Post. J. Draw AB; on AB descr. an equilat. Prop. 1. DAB; prod. DA, DB to E and F; from cr. B and dist. BC, descr. O CGH; from cr. D and dist. DG, descr. Post. 2. GKL. Then AL BC. Wherefore, from the given pt. A, a str. line AL has been drawn equal to the given str. line BC. From the greater of two given straight lines, to cut off a part equal to the less. Let AB and C be the given str. lines, whereof AB > C; it is required to cut off from AB a part = C. From A draw AD C; from cr. A and Prop. 2. dist. AD, descr. O DEF, cutting AB in E; Post. 3. then AE C. Wherefore, from AB, the greater of two str. lines, a part AE has been cut off = C, the less. Def. 15. Ax. 1. PROP. IV. THEOR. 4. 1 Eu. If two triangles have two sides of the one, equal to two sides of the other, each to each; and have likewise the angles contained by those sides equal to one another; they shall likewise have their bases, or third sides, equal, and the two triangles shall be equal; and their other angles shall be equal, each to each, viz., those to which the equal sides are opposite. Let ABC, DEF, be two As, of which AB =DE, AC = DF, and ▲ BAC = ▲ EDF. Then BC = EF, ▲ ABC=^ DEF, L ABCDEF, and ACB = ≤ DFE. For if BC do not coincide with EF, then two Ax. 10. str. lines enclose a space, which is impossible. BC coincides with and = EF, ABC coincides with and =▲ DEF, ZABC coincides with and = /DEF, ZACB coincides with and = LDFE. Therefore, if two triangles have, &c. PROP. V. THEOR. 5. 1 Eu. The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles upon the other side of the base shall be equal. Let ABC be an isosc., having AB AC. Let AB and AC be prod. to D and E. Then 2 ABC= ACB, and CBD = ▲ BCE. In BD take any pt. F; from AE, the greater, Prop. 3. cut off AG-AF; :. Therefore, the angles, &c. COR.-Hence every equilateral is also equiangular. PROP. V. THEOR. OTHERWISE DEMONSTRATED. The angles at the base of an isosceles triangle are equal to one another. Let ABC be an isosc. A, having the side AB side AC; then will ABC = Z ACB. Let the str. line AD divide the BAC into two parts. (AD is common to both As ABD, ACD, Wherefore the angles at the base, &c. Note. It is evident that some line, as AD, will bisect the BAC; and although the method of bisection is not known until Pro. 8 be solved, yet this does not affect the truth of |