268 
TRIGONOMETRY 
But, by the propositions proved above, 
GK _ sin GE GL _ sin OF q DT _ sin DR 
KA ~ slh^T LD ~ sпГDZ} , and TA = ABBA ’ 
therefore, by substitution, we have 
sin CE _ sin OF sin JDB 
sin EA sin FD sin BA 
Menelaus apparently also gave the proof for the .cases in 
which AD, GB meet towards A, 0, and in which AD, GB are 
parallel respectively, and also proved that in like manner, in 
the above figure, 
sin CA _ sin CD sin FB 
sin AD sinDD sin BE 
(the triangle cut by the transversal being here GFE instead of 
ADC). 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' 
are either equal or supplementary, sin c/ sin a = sin e'/sin o! 
and conversely. The particular case in which G, G' 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 5 or ‘ shadow-rule ’ of the 
Arabs. If ABG, A'B'G' be triangles right-angled at A, A', and 
G, G' are equal and both either > or < 90°, and if P, P' be 
the poles of AG, A'C', then 
sin AB __ sin A'B' sin BP 
sin AG sin A'C' sin B'P' 
Apply the triangles so that C' falls on G, G'B' on GB as CE, 
and G'A' on GA as CD; then the result follows directly from 
III. 1. Since sin BP = cos AB, and sin B'P' = cos A'B', the 
result becomes 
sin GA tan AB 
sin G'A' = tan A'B' 3 
, which is the ‘ tangent-rule ’ of the Arabs. 3 
1 Ptolemy, Syntaxis, i. 18, vol. i, p. 76. 
2 See Braunmiihl, Gesch. cler Trig, i, pp. 17, 47, 58-60, 127-9. 
3 Cf. Braunnuihl, op. cit. i, pp. 17-18, 58, 67-9, &c.