SCIENCE ABSOLUTE OF SPACE. 



FIG. 



BQ cuts ray AM (as BNllAMu 

 Therefore in revolving the hemi-plane 

 BCD around BC until it begins to 

 leave the ray AM, the hemi-plane 

 BCD at last will fall upon the hemi- 

 plane BCN. For the same reason this 

 c same will fall upon hemi-plane BCP. 

 Therefore BN falls in BCP. More 

 over, if BR II CP; then (because also AM II CP) 

 by like reasoning, BR falls in BAM, and also 

 (since BRIICP) in BCP. Therefore the 

 straight BR, being common to the two planes 

 MAB, PCB, of course is the straight BN, and 

 hence BN II CP.* 



If. therefore CP II AM, and B exterior to the 

 plane CAM; then the intersection BN of the 

 planes BAM, BCP is II as well to AM as to CP. 

 8. If BN II and ^ CP (or more briefly BN 

 II =^CP^, and AM (in NBCP) bisects 

 J_ the sect BC; then BN II AM. 



For if ray BN cut ray AM, also 

 ray CP would cut ray AM at the 

 same point (because MABN^ 

 MACP), and this would be common 

 to the rays BN, CP themselves, al- 



* The third case being put before the other two, these can be 

 demonstrated together with more brevity and elegance, like case 

 2 of 10. [Author s note.J 



