5. One arithmetical proportion is the _same_ with another arithmetical proportion, when one of the antecedents exceeds its consequent, or is exceeded by it, as much as the other antecedent exceeds its consequent, or is exceeded by it. And therefore, in four magnitudes, arithmetically proportional, the sum of the extremes is equal to the sum of the means. For if A. B :: C. D be arithmetically proportional, and the difference on both sides be the same excess, or the same defect, E, then B + C (if A be greater than B) will be equal to A - E + C; and A + D will be equal to A + C - E; but A - E + C and A + C - E are equal. Or if A be less than B, then B + C will be equal to A + E + C; and A + D will be equal to A + C + E; but A + E + C and A + C + E are equal. Also, if there be never so many magnitudes, arithmetically proportional, the sum of them all will be equal to the product of half the number of the terms multiplied by the sum of the extremes. For if A. B :: C. D :: E. F be arithmetically proportional, the couples A + F, B + E, C + D will be equal to one another; and their sum will be equal to A + F, multiplied by the number of their combinations, that is, by half the number of the terms. If, of four unequal magnitudes, any two, together taken, be equal to the other two together taken, then the greatest and the least of them will be in the same combination. Let the unequal magnitudes be A, B, C, D; and let A + B be equal to C + D; and let A be the greatest of them all; I say B will be the least. For, if it may be, let any of the rest, as D, be the least. Seeing therefore A is greater than C, and B than D, A+B will be greater than C + D; which is contrary to what was supposed. If there be any four magnitudes, the sum of the greatest and least, the sum of the means, the difference of the two greatest, and the difference of the two least, will be arithmetically proportional. For, let there be four magnitudes, whereof A is the greatest, D the least, and B and C the means; I say A + D. B + C :: A - B. C - D are arithmetically proportional. For the difference between the first antecedent and its consequent is this, A + D - B - C; and the difference between the second antecedent and its consequent this, A - B - C + D; but these two differences are equal; and therefore, by this 5th article, A + D. B + C :: A - B. C - D are arithmetically proportional. If, of four magnitudes, two be equal to the other two, they will be in reciprocal arithmetical proportion. For let A + B be equal to C + D, I say A. C :: D. B are arithmetically proportional. For if they be not, let A. C :: D. E (supposing E to be greater or less than B) be arithmetically proportional, and then A + E will be equal to C + D; wherefore A + B and C + D are not equal; which is contrary to what was supposed. [Sidenote: The definition and transmutations of analogism, or the same geometrical proportion.]