Prof. De Morgan on Anharmonic Ratio. 169 



(10.) The three sides, AB_, BC, CA, are anharmonically* 

 divided in the same manner, giving (AC'BC")='(BA'CA") 

 = (CB'AB"). The transversals passing through A' show that 

 CA': A'B is both (CB" : B"A)(AC' : C'B) and (CB': B'A) 

 (AC": C"B). A very easy form of the proof is this : each 

 of the ratios (AC'BC") &c. is nothing but the compound 

 (AC : C'B)(BA': A'C)(CB' : B'A). Converses may be made. 

 (11.) If two lines meeting in Z, be cut by three lines meeting 

 in O, in A, B, C, A', B', C', then they have similar anharmonic 

 divisions, and (ZABC) = (ZA'B'C'). Draw AB', A'B, &c. 

 and there are six cases of inscription ; A'BO and AB'O are 

 inscripts of ZCC; B'CO and BCD of ZAA'; C'AO and 

 AC'Oof ZBB': from which the theorem follows instantly, 

 and, of course, its extension to any pencil of four lines and 

 any pair of transversals. The particular case, where the an- 

 harmonic ratios become those of equality, must of course be 

 noted. Also, that if two similar anharmonic systems have one 

 point in common, the lines joining the points of the three cor- 

 responding couples meet in one point. (12.) Notice that 

 A"B"C" is an inscript both of ABC and A'B'C; and call it 

 the common inscript; or, when A"B"C" are in one straight 

 line, an evanescent inscript. (13.) When the inscript is har- 

 monic, that is when the sides of ABC are harmonically divided 

 by those of A'B'C, then AA', BB', CC meet in one point: 

 and the converse. (14-.) When the inscript is harmonic, the 

 common inscript is evanescent ; and the converse. 



(15.) The complete quadrilateral is a triangle and a trans- 

 versal, in four different ways. If, as before, ABC be the tri- 

 angle, and A'B'C the transversal, then A, A' are opposite 

 points, as are B, B', and C, C. The /r/nfrf of quadrilaterals is 

 ABA'B', BCB'C, CAC'A'. Draw the diagonals A A', BB', 

 CC, and let AA', BB' meet in C" ; BB', CC, in A"; CC, 

 A A', in B". In the figure are seven lines, and ABA'B' and 

 its attendants are complete. But there are six other sy- 

 stems of quadrilaterals, each of which is incomplete, wanting 

 one diagonal, and dispensing with one of the seven lines. 

 Thus AA'BB' has the diagonals AB, A'B', wants CC", and 

 has nothing to do with CC. Throw out the line CC, replace 



* I take this theorem to be new, which I should not have done from its 

 being new to me when 1 found it: but on communicating it to Mr. T. S. 

 Davies, whose research in this subject is far above mine, I found that it was 

 equally new to him. But I shall not definitively call it new till I hear what 

 M. Chasles says on the subject. I may add, that the substitution of (10.) 

 for the celebrated proposition of Pappus in (11.) is the joint work of Mr. 

 Davies and myself. On my communicating (10.), he returned me a de- 

 monstration of it by means of (11.), from which I saw that (11.) was really 

 a case of (10.) 



