69 
of the Equinoxes . 
1. Let C, (Plate II, Fig, 1.) be the centre of two circular 
arcs AB, EF, which are the measures of the angles ACB, 
ECF ; and let CB cut EF in D. Then, as the sectors ACB, 
ECD are similar, CA : CE : : AB : ED = But (Eu. 33. 
VI. ) ED : EF : : the angle ACB :the angle ECF ; and therefore 
: EF : : < ACB : <ECF. Consequently CExABx<ECF 
= CA x EF x < ACB ; and therefore AB : EF : : CA x < ACB 
: CE x < ECF. 
2. Let ACB, GEF (Fig. 2.) be two angles, of which the 
arcs AB, GF are the measures ; and the radii CA, EG not 
being necessarily equal, let the sines BK, 10 be equal to one 
another. Let BH, FM be tangents to the curves ; and let 
HD, MN be parallel to CA, EG respectively, and meet BK, 
FQ in D, N, as represented. Then as CBH, BDH are right 
angles, the triangles CBK, BHD are equiangular, and CB . 
BK : : BH : HD or its equal KL, if HL be drawn parallel to 
BK, and meet CA in L. Consequently BK = For 
the same reasons, if MP be parallel to FQ and meet EG in 
P, FQ = ’ and therefore as BK, FQ are, by hypothesis, 
equal, 2 |^ = Et*S£. Hence CB x KL x FM = EF x QP 
x BH, and BH : FM : : CB x KL : EF x QP. 
3. If, therefore, we suppose straight lines CH, EM to be 
drawn, and that the angles BCH, FEM are indefinitely small, 
and generated in the same time by the revolution of CB, EF 
respectively, then BH, FM may be considered as circular 
arcs, and by article 1, BH : FM : : CB x < BCH : EF x < FEM. 
Hence by article 2, ( and 1 1. V. ) CB x < BCH : EF x < FEM 
: : CB x KL : EF x QP, and therefore < BCH : < FEM : : KL 
I 2 
