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Euclid: Elementa

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Click to Expand/Collapse OptionTitle
Click to Expand/Collapse OptionPreface
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Proposition 11. 
PROBL. 6. PROPOS. 11. 
第十一題 
To draw a straight line at right angles to a given straight line from a given point on it. 
DATA recta linea, a puncto in ea dato, rectam lineam ad angulos rectos excitare. 
一直線。任於一點上求作垂線。 
Let AB be the given straight line, and C the given point on it.  Thus it is required to draw from the point C a straight line at right angles to the straight line AB. 
RECTA linea data sit A B; & in ea punctum C,   a quo iubemur erigere super A B, lineam ad angulos rectos, seu perpendicularem. 
法曰:甲乙直線,任指一點於丙。 
Let a point D be taken at random on AC; let CE be made equal to CD; [I. 3]  on DE let the equilateral triangle FDE be constructed, [I. 1]  and let FC be joined;  I say that the straight line FC has been drawn at right angles to the given straight line AB from C the given point on it. 
A puncto C, sumatur recta C D, cui æqualis auferatur C E.   Deinde super D E, constituatur triangulum æquilaterum D E F,  atque ex F, ad C, ducatur recta F C,  quam dico esse perpendicularem ad A B.  
For, since DC is equal to CE, and CF is common,  the two sides DC, CF are equal to the two sides EC, CF respectively;  and the base DF is equal to the base FE;  therefore the angle DCF is equal to the angle ECF; [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; [Def. 10]  therefore each of the angles DCF, FCE is right. 
Quoniam latera D C, C F, trianguli D C F, æqualia sunt lateribus E C, C F, trianguli E C F, utrumque utrique, nempe D C, ipsi E C, per constructionem, & C F, commune;    Est vero & basis D F, basi E F, æqualis, ob triangulum æquilaterum:   Erunt anguli ad C, contenti dictis lateribus, æquales.       Quare dicetur uterque rectus, 
Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given point C on it.  Q. E. F. 
atque adeo F C, recta, ad A B, perpendicularis. Data igitur recta linea a puncto in ea dato, &c.   Quod faciendum erat.

PRAXIS
EX puncto C, abscindantur utrinque lineæ æquales C D, C E, & ex D, & E, describantur duo arcus secantes sese in F. Recta namque ducta F C, erit perpendicularis. Demonstratio eadem est, quæ Euclidis, si modo ducantur rectæ D F, E F, quæ æquales erunt, propter æquales circulos ex D, & E, descriptos, qui se intersecant in puncto F. Quod si punctum datum in linea recta fuerit extremum, producenda erit linea in rectum & continuum, ad partes puncti dati, ut ex illo erigatur secundum praxim datam linea perpendicularis. Ut si linea data fuerit A C, & punctum datum C, extremum; protrahenda erit A C, in B, & sumendæ æquales C D, C E, &c. Si vero ad aliquam lineam constituenda sit linea perpendicularis, non quidem in puncto assignato, sed utcunque, id efficietur hac methodo. Ex duobus punctis A, & B, quibuscunque lineæ propositæ describantur tam superne, quam inferne duo arcus sese intersecantes in C, & D. Nam recta ducta C D, erit perpendicularis ad A B, hoc est, faciet duos angulos ad E, rectos, seu æquales. Quod non aliter probabis, quam supra praxim, qua lineam in duas æquales diuisimus partes, demonstrauimus. Nam per 4. propos. erunt anguli ad E, æquales, quippe qui super æquales bases A E, B E, consistant, opponanturque æqualibus lateribus A C, B C, quæ ex C, ad puncta A, & B, ducerentur.

EX PROCLO
SI punctum in linea datum fuerit extremum, & linea commods produci nequiuerit, poterimus ex puncto dato educere lineam perpendicularem, linea non producta, hac ratione. Sit recta A B, & punctum A. Ex C, puncto quolibet intra lineam educatur perpendicularis C D, ut docuit Euclides; & abscindatur C E, æqualis ipsi A C: Deinde diuidatur angulus C, bifariam, ducta recta C F: Et ex E, rursus, ut docuit Euclides, educatur E G, perpendicularis ad C D, secans rectam C F, in G. Ducta enim recta G A, perpendicularis erit ad A B. Quoniam eum latera A C, C G, triangula A C G, æqualia sint lateribus E C, C G, trianguli E C G, utrumque utrique, & anguli hisce lateribus contenti æquales quoque, per constructionem: Erunt anguli A, & E, oppositi communi lateri C G, æquales; Sed E, est rectus per constructionem; igitur & A, rectus erit, ideoque A C, ad A B, perpendicularis.

SCHOLION
BREVIVS lineam perpendicularem erigemus ex puncto dato, sive extremum illud sit, sive non, hoc modo. Sit data linea A B, punctumque in ea A. Ex centro C, extra lineam assumpto, ubi libuerit (dummodo recta A B, producta cum ipso non conueniat) interuallo vero accepto usque ad A, describatur arcus circuli secans A B, in D. Et ex D, per C, recta ducatur secans arcum in E. Recta enim ducta E A, erit perpendicularis ad A B. Nam angulus A, est rectus, cum sit in semicirculo D A E, ut ostendemus propositione 31. lib. 3.
 
 
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