178 



G. B. AIRY, ESQ., ON THE SUBSTITUTION OF METHODS, &c. 



parallelograms is double of the triangle BMF, and the latter is double of the triangle 

 MNF. Therefore the triangle BMF is equal to the triangle MNF, and therefore MF is 

 parallel to NB. 



Secondly. To prove that FI is parallel to NB. 



It will be shewn in exactly the same way that the parallelogram under AF, BI, with the 

 angle FAI, is equal to the parallelogram under AI, NF, with the angle FAI. But the 

 parallelogram under AF, BI, with the angle FAI, is the excess of the parallelogram under 

 AF, AI, with the angle FAI, above the parallelogram under AF, AB, with the same 

 angle; or is the excess of double the triangle AFI above double the triangle AFB, or is 

 double the triangle BFI. Similarly the parallelogram under AI, NF, with the angle FAI, 

 is double the triangle NFI. Therefore the triangles BFI, NFI, are equal; therefore FI is 

 parallel to NB. 



And as MF and FI are both parallel to NB, MF and FI are in the same straight line. 



a. E. D. 



' ADDENDUM. 



I AM permitted by Professor De Morgan to transcribe the simple process for demon- 

 strating the theorem of ex cequali in ordine perturhata, to which allusion is made above. 



li a : b ■.: B ; C, 



and b : c :: A : B, 



Then will a : c :: A : C. 



i 



then a 



A : C. 

 :: c : d. 



To exhibit the process more clearly to the eye, use the connecting mark /- — v for one 

 ratio and ^^i^ for the other; then the theorem stands thus, 



If a ^-N 6 ; 



and A^:=:::B^—^c, 



To prove it, take a fourth quantity d, such that a : b 



Then b ^p^:::: c ^— >. d. 

 But A^:=:::By-~>.C. 



Therefore, ex eequali, b : d :: A : C. 

 But, because a : b :: c : d, therefore alternando, a 

 fore the ratio a : c for 6 : d in the analogy just found, 



a : c :: A : C. q. k. d. 



c '.: b : d. Substituting there- 



RoTAL Observatoby, Greenwich. 

 September 2, 1867. 



G. B. AIRY. 



