Vignettes of Ancient Mathematics

(general diagram)

(diagram 1) For I say that if in a circle two unequal straight-lines are drawn, the larger to the smaller has a smaller ratio than the circular-arc on the larger straight-line to the one on the smaller.

(diagram 2) For let there be a circle, ABGD, and let two unequal lines be drawn on it with AB the smaller and BG the larger. I say that straight-line GB to straight-line BA has a smaller ratio than circular-arc BG to circular-arc BA. (diagram 3) For let angle ABG be bisected by BD, (diagram 4) and let AEG be joined as well as AD and GD.(diagram 5) And since angle ABG is bisected by straight-line BED, straight-line GD is equal AD [equal chords subtend equal angles], while GE is larger than EA.(diagram 6) Let there be drawn from D a perpendicular, DZ, to AE. (diagram 7) Then since AD is larger than ED, while ED is larger than DZ, therefore the circle inscribed with center D and distance DE cuts AD, and falls beyond DZ. Let there be inscribed [circle] HEQ, (diagram 8) and let DZQ be extended.

(diagram 9) And since section DEQ is larger than triangle DEZ, while triangle DEA is larger than section DEH, (diagram 10) therefore triangle DEZ to triangle DEA has a smaller ratio than section DEQ to DEH. (diagram 11) But as triangle DEZ is to triangle DEA, so is straight-line EZ to EA, (diagram 12) while as section DEQ is to section DEH, so is angle ZDE to EDA. (diagram 13) Therefore, straight-line ZE to EA has a smaller ratio angle ZDE to EDA. (diagram 14) Therefore, by composition straight-line ZA to EA also has a smaller ratio than angle ZDA to ADE. (diagram 15) The doubles of the leading terms, straight-line GA to AE has a smaller ratio than angle GDA to EDA. (diagram 16 = general diagram) And by division, straight-line GE to EA has a smaller ratio than angle GDE to EDA. But as straight-line GE is to EA, so is straight-line GB to BA, while as angle GDB is to BDA, so is circular-arc GB to BA. Therefore, straight-line GB to BA has a smaller ratio than circular-arc GB to circular-arc BA.

The theorem states the rule: chord GB : chord BA < arc GB : arc BA. If we keep in mind that a chord() = 2 radius Sin(/2), then rule is (anachronistically):
let a be the larger chord on angle , and b on angle
, with a > b, so that > . Then a : b < :
Chord() : Chord() < :
2 radius Sin(/2) : 2 radius Sin(/2) < :
Sin(/2) : Sin(/2) < :

Sin(/2) : Sin(/2) < /2 : /2

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