B Let the plane ABC be perpendicular to DBC, and let any straight line AB be drawn in the plane ABC perpendicular to the common section BC: the straight line AB shall be perpendicular to the plane D B C. From the point B, in the plane DB C, let BD be drawn at right angles to BC. Then, because the planes are at right angles to one another, the angle ABD is a right angle (17. Cor.): but ABC is likewise a right angle; therefore (3.) A B is perpendicular to the plane D B C. Next, let the straight line AB be perpendicular to the plane B CD; and let ABC be any plane passing through A B: the plane A B C shall be perpendicular to BCD. Let BC be the common section of the two planes; and, from the point B, in the plane BCD, draw BD at right angles to BC. Then, because AB meets the line BD drawn in the plane to which A B is perpendicular, the angle ABD is a right angle (def. 1.). But the angle AB D is contained by straight lines drawn in the two planes perpendicular to their common section B C. Therefore (17. Cor.) the dihedral angle A B C D is likewise a right dihedral angle, and the plane A B C is perpendicular to BCD. Therefore, &c. Cor. 1. If through the same point there pass any number of planes perpendicular to the same plane, they shall all of them pass through the same straight line, viz. the perpendicular which is drawn from the point to the plane. Cor. 2. Euc. xi. 19. If two planes, which cut one another, be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane. PROP. 19. (EUc. xi. 20.) If a solid angle be contained by three plane angles, any two of these shall be together greater than the third. Let the solid angle at A be contained by the three plane angles BAC, BAD, CAD: any A two of them, as BAC, BAD, shall be together greater than the third CAD. If one of the angles BAC, BAD be greater than CAD, or equal to it, the proposition is evident; for this, together with the other, must be greater than CAD. But, if the angles BAC, BAD be each of them less than C A D, from the angle CAD cut off the angle CAE equal to BAC; in A B take any point B, make A E equal to A B; through E draw any line CED, not parallel to A C or AD, and therefore cutting A C, A D in the points C, D respectively; and join BC, BD. Then, because the triangles ABC, AEC have two sides of the one equal to two sides of the other, each to each, and the included angles BAC, EAC equal to one another, the base BC is equal to the base EC (I. 4.): but BD and BC are together greater than D C, (I. 10.) that is, than DE and E C: therefore B D is greater than E D. And, because in the triangles ABD, AED the two sides A B, AD of the one are equal to the two sides AE, AD of the other, each to each, but the base B D greater than the base E D, the angle BAD is greater than the angle E A D (I. 11.). Therefore, B A C, BAD together are greater than EAC, EAD together, that is, than CAD. Therefore, &c. Cor. If three plane angles form a solid angle, any one of them shall be greater than the difference of the other two. of these triangles are (I. 19.) together equal to twice as many right angles as the figure has sides. But the angles of these triangles are those which contain the solid angle at A, together with the pairs of angles at the points B, C, D, &c.; and twice as many right angles as the figure has sides are equal to four right angles together with the angles of the figure (I. 20.). Therefore, the angles at A, together with the pairs of angles at B, C, D, &c. are equal to four right angles together with the angles of the figure BCDEF. But, because the solid angles at B, C, D, &c. are each of them contained by three plane angles, any two of which are (19.) together greater than the third, the pairs of an gles at B, C, D, &c. are together greater than the angles of the figure B CDEF. Therefore, the angles at A are together less than four right angles. Therefore, &c. Cor. It follows from this proposition that there cannot be more than five regular solids. For the solid angles of a regular solid are contained by the plane angles of equilateral triangles, or of squares, or of regular pentagons, &c. (def. 9.). Now, six angles of equilateral triangles are together equal to four right angles (I. 19.): therefore, only three, or four, or five, of such angles, may be taken to form a solid angle; and, accordingly, there cannot be more than three different regular solids, whose faces are equilateral triangles. Again, three angles of squares may be taken to form a solid angle; but four angles of squares are equal to four right angles (1. def. 20.). ↑ Lastly, three angles of regular pentagons may be taken to form a solid angle, for these are together less (I. 20.) than four right angles. But, because the angle of a regular pentagon is greater (I. 20.) than that of a square, four angles of a regular pentagon are greater than four right angles, so that not more than three of them can be taken to form a solid angle. With regard to the regular hexagon, and other regular figures that have a still greater number of sides, three angles of a regular hexagon are equal to four right angles, and those of the others still greater (I. 20.). It appears, therefore, that there cannot be more than five regular solids; of which, if there be so many, three will be included by equilateral triangles, one by squares, and one by regular pentagons. It will be seen in the section of problems, that all of these may be constructed, and that they are contained by 4, 8, 20, 6, and 12 faces respectively. The further consideration of solid angles is deferred to Book vi., where we shall find that the investigation of their properties is greatly facilitated by means of the spherical surface described about the angular point, and that they are, indeed, perfectly analogous to the properties of triangles upon such a surface. SECTION 3.-Of Solids contained by PROP. 21. If two triangular prisms have a principal edge of the one equal to a principal edge of the other, and the dihedral angles at those edges equal to one another, and likewise the sides which contain those dihedral angles equal, each to each; the two prisms shall be equal to one another. angular prisms which have the edge A B Let ABCDEF, abcdef be two triequal to the edge a b, the dihedral angle dabe, and the sides AC, A F equal to DABE equal to the dihedral angle the sides ac, af each to each: the prism ABCDEF shall be equal to the prism a b c d e f. First, let the angle ABC be equal to the angle abc, and the angle AB F to the angle abf, so that the sides AF, AC are similar and similarly placed to the sides af, ac, each to each. Then, if the straight line A B be made to coincide with the straight line ab, which is equal to it, and the plane A F with the plane af, the plane A C will also coincide with the plane ac, because the dihedral angle DABE is equal to the dihedral angle dabe: also the lines BC, B F will coincide with the lines bc, bf respectively; and therefore the lines AD, AE, which are parallel and equal to the former, each to each, will coincide respectively with the lines ad, ae, which are parallel and equal to the latter: therefore the prism A B C D E F will coincide with the prism abcdef. And because the two prisms may be made to coincide, they are equal to one another (I. ax. 11.). Next, let the sides A C, ac be equal and similar, and the sides A F, af equal but dissimilar. Make the angle ABG, in the plane AB F, equal to the angle abf; complete the parallelogram AB GH; and join CG, D H. Then because GH is equal and parallel to AB (I. 22.), which is equal and parallel to CD, the three, AB, CD, GH are equal and parallel, and, therefore, (def. 14.) ABCDHG is a triangular prism. Also, the parallelogram AHGB is equal to the parallelogram AEFB (I. 24.), that is, to the parallelogram a efb; and the parallelogram A D C B is equal to the parallelogram a dc b, by the supposition. Therefore, because the sides AC, AG of this new prism are respectively equal, similar, and similarly placed to the sides ac, af of the prism abcdef, it is equal to the prism abcdef, by the former case. Again, because H G is equal to DC, that is to EF, HE is equal to G F, and may be made to coincide with it; also the dihedral angle AEHD is equal to the dihedral angle BFGC, and may be made to coincide with it; in which case, also, the angles HED, HEA will coincide with the angles GFC, GFB, which (I. 15.) are equal to them respectively, and the lines E D, E A with the lines FC, FB, which are equal to them respectively (I. 22.). Therefore the solid HEAD, which is bounded by the four triangular planes HD E, HD A, HEA, EAD, may be made to coincide with the solid G F B C, which is bounded by the four triangular planes GCF, GCB, GFB, FB C, and (I. ax. 11.) is equal to it; and, therefore, adding the solid ABCDHF to each, the prism ABCDEF is equal to the prism A B C DH G, that is, to the prism abcdef. Lastly, let the sides A C, ac, as also A F, af, be equal, but dissimilar. Make the parallelogram AG similar to a f, and complete the triangular prism A B CDH G, as in the preceding case. Then, it is evident that the prism A B C DHG is equal to each of the prisms A B C D EF, abcdef; to the former, because the side AC is common to both, and the other sides AG, AF are equal to one another; and to the latter, because the sides AG, af are equal and similar (I. 24. and I. ax. 1.), and the other sides AC, a c equal to one another. Therefore (I. ax, 1.) the prism ABCDEF is equal to the prism abcdef. Therefore, &c. PROP. 22. (EUc. xi. 24 and 28.) The opposite faces of a parallelopiped are similar and equal parallelograms: also any two of its opposite edges are parallel to one another; and the plane which passes through them bisects the parallelopiped. A b B Let A a be a parallelopiped, and CD, cd any two opposite faces; they shall be similar and equal parallelograms. They are parallelograms, for the opposite sides of each, as A C, Db, being sections of parallel planes by the same plane, are parallel (12). Again, their corresponding sides, as AD, Bc, are equal to one another, because they are opposite sides of a parallelogram ABCD (I. 22.); and their corresponding angles, as CAD, d B c are equal to one another, because they are intercepted upon parallel planes CAD, dBc by the planes of the same dihedral angle CABD (15. Cor.): therefore the parallelograms CD, cd are equal and similar. Also, any two of the opposite edges of the parallelopiped, as A C, a c, are parallel to one another; for they are the common sections of parallel planes, namely, the opposite faces of the parallelopiped (12. Cor.). Lastly, therefore, let the parallelopiped be divided by the plane of any two opposite edges AC, ac, into the two triangular prisms dB AC ac, bDAC ac: these prisms shall be equal to one another. For the edges b D, dB, being each of them equal to CA (I. 22.), are equal to one another; and the dihedral angles at those edges are likewise equal (17. Scholium, 3.) and the sides which contain them are equal, each to each, because they are opposite faces of the parallelopiped. Therefore the prisms are equal to one another (20.). Therefore, &c. a Cor. 1. If a triangular prism be completed into a parallelopiped (which may be done by completing its triangular bases into parallelograms), the prism shall be equal to half theparallelopiped. Cor. 2. If a parallelopiped be cut by two planes which are drawn parallel respectively to two of its adjoining faces, the line of their in- A tersection will be parallel to the edge made by those faces (12. Cor.); and if that line be in the diagonal plane, and the parallelopiped be thus divided into four parallelopipeds, two of which are about the diagonal plane; the other two which, together with the former, make up the whole parallelopiped, shall be equal to one another. For, as in I. 23, the whole parallelopiped, and the two parts which are about the diagonal plane, are bisected by that plane. 'Scholium. It is worthy of remark, that "the four diagonals of a parallelopiped pass through the same point, and that the sum of their squares is equal to the sum of the squares of the twelve edges." For, if the diagonal planes which pass through A C, a c and BD, bd intersect one another in the line E F, the figures A Cac and BD bd will be parallelo D B d C a grams, having their opposite sides bisected by the line E F, (I. 22.); wherefore, the diagonals of each (which are the same with the diagonals of the parallelopiped), as, for instance, A a, will bisect EF; because the triangles AEG, a FG having the angles of the one equal to the angles of the other, each to each (I. 15.), and the side A E equal to the side a F, the side E G is likewise equal to the side FG (I. 5.). Therefore, the four diagonals of the parallelopiped pass through the same point, viz. the middle point of EF. And again, because A Cac and BDbd are parallelograms, the squares of their four diagonals are (1.41. Cor.) together equal to the squares of A C, a c, Ac, a C, and BD, bd, Bd, bD, that is to say, to the squares of the four edges AC, ac, Bd, bD, together with the squares of Ac, BD, and a C, bd, which are the diagonals of the parallelograms ABCD and ab Cd respec tively; that is, to the squares (I. 41. Cor.) of the twelve edges, which are made up of the sides of these parallelograms, and the four edges before mentioned. PROP. 23. (Etc. xi. 29 and 30.) Parallelopipeds upon the same base, and between the same parallel planes, are equal to one another. Let ABCD be the common base of two parallelopipeds, which have their opposite bases EFGH and KLMN lying in the same plane: the two parallelopipeds shall be equal to one another. Since EF and KL are both parallel to A B, either they are parts of the same straight line, or they are parallel to one another (6.). And first let them lie in the same straight line. Then, because LM N MH A G E P B C and EH are equal and parallel to B C, and therefore to one another (I. ax. 1. and 6.), M H is parallel to LE or KF (I. 21.), and MN, G H lie in the same straight line with MH and with each other. Therefore, the sides AL, DM, of the one parallelopiped, are in the same planes with the sides AF, DG of the other, each with each. And because the bases K M, EG are equal each to the base AC (22.), they are equal to one another. Therefore, also, the parallelogram K H is equal to the parallelogram LG. And because the triangular prisms KNHEAD, LMGFBC have the edges KN, LM equal to one another, and the dihedral angles at those edges equal (17. Scholium, 2.), and the sides containing those angles equal (22.), each to each, they are equal to one another (21.): but, if the first of these prisms be taken from the whole solid ADN KFGCB, there remains the parallelopiped AG; and if the other be taken from the same solid, there remains the parallelopiped AM: therefore the parallelopiped A G is equal to the parallelopiped A M. (I. ax. 3.) Next, let E F, K L not be in the same straight line. Let the planes ADHE, BCGF be produced to meet the planes ABLK, DCMN, themselves produced if necessary, and let the parallelopiped AP be completed* by producing * In this case, the sides of the parallelopiped to be completed are already drawn, viz. by producing the plete the parallelopiped, it is only necessary to draw sides of the parallelopipeds A M, A G; and, to comthe upper base by producing the plane of EFGH and straight line; and, for the like reason, the parallelopiped A M is equal to the same A P. Therefore (I. ax. 1.) the parallelopipeds AG, A M are equal to one another. Therefore, &c. PROP. 24. (Euc. xi. 31.) Parallelopipeds upon equal bases, and between the same parallel planes, are equal to one another. Let A G, KQ be two parallelopipeds upon equal bases ABCD, KLM N, and between the same parallel planes; the parallelopipeds AG, KQ shall be equal to one another. The bases ABCD, KLMN are either similar, or equiangular, or not equiangular. In the first case, since they are equal as well as similar, they may be made to coincide; and the parallelopipeds will then stand upon a common base, and between the same parallel planes: therefore they are equal to one another. Secondly, let them be equiangular, and not similar, having the angle at B equal to the angle at L; and therefore, also, their other angles equal, each to each, (I. 15. and I. ax. 3.). Pro KLMN to cut the sides. It may be observed, however, that a parallelopiped can always be completed when its base ABCD and one of its principal edges AE are given, viz. by drawing the side-planes EAB, EAD, and then through BC, CD the side-planes BG, CH parallel to EAD and EAB respectively, and, lastly, through the point E, the upper base E G parallel to ABCD. This operation is analogous to that by which a parallelogram is completed from two ad joining sides (see note, page 17); and will be sometimes understood when it is directed to complete a parallelopiped in future propositions. duce AB to R, so that BR may be equal to KL; complete the parallelogram BCSR; join S B, and produce it to meet D A produced in T; through T draw TU parallel to AB, to meet CB and S R produced in the points V and U; complete the parallelopiped DX upon the base DTUS, and with the edge DH; produce the planes ABFE, CBFG, in order to complete the parallelopiped BX, and draw the diagonal plane STY Z of the parallelopiped DX. Then the parallelopipeds BH and BX are equal to one another, because, together with the parallelopipeds BY and BZ about the diagonal plane, they complete the whole parallelopiped DX (22. Cor. 2.). Again, because the bases BU and BD are complements of the parallelogram DU, BU is equal to BD (I. 23.), that is, to LN; but BR is equal to KL, and the angle RBV to the angle ABC (I. 3.), that is, to KLM: therefore, also, B V is equal to L M, and the bases BU, LN are both equal and similar. Therefore, by the first case, the parallelopiped KQ is equal to the parallelopiped B X, that is, to BH or AG. Lastly, let the bases A B C D, KLM N be equal, but not equiangular. Make the angle ABS, in the plane ABCD, equal to the angle K L M, and complete the parallelogram A B ST, and the parallelopiped AX, upon the base A B ST Then it is and with the edge B F. evident that the triangular prisms D H EATY, CGFBSX have the edges DH, CG equal to one another (22.), and the dihedral angles at those edges equal (17. Scholium 2.), and the sides containing those angles equal, each to each, for the sides AH, B G are opposite faces of the parallelopiped AG (22.), and the sides TH, SG stand upon equal bases (I. ax. 1. and ax. 2. or 3.) TD, SC, and between the same parallels (I. 25.); therefore, the prisms are equal to one another (21.); and, these being taken from the whole solid AEYTCGFB, there remains the parallelopiped AG, equal to A X. But the parallelopiped KQ is equal to AX by the preceding case; because their bases are equal (I. 24. and |