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Euclid: Elementa
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Proposition 23. 
PROBL. 9. PROPOS. 23. 
第二十三題 
On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle. 
AD datam rectam lineam, datumque; in ea punctum, dato angulo rectilineo æqualem angulum rectilineum constituere. 
一直線。任於一點上、求作一角。與所設角等。 
Let AB be the given straight line, A the point on it, and the angle DCE the given rectilineal angle;  thus it is required to construct on the given straight line AB, and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE. 
DATA recta sit A B, datumque in ea punctum C, & datus angulus D E F.  Oportet igitur ad rectam A B, in puncto C, angulum constituere æqualem angulo E. 
.  . 
On the straight lines CD, CE respectively let the points D, E be taken at random; let DE be joined,  and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed  in such a way that CD is equal to AF, CE to AG, and further DE to FG. 
Sumantur in rectis E D, E F, duo puncta utcunque G, H, quæ recta G H, connectantur:  Deinde constituatur triangulum C I K, habens tria latera æqualia tribus rectis E G, G H, H E,  ita ut C I, æquale sit ipsi E G; & C K, ipsi E H; & I K, ipsi G H. (Quod facile fiet, si C I, sumatur æqualis ipsi E G, & C L, ipsi E H, & I M, ipsi G H. Deinde ex centris C, & I, interuallis vero C L, & I M, circuli describantur secantes sese in K, &c.) Dico angulum C, æqualem esse angulo E. 
.  .  . 
Then, since the two sides DC, CE are equal to the two sides FA, AG respectively, and the base DE is equal to the base FG, the angle DCE is equal to the angle FAG. [I. 8] 
Quoniam enim duo latera C I, C K, æqualia sunt duobus lateribus E G, E H, utrumque utrique, & basis I K, basi G H, per constructionem; erit angulus C, angulo E, æqualis: 
. 
Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE.  Q. E. F. 
Effecimus igitur angulum ad C, æqualem angulo F, &c.  Quod facere oportebat.

PRAXIS
NON differt huius problematis praxis ab illa, quam in præcedente problemate tradidimus; propterea quod triangulum constituere oportet æquale alteri triangulo, ut angulus dato angulo æqualis exhibeatur, ut perspicuum est. Facilius tamen hac arte problema efficies. Sit linea data A B, punctumque in ea C, & angulus datus E. Centro igitur E, & interuallo quouis arcus describatur G H: Eodemque interuallo ex centro C, arcus describatur I K, sumaturque beneficio circini arcus I K, arcui G H, æqualis. Recta enim ducta C K, faciet angulum ad C, æqualem angulo E. Nam si ducerentur rectæ I K, G H, essent ipsæ æquales, propterea quod circino non variato utramque distantiam I K, G H, accepimus. Cum ergo & duo latera I C, C K, æqualia sint duobus G E, E H, ob æqualia interualla, quibus arcus sunt descripti; erunt anguli I C K, G E H, æquales. 
.  . 
 
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