1842.] On Equations of Condition for a Quadrilateral. 31 



This result might have been deduced from (E), by dividing any of 

 the equations by another as they stand : thus : — 



Sin B3 • sin Ci • s in Di _ sin D3 * sin C2 ■ sin B2 and rejecting 

 SinC3 " sinDi * sin Ai sin A3 * sin D2 • sin C2 common factors. 

 Sin B3 sin Ci sin D3 sin B2 

 Sin C3 sin Ai sin A3 sin D2 

 .*. Sin A3 • sin B3 * sin Ci * sin D2 = sin C3 * sin D3 * sin Ai * sin B2 

 Again multiplying vertically all the equations, (E) rejecting common 

 factors, and taking the square root, we have 



Sin Ai * sin Bi • Ci sin Di = sin A2 ; sin B2 ' sin C2 * sin D2 ..(g) 

 In like manner in the 2d figure * we have the following equations : 

 Sin A3 : sin Bi * sin D3 = sin B3 * sin A2 ' sin D2 } Which may 



(H) Sin B 3 • sin Ci ■ sin Di = sin C 3 ' sin B 2 • sin D3 V be . te 1 rme i d 

 7 . . (external al- 



Sm C3 * sin Ai * sin D 2 === sin A3 * sin C2 " sin Di 1 ternate. 



Also 



Sin A3 ' sin Bi * sin Ci * sin Di — sin C3 * sin B2 * sin A2 ' sin D2 



(I) Sin B3 ' sin Ci * sin Ai • sin D2 = sin A3 * sin C2 * sin B2 * sin D3 



Sin C3 • sin Ai • sin Bi • sin D3 — sin B3 • sin A * sin C2 ' sin Di 



which may be termed medial alternate. 



And 



(K) Sin Ai * sin Bi * sin Ci = sin A2 * sin B2 ' sin C2 which may be 



termed internal alternate. 



Demonstration. 



sin U3 



B C = A C £** = CD S 4^?V A C ^t • S 4^1 



sin B3 sin sm d± sin D<2 sin Bi 



sin A3 _ sin A2 sin D2 

 sin B 3 sin D3 sin Bi 



.*. Sin A3 sin Bi sin D3 = sin B3 sin A2 sin D2 fhj 



Multiplying vertically pairs of (H), and rejecting common factors, we 



have the equations, (i) 



and multiplying vertically the whole of (H) in like manner, we have the 



equation, fkj 



I found at first some difficulty in trying to express in a convenient 

 form of words, the properties in the equations (H) and (I) ; but after 



