Then, since the straight line AD falling on the two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another, therefore EAF is parallel to BC. [I. 27]
Et quoniam in duas rectas BG et EZ recta incidens AD eos qui permutatim angulos EAD et ADG equales alternis fecit, equidistans ergo est recta (E) AZ recte BG.
Probatio huius: Quia super duas rectas lineas EZ; BG iam cecidit recta linea que est AD et fiunt duo anguli EAD; ADG equales qui sunt coalterni, erit linea EZ linee GB equidistans.
Rationis causa: Cum enim linea AD supra lineas HZ et BG ceciderit, fuerintque duo anguli coalterni HAD et ADG equales, erunt linee HZ et BG inconiunctive.
Title
Book I
