MENELAUS’S SPHAERICA 
267 
For, if AM, BN are perpendicular to OC, we have, as before, 
AT: TB = AM: BN 
= |(crd. 2 AG):\ (crd. 256") 
= sin AC': sin 
Now let the arcs of great circles APB, A EC be cut by the 
arcs of great circles DEC, BFE which themselves meet in F. 
Let G be the centre of the sphere and Join GB, GF, GE, AD. 
Then the straight lines AD, GB, being in one plane, are 
either parallel or not parallel. If they are not parallel, they 
will meet either in the direction of D, B or of A, G. 
Let AD, GB meet in T. 
Draw the straight lines AKG, DLG meeting GE, GF in K, L 
respectively. 
Then K, L, T must lie on a straight line, namely the straight 
line which is the section of the planes determined by the arc 
EFB and by the triangle A CD. 1 
A 
Thus we have two straight lines AC, AT cut by the two 
straight lines CD, TK which themselves intersect in L. 
Therefore, by Menelaus’s proposition in plane geometry, 
CK _ CL DT 
KA " LD'TA 
1 So Ptolemy. In other words, since the straight lines GB, GE, GF, 
•which are in one plane, respectively intersect the straight lines AD, AC, 
CD which are also in one plane, the points of intersection T, K, L are in 
both planes, and therefore lie on the straight line in which the planes 
intersect.