248 



ELEMENT AR* DEMONSTRATION, &C. 



S. Let BAD be an angle in- FIG. 4? 



commensurable with a right 

 angle. The proposition is true 

 in relation to BAC, the multiple 

 of (a) next less than BAD, and 

 of BAE, the multiple of (a) next u 

 higher, the difference between 

 which (zn a) may be less than any 

 assigned angle. But the equivalent of AB and BD must, in 

 respect of direction, be always intermediate between the equi- 

 valent of AB, BC, and that of AB, BE. It must, therefore, 

 pass through D ; and this is evidently true in the case of any 

 similar rectangle ; that is, so long as the angle BAD remains 

 the same. Hence it must be represented by AD, also in quan- 

 tity {vid. Lemma). 



From what is said above in § 2. it is manifest that what has 

 now been proved universally of the rectangle, may be extend- 

 ed to the oblique-angled parallelogram. 



XIII. 



