PROP. XLIII. THEOR. The angle formed by perpendiculars let fall from a point within a dihedral angle, upon its faces, is the supplement of the measure of the dihedral angle. Let OC, OD (fig. 48) be perpendiculars from the point O within the dihedral angle MABN, on its faces MB, NB; the angle COD is the supplement of the inclination of the faces MB, NB. Through C, O, D let a plane pass, cutting the planes MB, NB in CE, DE; then, as in the last proposition, the angle CED is the measure of the dihedral angle MABN. And because the four angles of the quadrilateral CEDO are together equal to four right angles (I. 32); and that the angles OCE, ODE are right angles; the angles COD and CED are equal to two right angles: that is, the COD is the supplement of the angle CED which measures the dihedral angle MABN: which was to be proved. PROP. XLIV. THEOR. If a solid angle be contained by three plane angles, any two of them are together greater than the third. Let the solid angle at A (fig. 49) be contained by the three plane angles BAC, CAD, DAB: any two of them together shall be greater than the third. If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are together greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A in the straight line AB, make, in the plane which passes through BA, AC, the angle BAE equal to the angle DAB (I. 23); and make AE equal to AD, and through E draw BEC cutting AB, AC, in the points B, C; and join DB, DC. And because DA is equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, each to each; and the angle DAB is equal to the angle EAB: therefore the base DB is equal to the base BE (I. 4). And because DB, DC are greater than CB (I. 20), and one of them BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC (I. Ax. 3). Again, because DA is equal to AE, and AC common, but the base DC greater than the base EC; therefore the angle DAC is greater than the angle EAC (I. 25); and, by the construction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater (I. Ax. 4) than BAE, EAC, that is, than the angle BAC: but BAC is not less than either of the angles DAB, DAC: therefore BAC with either of them is greater than the other. Wherefore, if a solid angle, &c.: which was to be proved. PROP. XLV. THEOR. Every solid angle is contained by plane angles, which together are less than four right angles. First, let the solid angle at A (fig. 50) be contained by three plane angles BAC, CAD, DAB: these three together shall be less than four right angles. Take in each of the straight lines AB, AC, AD, any points B, C, D, and join BC, CD, DB. Then because the solid angle at B is contained by the three plane angles CBA, ABD, DBC, any two of them are greater than the third (Prop. 44); therefore the angles CBA, ABD are greater than the angle DBC: for the same reason, the angles BCA, ACD are greater than the angle DCB; and the angles CDA, ADB greater than BDC: wherefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles DBC, BCD, CDB: but the three angles DBC, BCD, CDB are equal to two right angles (1.32); therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles: and because the three angles of each of the triangles ABC, ACD, ADB are equal to two right angles, therefore the nine angles of these three triangles, viz. the angles CBA, BAC, ACB, ACD, CDA, DAC, ADB, DBA, BAD, are equal to six right angles: of these the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles; therefore the remaining three angles BAC, CAD, DAB, which contain the solid angle at A, are less than four right angles. Next, let the solid angle at A (fig. 51) be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB: these shall together be less than four right angles. Let the planes in which the angles are be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB. And because the solid angle at B is contained by three plane angles CBA, ABF, FBC, of which any two are greater than the third (Prop. 44), the angles CBA, ABF are greater than the angle FBC: for the same reason, the two plane angles at each of the points C, D, E, F, viz. those angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the angles of the polygon BCDEF: therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon: and because all the angles of the triangles are together equal to twice as many right angles as there are triangles (I. 32); that is, as there are sides in the polygon BCDEF; and that all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon (I. 32. Cor. 1); therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles (I. Ax. 1): but all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved; wherefore the remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at A, are less than four right angles. Therefore, every solid angle, &c.: which was to be proved. PROP. XLVI. THEOR. If from a point taken within a trihedral angle perpendiculars are let fall upon the faces of the angle, that is upon the planes which form it, these perpendiculars will be the edges or intersections of the planes of a new trihedral angle, the plane angles forming which are the supplements of the measures of the dihedral angles at the edges about the original solid angle; and the measures of the dihedral angles about which are the supplements of the plane angles forming that solid angle. From the point O (fig. 52) within the trihedral angle formed by the three plane angles ASB, ASC, BSC, let the perpendiculars OP, OQ, OR upon the faces ASB, ASC, BSC be drawn; these perpendiculars are the edges of a trihedral angle, the plane angles of which, POQ, POR, QOR, are respectively the supplements of the measures of the dihedral angles at the edges SA, SB, SC, about the solid angle at S. Let PT, QT be the intersections of the plane POQ with the planes ASB, ASC; PV, RV the intersections of the plane POR with the planes ASB, BSC; and RU, QU the intersections of the plane QOR with the planes BSC, ASC. Then since OP and OQ are perpendiculars let fall from the point O upon the planes ASB, ASC which are the faces of the dihedral angle whose edge is SA, the angle POQ, one of the plane angles forming the solid angle at O, is the supplement of the angle PTQ which measures the corresponding dihedral angle about the solid angle at S (Prop. 43). In the same manner it may be shown that the other plane angles POR, QOR, forming with POQ the solid angle at O, are the supplements of PVR, QUR which measure the other dihedral angles about the solid angle at S. Also, the measures of the dihedral angles at the edges OP, OQ, OR, about the solid angle at O, are respectively the supplements of the plane angles ASB, ASC, BSC forming the solid angle at S. Since OP is at right angles to the plane ASP, it is at right angles to TP and PV (Def. 3), and therefore the angle TPV measures the dihedral angle whose faces are the planes OPTQ, OPVR, and whose edge is PO. And because ST and SV are perpendiculars from the point S upon the planes POQ, POR (Prop. 34), which are the faces of the dihedral angle whose edge is OP, the angle TSV is the supplement of the angle TPV, which measures the dihedral angle TPOV whose edge is OP (Prop. 43) ; and consequently the measure of the dihedral angle TPOV is the supplement of the angle ASB. In the same manner it may be shown, that the angle TQU, which measures the dihedral angle TQOU whose edge is OQ, is the supplement of the angle ASC; and that the angle URV, which measures the dihedral angle VROU whose edge is OR, is the supplement of the angle BSC. Therefore, if from a point taken within a trihedral angle, &c.: which was to be proved. Scholium. On account of these properties the trihedral angles whose vertices are S and O are called supplemental to each other. If two trihedral angles have the three faces or plane angles of the one equal to the three faces of the other, each to each; the dihedral angles contained by the faces of the one shall be equal to the dihedral angles contained by the faces equal to them of the other. Let the trihedral angles at S and V (fig. 53) have the three faces ASB, ASC, BSC, at S, equal to the three faces DVE, DVF, EVF, at V, each to each; the dihedral angle contained by the faces ASB and ASC shall be equal to the dihedral angle contained by the faces DVE and DVF; the angle contained by ASB and BSC equal to the angle contained by DVE and EVF; and the angle contained by ASC and BSC equal to the angle contained by DVF and EVF. In SA, SB, SC and VD, VE, VF take SG, SH, SI and VK, VL, VM all equal to each other; join GH, HI, IG and KL, LM, MK. And because SG is equal to SH, the angle SHG is equal to the angle SGH (I. 5), and therefore each of these angles is less than a right angle (I. 17). In the same manner it may be shown, that each of the angles SGI, SIG is less than a right angle; and that each of the angles VKL, VKM is less than a right angle. In GS take any point N, and from it, in the plane ASB, draw NP at right angles to SA, meeting GH, or GH produced (because the angles NGH, GNP are less than two right angles) in P; and in the plane ASC draw NO at right angles to SA, meeting GI, or GI produced (I. Ax. 12) in O: join OP. In KV take KQ equal to GN, and from Q draw, in the planes DVE, DVF, QT, QR at right angles to VD, meeting KL, KM, or these produced (I. Ax. 12) in T, R: and Join RT. Because in the two triangles GSH, KVL, the two sides GS, SH are equal to the two KV, VL, each to each, and their contained angles GSH, KVL are equal (by hypothesis), the base GH is equal to the base KL, and the angles SGH, SHG equal to the angles VKL, VLK, each to each (I. 4). In like manner it may be shown that GI is equal to KM, the angles SGI, SIG equal to VKM, VMK, each to each; that IH is equal to ML, and the angles SIH, SHI equal to VML, VLM, each to each. And because, in the two triangles PNG, TQK, the angles PNG, PGN are equal to TQK, TKQ, each to each, and GN is equal to KQ, GP is equal to KT and NP to QT (I. 26). In the same manner it may be shown, in the two triangles NGO, QKR, that GO is equal to KR, and NO to QR. Again, in the two triangles HGI, LKM, the sides HG, GI are equal LK, KM, each to each, and IH is equal to ML, therefore the angle HGI is equal to the angle LKM (I. 8). And because, in the two triangles PGO, TKR, the sides PG, GO are equal TK, KR, each to each, and the angle PGO has been shown to be equal to the angle TKR, the base PO is equal to the base TR (I. 4); and it has been shown that the sides PN, NO in the triangle PNO are equal to the sides TQ, QR in the triangle TQR, therefore the angle PNO is equal to the angle TQR (I. 8). But the angle PNO is the inclination of the planes ASB and ASC, because NP and NO are drawn, in these planes, perpendicular to SA (Def. 7); and for the same reason the angle TQR is the inclination of the planes DVE, DVF; therefore the inclination of the planes ASB, ASC is equal to the inclination of the planes DVE, DVF, that is, the dihedral angle contained by the faces ASB, ASC is equal to the dihedral angle contained by the faces DVE, DVF (Prop. 38). In like manner it may be shown, that the dihedral angle contained by the faces ASB, BSC is equal to that contained by the faces DVE, EVF, and the angle contained by the faces ASC, BSC equal to that contained by the faces DVF, EVF. Wherefore if two trihedral angles, &c.: which was to be proved. Scholium. The demonstration is precisely the same whether the faces which are equal follow the same order, in the same direction, about the two solid angles, or not; that is, supposing the intersection SC of the two faces ASC, BSC to be in front of the face ASB, whether the intersection VF of the two faces DVF, EVF be in front of the face DVE, or behind it. In the former case, the solid angles at S and V are equal, because if the solid angle at S be applied to the solid angle at V so that the face ASB coincide with the face DVE, the faces ASC and BSC will coincide with the faces DVF and EVF, the dihedral angles contained by the former faces with ASB being equal to those contained by the latter faces with DVF, and also in the same direction; in the latter case the solid angles at S and V are only symmetrical, because their faces will not coincide, the dihedral angles though equal not being in the same direction. Cor. It follows from this and Prop. 46, that if the three dihedral angles about two trihedral angles be equal, each to each, the faces or plane angles which form the trihedral angles will be equal, each |