SEVENTY-TWO CONSECUTIVE PROPOSITIONS IN TRANSVERSALS. 



39 



the locus of the anapole of two points which should move away from each other 

 in a straight line with uniform velocities, I was delighted to receive demonstra- 

 tions, perfectly simple and elegant, to the effect that, in the former case, the ana- 

 pole moves in a straight line, let the planet move as it may ; and that, in the latter 

 case, the locus of the anapole is a conic section, and becomes a straight line if the 

 uniform velocities are equal ; and farther, that the anapole of any two points 

 in an ellipse circumscribing the triangle of reference is invariable. Geometers, I 

 am sure, will admire these theorems of a rising young mathematician, and will 

 recognise the vein thus struck as promising to be a fertile one. Mr Smith added 

 another very beautiful property of the anapole, which turned out, on investigation, 

 to be identical with Theorem LXV. of the following series. 



Instead of defining a point by the equations y=£-=^, we shall say the point 



is — (/, m, n). Instead of defining a straight line by the equation la+mj3 + ny—0, 

 we shall say the line is — (I, m, n). 



The straight line joining the points (l v m lt n^, (£ 2 , m 2 , n 2 ) is; — 



m 2 , n 2 



» 



n v l 2 



3 



l v m 1 



l 2 , m 2 



The intersection of the two straight lines (l v m ± , n t ), (l 2 , m 2 , n 2 ) is defined by 

 the same expression. This identity of form is, in reality, the earliest germ of the 

 doctrine of pole and polar ; and gives rise to what is usually regarded as the first 

 promise of that doctrine, namely, the identity of the condition that the three 

 points (l lt m v w x ), (l 2 , m 2J n 2 ), (l s , m 3 , n s ) shall range in a straight line, with the 

 condition that the three straight lines (Z 15 m x , ra x ), (l 2 , m 2 , n 2 ), (l s , m s , w 3 ) shall 

 intersect in a point ; which is, in both cases, 





= o. 



THEOREMS. 



Theorem 1. 



On the sides of the triangle A B C, as bases, are constructed three triangles, 

 A X B C, A BjC, A B C 15 similar to each other, and so placed that the angles 

 A 1 BC = AB 1 C = ABC 1 ; B,C A = B C^B C A, ; and C 1 AB = C A^C AB r 

 Then A A v B B x , C C x meet in a point. 



The sides of ABC, taken as the triangle of reference, being a, b, c, the per- 

 pendiculars from Aj on a, b, c are, respectively, 



a . sin B x . sin C x a . sin B x . sin (C + C x ) a . sin C t . sin (B + B 2 ) 



sin A, 



sin A, 



sin A, 



