Bibliotheca Polyglotta

Let ABC be a circle, let the angle BEC be an angle at its centre, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base; I say that the angle BEC is double of the angle BAC.

IN circulo ABC, cuius centrum D, superbasin BC, consticuatur angulus BDC, ad centrum; et super eandem basin angulus BAC, ad peripheriam. Dico angulum BDC, duplum esse anguli BAC.

解曰。甲乙丙圜。其心丁。有乙丁丙分圜角。乙甲丙負圜角。同以乙丙圜分為底。題言乙丁丙角。倍大於乙甲丙角。

For let AE be joined and drawn through to F.

Includant enim primum duae AB, AC, duas DB, DC; et per centrum D, recta extendatur AE. Quoniam igitur rectae DA, DB, aequales sunt,¹ erunt anguli DAB, DBA, aequales: Est autem externus angulus BDE,² aequalis duobus angulis internis DAB, DBA. Quare BDE, plus erit alterius corum, ut anguli DAB. Eodem modo duplus ostendetur angulus CDE, anguli DAC. Quapropter totus BDC, duplus erit totius BAC. Quando enim duae magnitudines duarum sunt duplae, singulae singularum, est quoque aggregatum ex illis aggregati ex his duplum. Constat ergo propositum.

1. 5. primi. 2. 32. primi.

先論分圜角、在乙甲、甲丙、之內者曰。如上圖。試從甲、過丁心、作甲戊線。其甲丁乙角形之丁甲、丁乙等。卽丁甲乙、丁乙甲、兩角等。一卷四而乙丁戊外角。與內相對兩角幷等。一卷卅二卽乙丁戊、倍大於乙甲丁矣。依顯丙丁戊、亦倍大於丙甲丁。則乙丁丙全角。亦倍大於乙甲丙全角。

Then, since EA is equal to EB, the angle EAB is also equal to the angle EBA; [I. 5] therefore the angles EAB, EBA are double of the angle EAB. But the angle BEF is equal to the angles EAB, EBA; [I. 32] therefore the angle BEF is also double of the angle EAB. For the same reason the angle FEC is also double of the angle EAC. Therefore the whole angle BEC is double of the whole angle BAC.

DEINDE non includant rectae AB, AC, rectas DB, DC, sed AB, per centrum extendatur. Quoniam igitur externus angulus BDC,³ aequalis est duobus internis DAC, DCA. Hiautem duo⁴ inter se sunt aequales, quod latera DA, DC sint aequalia; erit angulus BDC, duplus alterius corum, nempe anguli BAC. Quod est propositum.

3. 32. primi. 4. 5. primi.

次論分圜角、不在乙甲、甲丙之內、而甲乙線過丁心者、曰。如下圖。依前論、推顯乙丁丙外角。等於內相對之丁甲丙、丁丙甲、兩角幷。一卷卅二而丁甲、丁丙、兩腰等。卽甲、丙、兩角亦等。一卷五則乙丁丙角。倍大於乙甲丙角。

Again let another straight line be inflected, and let there be another angle BDC; let DE be joined and produced to G. Similarly then we can prove that the angle GEC is double of the angle EDC, of which the angle GEB is double of the angle EDB; therefore the angle BEC which remains is double of the angle BDC.

TERTIO recta AB, secet rectam DC, et per centrum D, extendatur recta AE. Quoniam igitur angulus EDC, ad centrum, et angulus EAC, ad peripheriam, habent eandem basin EC, et recta AE, extenditur per centrum; erit angulus EDC, duplus anguli EAC, ut ostensum est in secunda parte. simili modo erit angulus EDB, duplus anguli EAB; habent enim hianguli eandem basin EB. Reliquusigitur angulus BDC, duplus erit reliqui anguli BAC.⁵ Quando enim totum totius est duplum, et ablatum ablati; est et reliquum reliqui duplum. In circulo igitur angulus ad centrum duplex est, etc.

5. 20. primi.

後論分圜角在負圜角線之外、而甲乙截丁丙者、曰。如下圖。試從甲過丁心、作甲戊線。其戊丁丙分圜角。與戊甲丙負圜角。同以戊乙丙圜分為底。如前次論戊丁丙角。倍大於戊甲丙角。依顯戊丁乙分圜角。亦倍大於戊甲乙負圜角。次於戊丁丙角、減戊丁乙角。戊甲丙角。減戊甲乙角。則所存乙丁丙角。必倍大於乙甲丙角。

Therefore etc. Q. E. D.

Quod erat demonstrandum.

增若乙丁、丁丙。不作角於心。或為半圜。或小於半圜。則丁心外餘地亦倍大於同底之負圜角。

論曰。試從甲過丁心、作甲戊線。卽丁心外餘地。分為乙丁戊、戊丁丙、兩角。依前論推顯此兩角。倍大於乙甲丁、丁甲丙、兩角。

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