Commentary on Proposition VI.23¶
Back to Proposition VI.23.
The second definition of the $Data$ says that A ratio is said to be given if we can find (πορίσασθαι) [another ratio that is] the same with it. Accordingly VI. 23 not only proves that equiangular parallelograms have to one another a ratio which is compounded of two others, but shows that that ratio is “given” when its component ratios are given, or that it can be represented as a simple ratio between straight lines.
Just as VI. 23 exhibits the operation necessary for compounding two ratios, a proposition (8) of the $Data$ indicates the operation by which we may divide one ratio by another. The proposition proves that Things which have a given ratio to the same thing have also a given ratio to one another. Euclid’s procedure is of course to compound one ratio with the inverse of the other; but, when this is once done and the result of Prop. 8 obtained, he uses the result in the later propositions as a substitute for the method of composition. Thus he uses the division of ratios, instead of composition, in the propositions of the $Data$ which deal with the same subject-matter as VI. 23. The effect is to represent the ratio of two equiangular parallelograms as a ratio between straight lines one of which is one side of one of the parallelograms. Prop. 56 of the $Data$ shows us that, if we want to express the ratio of the parallelogram AC to the parallelogram CF in the figure
of VI. 23 in the form of a ratio in which, for example, the side BC is the antecedent term, the required ratio of the parallelograms is BC : $X$, where
DC : CE = CG : X,
or $X$ is a fourth proportional to DC and the two sides of the parallelogram CF.
Measure CK along CB, produced if necessary, so that
DC : CE = CG : CK
(whence CK is equal to $X$).
[This may be simply done by joining DG and then drawing EK parallel to it meeting CB in K.]
Complete the parallelogram AK.
Then, since
DC : CE = CG : CK,
the parallelograms DK, CF are equal. [VI. 14]
Therefore
(AC) : (CF) = (AC) : (DK) [v. 7] = BC : CK [VI. 1] = BC : X.
Prop. 68 of the $Data$ uses the same construction to prove that, If two equiangular parallelograms have to one another a given ratio, and one side have to one side a given ratio, the remaining side will also have to the remaining side a given ratio.
I do not use the figure of the $Data$ but, for convenience’ sake, I adhere to the figure given above. Suppose that the ratio of the parallelograms is given, and also that of CD to CE.
Apply to CD the parallelogram DK equal to CF and such that CK, CB coincide in direction. [I. 45]
Then the ratio of AC to KD is given, being equal to that of AC to CF.
And
(AC) : (KD) = CB : CK;
therefore the ratio of CB to CK is given.
250 [BOOK VI [VI. 23
But, since
KD = CF, CD : CE = CG : CK. [VI. 14]
Hence CG : CK is a given ratio. And CB : CK was proved to be a given ratio. Therefore the ratio of CB to CG is given. [Data, Prop. 8]
Lastly we may refer to Prop. 70 of the Data, the first part of which proves what corresponds exactly to VI. 23, namely that, If in two equiangular parallelograms the sides containing the equal angles have a given ratio to one another [i.e. one side in one to one side in the other], the parallelograms themselves will also have a given ratio to one another. [Here the ratios of BC to CG and of CD to CE are given.]
The construction is the same as in the last case, and we have KD equal to CF, so that
CD : CE = CG : CK. [VI. 14]
But the ratio of CD to CE is given; therefore the ratio of CG to CK is given. And, by hypothesis, the ratio of CG to CB is given. Therefore, by dividing the ratios [Data, Prop. 8], we see that the ratio of CB to CK, and therefore [VI. 1] the ratio of AC to DK, or of AC to CF, is given.
Euclid extends these propositions to the case of two parallelograms which have given but not equal angles.
Pappus (VII. p. 928) exhibits the result of VI. 23 in a different way, which throws new light on compounded ratios. He proves, namely, that a parallelogram is to an equiangular parallelogram as the rectangle contained by the adjacent sides of the first is to the rectangle contained by the adjacent sides of the second.
Let AC, DF be equiangular parallelograms on the bases BC, EF, and let the angles at B, E be equal.
Draw perpendiculars AG, DH to BC, EF respectively.
Since the angles at B, G are equal to those at E, H, > the triangles ABG, DEH are equiangular.
Therefore
BA : AG = ED : DH. [VI. 4]
But
BA : AG = (rect. BA, BC) : (rect. AG, BC),
and
ED : DH = (rect. ED, EF) : (rect. DH, EF). [VI. 1]
Therefore [V. 11 and V. 16]
(rect. AB, BC) : (rect. DE, EF) = (rect. AG, BC) : (rect. DH, EF) = (AC) : (DF).
Thus it is proved that the ratio compounded of the ratios AB : DE and BC : EF is equal to the ratio of the rectangle AB, BC to the rectangle DE, EF.
Since each parallelogram in the figure of the proposition can be divided into pairs of equal triangles, and all the triangles which are the halves of either parallelogram have two sides respectively equal and the angles included by them equal or supplementary, it can be at once deduced from VI. 23 (or it can be independently proved by the same method) that triangles which have one angle of the one equal or supplementary to one angle of the other are in the ratio compounded of the ratios of the sides about the equal or supplementary angles. Cf. Pappus VII. pp. 894—6.
23 also shows that rectangles, and therefore parallelograms or triangles, are to one another in the ratio compounded of the ratios of their bases and heights.
The converse of VI. 23 is also true, as is easily proved by reductio ad absurdum. More generally, if two parallelograms or triangles are in the ratio compounded of the ratios of two adjacent sides, the angles included by those sides are either equal or supplementary.