and AC = DF) and have the angle A equal to the angle D contained under the equal right lines; they ſhall have the baſe EC equal to the baſe EF; and the triangle BAC ſhall be equal to the triangle EDF; and the remaining angles B, C, ſhall be equal to the remaining angles E, F, each to each, under which the equal ſides are ſubtended.
If the point D be applied to the point A, and the right line DE plac'd upon the right line AB, the point E ſhall fall upon B,becauſe DE aa hyp. = AB, alſo the right line DF ſhall fall upon AC,becauſe the angle A a = D. moreover the point F ſhall fall upon the point C, becauſe AC ab 14. ax. = DF. Therefore the right lines EF, BC ſhall agree, becauſe they have the ſame Terms, and conſequently are equal. Wherefore the triangles BAC, DEF, and the angles B, E, as alſo the angles C, F, do agree, and are equal. Which was to be Demonſtrated.
PROP. V.
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The angles ABC, ACB, at the baſe of an Iſoſceles triangle ABC, are equalone to the other: And if the equal ſides AB, AC be produc'd, the angles CBD, BCE, under the baſe, ſhall be equal one to the other.
aa 3. 1.
b 1. post. Take AE= AD; and b join CD, and BE.
d conſtr.
e 4. 1. = AC, and AE d = AD, and the angle A common to them both, e therefore is the angle ABE = ACD,and the angle AEB e = ADC, and the baſe BE e = CD; alſo EC ff 3. ax.
g 4. 1. = DB. Therefore in the triangles BEC, BDC g ſhall be the angle ECB = DBC. Which was to be Dem. Alſo therefore the angle EBC = DCB. but the angle ABE hh before.
k 3. ax. = ACD; therefore the angle ABC k = ACB. Which was to be Dem.
