Commentary on Proposition XI.30¶
Back to Proposition XI.30.
This proposition completes the proof of the theorem that Two parallelepipeds on the same base and of the same height are equivalent.
Legendre deduced the useful theorem that Every parallelepiped can be changed into an equivalent rectangular parallelepiped having the same height and an equivalent base.
For suppose we have a parallelepiped on the base ABCD with EFGH for the opposite face.
Draw AI, BK, CL, DM perpendicular to the plane through EFGH and all equal to the height of the parallelepiped AG. Then, on joining IK, KL, LM, MI, we have a parallelepiped equivalent to the original one and having its lateral faces AK, BL, CM, DI rectangles.
If ABCD is not a rectangle, draw AO, DN in the plane AC perpendicular to BC, and IP, MQ in the plane IL perpendicular to KL.
Joining OP, NQ, we have a rectangular parallelepiped on AOND as base which is equivalent to the parallelepiped with ABCD as base and IKLM as opposite face, since we may regard these parallelepipeds as being on the same base ADMI and of the same height (AO).
That is, a rectangular parallelepiped has been constructed which is equivalent to the given parallelepiped and has (1) the same height, (2) an equivalent base.
The American text-books which I have quoted adopt a somewhat different construction shown in the subjoined figure.
The edges AB, DC, EF, HG of the original parallelepiped are produced and cut at right angles by two parallel planes at a distance apart A'B' equal to AB.
Thus a parallelepiped is formed in which all the faces are rectangles except A'H', B'G'.
Next produce D'A', C'B', G'F', H'E' and cut them perpendicularly by two parallel planes at a distance apart B''C'' equal to B'C'.
The points of section determine a rectangular parallelepiped.
The equivalence of the three parallelepipeds is proved, not by Eucl. XI. 29, 30, but by the proposition about a right section of a prism given above in the note to XI. 28 (3 in that note).