Commentary on Proposition XI.36

Back to Proposition XI.36.

[Folio 0362 | p. 358]

The edges of the parallelepiped HL being respectively equal to A, B, C, and those of the equiangular parallelepiped KE being all equal to B, we regard MN (not containing the edge OL equal to B) as the base of the first parallelepiped, and consequently FD, equiangular to MN, as the base of KE.

Then the solids have the same height. [XI. 35, Por.]

Hence

$$\text{(solid } HL\text{)} : \text{(solid } KE\text{)} = \mathbin{▱} MN : \mathbin{▱} FD. \qquad \text{[XI. 32]}$$

But, since A, B, C are in continued proportion,

$$A : B = B : C,$$

or

$$LM : EF = DE : LN.$$

Thus the sides of the equiangular ▱s MN, FD are reciprocally proportional, whence

$$\mathbin{▱} MN = \mathbin{▱} FD, \qquad \text{[VI. 14]}$$

and therefore

$$\text{(solid } HL\text{)} = \text{(solid } KE\text{)}.$$