Let A (in fig. 3) be the centre of the scale, and A B the beam; to which let the two ponderants C and D be applied at the points B and E. I say, the proportion of the moment of the ponderant C to the moment of the ponderant D, is compounded of the proportions of A B to A E, and of the weight C to the weight D; or, if C and D be of the same species, of the magnitude C to the magnitude D. Let either of them, as C, be supposed to be bigger than the other, D. If, therefore, by the addition of F, F and D together be as one body equal to C, the moment of C to the moment of F + D will be (by the last article) as B G is to E H. Now as F + D is to D, so let E H be to another E I; and the moment of F + D, that is of C, to the moment of D, will be as B G to E I. But the proportion of B G to E I is compounded of the proportions of B G to E H, that is, of A B to A E, and of E H to E I, that is, of the weight C to the weight D. Wherefore unequal ponderants, when they are applied, &c. Which was to be proved. 6. The same figure remaining, if I K be drawn parallel to the beam A B, and cutting A G in K; and K L be drawn parallel to B G, cutting A B in L, the distances A B and A L from the centre will be proportional to the moments of C and D. For the moment of C is B G, and the moment of D is E I, to which K L is equal. But as the distance A B from the centre is to the distance A L from the centre, so is B G, the moment of the ponderant C, to L K, or E I the moment of the ponderant D. [Sidenote: If two ponderants have their weights and distances from the centre of the scale in reciprocal proportion, they are equally poised; and contrarily.] 7. If two ponderants have their weights and distances from the centre in reciprocal proportion, and the centre of the scale be between the points to which the ponderants are applied, they will be equally poised. And contrarily, if they be equally poised, their weights and distances from the centre of the scale will be in reciprocal proportion.