Elements of Algebra

CHAP. VIII. OF ALGEBRA. 153 have only to divide both sides by $bd$, which gives us $\frac{ad}{bd} = \frac{bc}{bd}$, or $\frac{a}{b} = \frac{c}{d}$, and consequently $a : b :: c : d$. 466. The four terms of a geometrical proportion, as $a : b :: c : d$, may be transposed in different ways, without destroying the proportion; for the rule being always, that the product of the extremes is equal to the product of the means, or $ad = bc$, we may say, 1st. $b : a :: d : c$; 2dly. $a : c :: b : d$; 3dly. $d : b :: c : a$; 4thly. $d : c :: b : a$. 467. Beside these four geometrical proportions, we may deduce some others from the same proportion, $a : b :: c : d$; for we may say, $a + b : a :: c + d : c$, or the first term, plus the second, is to the first, as the third, plus the fourth, is to the third; that is, $a + b : a :: c + d : c$. We may farther say, the first, minus the second, is to the first, as the third, minus the fourth, is to the third, or $a - b : a :: c - d : c$. For, if we take the product of the extremes and the means, we have $ac - bc = ac - ad$, which evidently leads to the equality $ad = bc$. And, in the same manner, we may demonstrate that $a + b : b :: c + d : d$; and that $a - b : b :: c - d : d$. 468. All the proportions which we have deduced from $a : b :: c : d$ may be represented generally as follows: $ma + nb : pa + qb :: mc + nd : pc + qd$. For the product of the extreme terms is $mpac + npbc + mqad + nqbd$; which, since $ad = bc$, becomes $mpac + npbc + mqbc + nqbd$; also the product of the mean terms is $mpac + mqbc + npad + nqbd$; or, since $ad = bc$, it is $mpac + mqbc + npbc + nqbd$: so that the two products are equal. 469. It is evident, therefore, that a geometrical proportion being given, for example, $6 : 3 :: 10 : 5$, an infinite number of others may be deduced from it. We shall, however, give only a few: $3 : 6 :: 5 : 10$; $6 : 10 :: 3 : 5$; $9 : 6 :: 15 : 10$; $3 : 3 :: 5 : 5$; $9 : 15 :: 3 : 5$; $9 : 3 :: 15 : 5$. 470. Since in every geometrical proportion the product of the extremes is equal to the product of the means, we may, when the three first terms are known, find the fourth from them. Thus, let the three first terms be $24 : 15 :: 40$ to the fourth term: here, as the product of the means is $600$, the fourth term multiplied by the first, that is by $24$, must