draw perpendiculars to the planes in which are the bases EH, NP, meeting those planes in the points S, Y, V, T; Q, I, U, Z; and complete the solids FV, XU, which are parallelopipeds, as was proved in the last part of Prop. 31, of this book. K B R D Because the solid AB is equal to the solid CD, and that the solid AB is equal to the solid BT (h), for they are upon the same base FK, and of the same altitude; and that the solid CD is equal to the solid DZ (h), being upon the same base XR, and of the same alti tude; therefore the solid BT is equal to Н S E (h) XI. 29 or 30. the solid DZ: but the bases are reciprocally proportional to the altitudes of equal solid parallelopipeds of which the insisting straight lines are at right angles to their bases, as before was proved; therefore as the base FK to the base XR, so is the altitude of the solid DZ to the altitude of the solid BT: and the base FK is equal to the base EH, and the base XR to the base NP; wherefore, as the base EH is to the base NP, so is the altitude of the solid DZ to the altitude of the solid BT; but the altitudes of the solids DZ, DC, as also of the solids BT, BA, are the same; therefore as the base EH to the base NP, so is the altitude of the solid CD to the altitude of the solid AB; that is, the bases of the solid parallelopipeds AB, CD are reciprocally proportional to their altitudes. [2] Next, let the bases of the solids AB, CD be reciprocally proportional to their altitudes, viz. the base EH is to the base NP, as the altitude of the solid CD is to the altitude of the solid AB: the solid AB shall be equal to the solid CD. The same construction being made; because, as the base EH is to the base NP, so is the altitude of the solid CD to the altitude of the solid AB; and that the base EH is equal to the base FK, and NP to XR; therefore the base FK is to the base XR, as the altitude of the solid CD to the altitude of AB; but the altitudes of the solids AB, BT are the same, as also of CD and DZ; therefore as the base FK is to the base XR, so is the altitude of the solid DZ to the altitude of the solid BT: wherefore the bases of the solids BT, DZ are reciprocally proportional to their altitudes: and their insisting straight lines are at right angles to the bases; wherefore, as was before proved, the solid BT is equal to the solid DZ: but BT is equal to the solid BA (h), and DZ to the solid DC, because they are upon the same bases, and of the same altitude; therefore the solid AB is equal to the solid CD. PROPOSITION XXXV. THEOREM.-If, from the vertices (A and D) of two equal plane angles (BAC, EDF), there be drawn two straight lines (AG, DM) elevated above the planes in which the angles are, and containing equal angles with the sides of those angles, each to each (GAB to MDE, and GAC to MDF); and if in the lines (AG, DM) above the planes there be taken any points (G, M), and from them perpendiculars (GL, MN) be drawn to the planes in which are the first-named angles (BAC, EDF); and from the points (L, N) in which they meet the planes, straight lines (LA, ND) be drawn to the vertices of the angles first-named; these straight lines shall contain equal angles (GAL, MDN) with the straight lines which are above the planes of the angles. DEMONSTRATION. Make AH equal to DM, and through H draw HK parallel to GL: but GL is perpendicular to the plane BAC; wherefore HK is perpendicular to the same plane (a). From the points K, N, to the straight lines AB,AC, DE, DF, draw perpendiculars KB, KC, NE, NF, and join HB, BC, ME, EF. Because HK is perpendicular to the plane BAC, the plane HBK which passes through HK is at right angles to the plane BAC (); and AB is drawn in the plane BAC at right angles to the common section BK of the two planes; therefore AB is perpendicular to the plane HBK (c), and makes right angles with every straight line meeting it in that plane (d): but BHI meets it in that plane; therefore ABH is a right angle: for the same reason, DEM is a right angle, and is therefore equal to the angle ABH: and the angle HAB is equal to the angle MDE (e); therefore in the two triangles HAB, MDE, there are two angles in one, equal to two angles in the other, each to each, and one side equal to one side, opposite to one of the equal angles in each, viz. HA equal to DM; therefore the remaining sides are equal, each to each (ƒ), wherefore AB is equal to DE. In the same manner, if HC and MF be joined, it may demonstrated, be that AC is equal to DF: therefore, since AB is equal to DE, BA and AC are equal to ED and DF, each to each; and the angle BAC is equal to the angle EDF (e); wherefore the base BC is equal to the base EF (g), and the remaining angles to the remaining angles; therefore the angle ABC is equal to the angle DEF: and the right angle ABK is equal to the right angle DEN; whence the remaining angle CBK is equal to the remaining angle FEN: for the same reason, the angle BCK is equal to the angle EFN; therefore in the two triangles BCK, EFN there are two angles in one, equal to two angles in the other, each to each, and one side equal to one side adjacent to the equal angles in each, viz. BC equal to EF; therefore the other sides are equal to the other sides; BK then is equal to EN: but AB is equal to DE; wherefore AB, BK are equal to DE, EN, each to each; and they contain right angles; wherefore the base AK is equal to the base DN. And since AH is equal to DM, the square on AH is equal to the square on DM: but the squares on AK, KH are equal to the square on AH, because AKH is a right angle (h); and the squares on DN, NM are equal to the square on DM, for DNM is a right angle: wherefore the squares on AK, KH are equal to the squares on DN, NM: and of these the square on AK is equal to the square on DN; therefore the remaining square on KH is equal to the remaining square on NM; and the straight line KH to the straight line NM; and because HA, AK are equal to MD, DN, each to each, and the base HK to the base MN, as has been proved, therefore the angle HAK, that is, GAL, is equal to the angle MDN (i). COROLLARY. From this it is manifest, that if from the vertices of two equal plane angles, there be elevated two equal straight lines containing equal angles with the sides of the angles, each to each; the perpendiculars drawn from the extremities of the equal straight lines to the planes of the first angles, are equal to one another. SCHOLIUM. Of this Corollary another demonstration may be given, as follows: Let the plane angles BAC, EDF be equal to one another, and let AH, DM be two equal straight lines above the planes of the angles, containing equal angles with BA, AC; ED, DF, each to each, viz. the angle HAB equal to MDE, and HAC equal to the angle MDF; and from H, M let HK, MN be perpendiculars to the planes BAC, EDF: HK shall be equal to MN. Because the solid angle at A is contained by the three plane angles BAC, BAH, HAC, which are, each to each, equal to the three plane angles EDF, EDM, MDF, containing the solid angle at D; the solid angles at A and D are equal, and therefore coincide with one another; to wit, if the plane angle BAC be applied to the plane angle EDF, the straight line AH coincides with DM, as was shown in Prop. B, of this book: and because AH is equal to DM, the point H coincides with the point M: wherefore HK, which is perpendicular to the plane BAC, coincides with MN, which is perpendicular to the plane EDF, because these planes coincide with one another (a). Therefore HK is equal to MN. PROPOSITION XXXVI. THEOREM.-If three straight lines (A, B, C) be proportionals, the solid parallelopiped described from all three, as its sides, is equal to the equilateral parallelopiped described from the mean proportional (B), one of the solid angles of which is contained by three plane angles equal, each to each, to the three plane angles containing one of the solid angles of the other figure. DEMONSTRATION. Take a solid angle D, contained by three plane angles EDF, FDG, GDE; and make each of the straight lines ED, DF, DG equal to B, and complete the solid parallelopiped DH: make LK equal to A, and at the point K, in the straight line LK, make a solid angle contained by the three plane angles LKM, MKN, ÑKL, equal H N M K E A B (a) XI. 26. to the angles EDF, FDG, GDE, each to each (a); and make KN equal to B, and KM equal to C; and complete the solid parallelopiped KO. And because, as A is to B, so is B to C, and that A is A M N B equal to LK, and B is equal to each of the straight lines DE, DF, and C is equal to KM; therefore LK is to ED, as DF to KM; that is, the sides about the equal angles are reciprocally proportional; therefore the parallelogram LM is equal to EF (6): and because EDF, LKM are two equal plane angles, and the two equal straight lines DG, KN are drawn from their vertices above their planes, and contain equal angles with their sides; therefore the perpendiculars from the points G, N, to the planes EDF, LKM are equal to one another (c): therefore the solids KO, DH are of the same altitude; and they are upon equal bases LM, EF; and therefore they are equal to one another (d): but the solid KO is described from the three straight lines A, B, C, and the solid DH from the straight line B. PROPOSITION XXXVII. (b) VI. 14. THEOREM [1.]-If four straight lines (AB, CD, EF, GH) be proportionals, the similar solid parallelopipeds (AK, CL, FM, HN) similarly described from them shall also be proportionals and if the similar parallelopipeds similarly described from four straight lines be proportionals, the straight lines shall be proportionals. DEMONSTRATION [1.] Make AB, CD, O, P, continual proportionals, as also EF, GH, Q, R (a): and because as AB is to CD, so is EF to GH; and that CD is to O, as GH to Q (b), and O is to P, as Q to R; therefore, ex aquali, AB is to P, as EF to R (c): but as AB is to P, so is the solid AK to the solid CL (d); and as EF is to R, so is the solid FM to the solid HN (d): |