Return to Vignettes of Ancient Mathematics
Scholion 16 to Aristarchus, On the Sizes and Distances of the Sun and the
Moon, prop.4, ed. by Marquis de Fortia d'Urbain (Paris, 1810), pp. 120-23
(prop. 5 in this edition). I have altered the punctuation in several places.
The diagram may be reversed (the translator's xerox did not include the diagrams,
for which he apologizes).
(diagram 1) Let there be a triangle, ABG, with AB smaller than AG. I say that AB has a larger ratio to AG than angle G to angle B.
(diagram 2) Let an altitude, AD be drawn, and AQ parallel to BD. (diagram 3) And since AG is larger than AB, i.e., the square of AG than the square of AB, i.e., the squares of AD, DG than the squares of AD, DB. Therefore, by subtraction of what's common, the square of AD, line DG is larger than BD. (diagram 4) Let there be placed an equal, DE, to BD [DG > BD E is on DG]; and so let AE be joined. Therefore, AE is equal to AB. (diagram 5) And with center A and distance AE, let there be inscribed a circle, EZH. (diagram 6) And let EZ be joined and (diagram 7) let it be extended [to Q]. For it falls onto AQ, since it is also on its parallel BG [the sense is that if a line intersects one parallel, it will intersect the other when extended]
(diagram 8) And so, since triangle QAZ again has a larger ratio to section HAZ than triangle ZAE has to section ZAE (diagram 9 and diagram 10) (and alternately both the base and angle [an odd expression perhaps, but the sense is that triangles with equal height are as their bases, and the angles as the sections]), (diagram 11 and diagram 12) therefore QZ to ZE, i.e., AZ to ZG (diagram 13) (due to the similarity of triangles QAZ, ZEG [AQ||GE, so that the alternate angles are equal]), has a larger ratio than angle QAZ to angle ZAE. (diagram 14) By composition and conversion, and yet again, AZ has a larger ratio to AG than angle HAZ to HAE. (diagram 15) But AZ is equal to AE, i.e., to AB, (diagram 16) while angle HAG is equal to AGE, since they are alternate angles, (diagram 17) and angle EAH is equal to AED [also alternate angles], i.e. to ABE. For AE is equal to AB [forming isosceles triangle ABE]. (diagram 18) Therefore, BA has a larger ratio to AG than angle G to B.
A note on point Q. According to the order of the letters, Q is the last point constructed, but the author needs the parallel to BG, i.e., AQ, to demarcate point H and to mark the extension of EZ.
This is one of the basic trigonometric theorems. Anachronistically,
the theorem proves,
> Csc() : Csc() > :
Sin() : Sin() > :
: > Sin() : Sin()