170 



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



Apply the two triangles together, so that their right angles coincide, and their homony- 

 mous sides are in the same straight lines. In consequence of the equality of the remaining 

 angles, the hypothenuses EG, FH, will be parallel. Therefore the triangle FEG is equal 

 to the triangle HEG. To each add the triangle EDG, then the triangle FDG is equal to 

 the triangle EDH. But the rectangle under 6, C, is double of the triangle EDH ; and the 

 rectangle under c, B, is double of the triangle FDG. Therefore the rectangle under 6, C> 

 is equal to the rectangle under c, B. a.E.D. 



Pkoposition (F). If the rectangle contained under the lines a, B, is equal to the rect- 

 angle contained under the lines J, b ; the parallelogram contained under the lines a, B, will 

 be equal to the equiangular parallelogram contained under the lines J, b. 



[This is equivalent to the proposition, 



U a : b :: J : B, 

 Then a : b :: A.cosa : B.cosa.^ 



In the figure, produce the upper sides of the parallelograms to cut the vertical sides of 

 the rectangles in D and H. The rectangles DG, HL, are equal to the given parallelograms, 

 therefore it is to be proved that the rectangle DC? is equal to the rectangle HL, or that the 

 rectangle under a, EG^ is equal to the rectangle under b, IL. 



Since the parallelograms are equiangular, the right-angled triangles EGF, ILK, are equi- 

 angular ; and therefore by Proposition (C), the rectangle under EG, A, is equal to the 

 rectangle under IL, B. But by hypothesis, the rectangle under B, a, is equal to the rectangle 

 under A, b ; therefore by Proposition (B), the rectangle under EG, a, is equal to the rectangle 

 under IL, b. Or, the parallelogram under the lines B, a, is equal to the equiangular paral- 

 lelogram under the lines A, b. a.E.D. 



