Commentary on Proposition VI.31

Back to Proposition VI.31.

As we have seen (note on I. 47), this extension of I. 47 is credited by Proclus to Euclid personally.

There is one inference in the proof which requires examination. Euclid proves that

$$CB : BD = (\text{figure on } CB) : (\text{figure on } BA),$$

and that

$$BC : CD = (\text{figure on } BC) : (\text{figure on } CA),$$

and then infers directly that

$$BC : (BD + CD) = (\text{fig. on } BC) : (\text{sum of figs. on } BA \text{ and } AC).$$

Apparently V. 24 must be relied on as justifying this inference. But it is not directly applicable ; for what it proves is that, if

$$a : b = c : d,$$

and

$$e : b = f : d,$$

then

$$(a + e) : b = (c + f) : d.$$

Thus we should invert the first two proportions given above (by Simson’s Prop. B which, as we have seen, is a direct consequence of the definition of proportion), and thence infer by V. 24 that

(BD + CD) : BC = (sum of figs. on BA, AC) : (fig. on BC).

But $BD + CD$ is equal to $BC$; therefore (by Simson’s Prop. A, which again is an immediate consequence of the definition of proportion) the sum of the figures on $BA$, AC is equal to the figure on $BC$.

The MSS. again give an alternative proof which Heiberg places in the Appendix. It first shows that the similar figures on the three sides have the same ratios to one another as the squares on the sides respectively. Whence, by using I. 47 and the same argument based on V. 24 as that explained above, the result is obtained.

If it is considered essential to have a proof which does not use Simson’s Props. B and A or any proposition but those actually given by Euclid, no method occurs to me except the following.

Eucl. V. 22 proves that, if $a, b, c$ are three magnitudes, and $d, e, f$ three others, such that

$$a : b = d : e,$$

$$b : c = e : f,$$

then, ex aequali,

$$a : c = d : f.$$

If now in addition

$$a : b = b : c,$$

so that, also,

$$d : e = e : f,$$

the ratio $a : c$ is duplicate of the ratio $a : b$, and the ratio $d : f$ duplicate of the ratio $d : e$, whence the ratios which are duplicate of equal ratios are equal.

Now (fig. on AC) : (fig. on AB) = the ratio duplicate of AC : AB

$$= \text{the ratio duplicate of } CD : DA$$

$$= CD : BD.$$

Hence (sum of figs. on AC, AB) : (fig. on AB) = BC : BD. [V. 18]

But (fig. on BC) : (fig. on AB) = BC : BD (as in Euclid’s proof).

Therefore the sum of the figures on AC, AB has to the figure on AB the same ratio as the figure on BC has to the figure on AB, whence the figures on AC, AB are together equal to the figure on BC. [V. 9]