dence by superposition if the three parts of the one follow each other in the same order as in the figure. But if they follow in a reversed order, as when S. D'EF is substituted for S. DEF, then equality by coincidence can no longer be inferred; since coincidence itself cannot be made by any change of the faces of any one of the planes concerned. We are then either obliged to have recourse to other methods, or to adopt the principle of equality by symmetry. In general, symmetry may be defined as similarity of position in rerersed directions, estimated from a point, line, or plane. Thus, if two points A, A', be situated in a straight line, equi-distant from a point O in that line, and in opposite directions, A and A' are symmetrical with respect to the point 0. Or, again, if two points A, A' be situated on the opposite sides of a line X'X, and that perpendiculars from them meet X'X in the same point P, and are likewise equal: then A, A' are symmetrical with respect to the line X'X. Further, if there be two lines AB, AB' drawn from the same point A in the line X'X, making equal angles XAB, XAB', above and below X'X; these lines are symmetrically situated with respect to X'X. The same holds if for lines we substitute planes : thus, if AA' meet the plane MN at right angles in P, and AP, AP be equal: then A and A' are symmetrically situated with respect to MN. Other instances will occur to the student's mind which it will not be necessary to give formally here. It only remains to show the application of the principle to trihedral angles. (1.) Symmetry with respect to the vertex. Let S. ABC and s. abc be two trihedral angles whose corresponding equal parts lie in contrary directions, or circulate in contrary order. Produce AS, BS, CS, and make SA', SB', SC' equal respectively to sa, sb, sc; and join A'B', B'C', C'A'. Then the trihedral angle S. A'B'C' will be equal to s. abc, in all respects, if the three fixing conditions of equality be correctly given. It is the business of the investigator to ascertain that this is the case ; and to show that the other parts will be equal in magnitude in the two trihedrals whose common vertex is S. (2.) Symmetry with respect to an edge. Let S. ABC and s. abc be the two trihedral angles which have their pairs of equal magnitudes ranged in reverse order. Make the edges SA, sa coincide, and let also the plane asb coincide with the continuation of the plane ASB: then the plane asc will coincide with the continuation of ASC. The two trihedral angles are then symmetrical with one another; and the fixing conditions being correctly given, the parts of the one can be proved to be equal to the corresponding parts of the other. (3.) Symmetry with respect to a face. In this case the faces ASC, asc are placed in contact, the point a coinciding with A, or taking the place of A' in the symmetrical position ; and c taking the place C or C'. Then B and B' (that is b trans AA posed) is as in the figure; and the fixing conditions of equality being correctly given, it is to be shown that the remaining parts are equal each to each. Other forms of symmetry might be given: but these will be sufficient for all the ordinary investigations respecting trihedral angles. PROPOSITION I. If two trihedral angles have two faces of the one equal two faces of the other, and their included dihedral angles equal, the remaining face of the one will be equal to the remaining face of the other, and the remaining dihedral angles will be equal, each to each, namely, those to which the equal faces are opposite. Case 1. Let the two trihedral angles S. ABC, s. abc have the two faces BSC, CSA equal to the two bsc, csa, each to each, and the dihedral angle (CSB, CSA) equal to (csb, csa); and let the equals succeed each other in the same order; then the faces ASB, asb, will be equal ; and the dihedral angles (CBS, BSA) and (cbs, bsa) will be equal, and likewise the angles (CSA, ASB) and (csa, asb). For, let the trihedral angle s. abc be applied to S. ABC, so that s shall coincide with S, the line bs shall coincide with BS, and the plane csb with the plane CSB. Then, since the angle bsc is equal to BSC, cs will coincide with CS. Again, since the dihedral angle (csb, csa) is equal to (CSB, CSA) the plane cas will coincide with the plane CAS. Also, since the angle csa is equal to CSA, the line as will coincide with AS, and the plane asb with the plane ASB. Whence, as, sb coinciding with AS, SB, the angle asb is equal to the angle ASB. Also, since the planes asb, asc coincide with ASB, ASC, they contain equal dihedral angles; and similarly the dihedral angles (asb, bsc) and (ASB, BSC) are equal. Case 2. Let the trihedral angles S. ABC, s. abc have the same parts equal as before, but situated in contrary order. Produce the planes ASC, BSC; their intersection SC' will be in the prolongation of SC. In these planes respectively produce BS and AS, to B' and A', so that SA', SB', SC may be equal to sa, sb, sc. Then, since CS and BS are produced, the angle B'SC' is equal to BSC; and this latter by hypothesis, to bsc. Whence B'SC' is equal to bsc. In the same way A'SC' is equal to asc. Again, the dihedral angle (A'SC', B'SC') is made by the same planes as the angle (ASC, BSC) they being opposite dihedral angles: whence they are equal. Also (ASC, BSC) is equal to (asc, bsc) by hypothesis ; and hence (A'SC', B'SC'') is equal to the dihedral angle (asc, bsc). : Again, since ASB, A'SB' make opposite dihedral angles with the other two faces, those angles are respectively equal each to each ; viz., (ASB, CSB) to (A'SB', C'SB') and (ASB, ASC) to (A'SB', A'SC'); and the third face ASB is equal to the third A'SB', since they are opposite plane angles. All the parts of S. A'B'C' are, therefore, equal to the corresponding parts of S. ABC. But the parts of the trihedral angle S. A'B'C' follow in the same order as those of s. abc; and by the first case are respectively equal to them. Wherefore, also, the several remaining parts of S. ABC are equal to the corresponding ones of s. abc. That is to say: The third face ASB is equal to the third face asb ; the dihedral angle (ASB, BSC) to (asb, bsc); and the dihedral angle (BSA, ASC) to (bsa, asc). The test of equality laid down by Euclid as the primary one of equality, is capability of coincidence. This latter test by symmetry is equally valid ; and it may be stated generally that, under the circumstances, the two trihedral angles are equal. The analogous proposition in plano is Euo!id 1.4; and the species of symmetry employed in the proof is the first of those enumerated. PROPOSITION II. If two trihedral angles have two dihedral angles of the one equal to two dihedral angles of the other, and likewise their included faces equal : then the remaining dihedral angle of the one will be equal to that of the other; and the remaining faces of the one to those of the other, each to each ; riz., those to which the equal dihedral angles are oppusite. Case 1. Let the face ASB be equal to the face asb ; the dihedral angle (ASB, ASC) to (asb, asc); and the dihedral angle (BSA, BSC) to (bsa, bsc) the order of succession being the same in both : then will CS the faces ASC, BSC be equal to the faces asc, bsc; and the dihedral angle (ASC, BSC) to the dihedral angle (asc, bsc). For, place the point S upon s, the line SA upon sa, and the plane BSA upon bsa. Then, since the angles ASB, asb are equal, SB will coincide with sb. Again, since the dihedral angle (BSA, ASC) is equal to (bsa, asc), and the edge SA coincides with the edge sa, and the plane BSA with bsa, the plane ASC coincides with asc ; and, similarly, the plane BSC coincides with bsc. Wherefore their intersections SC, sc coincide; and the trihedral angles wholly coincide. Their corresponding parts, therefore, as detailed above, are equal, each to each. ? Case 2. Let the equal parts follow in a contrary order, the lettering of the figures remaining as before. In the plane ASB produce the lines SA, SB in a symmetrical direction, and likewise produce the planes ASC, BSC ; and let SC' be their intersection. This will be in the prolongation of SC. Also the angle A'SB' is equal to its opposite ASB, and therefore to asb. *Again the dihedral angle (A'SB', B'SC'') is opposite to (ASB, BSC), and therefore equal to it; and consequentiy to (asb, bsc). Similarly, the dihedral angle (A'SB', A'SC) is equal to (asb, asc). Whence, by the preceding case the trihedral angle s. abc is equal in all respects to S. A'B'C'. But the dihedral angle (A'SC", B'SC'') is opposite to (ASC, BSC), and hence equal to it; whence also (ASC, BSC) is equal to (asc, bsc), that is the remaining dihedral angle of the one to the remaining dihedral angle of the other. And finally, the faces ASC, BSC are equal to A'SC', B'SC'; that is, to the faces asc, bsc respectively. |