Commentary on Proposition VI.14

Back to Proposition VI.14.

De Morgan says upon this proposition: “Owing to the disjointed manner in which Euclid treats compound ratio, this proposition is strangely out of place. It is a particular case of VI. 23, being that in which the ratio of the sides, compounded, gives a ratio of equality. The proper definition of four magnitudes being reciprocally proportional is that the ratio compounded of their ratios is that of equality.”

It is true that VI. 14 is a particular case of VI. 23, but, if either is out of place, it is rather the latter that should be placed before VI. 14, since most of the propositions between VI. 15 and VI. 23 depend upon VI. 14 and 15. But is perfectly consistent with Euclid’s manner to give a particular case first and its extension later, and such an arrangement often has great advantages in that it enables the more difficult parts of a subject to be led up to more easily and gradually. Now, if De Morgan’s view were here followed, we should, as it seems to me, be committing the mistake of explaining what is relatively easy to understand, viz. two ratios of which one is the inverse of the other, by a more complicated conception, that of compound ratio. In other words, it is easier for a learner to realise the relation indicated by the statement that the sides of equal and equiangular parallelograms are “reciprocally proportional” than to form a conception of parallelograms such that “the ratio compounded of the ratio of their sides is one of equality.” For this reason I would adhere to Euclid’s arrangement.

The conclusion that, since DB, BE are placed in a straight line, FB, BG are also in a straight line is referred to I. 14. The deduction is made clearer by the following steps.

The angle DBF is equal to the angle GBE; add to each the angle FBE;
therefore the angles DBF, FBE are together equal to the angles GBE, FBE. [C. N. 2]
[I. 13]But the angles DBF, FBE are together equal to two right angles,
therefore the angles GBE, FBE are together equal to two right angles, [C. N. 1]
[I. 14]and hence FB, BG are in one straight line.

The result is also obvious from the converse of I. 15 given by Proclus (see note on I. 15).

The proposition VI. 14 contains a theorem and one partial converse of it; so also does VI. 15. To each proposition may be added the other partial converse, which may be enunciated as follows, the words in square brackets applying to the case of triangles (VI. 15).

A D F B E K H C G A D F B E C K G H
Equal parallelograms [triangles] which have the sides about one angle in each reciprocally proportional are equiangular [have the angles included by those sides either equal or supplementary].

Let AB, BC be equal parallelograms, or let FBD, EBG be equal triangles, such that the sides about the angles at B are reciprocally proportional, i.e. such that

DB : BE = GB : BF.

We shall prove that the angles FBD, EBG are either equal or supplementary.

Place the figures so that DB, BE are in one straight line. Then FB, BG are either in a straight line, or not in a straight line.

  1. If FB, BG are in a straight line, the figure of the proposition (with the diagonals FD, EG drawn) represents the facts, and

[I. 15]the angle FBD is equal to the angle EBG.

  1. If FB, BG are not in a straight line, produce FB to H so that BH may be equal to BG. Join EH, and complete the parallelogram EBHK.

Now, since

DB : BE = GB : BF and GB = HB, DB : BE = HB : BF,

and therefore, by VI. 14 or 15, the parallelograms AB, BK are equal, or the triangles FBD, EBH are equal.

But the parallelograms AB, BC are equal, and the triangles FBD, EBG are equal; therefore the parallelograms BC, BK are equal, and the triangles EBH, EBG are equal.

Therefore these parallelograms or triangles are within the same parallels: that is, G, C, H, K are in a straight line which is parallel to DE. [I. 39]

Now, since BG, BH are equal, the angles BGH, BHG are equal.

By parallels, it follows that

the angle EBG is equal to the angle DBH,

whence the angle EBG is supplementary to the angle FBD.