If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle.
SI in circuli peripheria duo quælibet puncta accepta fuerint; Recta linea, quæ ad ipsa puncta adiungitur, intra circulum cadet.
圜界、任取二點。以直線相聯。則直線全在圜內。
Let ABC be a circle, and let two points A, B be taken at random on its circumference;
I say that the straight line joined from A to B will fall within the circle.
IN circulo ABC, sumantur quaelibet duo puncta A, et C, in eius circumferentia.
Dico rectam ex A, in C, ductam cadere intra circulum, ita ut ipsum secet.
解曰。甲乙丙圜界上。任取甲、丙、二點。
作直線相聯。題言甲丙線、全在圜內。
For suppose it does not, but, if possible, let it fall outside, as AEB;
let the centre of the circle ABC be taken [III. 1], and let it be D;
let DA, DB be joined, and let DFE be drawn through.
Si enim non cadit intra, cadat extra, qualis est linea ADC, recta.
Invento igitur centro E,1
ducantur ab eo ad puncta assumpta A, et C, nec non ad quodvis punctum D, in recta ADC, lineae rectae EA, EC, ED, fecetque ED, circumferentiam in B.
Then, since DA is equal to DB, the angle DAE is also equal to the angle DBE. [I. 5]
And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE. [I. 16]
But the angle DAE is equal to the angle DBE;
therefore the angle DEB is greater than the angle DBE.
And the greater angle is subtended by the greater side; [I. 19]
therefore DB is greater than DE.
But DB is equal to DF;
therefore DF is greater than DE, the less than the greater: which is impossible.
Therefore the straight line joined from A to B will not fall outside the circle.
Similarly we can prove that neither will it fall on the circumference itself; therefore it will fall within.
Quoniam ergo duo latera EA, EC, trianguli, cuius basis ponitur recta ADC, aequalìa sunt, (e centro enim ducuntur2
) erunt anguli EAD, ECD, aequales:
Est autem angulus EDA, angulo ECD, maior, externus interno opposito,3
cum latus CD, in triangulo ECD, sit productum ad A. Igitur et angulo EAD, maior erit idem angulus EDA.
See two records above
Quare recta EA, maiori angulo opposita,4
hoc est, recta EB, sibi aequalis,
maior erit, quam recta ED, pars quam totum. Quod est absurdum.
Non igitur recta ex A, in C, ducta extra circulum cadet, sed intra.
Eodem enim modo demonstrabitur, rectam ductam ex A, in C, non posse cadere super arcum ABC, ita ut eadem sit, quae circumferentia ABC. Esset enim recta EA, maior, quam recta EB. Quod etiam ex definitione rectae lineae patet, cum ABC, arcus sit linea curua, non autem recta.
Itaque si in circuli peripheria duo quaelibet puncta, etc.
Quod erat ostendendum.
SCHOLION. IDEM hoc theorema demonstrari poterit affirmative, hoc modo, Recta AB, coniungat duo puncta A, et B, in circumferentia circuli AB, cuius centrum C. Dico rectam AB, intra circulum cadere, ita ut omnia eius puncta media intra circulum existant. Assumatur enim quodcunque eius punctum intermedium D, et ex centro educantur recta CA, CB, CD. Quoniam igitur duo latera CA, CB, trianguli CAB, aequalia sunt, erunt anguli CAB, CBA, aequales.5
Est autem angulus CDA, angulo CBA, maior, externus interno;6
Igitur idem angulus CDA, angulo CAD, maior erit. Quare cum CA, sit ducta a centro ad circumferentiam,7
usque, non perveniet recta CD, ad circumferentiam, ideoque punctum D, intra circulum cadet: Idem ostendetur de quolibet alio puncto assumpto. Tota igitur recta AB, intra circulum cadit. Quod est propositum.
COROLLARIUM. HINC est manifestum, lineam rectam, quae circulum tangit, ita ut cum non secet, in uno tantum puncto ipsum tangere. Si enim in duobus punctis eum tangeret, caderet pars rectae inter ea duo puncta posita intra circulum; Quare circulum secaret, quod est contra hypothesin.
Complete text
Book I