } BT\text{)} = \text{(solid } DX\text{),}$$
and, by the first part of this proposition,
$$\text{▱ } KF : \text{▱ } OR = MX : GT,$$
or
$$\text{▱ } HE : \text{▱ } NQ = MX : GT.$$
If
$$\text{▱ } HE : \text{▱ } NQ = MX : GT,$$
then
$$\text{▱ } KF : \text{▱ } OR = MX : GT,$$
so that, by the first half of the proposition, the solids BT, DX are equal, and consequently
$$\text{(solid } AB\text{)} = \text{(solid } CD\text{).}$$
The text of the second part of the proposition four times contains, after the words “of the same height,” the words “in which the sides which stand up are not on the same straight lines.” As Simson observed, they are inept, as the extremities of the edges may or may not be “on the same straight lines”; cf. the similar words incorrectly inserted at the end of XI. 31.
Words purporting to quote the result of the first part of the proposition are also twice inserted; but they are rejected as unnecessary and as containing an absurd expression—“(solids) in which the heights are at right angles to their bases,” as if the heights could be otherwise than perpendicular to the bases.