then the angle ADC bb 5. 1.
c ſuppoſ.
d 9. ax. = ACD; as alſo, becauſe BD c = BC, the angle FDC = b ACD. therefore is the angle FDC d ACD, that is, the angle FDC ADC. d Which is impoſſible.

3. Caſe. If D falls without the triangle ACB, let CD be joined.

Again the angle ACD ee 5. 1.
f 9. ax. = ADC, and the angle BCD e = BDC. f Therefore the angle ACD BDC, viz. the angle ADC BDC. Which is impoſſible. Therefore, &c.

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If two triangles ABC, DEF, have two ſides AB, AC equal to two ſides DE, DF, each to each, and the baſe BC equal to the baſe EF, then the angles contained under the equal right lines ſhall be equal, viz. A to D.

Becauſe BC aa hyp.
b ax. 8.
c hyp. = EF, if the baſe BC be laid on the baſe EF, b they will agree: therefore whereas AB c = DE, and AC = DF, the point A will fall on D (for it cannot fall on any other point, by the precedent propoſition) and ſo the ſides of the angles A and D are coincident; dd 8. ax. wherefore thoſe angles are equal. Which was to be Dem.

1. Hence, Triangles mutually equilateral, are alſo mutually ee 4. 1. equiangular.

2. Triangles mutually equilateral, e are equal one to the other.