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Euclid: Elementa
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Proposition 10. 
PROBL. 5. PROPOS. 10. 
第十題 
To bisect a given finite straight line. 
DATAM rectam lineam finitam bifariam secare. 
一有界線。求兩平分之。 
Let AB be the given finite straight line.  Thus it is required to bisect the finite straight line AB. 
SIT recta finita A B,   diuidenda bifariam, id est, in duas partes æquales. 
.  . 
Let the equilateral triangle ABC be constructed on it, [I. 1]  and let the angle ACB be bisected by the straight line CD; [I. 9]  I say that the straight line AB has been bisected at the point D. 
Describatur super A B, triangulum æquilaterum A B C,   cuius angulus C, per rectam C D, diuidatur bifariam, rectaque C D, rectam A B, secet in D.   Dico rectam A B, bifariam esse diuisam in D.  
.  .  . 
For, since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD respectively;  and the angle ACD is equal to the angle BCD;  therefore the base AD is equal to the base BD. [I. 4] 
Quoniam duo latera A C, C D, trianguli A C D, æqualia sunt duobus lateribus B C, C D, trianguli B C D, utrumque utrique, nempe A C, ipsi B C, cum sint ambo latera trianguli æquilateri, & C D, est commune;  Est autem & angulus A C D, angulo B C D, æqualis, per constructionem:   Erit basis A D, basi B D, æqualis.  
.  .  . 
Therefore the given finite straight line AB has been bisected at D.  Q. E. F. 
Datam ergo rectam A B, bifariam secuimus in D,   quod facere oportebat.

PRAXIS
EX centro A, ad quoduis intervallum, quod tamen dimidium linea A B, excedat, describantur duo arcus, unus superne, alter inferne; Et ex centro B, ad idem intervallum omnino alii duo arcus delinquentur, qui priores secent in C, & D. Recta enim ducta C D, secabit rectam A B, in E, bifariam. Si enim ex A, & B, ad C, & D, ducantur quatuor rectæ, erunt hæ omnes inter se æquales, cum ex centris ad circumferentias æqualium circulorum cadant; Nam arcus circulorum descripti sunt eodem intervallo. Quoniam igitur latera A B, C D, æqualia sunt lateribus B C, C D, utrumque utrique, & basis A D, basi B D, erit angulus A C D, angulo B C D, æqualis. Rursus quia latera A C, C E, æqualia sunt lateribus B C, C E, utrumque utrique, & angulus A C E, angulo B C E, ut ostensum fuit; erit basis A E, basi B E, æqualis.

SCHOLION
PERSPICVVM est, eodem modo dividi posse eandem lineam rectam A B, in 4. partes æquales, & in 8. in 16. in 32. &c. sicuti in propositione præcedenti diximus de diuisione trianguli rectilinei. Qua vero ratione quævis recta linea proposita diuidenda sit in quotcunque partes æquales, uberrime trademus ad propositionem 10. liber 6. ubi varias, & non iniucundas praxes in medium adducemus. Ibi enim videtur esse proprius huic rei locus, cum huiusmodi praxes fere omnes per linearum proportiones facilius demonstrentur. Neque vero unquam indigebimus diuisione lineæ in plures, quam in duas partes æquales, ad eum locum usque. 
.  . 
 
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