SCIENCE ABSOLUTE OF SPACE. 



FIG. 2. 



2. If BN II AM, we will have also CN II AM. 

 For take D anywhere in MACN. 

 If C is on ray BN, ray BD cuts 

 ray AM, since BN II AM, and so 

 also ray CD cuts ray AM. But 

 if C is on ray BP, take BQ II CD; 

 BQ falls within the Z ABN (1), 

 and cuts ray AM; and so also 

 ray CD cuts ray AM. Therefore 

 every ray CD (in ACN) cuts, in 

 p each case, the ray AM, without 



CN itself cutting ray AM. Therefore always 



CN II AM. 



3. (Fig. 2.) If BR and CS and each II AM, 

 and C is not on the ray BR, then ray BR and 

 ray CS do not intersect. For if ray BR and 

 ray CS had a common point D, then ( 2) DR 

 and DS would be each II AM, and ray DS ( 1) 

 would fall on ray DR, and C on the ray BR 

 (contrary to the hypothesis). 



4. If MAN>MAB, we will have for every 

 point B of ray AB, a point 

 p C of ray AM, such that 

 BCM=NAM. 



For (by 1) is granted 

 N BDM>NAM, and so that 

 FIG. 3. MDP-MAN, and B falls in 



