[ 9 2 4 ] 
Now, in the 
fpherical triangle 
ABC, through 
B, and to the 
poles A, C, let 
the femicircles 
EBF, GBD be 
defcribed , and 
the arch A C 
completed to a 
circle. Then the 
femicircles EBF, 
GBD will be 
both perpendi- 
cular to the plane of this circle, cut it in their 
diameters E F, G D, and their common interfedion 
B H be alfo perpendicular to it, and confequently 
perpendicular to G D, the diameter of the femicircle 
GBD, whereby G B, D B being joined, G D x G H 
is equal to G B q , and GDxDH equal to BD’. 
Again, A K being drawn parallel to G D, the arch 
ACK is double AC, GCD double CB, EAD 
the fum of all the three fides of the triangle, EG:- 
ED~DG(2CB),GF = AF(AE)4-CG(CD)-AC 
= E D — 2 A C, and FD = ED — EF (2 A B). 
Then EG, GF, FD, ED being joined; in the 
firft place, AKistoDGasAKxGH toDGxGH, 
alfo AK to D G as AK x DG to D G % but AK x G H 
= EG x GF by the preceding lemma, and DG x GH 
— G B q ; whereby AK x DG is to D G q asEGxGF 
toGBl Hence A K being twice the fine of the arch 
AC, D G twice the fine of the arch G C, equal to 
B C, EG twice the fine of half the arch E A G, G F 
* • 
twice 
