(baker] the foundations OF GEOMETRY 121 



Every sect is congruent to itself. The sect A B is congruent to 

 the sect B A. 



The assumption means that a given sect can be taken on a given 

 straight line, and on a given side of a given point, in one and only 

 one way. 



(2). Sects that are congruent to the same sect are congruent to 

 one another. 



(3). If AB, BC be two sects on the same straight line and 

 without common points except B, and likewise A' B', B' C two sects 

 on the same straight line and without common points except B', if 

 A B E A' B' and B C E B' C, theu is A C E A' C. 



(4). There is in a plane, and on the same side of B' C, only- 

 one ray B' A' such that the angle A' B' C is congruent to the angle 

 ABC. Written < A' B' C'E< ABC. 



Every angle is congruent to itself. Also <;ABCE<CBA. 



This assumption means that in a given plane every angle can be laid 

 off towards a given side, against a given ray, in one and only one way. 



(5). Angles that are congruent to the same angle are congruent to 

 one another. 



(6). If A, B, C be three points, and A' B' C three other points, and if 

 A B E A' B, A C E A' C, and < B A C E < B' A' C, 

 then < ABCE> ABC, and< ACBE< A'C'B'. 



Here again, in (6), we are startled by the assumption which, I 

 think, no one would speak of as a self-evident truth, just as we were 

 ptartled by Pasch^s assumption, and in the long ago by Euclid's eleventh 

 axiom. It is no more apparent that < A B C E < A' B'C, than that 

 B C E B' C. However, we are justified in assuming what is necessary, 

 but not in assuming more than is necessary ; and with the assumption of 

 (6) we shall see that we are able to prove B C E B' C. 



Theorem P.— In two triangles A B C, A' B' C, if A B - A' B' 

 ACe A'C and < BACE< B'A'C, then is BCE B'C. 



>?A >/^ ï'or by (6) < A B C 



E < A' B' C, Take then 



B' C" congruent to B C. 



Then in triangles ABC, 



A'B'C" we have A B= A'B', 



B C E B' C", < A B C E 



< A B' C". Therefore by 



(6) < BAG E < B'A'C". 



3 c S' C C But by hypotlies'is < B A C 



E < B' A'C. Hence [III, (5)] < B' A' C E < B' A' C", which by III, 



(4), is impossible. Hence BCE B'C. 



