ABC be applied to ▲ DEF, so B coincides with E. C coincides with F. BC coincides with EF. 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, DFE. 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. A, having AB = AC. Let AB and AC be prod. to D and E. Then ▲ ABC ACB, and CBD = A BCE. In BD take any pt. F; from AE, the greater, Prop. 3. cut off AG-AF; and FAG common to As AFC, AGB, base FC base GB, ▲ AFC = ▲ AGB. Again, whole AF whole AG, and part AB = part AC, .. remain. BF = remain. CG; Constr. Hyp. Prop. 4. Hyp. Ax. 3. Prop. 4. Ax. 3. Hyp. Prop. 4. .. 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., having the side AB = side AC; then will ▲ ABC = ▲ ACB. Let the str. line AD divide the BAC into two = parts. (AD is common to both AS ABD, ACD, and AB < BAD = ABD = AC, Z DAC; ▲ ACD, ZACB, which are the s at the base. ={< Z ABC = 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 the proposition; since we are at liberty to suppose any line to be drawn, or any figure to be constructed, the existence of which does not involve an absurdity. The first proof of this proposition is that of Euclid, who, to avoid the possibility of the taking for granted that which may imply a contradiction, never supposes a thing to be done, the manner of doing which has not been previously explained. The equality of the s on the oppo. side of the base will follow from Prop. 12. PROP. VI. THEOR. 6. 1 Eu. If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to the equal angles, shall be equal to one another. Let ABC have ABC = ▲ ACB; then AB AC. Prop. 4. BC common to both as; base DC = base AB, and DBC = ▲ ACB, or, the less { the greater, which is absurd; .. AB is not AC, i. e. AB AC. Wherefore, if two angles, &c. COR.-Hence every equiangular triangle is equilateral. PROP. VII. THEOR. If two triangles have three sides of the one respectively equal to the three sides of the other, each to each, the triangles are equal, and the angles are equal which are opposite to the equal sides. In As CBA, CDA, if AB, BC, CA = AD, DC, CA, respectively, ea. to ea. then CBÁ = ▲ CDA, ▲ CBA = ▲ CDA, ▲ BAC = ≤ CAD, ACB = ▲ ACD. |