other two. To establish this proposition, we have only to compare the greatest of the three angles with the sum of the other two. Suppose, then, BAC to be the greatest angle, and draw in its plane B the line AE, making the angle CAE equal to the angle CAD. On C E D AE, take any point, E, and through it draw the line CEB. Take AD, equal to AE, and draw BD and DC. Now, the two triangles, CAD and CAE, having two sides and the included angle of the one equal to the two sides and included angle of the other, each to each, are equal, and CE = CD; but in the triangle, BDC, BC< BD+ DC. Taking EC from the first member of this inequality, and its equal, DC, from the second, we have, BE BD. In the triangles, BAE and BAD, BA is common, and AE AD by construction; but the third side, BD, in the one, is greater than the third side, BE, in the other; hence, the angle BAD is greater than the angle BAE, (Th. 22, B. I); that is, LBAE <L\BAD; adding the EAC to the first member of this inequality, and its equal, the DAC, to the other, we have = |_ BAE+|_ EAC<\ BAD + \ DAC. And, as the BAC is made up of the angles BAE and EAC, we have, as enunciated, | BAC<L BAD + | DAC. THEOREM XIX. The sum of the plane angles forming any solid angle, 28 always less than four right angles. Let the planes which form the solid angle at A, be cut by another plane, which we may call the plane of the base, BCDE. Take any point, a, in this plane, and draw aB, aC, aD, aE, etc., thus making as many triangles on the plane of the base as there are triangular planes forming the solid angle A. Now, since the sum of the angles of every ▲ is two right angles, the sum of all the angles of the A's which have their vertex in A, is equal to the sum of all angles of the A's which have their vertex in a. But, the angles BCA +ACD, are, together, greater than the angles BCa+aCD, or BCD, by the last proposition. That is, the sum of all the angles at the bases of the 's which have their vertex in A, is greater than the sum of all the angles at the bases of the ▲'s which have their vertex in a. Therefore, the sum of all the angles at a is greater than the sum of all the angles at A; but the sum of all the angles at a is equal to four right angles; therefore, the sum of all the angles at A is less than four right angles. THEOREM XX. If two solid angles are formed by three plane angles respectively equal to each other, the planes which contain the equal angles will be equally inclined to each other. Let the ASC-the|___DTF, the ASB= the __DTE, and the BSC=the ETF; then will the inclination of the planes, ASC, ASB, be equal to that of the planes, DTF, DTE. Having taken SB at pleasure, draw BO perpendicular A T to the plane ASC; from the point 0, at which that perpendicular meets the plane, draw OA and OC, perpendicular to SA and SC; draw AB and BC; next take TE = SB, and draw EP perpendicular to the plane DTF; from the point P, draw PD and PF, perpendicular to TD and TF; lastly, draw DE and EF. = = = = = The triangle SAB, is right-angled at A, and the triangle TDE, at D, (Th. 5); and since the L ASB = the LDTE, we have |_ SBA =|_ TED; likewise, SB = TE; therefore, the triangle SAB is equal to the triangle TDE; hence, SA TD, and AB = DE. In like manner it may be shown that SC TF, and BC EF. That granted, the quadrilateral SAOC is equal to the quadrilateral TDPF; for, place the angle ASC upon its equal, DTF, and because SA TD, and SC TF, the point A will fall on D, and the point Con F; and, at the same time, AO, which is perpendicular to SA, will fall on PD, which is perpendicular to TD, and, in like manner, OC on PF; wherefore, the point 0 will fall on the point P, and AO will be equal to DP. But the triangles, AOB, DPE, are right angled at O and P; the hypotenuse AB = DE, and the side 40 DP; hence, those triangles are equal, AO = (Cor, Th. 39, B. I), and [_OAB=|_PDE. The angle OAB is the inclination of the two planes, ASB, ASC; the angle PDE is that of the two planes, DTE, DTF; consequently, those two inclinations are equal to each other. Hence the theorem. SCHOLIUM 1.—The angles which form the solid angles at S and T, may be of such relative magnitudes, that the perpendiculars, BO and EP, may not fall within the bases, ASC and DTF; but they will always either fall on the bases, or on the planes of the bases produced, and O will have the same relative situation to A, S, and C, as P has to D, T, and F. In case that O and P fall on the planes of the bases produced, the angles BCO and EFP, would be obtuse angles; but the demonstration of the problem would not be varied in the least. SCHOLIUM 2.—If the plane angles bounding one of the triedral angles be equal to those of the other, each to each, and also be similarly arranged about the triedral angles, these solid angles will be absolutely equal. For it was shown, in the course of the above demonstration, that the quadrilaterals, SAOC and TDPF, were equal; and on being applied, the point O falls on the point P; and since the triangles AOB and DPE are equal, the perpendiculars OB and PE are also equal. Now, because the plane angles are like arranged about the triedral angles, these perpendiculars lie in the same direction; hence the point B will fall on the point E, and the solid angles will exactly coincide. SCHOLIUM 3.-When the planes of the equal angles are not like disposed about the triedral angles, it would not be possible to make these triedral angles coincide; and still it would be true that the planes of the equal angles are equally inclined to each other. Hence, these triedral angles have the plane and diedral angles of the one, equal to the plane and diedral angles of the other, each to each, without having of themselves that absolute equality which admits of superposition. Magnitudes which are thus equal in all their component parts, but will not coincide, when applied the one to the other, are said to be symmetrically equal. Thus, two triedral angles, bounded by plane angles equal each to each, but not like placed, are symmetrical triedral angles. BOOK VII. SOLID GEOMETRY. DEFINITIONS. 1. A Polyedron is a solid, or volume, bounded on all sides by planes. The bounding planes are called the faces of the polyedron, and their intersections are its edges. 2. A Prism is a polyedron, having two of its faces, called bases, equal polygons, whose planes and homologous sides are parallel. The other, or lateral faces, are parallelograms, and constitute the convex surface of the prism. The bases of a prism are distinguished by the terms, upper and lower; and the altitude of the prism is the perpendicular distance between its bases. Prisms are denominated triangular, quadrangular, pentangular, etc., according as their bases are triangles, quadrilaterals, pentagons, etc. 3. A Right Prism is one in which the planes of the lateral faces are perpendicular to the planes of the bases. 4. A Parallelopipedon is a prism whose bases are parallelograms. |