PROP. XXXIV. THEOR. R D M X K B T L P The bases and altitudes of equal parallelepipeds are reciprocally proportional: and (1.) if the bases and altitudes be reciprocally proportional, the parallelepipeds are equal. 1. Let AB, CD be equal parallelepipeds; their bases are reciprocally proportional to their altitudes; that is, as the base EỈ is to the base NP, so is the altitude of CD to the altitude of AB. First, let the insisting lines AG, CM, &c., be at right angles to the bases. As the base EH to NP, so is CM to AG. If EH be equal to NP, then because AB is likewise equal to CD, CM is equal to AG. For, if EH, NP be equal, but AG, CM be not equal, neither is AB equal to CD. But (hyp.) the solids are equal. Therefore CM is not unequal to AG; that is, they are equal. Wherefore, as EH to NP, SO is CM to AG. Next, let the bases EH, NP not be equal, but EH greater than the other. Since then the solid AB is equal to the solid CD, CM is greater than AG. For, if it be not, neither also, in this case, would the solids AB, CD be equal, which (hyp.) are equal. Make then CT equal to AG, and complete the parallelepiped CV, of which the base is NP and altitude CT. Because AB is equal to CD, therefore (V. 7.) AB is to CV, as CD to CV. But (XI. 32.) as AB to CỤ, so is EH to NP; for the solids AB, CV are of the same altitude ; and (XI. 25.) as CD to CV, so is the base MP to PT, and (VI. 1.) so is the straight line MC to CT; and CT is equal to AG. Therefore, as EH to NP, so is MC to AG: wherefore (VI. def. 2.) the bases of the parallelepipeds AB, CD are reciprocally proportional to their altitudes. 2. Let now the bases of the parallelepipeds AB, CD be reciprocally proportional to their altitudes ; viz., as EH to NP, so let the altitude of CD be to the altitude of AB; AB is equal to CD. Let the insisting lines be, as before, at right angles to the bases. Then, if EH be equal to NP, since EH is to NP, as the altitude of CD is to the altitude of AB, therefore (V. A.) the altitude of CD is equal to the altitude of AB. But (XI. 31.) parallelepipeds upon equal bases, and of the same altitude, are equal to one another; therefore AB is equal to CD. But let the bases EH, NP be unequal, and let EH be the greater. Therefore, since EH is to NP as CM, the altitude of А E с N D B F X P A E с N CD, is to AG, the altitude of AB, CM is greater (V. A.) than AG. Take CT equal to AG, and complete, as before, the solid CV. Then, because EH is to NP, as CM to AG, and that AG is equal to CT, therefore EH is to NP, as MC to CT. But as EH is to NP, so is AB to CV: for AB, CV are of the same altitude ; and as MC to CT, so (VI. 1.) is the base MP to PT, and (XI. 25.) the solid CD to CV: and therefore as AB to CV, so is CD to CV; wherefore (V. 9.) AB is equal to CD. Second general case. 1. Let the insisting lines FE, XN, &c., not be at right angles to the bases of the solids. Then, if a parallelepiped be described on the base EH having the same altitude as AB, and having its insisting lines perpendicular to its base, it will (XI. 29. or 30.) be equal to AB. For the same reason, if a parallelepiped be described on the base NP of the same altitude as CD, and having its insisting lines perpendicular to its base, it will be equal to CD. These parallelepipeds, therefore, are equal, if AB, CD be equal : and, by the first general case, their bases and altitudes, which are also the bases and altitudes of AB, CD, are reciprocally proportional. 2. Next, let the bases EH, NP be reciprocally proportional to the altitudes ; AB is equal to CD. For, if the same construction were made, the bases and altitudes of the parallelepipeds so constructed would be reciprocally proportional to one another, being the same as those of AB, CD: and therefore, by the first general case, those parallelepipeds are equal; and so also are AB, CD, which are equal to them. Therefore the bases, &c. PROP. XXXV. THEOR. IF, from the vertices of two equal plane angles, there be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the sides of those angles, each to each ; and if from any points in the lines above the planes perpendiculars be drawn to those planes ; and if from the points in which they meet the planes, straight lines be drawn to the vertices of the first mentioned angles; these straight lines make equal angles with the straight lines above the planes. Let BAC, EDF be two equal plane angles ; and from A, D let A the straight lines AG, DH be elevated above the planes of the angles, making the angle GAB equal to HDE, and GAC to HDF; from any points G, H, in those lines let perpendiculars GL, HK be drawn (XI. 11.) to the planes BAC, EDF meeting these planes in L, K, and join LA, KD: the angles GAL, HDK, are equal. If AG, DH be equal, then the angle GAL is equal to HDK, as was shown in the corollary to : proposition B. of this book. But if AG, DH be unequal, make AM equal to DH, and (I. 31.) draw MN parallel to GL, meeting AL, the common section of the planes BAC, GAL, in N. Then (XI. 8.) MN is perpendicular to the plane BAC, because GL, which is parallel to MN, is perpendicular to BAC: and therefore, by the first case, the angle MAN, that is, GAL, is equal to HDK. * If, therefore, &c. E B M N с F H G L PROP. XXXVI. THEOR. IF three straight lines be proportionals, any parallelepiped described from all three as its sides, is equal to the equilateral parallelepiped described from the mean proportional, 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. Let A : B:: B: C: any parallelepiped described from A, B, C, is equal to the equilateral parallelepiped described from B, equiangular to the other. Take a solid angle D contained by three plane angles EDF, FDG, GDE; and make ED, DF, DG, each equal to B, and complete the parallelepiped DH. Make LK equal to A, and (XI. 26.) at the point K in the straight line LK make a solid angle contained by the three plane angles LKM, MKN, NKL equal to the angles EDF, FDG, GDE, each to each : make also KN equal to B, and KM equal to C; and complete the parallelepiped KO. Then, because A: B::B:C, and that A is equal to LK, and B to each of the lines DE, DF, and C to KM; therefore LK : ED :: H N M L E А B с * This short and easy demonstration is substituted for the very tedious one given by Euclid. DF: KM; that is, the sides about the equal angles are reciprocally proportional; therefore (VI. 14.) the parallelogram LM is equal to EF. Also, 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 (XI. B. cor.) the perpendiculars from G, N, to the planes EDF, LKM, are equal." The solids KO, DH, therefore, are of the same altitude ; and since they are upon equal bases LM, EF, they are equal (XI. 31.) to one another. "If, therefore, &c. PROP. XXXVII. THEOR. If four straight lines be proportionals, the similar parallelepipeds similarly described from them are also proportionals: and (2.) if the similar parallelepipeds similarly described from four straight lines be proportionals, the lines are proportionals. 1. Let the four straight lines AB, CD, EF, GH be proportionals, viz., as AB to CD, so EF to GH; and let the similar parallelepipeds AK, CL, EM, GN be similarly described from them: AK is to CL, as EM to GN. Make (VI. 11.) AB, CD, O, P continual proportionals, as also EF, GH, Q, R. Then, because AB: CD :: EF:GH; and that (V. 11.) CD:0:: GH: Q, and 0 : P::Q:R; therefore, ex æquo, AB:P :: EF: R. But (XI. 33. cor.) as AB to P, so is AK to CL ; and as EF to R, so is EM to GN: therefore (V. 11.) as AK to CL, so is EM to GN. 2. Next, let AK be to CL, as EM to GN: the straight line AB is to CD as EF to GH. Take (VI. 12.) AB to CD, as EF to ST, and (XI. 27.) from ST describe a parallelepiped SV similar and similarly situated to either EM or GN. Then, by the preceding part, AK : CL :: EM:SV. But (hyp.) AK : CL :: EM:GN: therefore (V. 9.) GN is equal to SV : it is likewise similar and similarly situated to SV; therefore the planes which contain the solids GN, SV are similar and equal, and their homologous sides GH, ST equal to one another : and because AB : CD:; EF: ST, and that ST is equal to GH; AB : CD :: EF: GH. If, therefore, &c. PROP. XXXVIII. THEOR. с If one plane be perpendicular to another, a straight line drawn from a point in one of them perpendicular to the other, falls on their common section. Let the plane CD be perpendicular to AB, and let AD be their common section ; if any point E be taken in CD, the perpendicular drawn from E to AB will fall on AD. From E draw EG in the plane CD perpendicular to AD, and in AB draw GF also perpendicular to AD. Then (XI. def. 3.) EG is perpendicular to GF; and (const.) it is perpendicular to AD: it is therefore (XI. 4.) perpendicular to the plane AB : and (XI. 13.) through E no other straight line can be drawn perpendicular to AB. Therefore the perpendicular from E to AB falls on AD. Wherefore, if one plane, &c. E A D F B PROP. XXXIX. THEOR. D K In a parallelepiped, if the sides of two of the opposite planes be each bisected, the common section of the planes passing through the points of division, and any diagonal of the parallelepiped bisect each other. Let the sides of the opposite planes CF, AH of the parallelepiped AF, be bisected in the points K, L, M, N; X, O, P, R; and join, KL, MN, XO, PR. Then, (I. 33.) because DK, CL are equal and parallel, KL is parallel to DC. For the same reason MN is parallel to BA. But BA is parallel to DC; therefore (XI. 9.) KL is parallel to BA. Also (XI. 9.) because KL, MN are each of them parallel to BA, KL is parallel to MN; wherefore (I. def. 11.) KL, MN are in one plane. In like manner, it may be proved, that x0, PR are in one plane. Let YS be the common section of the planes KN, XR; and DG a diagonal of the parallelepiped AF: YS and DG meet and bisect one another. Join DY, YE, BS, SG. Because DX is equal to OE, and E L T M H B R P S N |