SCIENCE ABSOLUTE OP SPACE. 



7 



NADP. If therefore NAM is carried along 

 AM until ray AN arrives on ray DP, ray 

 AN will somewhere have necessarily passed 

 through B, and some BCM=NAM. 



5. If BN II AM, there is on the straight 

 iN AM a point F such that FM^=BN. 

 For by 1 is granted BCM&amp;gt;CBN; 

 and if CE=CB, and so EOBC; 

 evidently BEM&amp;lt;EBN. The point 

 P is moved on EC, the angle BPM 

 always being called u, and the an 

 gle PBN always v; evidently u is 

 at first less than the corresponding 

 v, but afterwards greater. Indeed 

 u increases continuously from 

 BEM to BCM; since (by 4) there 

 exists no angle &amp;gt;BEM and &amp;lt;BCM, to which 

 u does not at some time become equal. Like 

 wise v decreases continuously from EBN to 

 CBN. There is therefore on EC a point F 

 such that BFM=FBN. 



6. If BNIIAM ;md E anywhere in the 

 straight AM, and G in the straight BN; then 

 GN II EM and EM II GN. For (by 1) BN II EM, 

 whence (by 2) GN II EM. If moreover FM^ 

 BN (5); then MFBN^NBFM, and conse 

 quently (since BN II FM) also FM II BN, and 

 (by what precedes) EM II GN. 



FIG. 4. 



