On Reduction, 2 of 11

Geometry of Reduction and Protraction

     Consider this diagram, the “fish-tail”:

           *     *

           |\   /|

           | \ / |

         x |  *  | y    

           | /|\ |

           |/ |z\|

           *--*--*

             a  b      

     The line segments x and y need not be equal; they and the z segment are assumed only to be parallel; so here is a theorem in affine geometry:

     Theorem:     x <+> y     =     z

     Proof:

     By similar triangles:

          

           x/(a+b)  =  z/b

           y/(a+b)  =  z/a

     Reducing equals with equals, we get:

           x/(a+b)  <+>  y/(a+b)  =   z/b  <+>  z/a

     Therefore:

           (x <+> y)/(a+b)  =   z/(b + a)

     The left side follows from the common denominators law, the right side follows from the common numerators law. Cancelling (a+b), we get:

           x <+> y   =   z

     QED. 

     From this it follows that  y = z <-> x   and   x = z <-> y .