A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

MENELAUS'S SPIIAERICA 267
For, if AM, BN are perpendicular to OC, we have, as before,
AT:TB = AM:BN
= !(crd. 2^6'):i(crd. 2BC)
= sin.J_C':sini?G Y .
Now let the arcs of great circles ADB, AEC be cut by the
arcs of great circles DFC, 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 nieet in T.
Draw the straight lines ARC, DLC 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 AGD. 1
Thus we have two straight lines AC, AT cut by the two
straight lines CD, TK which themselves intersect in L.
1
Therefore, by Menelaus's proposition in plane geometry,
CK CL DT
KA~ LD'TA
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.
268 TRIGONOMETRY
But, by the propositions proved above,
GK sin GE GL sin GF DT _ sin DB
KA ~ sin EA' LD ~~ sin FD* YA ~ sJn~BA '
therefore, by substitution, we have
sin CE __ sin (LP sin DB
sin EA " sin FD ' sin BA "
Menelaus apparently also gave the proof for the cases in
which J.Z),
(ri? meet towards A, G, and in which AD, GB are
parallel respectively, and also proved that in like manner, in
the above figure,
sin GA sin CD sin FB
sin AE sin DF sin BE
(the triangle cut by the transversal being here CFE instead of
ADG). Ptolemy 1 gives the proof of the above case only, and
dismisses the last-mentioned result with a similarly '
'.
(/3) Deductions from Menelaus s Theorem.
III. 2 proves, by means of I. 14, 10 and III. 1, that, if ABC,
A'B'G' be two spherical
triangles in which A — A', and G, G f
are either equal or supplementary, sin c/sin a = sin c'/sin a'
and conversely. The particular case in which G, (7 are right
angles gives what was afterwards known as the regula
'
quattuor quantitatum ' and was fundamental in Arabian
trigonometry. 2 A similar association attaches to the result of
III. 3, which is the so-called tangent ' ' or shadow-rule of the
' '
Arabs/ If ABC, A f B'G' be triangles right-angled at A, A', and
G, G f are equal and both either > or < 90°, and if P, P f be
the poles of AG, A'C, then
sin AB _ sinA'B' sin BP
sin AG ~ sin A'G' '
sin B'P' '
Apply the triangles so that G' falls on C, C'B' on GB as GE,
then the result follows directly from
and C A' on GA as GD
;
III. 1.
result becomes
Since sin BP — cos AB, and sin B'P' = cos A'B\ the
sin GA
tan AB
sin C'A' " ta^rZ 7^ 5
which is the ' tangent-rule ' of the Arabs. 3
1
Ptolemy, Syntax-is, i. 13, vol. i, p. 76.
2
See Braunmuhl, Gesch. der Trig, i, pp. 17, 47, 58-60, 127-9.
3
Cf. Braunmuhl, op. cit. i, pp. 17-18, 58, 67-9, &c.