Either of the two anharmonic ratios expresses a relation between the parts into which the segment AB is divided by the points C and D. Evidently the two anharmonic ratios have the same sign, and when one of them is zero the other is infinite, and vice versa. The last form is to be preferred, other things being convenient, on account of its symmetry with respect to A and B, the end-points of the divided segment. = 299°. The following results readily follow. I. Let AD AC. BD AC be +. Then and have like signs, AD. BC BC BD and therefore C and D both divide AB internally or both externally. (298°) In this case the order of the points must be some one of the following set, where AB is the segment divided, and the letters C and D are considered as being interchangeable : CDAB, ACDB, CABD, ABCD. 2. Let and have opposite AC. BD AC be AD. BC BC signs, and one point divides AB internally and the other externally. The order of the points is then one of the set CADB, ACBD. or Then or +I. AC. BD 4. Let AD. BC both internal or both external. 3. When either of the two anharmonic ratios is 1, these ratios are equal. AC AD AD BD and C and D are Also AC-BC BC AD-BD or AB AB BC BD' = and C and D coincide. Hence, when C and D are distinct points, the anharmonic ratio of the parts into which C and D divide AB cannot be positive unity. 5. Let AC.BD == I. Then And since C and D are now one external and one internal (2), they divide the segment AB in the same ratio internally and externally, disregarding sign. Such division of a line segment is called harmonic. (208°, Cor. 1) Harmonic division and harmonic ratio have been long employed, and from being only a special case of the more general ratio, this latter was named "anharmonic" by Chasles, "who was the first to perceive its utility and to apply it extensively in Geometry.” AC AD BC BD 300°. Def.-When we consider AB and CD as being two segments of the same line we say that CD divides AB, and that AB divides CD. Now the anharmonic ratios in which CD divides AB are and AD. BC And the anharmonic ratios in which AB divides CD are and CB. DA CA. DB .. the anharmonic ratios in which CD divides AB are the same as those in which AB divides CD. Or, any two segments of a common line divide each other equianharmonically. 301°. Four points A, B, C, D taken on a line determine six segments AB, AC, AD, BC, BD, and CD. These may be arranged in three groups of two each, so that in each group one segment may be considered as dividing the others, viz., AB, CD; BC, AD; CA, BD. Each group gives two anharmonic ratios, reciprocals of one another; and thus the anharmonic ratios determined by a range of four points, taken in all their possible relations, are six in number, of which three are reciprocals of the other three. These six ratios are not independent, for, besides the reciprocal relations mentioned, they are connected by three relations which enable us to find all of them when any one is given. Denote by P, by Q, CB. AD by R. AC. BD BA.CD AD. BC BD. CA Then P, Q, R are the anharmonic ratios of the groups ABCD, BCAD, and CABD, each taken in the same order. But in any range of four (233°) we have AB. CD + BC. AD+CA. BD=0. And dividing this expression by each of its terms in succession, we obtain Q + 1 = R + 1 = P + 1 = 1 I I I 1. Р Q From the symmetry of these relations we infer that any general properties belonging to one couple of anharmonic ratios, consisting of any ratio and its reciprocal, belong equally to all. Hence the properties of only one ratio need be studied. The symbolic expression {ABCD} denotes any one of the anharmonic ratios, and may be made to give all of them by reading the constituent letters in all possible orders. Except in the case of harmonic ratio, or in other special cases, we shall read the symbol in the one order of alternating the letters in the numerator and grouping the extremes and means in the denominator. Thus (ABCD) denotes AC. BD It is scarcely necessary to say that whatever order may be adopted in reading the symbol, the same order must be employed for each when comparing two symbols. 302°. Theorem.-Any two constituents of the anharmonic symbol may be interchanged if the remaining two are interchanged also, without affecting the value of the symbol. Proof.- {ABCD)=AC.BD: AD. BC. Interchange any two as A and C, and also interchange the remaining two B and D. Then {CDAB} = CA. DB:CB.DA Similarly it is proved that {ABCD} = {BADC} = {CDAB} = {DCBA}. q.e.d. 303°. If interchanging the first two letters, or the last two, without interchanging the remaining letters, does not alter the value of the ratio, it is harmonic. For, let {ABCD}={ABDC}. AD. BC AC. BD' Then or, multiplying across and taking square roots, AC. BD=±AD. BC. But the positive value must be rejected (299°, 4), and the negative value gives the condition of harmonic division. A 304°. Let ABCD be any range of four and O any point not on its axis. B Thus The anharmonic ratio of the pencil O. ABCD corresponding to any given ratio of the range is the same function of the sines of the angles as the given D ratio is of the corresponding segments. sin AOC. sin BOD sin AOD. sin BOC с corresponds to AC.BD or, symbolically, O{ABCD} corresponds to {ABCD}. To prove that the corresponding anharmonic ratios of the range and pencil are equal. AC_▲AOC_OA. OC sin AOC_OA sin AOC BC ABOC OB sin BOD OA sin AOD' Similarly, BD AD AC. BD sin AOC. sin BOD AD. BC sin AOD. sin BOC Hence, symbolically {ABCD}=0(ABCD}; and, with necessary formal variations, the anharmonic ratio of a range may be changed for that of the corresponding pencil, and vice versa, whenever required to be done. Cor. I. Two angles with a common vertex divide each other equianharmonically. (300°) Cor. 2. If the anharmonic ratio of a pencil is +1, two rays coincide, and if - 1, the pencil is harmonic. (299°, 4, 5) Cor. 3. A given range determines an equianharmonic pencil at every vertex, and a given pencil determines an equianharmonic range on every transversal. Cor. 4. Since the sine of an angle is the same as the sine of its supplement (214°, 1), any ray may be rotated through a straight angle or reversed in direction without affecting the ratio. Corollaries 2, 3, and 4 are of special importance. 305. Theorem. If three pairs of corresponding rays of two equianharmonic pencils intersect collinearly, the fourth pair intersect upon the line of collinearity. Proof.-Let O{ABCD}=0'{ABCD'}, and let the pairs of corresponding rays OA and O'A, OB and O'B, OC and O'C intersect in the three collinear points A, B, and C. Let the fourth corresponding rays meet the axis of ABC in D and D' respectively. Then R {ABCD} = {ABCD'}, (304°) |