Bibliotheca Polyglotta

For let the two circles ABC, ADE touch one another externally at the point A, and let the centre F of ABC, and the centre G of ADE, be taken I say that the straight line joined from F to G will pass through the point of contact at A.

CIRCVLI duo ABC, DBE, tangant se exterius in B, et centrum circuli ABC, sit F, circuli vero DBE, centrum sit G. Dico rectam extensam per F, et G, transire per contactum B.

解曰。甲乙丙、丁乙戊、兩圜。外相切於乙。其甲乙丙心為己。丁乙戊心為庚。題言作己庚直線。必過乙。

For suppose it does not, but, if possible, let it pass as FCDG, and let AF, AG be joined.

Si enim non transit,

論曰。如云不然。

Then, since the point F is the centre of the circle ABC, FA is equal to FC. Again, since the point G is the centre of the circle ADE, GA is equal to GD. But FA was also proved equal to FC; therefore FA, AG are equal to FC, GD, so that the whole FG is greater than FA, AG; but it is also less [I. 20]: which is impossible. Therefore the straight line joined from F to G will not fail to pass through the point of contact at A; therefore it will pass through it.

secet circunferentias in C, et E, ducanturque a centris F, G, ad B, contactum rectae FB, GB. Quoniam igitur in triangulo FBG, latera duo BF, BG, ¹ maiora sunt latere FG: Est autem recta BF, rectae FC, aequalis: (quod F, ponatur centrum circuli ABC,) et recta GB, rectae GE, aequalis; (quod G, ponatur centrum circuli DBE,) erunt et rectae FC, GE, maiores quam recta FG, pars quam totum, (cum FG, contineat praeter FC, GE, rectam adhuc CE,) quod est absurdum.

1. 20. primi.

而己庚線、截兩圜界於戊、於丙。令於切界作乙己、乙庚、兩線。其乙己庚角形之己乙、乙庚、兩邊幷。大於己庚一邊。而乙庚與庚戊。乙己、與己丙。俱同心所出線。宜各等。卽庚戊、丙己、兩線幷。亦大於庚己一線矣。一卷二十夫庚己線。分為庚戊、丙己。尚餘丙戊。而云庚戊、丙己。大於庚己。則分大於全也。故直線聯己庚。必過乙。

Therefore etc. Q. E. D.

Si igitur duo circuli sese exterius contingant, etc. Quod erat demonstrandum.

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