The sum of any two of the plane angles formed by the edges of a triedral angle, is greater than the third. Let SA, SB, and SC, be the edges of a triedral angle then will the sum of any two of the plane angles formed by them, as ASC and CSB, be greater than the third ASB. If the plane angle ASB is equal to, or less than, either of the other two, the truth of the proposition is evident. Let us suppose, then, that ASB is greater than either. In the plane ASB, construct the angle BSD equal to BSC; draw AB in that plane, at plea S mon, and the included angles BSD and BSC equal, by construction; the triangles are therefore equal in all their parts hence, BD is equal to BC. But, from Proposition VII., Book I., we have, BC + CA > BD + DA. Taking away the equal parts BC and BD, we have, CA > DA; hence (B. I., P. IX., C.),. we have, angle ASC > angle ASD ; and, adding the equal angles BSC and BSD, or, angle ASC + angle CSB > angle ASD + angle DSB ; angle ASC + angle CSB > angle ASB ; which was to be proved. PROPOSITION XX. THEOREM. The sum of the plane angles formed by the edges of any polyedral angle, is less than four right angles. Let S be the vertex of any polyedral angle whose edges are SA, SB, SC, SD, and SE; then will the sum of the angles about S be less than four right angles. For, pass a plane cutting the edges in the points A, B, C, D, and E, and the faces in the lines AB, BC, CD, DE, and EA. From any point within the polygon thus formed, as 0, draw the straight lines OA, OB, OC, OD, and OE. B D both having comNow, in the set We then have two sets of triangles, one set having a common vertex S, the other having a common vertex 0, and mon bases AB, BC, CD, DE, EA. which has the common vertex S, the sum of all the angles is equal to the sum of all the plane angles formed by the edges of the polyedral angle whose vertex is S, together with the sum of all the angles at the bases: viz., SAB, SBA, SBC, &c.; and the entire sum is equal to twice as many right angles as there are triangles. In the set whose common vertex is O, the sum of all the angles is equal to the four right angles about 0, together with the interior angles of the polygon, and this sum is equal to twice as many right angles as there are triangles. Since each set, is the same, it follows But in the triedral angle whose the number of triangles, in ABS+ SBC > ABC; A S D and the like may be shown at each of the other vertices, C, D, E, A: hence, the sum of the angles at the bases, in the triangles whose common vertex is S, is greater than the sum of the angles at the bases, in the set whose common vertex is 0: therefore, the sum of the vertical angles about S, is less than the sum of the angles about 0: that is, less than four right angles; which was to be proved. B Scholium. The above demonstration is made on the supposition that the polyedral angle is convex, that is, that the diedral angles of the consecutive faces are each less than two right angles. If the plane angles formed by the edges of two triedral angles are equal, each to each, the planes of the equal angles are equally inclined to each other. Let S and T be the vertices of two triedral angles, and let the angle ASC be equal to DTF, ASB to DTE and BSC to ETF: then will the planes of the equal angles be equally inclined to each other. For, take any point of SB, as B, and from it draw in the two faces ASB and CSB, the lines BA and BC, respectively perpendicular to SB: then will the angle ABC measure the inclination of these faces. Lay off TE equal to SB, and from E draw in the faces DTE and FTE, the lines ED and EF, respectively perpendicular to TE; then will the will the angle DEF measure the inclination of these faces. Draw AC and DF The right-angled triangles SBA and TED, have the side SB equal to TE, and the angle ASB equal to Ꭲ . A A DTE; hence, AB is equal to DE and AS to TD. In like manner, it may be shown that BC is equal to EF, and, CS to FT. The triangles ASC and DTF, have the angle ASC equal to DTF, by hypothesis, the side AS equal to DT, and the side CS to FT, from what has just been shown; hence, the triangles are equal in all their parts, and consequently, AC is equal to DF. Now, the triangles ABC and DEF have their sides equal, each to each, and consequently, the corresponding angles are also equal; that is, the angle ABC is equal to DEF: hence, the inclination of the planes ASB and CSB, is equal to the inclination of the planes DTE and FTE In like manner, it may be shown that the planes of the other equal angles are equally inclined; which was to be proved. Scholium. If the planes of the equal plane angles are like placed, the triedral angles are equal in all respects, for they may be placed so as to coincide. If the planes of the equal angles are not similarly placed, the triedral angles are equal by symmetry. In this case, they may be placed so that two of the homologous faces shall coincide, the triedral angles lying on opposite sides of the plane, which is then called a plane of symmetry. In this position, for every point on one side of the plane of symmetry, there is a corresponding point on the other side. BOOK VII. POLY EDRONS. DEFINITIONS. 1. A POLYEDRON is a volume bounded by polygons. The bounding polygons are called faces of the polyedron; the lines in which the polygons meet, are called edges of the polyedron; the points in which the edges meet, are called vertices of the polyedron. 2. A PRISM is a polyedron, two of whose faces are equal polygons having their homologous sides parallel, the other faces being parallelograms The equal polygons are called bases of the the other the prism; lower base; the parallelograms taken together one the upper, and the make up the lateral or convex surface of the prism; lines in which the lateral faces meet, are called lateral edges of the prism. 3. The ALTITUDE of a prism is the perpendicular dis tance between the planes of its bases. 4. A RIGHT PRISM is one whose lateral edges are perpendicular to the planes of the bases, In this case, any lateral edge is equal to the altitude. |