[bakek] the foundations OF GEOMETRY 119 



Then [II. (4)] A C, through E, must have a point in common with 

 D B or F B. That is, A C meets B C or B A in two points, which is 

 impossible [Thm. 1]. 



And now we reach a theorem from which we 

 first realize that a line has "sides," and that it 

 divides a plane in the way in which in Euclid, 

 without hesitation, we assume, it divides the plane. 

 Theorem 7. — Every straight line a in a plane 

 divides the plane into two regions such that all 

 points P, P', . . . of one region determine sects 

 that have no points in common with a; and all points Q, Q', ... of the 

 other region determine sects that have no point in common with a ; but 

 every one of the points P, P', . . . of one region determines with 

 every one of the points Q, Q', . . . of the other region a sect that has a 

 point in common with a. 



Let A be a point not in a ; and let 

 P, P' . . . be points such that the sects 

 A P, A P', . . . have no point in com- 

 mon with a; also let Q, Q', . . . be 

 points such that A Q, A Q', . . . have 

 points in common with a. Then if 

 a had a point in common with P P', 

 it would have a point in common 

 with A P or A P' [II., (4)], which by 

 hypothesis it has not. Hence P P' 

 has no point in common with a. Again, since A Q, A Q' have points in 

 common with a, therefore, Q Q' has no point in common with a [Thm. 6], 

 Also since A Q has a point in common with a, and A P no point in com- 

 mon with a ; therefore, P Q has a point in common with a [[I., (4)]. 



It is here assumed that A being talcen not in a, there are points 

 giving sects with points in common with a, and others not having 

 This can readily be proved: 



For B C [I, (3)] being points on a, 

 there exists a point P in B A 

 such that A lies between P and B 

 [II, (2)] ; and, therefore, A P has no 

 point in common with a. Again, 

 between B, C there exists on a a 

 point D [II, (2)] ; and on the straight 

 line determined by A, D there exists 

 a point Q such that D lies between A and Q [II, (2)]. Hence A Q has a 

 point in common with a. 



such common points. 



7>y 



