SCIENCE ABSOLUTE OF SPACE. 9 



BQ cuts ray AM fas BNHAM>. 

 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 

 same will fall upon hemi-plane BCP. 

 FIG. 7. 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 (inNBCP) 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= 

 c 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 ease 

 2 of 10. [Author's note.] 



