Commentary on Proposition V.16

Back to Proposition V.16.

Algebraically, if

$$\begin{aligned} a : b &= c : d, \\ \text{then}\qquad a : c &= b : d. \end{aligned}$$

Taking equimultiples $ma$, $mb$ of $a$, $b$, and equimultiples $nc$, $nd$ of $c$, $d$, we have, by V. 15,

$$\begin{aligned} a : b &= ma : mb, \\ c : d &= nc : nd. \end{aligned}$$

And, since $a : b = c : d$, we have [V. 11] $ma : mb = nc : nd$.

Therefore [V. 14], according as $ma > = < nc$, $mb > = < nd$, so that $a : c = b : d$.

Aristotle tacitly uses the theorem in $Meteorologica$ III. 5, 376 a 22—24.

The four magnitudes in this proposition must all be of the same kind, and Simson inserts “of the same kind” in the enunciation.

This is the first of the propositions of Eucl. V. which Smith and Bryant (Euclid’s Elements of Geometry, 1901, pp. 298 sqq.) prove by means of VI. 1 so far as the only geometrical magnitudes in question are straight lines or rectilineal areas ; and certainly the proofs are more easy to follow than Euclid’s. The proof of this proposition is as follows.

To prove that, If four magnitudes of the same kind [straight lines or rectilineal areas] be proportionals, they will be proportionals when taken alternately.

Let P, Q, R, S be the four magnitudes of the same kind such that

$$P : Q = R : S ;$$

then it is required to prove that

$$P : R = Q : S.$$

$First$, let all the magnitudes be areas.

Construct a rectangle abcd equal to the area P, and to bc apply the rectangle bcef equal to Q.

Also to ab, bf apply rectangles ag, bk equal to R, S respectively.

Then, since the rectangles ac, be have the same height, they are to one another as their bases. [VI. 1]

d c e a b f h g k

Hence P : Q = ab : bf. But P : Q = R : S. Therefore R : S = ab : bf, [V. 11] i.e. rect. ag : rect. bk = ab : bf.

Hence (by the converse of VI. 1) the rectangles $ag, bk$ have the same height, so that k is on the line hg.

Hence the rectangles ac, ag have the same height, namely ab; also be, bk have the same height, namely bf.

$$\begin{aligned} \text{Therefore} \qquad \text{rect. } ac : \text{rect. } ag &= bc : bg, \\ \text{and} \qquad \text{rect. } be : \text{rect. } bk &= bc : bg. && [\text{VI. 1}] \\ \text{Therefore} \qquad \text{rect. } ac : \text{rect. } ag &= \text{rect. } be : \text{rect. } bk. && [\text{V. 11}] \\ \text{That is,} \qquad P : R &= Q : S. \end{aligned}$$

Secondly, let the magnitudes be straight lines AB, BC, CD, DE. Construct the rectangles Ab, Bc, Cd, De with the same height.

a b c d e A B C D E

Then Ab : Bc = AB : BC, and Cd : De = CD : DE. [VI. 1] But AB : BC = CD : DE. Therefore Ab : Bc = Cd : De. [V. 11]

Hence, by the first case,

$$Ab : Cd = Bc : De,$$

and, since these rectangles have the same height,

$$AB : CD = BC : DE.$$