K А B с sides shall be = A, B, C, each to each. Const.—Take a line DE, D terminated towards D, but unlimited towards E, and make DF = A, FG = B, GH = c. (Prop. III.) From cr. F, with rad. FD, describe a O DKL. From cr. G, with rad. gh, describe a o HLK (Post. 3). Join KF, KG (Post. 1). Then the A KFG is the triangle required. DEMONST.—By the property of a o (Def. 15), KF = FD = A; and KG = GH = c; also FG = B. (Const.): .. the A KFG has its three sides, KF, FG, GK = to the three gn. st. lines A, B, C respectively.-Q. E. F. That the three straight lines must answer the conditions indicated in the enunciation is manifest from Prop. XX. ; and, such being the case, the two Os must necessarily cut each other; for since FG + gh are > DF, the must meet FE between F and h; and DF + FG being > gh, the O HLK must meet gd between G and D; so that one of these Os cannot be wholly within the other. Also, because DF + Gh is > FG, the two O cannot lie wholly without each other. It is clear that if DF the A will be isosceles, and the Proposition will coincide with Prop. B; and if the three lines be all equal, with Prop. I. Another mode of solving the Proposition is as follows : Taking the longest line A, or de = A, to form the base of the A, make DF = B, and Eg = C. (Prop. III.) From cr. D, with rad. DF, describe the opku. From cr. E, with rad. EG, describe the O GKH. Join DH, EH. Then deh is the A required. For by property of O, DH = DF = B; EH = EG = c; and DE A; .:. the sides of the A DEH = A, B, C respectively.-Q. E. F. DKL GH, н, D E A similar A may of course be constructed on the other side of DE. PROP. XXIII. PROB. G Gen. Enun.—At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. PART. Enun.—Let AB be the gn. st. line; A the gn. pt. in it; dce the gn. rectilineal Z. Then it is required to make an 2 at the gn. pt. A = ZDCE. Const.—Take in CD, CE any pts. D, E, and join DE; make the a AFG, of which the sides shall be = the st. lines CD, DE, CE, each to each ; viz. AF = CD, AG = CE, FG = DE. (Prop. XXII.) Then the Fag is the required. DEMONST.—Because DC, CE = FA, AG, each to each, and the base De = base fg (Const.), ... the Z FAG ZDCE. (Prop. VIII.) Wherefore, at the gn. pt. A, in the gn. st. line AB, the < FAG has been made = the gn. ZDCE.-Q. E. F. 'B PROP. XXIV. THEOR. GEN. ENUN.If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides, equal to them, of the other, the base of that which has the greater angle shall be greater than the base of the other. Part. Enun.--Let ABC, DEF be two triangles, which have the two sides AB, AC = to the two sides DE, DF, each to each, viz. AB = DE, and AC = DF; but let the < BAC be > the L EDF: then the base bc shall be > the base EF. Const. Of the two sides DE, DF, let de be not > DF; and at the pt. D, in the st. line DE, make Z EDG = { BAC. (Prop. XXIII.) Make also DG = AC = DF (Prop. III.), and join EG, GF (Post. 1). DEMONST.-1. Because AB = DE (Hyp.), and ac = DG (Const.), :the two sides AB, AC = DE, DG, each to each : and the < BAC = 4 EDG (Const.); .. the base bc = base EG. (Prop. IV.) 2. Again, because ng = DF, :. DFG = DGF (Prop. V.); but the Z DGF is > Z EGF (Ax. 9); :: Zorg is > Z EGF; .., a fortiori, 2 EFG is > < EGF (Ax. 9). A D B CE G K 3. Now the > < of a A is subtended by its greater side (Prop. XIX.); .. in the EGF, the side Eg is > EF; but EG = BC; : also BC is > EF. Wherefore, if two As have two sides &c.Q. E. D. There is clearly another case of this Theorem, when DE is > than DF, of which the proof is somewhat different, and is therefore subjoined. Proceed as before, and produce df, DG to , K: then, as before, EG = BC; and because DF = DG, the Zs on the other side of the base of the A DFG are = (Prop. V.); .. the L GFH FGK; but / fok is > | FGE (Ax. 9); .. 1 GFh is > LFGE; .., a fortiori, Gre is > L FGE; .. side eg is > EF (Prop. XIX.); but EG = BC; .. BC is > EF.-Q. E. D. The reason why Euclid did not adopt this proof may be, that it is, in point of fact, superfluous ; for if dE be > DF, DF is < than DE; and by an inverted construction of the figure, the same demonstration will suffice. Of this the student will do well to satisfy himself. That in this case the pt. F falls below the line EG is manifest; for since DG = DF, the pt. G is in the Oce of a O described from the cr. D, with rads. DF, and must be above DF, because the 2 EDG is > than LEDF. It is also worthy of observation that the conclusion, derived from the inequality of the 2$ contained by the equal sides, is not coextensive with that derived from their equality in Prop. IV. It is not necessary that the area or the remaining 28 of that which has the greater base should be > than those of the other. Without proceeding further, at least for the present, it will be seen at once that, in the first case, as given in the Proposition, the L ABC will always be <, and, in the second, always > the DEF. PROP. XXV. THEOR. GEN. ENUN.-If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other, the angle contained by the sides of that which has the greater base shall be greater than the angle contained by the sides, equal to them, of the other. PART. Enun.—Let ABC, DEF be two AS which have the two sides AB, AC equal to the two sides DE, DF, each to each ; i. e. AB = DE, and ac = DF: but let the base bc be > the base EF. Then the < BAC shall be > the 2 EDF. DEMONST.-For, if not >, the < BAC is either = or < the 2 EDF. 1. Now it is not = to it, for then base BC would = base EF (Prop. IV.); but it is not = to it. (Hyp.) 2. Neither is it <, for then BC would be < EF (Prop. XXIV.); but it B is not. (Hyp.) 3. ... the < BAC is neither = nor < the Z EDF; i. e. Z BAC = 4 EDF. Wherefore, if two As have two sides &c.Q. E. D. This Proposition is the converse of the preceding. А E с |