lo SCIENCE ABSOLUTE OF SPAU . 



though BN II CP. But every ray BO (in CBN) 

 cuts ray CP; and so ray P&amp;gt;O cuts also ray AM. 

 Consequently BX AX. 



If BN II AM, and MAPj_MAB, and the 

 Z, which NBD makes with 

 NBA (on that side of MABX, 

 \ where MAP is) is &amp;lt;rt.Z; then 

 MAP and NBD intersect. 



For let ZBAM=rt.Z, and 

 AClBN (whether or not C 

 falls on B), and CEj_BN (in 

 NBD); by hypothesis ZACE 

 &amp;lt;rt.Z, and AF (iCE) will fall in ACE. 



Let ray AP be the intersection of the hemi- 



planes ABF, AMP (which have the point A 



common); &amp;gt; i nee BAM l MAP, ZBAP=ZBAM 



-rt.Z. 



If finally the hemi-plane ABF is placed upon 



the heini-plane ADM (A and 1&amp;gt; remaining), ray 

 AP will fall on ray AM; and since AC 1 1 iX. 

 and sect AF&amp;lt;sect AC, evidently sect AF will 

 terminate within ray BX, and &amp;gt;o HP falls in 

 ABX. lint in /his position, ray \\V cuts ray AP 

 (because BN II AMi: and so ray AP and ray BF 



. intersect also in the original position; and the 



point of section is common to the hemi-planes 

 MAP and NBD. Therefore tin- hemi-planes 

 MAP and XBD intersect. Hence follows eas- 



