Commentary on Proposition XI.29

Back to Proposition XI.29.

As usual, Euclid takes one case only and leaves the reader to prove for himself the two other possible cases shown in the subjoined figures. Euclid’s proof holds with a very slight change in each case. With the first figure, the

D H E K F M G N C B A L D H E K F M G N C B A L

only difference is that the prism of which the $\Delta\text{s}$ GAL, ECB are the bases takes the place of “the solid of which the parallelogram AB is the base and GEHM its opposite”; while with the second figure we have to subtract the prisms which are proved equal successively from the solid of which the parallelogram AB is the base and FDKN its opposite.

Simson, as usual, suspects mutilation by “some unskilful editor,” but gives a curious reason why the case in which the two parallelograms opposite to AB have a side common ought not to have been omitted, namely that this case “is immediately deduced from the preceding 28th Prop which seems for this purpose to have been premised to the 29th.” But, apart from the fact that Euclid’s Prop. 28 does not prove the theorem which it enunciates (as we have seen), that theorem is not in the least necessary for the proof of this case of Prop. 29, as Euclid’s proof applies to it perfectly well.