Bibliotheca Polyglotta

If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at right angles; and if it cut it at right angles, it also bisects it.

SI in circulo recta quaedam linea per centrum extensa quandam non per centrum extensam bifariam secet; et ad angulos rectos ipsam secabit. Et si ad angulos rectos eam secet, bifariam quoque eam secabit.

直線過圜心。分他直線為兩平分。其分處必為兩直角。為兩直角。必兩平分。

Let ABC be a circle, and in it let a straight line CD through the centre bisect a straight line AB not through the centre at the point F; I say that it also cuts it at right angles.

PER A, centrum circuli BCD, recta CE, extensa dividat rectam BD, non per centrum extensam, bifariam in F. Dico rectam AF, esse ad angulos rectos ipsi BD.

解曰。乙丙丁圜。有丙戊線。過甲心。分乙丁線為兩平分於己。題言甲己必是垂線。而己旁為兩直角。又言己旁旣為兩直角。則甲己分乙丁。必兩平分。

For let the centre of the circle ABC be taken, and let it be E; let EA, EB be joined.

Ductis enim rectis AB, AD,

先論曰。試從甲作甲乙、甲丁、兩線。

Then, since AF is equal to FB, and FE is common, two sides are equal to two sides; and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE. [I. 8] But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10] therefore each of the angles AFE, BFE is right. Therefore CD, which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles.

erunt duo latera AF, FB, trianguli AFB, duobus AF, FD, trianguli AFD, aequalia; et bases AB, AD, aequales. Igitur anguli AFB, AFD, aequales erunt,¹ hoc est, recti. Quod erat primo propositum. SIT iam AF, ad angulos rectos ipsi BD. Dico rectam BD, bifariam secari in F, a recta CE.

1. 8. primi.

卽甲乙己角形之乙己。與甲丁己角形之丁己。兩邊等。甲己同邊。甲乙、甲丁、兩線、俱從心至界。又等。卽兩形等。則其對等邊之甲己乙、甲己丁。亦等。一卷八而為兩直角矣。

Again, let CD cut AB at right angles; I say that it also bisects it, that is, that AF is equal to FB.

For, with the same construction, since EA is equal to EB, the angle EAF is also equal to the angle EBF. [I. 5] But the right angle AFE is equal to the right angle BFE, therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore AF is equal to FB.

Ductis enim iterum rectis AB, AD; cum latera AB, AD, trianguli ABD, sint aequalia, erunt anguli ABD, ADB, aequales.² Quoniam igitur duo anguli AFB, ABF, trianguli ABF, aequales sunt duobus angulis AFD, ADF, trianguli ADF; et latera AB, AD, quae rectis angulis aequalibus opponuntur, aequalia quoque: erunt latera FB, FD, aequalia. Quod secundo proponebatur.³

2. 5. primi. 3. 26. primi.

後論曰。如前作甲乙、甲丁、兩線。甲乙丁角形之甲乙、甲丁、兩邊旣等。則甲乙丁、甲丁乙、兩角亦等。一卷五又甲乙己角形之甲己乙、甲乙己、兩角。與甲丁己角形之甲己丁、甲丁己、兩角。各等。而對直角之甲乙、甲丁、兩邊又等。則己乙、己丁、兩邊亦等。一卷廿六

Therefore etc. Q. E. D.

Si igitur in circulo recta quaedam linea per centrum extensa, etc. Si igitur in circulo recta quaedam linea per centrum extensa, etc. Quod demonstrandum erat.

FACILE quoque demonstrari poterat secunda haec pars; quae quidem conuersa est primae partis, hac ratione. Si enim AF, perpendicularis est ad BD, erit tam quadratum rectae AB, aequale quadratis rectarum AF, FB, quam quadratum rectae AD, quadratis rectarum AF, FD.⁴ Cum igitur quadratum rectae AB, aequale sit quadrato rectae AD; erunt et quadrata rectarum AF, FB, aequalia quadratis rectarum AF, FD. Quare dempto communi quadrato rectae AF, remanebunt quadrata rectarum FB, FD, aequalia; atque idcirco et rectae FB, FD, aequales erunt.

4. 47. primi.

欲顯次論之旨。

又有一說。如甲丁上直角方形。與甲己、己丁、上兩直角方形幷、等。一卷四七而甲乙上直角方形。與甲己、乙己、上兩直角方形、幷。亦等、卽甲己、己乙、上兩直角方形、幷。與甲己。己丁、上兩直角方形幷、亦等。此二率者。每減一甲己上直角方形。則所存乙己、己丁、上兩直角方形、自相等。而兩邊亦等。

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