68 PROCEEDINGS OF THE AMERICAN ACADEMY. 



where k is a function of A and s. This gives on applying formulae 

 (11) and (18) after replacing s by [B C] 



{F A B C) 



ABC = k 



(0 A) (0 B) {<!> C) 



The areas of two triangles having the same vertex and base line are 

 then proportional to the quantities 



{FABC) 



{4> A) ( (0 B) (4> C) 



By a series of operations consisting of moving one side of the 

 triangle along its line and keeping the opposite vertex fixed we can 

 move the triangle into coincidence with any other having the same 

 area. Under each of these operations the area is proportional to the 

 above quantity. Hence any two areas are to each othei as those 

 quantities. Then by a proper choice of unit we have 



{FABC) 



^^^-(0.4)(0^)(0C) • • • ^-'^ 

 Writing F = (0 p) we have 



{<f>p-ABC) _ (0.1) (pBC) + {4>B) (p CA) + (0C) (pAB) 



ABC = 



(0.4) (05) (0C) ■ (0 yl) (0 5) (0 C) 



= BC-i-CA + AB (22) 



Thus the line area of a triangle is equal to its perimeter. 



Dually we can find as the trihedral angle between three planes 



a, t^, 7, 



^— ^ (0 g /^ 7) 



""'^ {F a) {F f:l) (F y) 



^aT + z^T + Ya . . . (23) 



19. Volume of a tetrahedron. Similarly we define the volume 

 of a tetrahedron A B C D as such a function A B C D of the four 

 points that given the vertex and plane of the base, the volume is 

 proportional to the area of the base. From the definition we have 



ABCD = IcBC D-Aa 



k {F BCD)- {A a) _ , {ABCD) 



— A* 



(0 B) (0 C) (0 D) (0 A) {F a) (0 .4) (0 B) (0 C) (0 D)' 



