168 Prof. De Morgan on Anharmonic Ratio. 



less is entirely within, or entirely without, the greater. Hence, 

 Pbeingany point, CPDand C'PD' cannot both be right angles; 

 from which, by a very easy reductio ad absurdum, it follows 

 that if CPD be a right angle, PC bisects the angle APB. For 

 an easy* mode of finding any number of pairs of conjugates, 

 with a given line AB, proceed as follows. Take BAP iso- 

 sceles at A, and bisect BP in Q. Having a point C in AB, 

 take in AP AC'= AC, and draw C'Q meeting AB in D : then 

 C and D are conjugate. 



(7.) If two triangles, ABC, A'B'C, have their bases, AB 

 A'B', in the same line, which cuts CC in Z, the areas are in 

 the ratio compounded of the bases AB, A'B', and the segments 

 of the line joining the vertices, CZ, C'Z. Join BC : then 

 CAB: C'A'B' is (CAB: CZB)(CZB : ZBC')(ZBC': C'A'B'), 

 or (AB:ZB)(ZC:ZC')(ZB:A'B'), or (AB : AB')(CZ: C'Z). 

 If the bases be equal, the areas are as CZ:C'Z. Hence 

 the fundamental proposition on transversals is most easily 

 proved. If the sides of ABC be cut by a line cutting BC in 

 A', CA in B', AB in C, then the number of external sectious 

 is odd, and (AC: C'B)(BA': A'C)(CB': B'A)=1 : 1. Join 

 A A', CC by vi. 1, and (7 ) : the compound just mentioned is 

 (AA'C : BA'C) (BA'C : CA'C) (CA'C : AA'C) or 1:1. 

 Many writers prove the converse incorrectly, both in this pro- 

 position and others. If there be A' B' Con the three sides, 

 with an odd number external, and (AC: CB)(BA': A'C) 

 (CB': B'A)=1 : 1, then A', B', C are in one line. If not, 

 let A'B' meet AB in C" : then C" may be written for Cin the 

 hypothesis, whence AC : C'B : : AC" : C" B, and either C and 

 C" coincide, or they are conjugate points in AB. In the last 

 alternative C" is internal or external, according as C is ex- 

 ternal or internal : but A', B', C have by hypothesis, an odd 

 number of externals, therefore A', B', C" (in a straight line) 

 have an even number or none, which is absurd. 



(8.) If O be any point, and OA, OB, OC, cut BC, CA, AB 

 in A', B', C, then an odd number of sides is cut interrially, 

 (AC':C'B)fBA': A'C)(CB' : B'A) = 1 : 1. This compound, 

 by (7.), is (AOC : COB) (BOA : AOC)(COB : BOA), or 1 : I. 

 Treat the converse as before. 



(9.) Let A'B'C be a triangle having its vertices on the 

 sides BC, CA, AB of the triangle ABC, and let A'B', B'C, 

 CA' cut AB, BC, CA, in C", B", A"; so that BC is cut in 

 A' and A", &c. Then the division is internal and external in 

 all, or in none. Let A'B'C be called an inscript of ABC, 

 and ABC a descript of A'B'C. 



* Even in drawing without ruler and compasses, this method will be 

 found a useful assistant to the eye. 



