Commentary on Proposition VI.1¶
Back to Proposition VI.1.
The proof assumes—what is however an obvious deduction from I. 38—that, of triangles or parallelograms on unequal bases and between the same parallels, the greater is that which has the greater base.
It is of course not necessary that the two given triangles should have a common side, as in the figure; the proof is just as easy if they have not. The proposition being equally true of triangles and parallelograms of equal heights, Simson states this fact in a corollary thus:
“From this it is plain that triangles and parallelograms that have equal altitudes are to one another as their bases.
Let the figures be so placed as to have their bases in the same straight line; and, if we draw perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are, because the perpendiculars are both equal and parallel to one another [I. 33]. Then, if the same construction be made as in the proposition, the demonstration will be the same.”
The object of placing the bases in one straight line is to get the triangles and parallelograms within the same parallels. Cf. Proclus’ remark on I. 38 (p. 405, 17) that having the same height is the same thing as being in the same parallels.
Rectangles, or right-angled triangles, which have one of the sides about the right angle of the same length can be placed so that the equal sides coincide and the others are in a straight line. If then we call the common side the base, the rectangles or the right-angled triangles are to one another as their heights, by VI. 1. Now, instead of each right-angled triangle or rectangle, we can take any other triangle or parallelogram respectively with an equal base and between the same parallels. Thus
Triangles and parallelograms having equal bases are to one another as their heights.
Legendre and those authors of modern text-books who follow him in basing their treatment of proportion on the algebraical definition are obliged to divide their proofs of propositions like this into two parts, the first of which proves the particular theorem in the case where the magnitudes are commensurable, and the second extends it to the case where they are incommensurable.
Legendre (Éléments de Géométrie, III. 3) uses for this extension a rigorous method by reductio ad absurdum similar to that used by Archimedes in his treatise On the equilibrium of planes I. 7. The following is Legendre’s proof of the extension of VI. 1 to incommensurable parallelograms and bases.
The proposition having been proved for commensurable bases, let there be two rectangles ABCD, AEFD as in the figure, on bases AB, AE which are incommensurable with one another.
To prove that rect. ABCD : rect. AEFD = AB : AE.
For, if not, let rect. ABCD : rect. AEFD = AB : AO, ………………..(1)
where AO is (for instance) greater than AE.
Divide AB into equal parts each of which is less than EO, and mark off on AO lengths equal to one of the parts; then there will be at least one point of division between E and O.
Let it be I, and draw IK parallel to EF.
Then the rectangles ABCD, AIKD are in the ratio of the bases AB, AI, since the latter are commensurable.
Therefore, inverting the proportion, rect. AIKD : rect. ABCD = AI : AB ………………………(2).
From this and (1), ex aequali, rect. AIKD : rect. AEFD = AI : AO.
But AO > AI; therefore rect. AEFD > rect. AIKD.
But this is impossible, for the rectangle AEFD is less than the rectangle AIKD.
Similarly an impossibility can be proved if AO < AE.
Therefore rect. ABCD : rect. AEFD = AB : AE.
Some modern American and German text-books adopt the less rigorous method of appealing to the theory of limits.