182 PROCEEDINGS OF THE [Sept., 1887. 



radius of the escribed circle which, at L touches the side BC, at 1 touches the 

 side AB prolonged, and at M touches the side AC prolonged. Also in this sec- 

 ond rectangle, AE is, by Prop. IV, the difference between the half -perimeter 

 and the side BC, which is tangent to the escribed cu-clo whose centre is K. 



13. Prop. X, If a rectangular parallelopipedon be formed by taking the 

 three sides of any triangle as the three edges of one of its soHd angles, and if a 

 right triangular prism be formed, having that triangle for its base, and having 

 for its height twice the diameter of circle circumscribing that triangle ; then 

 will the volumes of the parallelopipedon and of the prism be equal. 



By a well known theorem of Legendre's fourth Book, (American edition, ) the 

 area of the rectangle formed by any two sides of a triangle, as ABX AC, equals 

 that of the rectangle formed by the diameter of the circumscribing circle, and 

 the perpendicular from the vertex A, on the remaining side BC ; put AP for 

 this perpendicular, and AE for the diameter, then rectangle ABxAC=rectan- 

 gle ARXAP. 



If now we take rectangle AB X AC as the base of a rectangular parallelopipe- 

 don, and the remaining side BC as its height, we shall have the parallelopipedon 

 ABx ACxBC, given in enunciation above ; and if we take rectangle AEx AP as 

 the base of a right rectangular prism, and the side BC as its height also, we 

 shall have the prism ARxAPxBC, and the volume of the prism will equal 

 that of the parallelopipedon, since they have equal bases and equal heights. 



But the prism may also be regarded as having AB for its height, and the rect- 

 angle AP X BC for its base, and as this base is evidently double the area of the 

 triangle, it follows that if we form a right triangular prism, whose base is the 

 given triangle ABC, and its height equal to twice AR, its volume equals that of 

 the right rectangular prism, since its base is half that of the last and its height 

 is double that of the last. 



The triangular prism is therefore equal to the parallelopipedon, which was. to 

 be proved. 



14. The numerical results derivable from the preceding propositions may be 

 expressed algebraically thus : Let 



a, b, c, = the sides opposite to the angles A, B, C, respectively. 

 8=hfa+b+c) =:th.e half -perimeter. 

 ^=the area of the triangle, ABC. 

 r=the radius of the inscribed circle, 

 r', r", r'", =the radii of the three escribed circles, touching externally respect- 

 ively, the sides a, b, c. 

 Ii= the radius of the circumscribed circle. 

 P; P\ ^3",= perpendiculars from angles on sides a, b, c, respectively. 



Then, from what has been demonstrated, we can derive the following equa- 

 tions applicable to the numerical values of the symbols, in any given case : 



ti=s{s-a) (s-b) (s-c) (1; 



t=rs=r'is-a)=r"{s-b)=r"'(s-c)=^ (2) 



^t=pa=p'b=p"c (3) 



