Then, since the straight line BF has been cut into equal segments at G, and into unequal segments at E, the rectangle contained by BE, EF together with the square on EG is equal to the square on GF. [II. 5]
Quoniam ergo recta BZ divisa est in equalia quidem secundum I, in inequalia vero secundum E, sub rectis ergo BE, EZ contentum orthogonium cum eo quod ab EI tetragono equale est ei quod ab IZ tetragono.
Probatio huius: Et quoniam linea recta BZ iam est divisa in duas equales sectiones in puncto H et duas inequales in puncto E, erit superficies rectorum angulorum, que ab his duabus comprehenditur lineis BE; EZ, cum quadrato quod fit ex linea EH quadrato quod fit ex linea HZ equalis.
فلان خط (د ط) قد قسم بنصفين على نقطة (ك) وبقسمين مختلفين على نقطة (ه) فبحسب برهان [ه] من [ب] فان السطح الذى يحيط به خطا (د ه) (ه ط) مع المربع الكائن من خط (ك ه) مساو للمربع الكائن من خط (ك ط)
Title
Book I
