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Euclid: Elementa

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eng
Q. E. F.
lat Clavius
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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