Commentary on Proposition XI.33

Back to Proposition XI.33.

The proof may be summarised as follows.

The three edges AE, GE, HE of the parallelepiped AB which meet at E, the vertex corresponding to R in the other parallelepiped, are produced, and lengths EK, EL, EM are marked off equal respectively to the edges CF, FN, FR of CD.

The parallelograms and solids are then completed as shown in the figure.

Euclid first shows that the solid CD and the new solid PK are equal and similar according to the criterion in XI. Def. 10, viz. that they are contained by the same number of equal and similar planes. (They are arranged in the same order, and it would be easy to prove equality by proving the equality of a pair of solid angles and then applying one solid to the other.)

We have now, by hypothesis,

$$AE : CF = EG : FN = EH : FR\text{;}$$

that is,

$$AE : EK = EG : EL = EH : EM\text{.}$$

But

$$\begin{align*} AE : EK &= \square AG : \square GK, & \text{[VI. 1]} \\ EG : EL &= \square GK : \square KL, & \\ EH : EM &= \square HK : \square KM\text{.} & \end{align*}$$

Again, by XI. 25 or 32,

$$\begin{align*} \square AG : \square GK &= (\text{solid } AB) : (\text{solid } EO)\text{,} \\ \square GK : \square KL &= (\text{solid } EO) : (\text{solid } QL)\text{,} \\ \square HK : \square KM &= (\text{solid } QL) : (\text{solid } KP)\text{.} \end{align*}$$

Therefore

$$(\text{solid } AB) : (\text{solid } EO) = (\text{solid } EO) : (\text{solid } QL) = (\text{solid } QL) : (\text{solid } KP)\text{,}$$

or the solid AB is to the solid KP (that is, CD) in the ratio triplicate of that which the solid AB has to the solid EO, i.e. the ratio triplicate of that which AE has to EK (or CF).

Heiberg doubts whether the Porism appended to this proposition is genuine.

Simson adds, as Prop. D, a useful theorem which we should have expected to find here, on the analogy of VI. 23 following VI. 19, 20, viz. that Solid parallelepipeds contained by parallelograms equiangular to one another, each to each, that is, of which the solid angles are equal, each to each, have to one another the ratio compounded of the ratios of their sides.

The proof follows the method of the proposition XI. 33, and we can use the same figure. In order to obtain one ratio between lines to represent the ratio compounded of the ratios of the sides, after the manner of VI. 23, we take any straight line $a$, and then determine three other straight lines $b$, $c$, $d$, such that

$$\begin{align*} AE : CF &= a : b\text{,} \\ EG : FN &= b : c\text{,} \\ EH : FR &= c : d\text{,} \end{align*}$$

whence $a : d$ represents the ratio compounded of the ratios of the sides. We obtain, in the same manner as above,

$$\begin{align*} (\text{solid } AB) : (\text{solid } EO) &= \square AG : \square GK = AE : EK = AE : CF \\ &= a : b\text{,} \\ (\text{solid } EO) : (\text{solid } QL) &= \square GK : \square KL = GE : EL = GE : FN \\ &= b : c\text{,} \\ (\text{solid } QL) : (\text{solid } KP) &= \square HK : \square KM = EH : EM = EH : FR \\ &= c : d\text{,} \end{align*}$$

whence, by composition [V. 22],

$$\begin{align*} (\text{solid } AB) : (\text{solid } KP) &= a : d\text{,} \\ \text{or } (\text{solid } AB) : (\text{solid } CD) &= a : d\text{.} \end{align*}$$