790 PROCEEDINGS OF THE AMERICAN ACADEMY. 



geometry in one dimension. In fact, if a point xV is represented as a 

 linear function of three points, one of which is F, the multiple of F is 

 lost in the product 



AB =: (ABF). 



Thus, so far as concerns their values in products of the form AB, 

 points of the plane are linear functions of two points. In respect to 

 this multiplication the plane is then one dimensional. Thus any homo- 

 geneous identical relation between products AB along a line will hold 

 for distances AB in the plane. For example, along a line 



AB CD = AC BD - AD BC. 



Therefore 



AB-CD = AC-BD-A5-BC (48) 



where A, B, C, D are any four points in the plane. This can be proved 

 directly by using the equation 



(ABF) C - (ABC) F -f (AFC) B - (BFC) A = 



which is obtained from (43) by changing the notation. Multiplying 

 by DF and transposing we get 



(ABF) (CDF) = (ACF) (BDF) - (ADF) (BCF) 



which is equivalent to (48). 



Massachusetts Institute op Technology. 



