Bibliotheca Polyglotta

Let AB be the given infinite straight line, and C the given point which is not on it; thus it is required to draw to the given infinite straight line AB, from the given point C which is not on it, a perpendicular straight line.

Sit recta A B, interminatæ quantitatis, & extra ipsam punctum C, a quo oporteat lineam perpendicularem deducere ad rectam A B.

For let a point D be taken at random on the other side of the straight line AB, and with centre C and distance CD let the circle EFG be described; [Post. 3] let the straight line EG be bisected at H, [I. 10] and let the straight lines CG, CH, CE be joined. [Post. 1] I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

¹ Centro C, interuallo vero quolibet circulus describatur secans A B, in D, & E. (quoniam interuallum assumptum tantum esse debet, ut transcendat rectam A B; alias eam non secaret.) Diuisa autem recta D E, bifariam in F, ducatur recta C F, quam dico perpendicularem esse ad A B.

1. Clavius ommitts the point on the opposite side of the line, and instead utilizes a random point on the line itself. ()

. . . . .

For, since GH is equal to HE, and HC is common, the two sides GH, HC are equal to the two sides EH, HC respectively; and the base CG is equal to the base CE; therefore the angle CHG is equal to the angle EHC. [I. 8] And they are adjacent angles. 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, and the straight line standing on the other is called a perpendicular to that on which it stands. [Def. 10]

Si enim ducantur C D, C E, erunt duo latera D F, F C, trianguli D F C, æqualia duo bus lateribus E F, F C, trianguli E F C, utrumque utrique, per constructionem; est autem & basis C D, basi C E, æqualis, cum hæ sint ex centro C, ad circumferentiam. Quare erit angulus D F C, angulo E F C, æqualis, & propterea uterque rectus.

. . . . . . .

Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it. Q. E. F.

Ducta est igitur C F, perpendicularis, quod faciendum erat.

SCHOLION
PROBE apposuit Euclides hanc particulam: infinitam. Si enim linea esset infinita, non posset semper a puncto dato extra ipsam perpendicularis ad eam deduci. Ut si linea finita esset B E, & punctum C, non posset ex C, describi circulus secans B E, in duobus punctis, quare neque ex C, perpendicularis duci ad B E. Hac igitur de causa vult Euclides, rectam datam esse infinitam, hoc est, non habere magnitudinem determinatam, ut saltem ad ipsam productam perpendicularis possit deduci. Ita enim fiet hic, si B E, producatur, donec circulus ex C, descriptus secet totam B A, productam in D, & E, &c.

PRAXIS
CENTRO facto C, & interuallo quovis eodem, describantur duo arcus secantes rectam datam in A, & B. Deinde ex A, & B, eodemque interuallo, vel alio, si placuerit, alii duo arcus describantur secantes se in D. Nam ducta recta C D secans A B, in E, erit perpendicularis ad A B. Demonstratio huius operationis non differt a demonstratione tradita in praxi propositionis 10. Nam anguli ad E, erunt recti, nempe inter se æquales.
IDEM officiemus hoc modo. Ex quovis puncto A, in linea data, & interuallo quolibet usque ad C, assumpto, arcus circuli describatur: Deinde ex quolibet alio puncto B, interualloque usque ad idem C, alius arcus describatur priorem secans in C, & D; Eritque recta C D, secans A E, in E, perpendicularis ad A B. Demonstratio eadem est, quæ prior. Non est autem necesse, ut intervallum B C, æquale sit intervallo A C, ut in hac figura apparet. Facilior tamen erit, & breuior operatio, si idem semper intervallum accipiatur.

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