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But here DK alfo will be perpendicular to AB, 
and A B meeting the earth’s orbit in L and M, the 
2 redangle under K AG will be equal to the 
fquare of A M. But B G being to B K as 
A G to I, if B N be taken equal to I, B G will be to 
B K as A G to B N, and A B to KN alfo as A G to 
B N, and the redangle under NK, AG equal to 
that under A B and I : therefore the redangle under 
K A G being equal to the fquare of A M, N K will 
be to K A as the redangle under A B, I to the fquare 
of A M, that is, in a given ratio, and K D with the 
point D will be given in polition. 
Again, when C B is not perpendicular to L M, 
let D O be perpendicular to L M. Then the 
redangle under OAG will be equal to the fquare 
p io . 2 „ of AM. But BN being taken equal 
to I, as before, the redangle under N K, 
A G will be equal to that under AB, I; whence 
N K will be to AO in the given ratio of the 
redangle under AB, I to the fquare of AM, There- 
fore N P being taken to P A in that ratio, the point 
P will be given, and K P, the excefs of N P above 
N K, will be to P O, the excefs of A P above 
A O, in the fame ratio. Hence, as D K is parallel to 
C B and D O perpendicular to LM, the triangle K O D 
is given in fpecies, and if P D be drawn, the angle 
O P D will be given ; for the co-tangent of the angle 
O K D will be to the co-tangent of the angle O P D, 
as KO to OP, that is, as the redangle under A B, I 
together with the fquare of A M to the fquare of A M, 
and hence the point D is given by the line P D drawn 
from a given point P in a given angle APDj and if 
A D be drawn, A D will be to A P as the line of the 
angle A P D to the line of the angle PDA; this angle 
therefore 
