Conceive the polyedral angle S to be cut by any plane ABCDE; from O a point in that plane, draw to the several angles straight lines AO, OB, OC, OD, OE. The sum of the angles of the triangles ASB, BSC, etc., formed about the vertex S, is equiva lent to the sum of the angles of an equal number of triangles AOB, BOC, etc., formed about the point O. But at the point B the angles ABO, OBC, taken together, make the angle ABC (T. XIX.) less than the sum of the angles ABS, SBC. In the same manner, at the point C we have BCO + OCD <BCS+ SCD, and so with all the angles of the polygon ABCDE. Whence it follows that the sum of all the angles at the bases of the triangles whose common vertex is in O, is less than the sum of all the angles at the bases of the triangles whose common vertex is in S; hence, to make up the deficiency, the sum of the angles formed about the point O, is greater than the sum of the angles about the point S. But the sum of the angles about the point O is equal to four right angles (B. I., T. I., C. III.); therefore the sum of the plane angles, which form the polyedral angle S, is less than four right angles. Scholium. This demonstration is founded on the supposition that the polyedral angle is convex, or that the plane of no one surface produced can ever meet the polyedral angle. If it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude. THEOREM XXI. If two polyedral angles are composed of three plane angles respectively equal to each other, the planes which contain the equal angles will be equally inclined to each other. Let the angle ASC DTF, the angle ASB = DTE, and the angle BSC 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 T S F E B A D Р BO perpendicular to the plane ASC; from the point O at which that perpendicular meets the plane, draw OA, OC perpendicular to SA, SC; join AB, BC; next take TE=SB; draw EP perpendicular to the plane DTF; from the point P draw PD, PF perpendicular to TD, TF; lastly, join DE, EF. The triangle SAB is right-angled at A, and the triangle TDE is right-angled at D (T. VI.), and since the angle ASB = DTE, we have SBA = TED. Likewise SBTE, therefore the triangle SAB is equal to the triangle DTE; therefore 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, because SA=TD, and SC TF, the point A will coincide with D, and the point C with F; and at the same time AO, which is perpendicular to SA, will coincide with PD, which is perpendicular to TD, and in like manner CO with FP; wherefore the point O will coincide with the point P, and AO will be equal to DP. But the triangles AOB, DPE are rightangled at O and P; the hypotenuse AB = DE, and the side AO=DP; hence those triangles are equal-therefore the angle OAB = PDE. The angle OAB is the inclination of the two planes ASB, ASC, and the angle PDE is that of the two planes DTE, DTF; hence these two inclinations are equal to each other. It must, however, be observed, that the angle A of the rightangled triangle AOB is properly the inclination of the two planes ASB, ASC, only when the perpendicular BO falls on the same side of SA as SC falls; for if it fell on the other side, the angle of the two planes would be obtuse, and, added to the angle A of the triangle OAB, it would make two right angles. But, in the same case, the angle of the two planes TDE, TDF would also be obtuse, and, added to the angle D of the triangle PDE, it would make two right angles; and the angle A being thus always equal to the angle at D, it would follow, in the same manner, that the inclination of the two planes ASB, ASC must be equal to that of the two planes DTE, DTF. SIXTH BOOK. BODIES BOUNDED BY PLANES. DEFINITIONS. I. A solid, bounded by polygons, is called a Polyedron. The bounding polygons are called faces, the straight lines formed by the intersection of any two faces are called edges, and the points where three or more faces meet are called corners. A straight line joining any two corners not situated in the same face is called a diagonal. II. A polyedron, having two of its faces equal and parallel, and all the other faces parallelograms, is called a Prism. The two equal and parallel faces are called the bases of the prism, usually referred to as the lower base and the upper base; and the parallelograms which make up the other faces, taken together, constitute the lateral surface of the prism. III. A prism is triangular, quadrangular, pentagonal, etc., according as the bases are triangles, quadrilaterals, pentagons, etc. IV. When the edges formed by the intersection of the lateral faces of a prism are perpendicular to its bases, it is a right prism: each edge is then equal to its altitude. In all other cases the prism is said to be oblique, and the edges are greater than the altitude. 7. V. A prism, whose bases are parallelograms, is called a Parallelopipedon. When all the faces are rectangles it is called a Rectangular Parallelopipedon. When all the faces are squares it is called a Cube, or regular hexaedron. VI. A Pyramid is a solid formed by several triangular planes meeting in a point, called the vertex, and terminating in the sides of a polygon, forming its base. The triangles meeting at the vertex, taken together, constitute the lateral surface of the pyramid. The perpendicular distance from the vertex to the base is called its altitude. VII. A pyramid is triangular, quadrilateral, pentagonal, etc., according as the base is a triangle, quadrilateral, pentagon, etc. VIII. When the base of a pyramid is a regular polygon, and the triangles forming the lateral surface are isosceles, it is a right pyramid. The perpendicular drawn from the vertex to the base will, in this case, pass through its centre, and it is then called the axis of the pyramid. A line drawn from the vertex of a right pyramid to the middle point of one of the sides of the polygon constituting its base, is called its slant height. IX. A triangular pyramid, having in all four triangular faces, is called a Tetraedron. X. If a pyramid is cut by a plane parallel to its base it will be divided into two portions, one of which will be a second pyramid having the same vertex as the first. The remaining portion, having two parallel bases, is called a truncated pyramid, or frustum of a pyramid. The altitude of a frustum of a pyramid is the perpendicular distance between its bases. The slant height of a frustum of a right pyramid is the portion of its slant height included between its bases. XI. Two tetraedrons are similar when they have their edges proportional and arranged in the same order. Consequently the triangular faces of two similar tetraedrons are similar triangles each to each (B. III., D. II.). Any two polyedrons whatever, are similar when they are capable of being decomposed into the same number of similar tetraedrons each to each, having the same order of arrange ment. XII. A regular Polyedron is one having all its faces equal and regular polygons; and all its diedral angles equal. THEOREM I. The lateral surface of a right prism is equal to the perimeter of its base multiplied by its altitude. For this surface is equal to the sum of the rectangles AFGB, BGHC, CHID, etc., which compose it. Now the altitudes AF, BG, CH, etc., of those rectangles, are equal to the altitude of the prism; their bases AB, BC, CD, etc., taken together, make up the perimeter of the prism's base: hence the sum of these rectangles, or the lateral surface of the prism, is equal to the perimeter of its base multiplied by its altitude. F G K H I E A D B C Cor. If two right prisms have the same altitude, their lateral surfaces will be to each other as the perimeters of their bases. THEOREM II. Two prisms are equal when a polyedral angle in each is contained by three planes, which are respectively equal in both, and similarly situated. Let the base ABCDE be equal to the base abcde; the parallelogram ABGF equal to the parallelogram abgf, and the parallelogram BCHG equal to the parallelogram behg; then will the prism ABCI be equal to the prism abci. For, apply the base ABCDE upon its equal abcde, so that |