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Scholion to Theodosius, Sphaerica iii 11 (p. 158.12-5),
(diagram 1) Let there be a right-angled triangle ABG, and let some line, AD, be drawn. To show that BG has a larger ratio to BD than angle ADB to angle AGB.
(diagram 2) For let DE be drawn through D parallel to AG. And since DE is larger than BD [since it subtends a larger angle; for it is right while D is acute; therefore ADG is obtuse--del. Tannery], (diagram 3) while AD is larger than ED, (diagram 4) therefore the circle inscribed with center D and distance DE cuts AD but falls beyond BD. Let it come as EQZ. (diagram 5 with pt H)
(diagram 6) Therefore, triangle AED has a larger ratio to section EDZ than triangle EBD to section EHD.
(diagram 7) And by alternation triangle AED has a larger ratio to triangle EBD than section EDZ to section EHD. (diagram 8) But as triangle AED is to triangle EBD, so is AE to BE, (diagram 9) while as section EDZ to section EHD, so is angle ZDE to angle EDB. (diagram 10) And by composition, AB has a larger ratio to BE than angle ZDH to angle EDB. (diagram 11) But angle EDB is equal to angle AGB since ED is parallel to one of the sides of triangle ABG, AG. (diagram 12) Therefore AB has a larger ratio to BE than angle ZDB to angle AGB. (diagram 13) Therefore, GB has a larger ratio to BD than angle ZDB to angle EDB.
(diagram 14) For ED cuts the sides proportionally, and it comes to be that AB is to BE so is GB to BD. (Euclid, Elements vi 2)
This theorem is one of two basic trigonometic theorems mentioned by Apollonius and Archimedes. Ptolemy, Almagest i 10, also has a proof of this theorem.
(diagram 15) The theorem is ADB
GB : BD > ADB
but it may be anachronistically rewritten
ADB= and AGB = ,
> Cot() : Cot() > :
> Tan() : Tan() > :