DH with NR, and the point H with R. Also, because the solid angles at B and L are equal, it may be proved, in the same manner, that the figure BG coincides with LQ, the straight line CG with MQ, and the point G with Q. Since, therefore all the planes and sides of AG coincide with those of KQ, AG is equal and similar (XI. def. 8.) to KQ. And, in the same manner, any other solid figures whatever, contained by the same number of equal and similar planes, alike situated, and having none of their solid angles contained by more than three plane angles, may be proved to be equal and similar to one another. Therefore solid figures, &c. PROP. XXIV. THEOR. If a solid be contained by six planes, two and two of which are parallel, the opposite planes are similar and equal parallelograms. Let the solid BE be contained by the parallel planes AC, GF; BG, CE; BF, AE: its opposite planes are similar and equal parallelograms. Because the parallel planes BG, CE are cut by the plane AC, their common sections AB, CD are (XI. 16.) parallel. Also, because the parallel planes BF, AE are cut by the plane AC, their common sections AD, BC are parallel : and AB, CD are parallel ; therefore AC is a parallelogram. In like manner, it may be proved, that CE, FG, GB, BF, AE are parallelograms. Join AH, DF; and because AB is parallel to DC, and BH to CF, the angle ABH is equal (XI. 10.) to DCF. Then, because AB, BH are equal to DC, CF, and the angle ABH to DCF; the triangle ABH is equal (i. 4.) to DCF. But (I. 34.) the parallelogram BG, is double of the triangle ABH, and CE is double of DCF ; therefore the parallelogram BG is equal and similar to the parallelogram CE. In the same manner it may be proved, that AC is equal and similar to GF, and AE to BF. Therefore, if a solid, &c. B н G E PROP. XXV. THEOR. A PLANE cutting a parallelepiped, and parallel to two of its opposite planes, divides the whole into two solids, the base of one of which is to the base of the other, as the one solid is to the other. Let the parallelepiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole X B G T z Р V D U into the two solids AV, ED; as the base A EFY of the first is to the base EHCF of the other, so is AV to ED. Produce AH both ways, and take any number of straight lines HM, MN, each equal to EH, and any number AK, KL, each equal to EA, and complete the parallelograms LO, KY, HQ, Ms, and the solids LP, KR, HU, MT. Then, because LK, KA, AE are all equal, the parallelograms LO, KY, AF (I. 36.) are equal : and likewise the parallelograms KX, KB, AG; as also (XI. 24.) the parallelograms LZ, KP, AR, because they are opposite planes. For the same reason, the parallelograms EC, HQ, MS, are equal ; and the parallelograms HG, HI, IN, as also HD, MU, NT. Therefore three planes of the solid LP, are equal and similar to three planes of KR, as also to three planes of AV. But (XI. 24.) the three planes opposite to these three are equal and similar to them in the several solids, and none of their solid angles are contained by more than three plane angles : therefore (XI. C.) the solids LP, KR, AV are equal to one another. For the same reason, ED, HU, MT are equal to one another. Therefore whatever multiple the base LF is of AF, the same multiple is the solid LV of AV. For the same reason, whatever multiple the base NF is of HF, the same multiple is the solid TE of ED. And if the base LF be equal to NF, the solid LV is equal (XI. C.) to ET; if greater, greater ; and if less, less. Therefore (V. def. 5.) as the base AF is to the base FH, so is the solid AV to the solid ED: wherefore a plane, &c. A R E K H н M N 1 Y PROP. XXVI. PROB. At a given point in a given straight line, to make a solid angle equal to a given solid angle contained by three plane angles. Let A be a given point in a given straight line AB, and D a given solid angle contained by the three plane angles EDC, EDF, FDC; it is required to make at A in the straight line AB a solid angle equal to the solid angle D. În DF take any point F, from which draw (XI. 11.) FG perpendicular to the plane EDC, meeting that plane in G; join DG, and (I. 23.) make the angle BAL equal to EDC, and in the plane BAL make the angle BAK equal to EDG; then make AK equal to DG, and (XI. 12.) draw KH perpendicular to the plane BAL, and equal to GF, and join AH.' Then the solid angle at A, which is contained by the plane angles, BAL, BAH, HAL, is equal to the given solid angle at D. Take AB, DE equal to one another; and join HB, KB, FE, GE: and (XI. def. 1.) because FG is perpendicular to the plane EDC, FGD, FGE are right angles. For the same reason, HKA, HKB are right angles : and because KA, AB are equal to GD, DE, each to each, and contain equal angles, BK is equal (I. 4.) to EG ; also KH is equal to GF, and HKB, FGE are right angles ; therefore HB is equal to FE. Again, because AK, KH are equal to DG, GF, and contain right angles, AH is equal to DF; also AB is equal to DE, and HB to FE; therefore (I. 8.) the angles BAH, EDF are equal. Again, since (const.) the angle BAL is equal to EDC, and BAK to EDG, the maining angles KAL,GDC are(I.ax.3.) equal to one another; and, by taking AL and DC equal, and joining LH, LK, CF, CG, it would be proved, as in the foregoing part, that the angle HAL is equal (I. 8.) to FDC. Therefore, because the three plane angles BAL, BAH, HAL, which contain the solid angle at A, are equal to the three EDC, EDF, FDC, which contain the solid angle at D, each to each, and are situated in the same order, the solid angle at A is equal (XI. B.) to the solid angle at D. Therefore what was required has been done. re E B L с H PROP. XXVII. PROB. To describe from a given straight line a parallelepiped similar, and similarly situated to one given. L D H Let AB be the given straight line, and CD the given parallelepiped. It is required from AB to describe a parallelepiped similar and similarly situated to CD. At the point A of the straight line AB, make (XI. 26.) a solid angle equal to the solid angle at C, and let BAK, KAH, HAB be the plain angles which contain it, so that BAK is equal to ECG, KAH to GCF, and HAB to FCE: also as EC to CG, so make (VI. 12.) BA to AK; and as GC to CF, so make KA to AH; wherefore, ex æquo, as EC: CF:: BA: AH. Complete the parallelogram BH, and the solid AL. Then, because EC: CG:: BA : AK, the sides about the equal angles ECG, BAK are proportionals; therefore the parallelogram BK is similar to EG. For the same reason KH is similar to GF, and HB to FE. Therefore three parallelograms of the solid AL are similar to three of the solid CD; and. (XI. 24.) the three opposite ones F A B с B in each are equal and similar to these, each to each. Also, because the plane angles which contain the solid angles of the figures are equal, each to each, and situated in the same order, the solid angles (XI. B.) are equal, each to each. Therefore (XI. def. 8.) AL is similar to CD. Wherefore, from a given straight line AB a parallelepiped AL has been described similar, and similarly situated to the given one CD: which was to be done. PROP. XXVIII. THEOR. If a parallelepiped be cut by a plane passing through the diagonals of two of the opposite planes; it is bisected by that plane. с B G D H Н Let AB be a parallelepiped, and DE, CF the diagonals of the opposite parallelograms AH, GB, viz., those which join the equal angles in each. Then (XI. 9.) CD, FE are parallels, because each of them is parallel to GA; wherefore (XI. 7.) the diagonals CF, DE are in the plane in which the parallels are, and (XI. 16.) are themselves parallels. Again, because (I. 34.) the triangle CGF is equal to CBF, and DAE to DHE, and that (XI. 24.) the parallelogram CA is equal and similar to the opposite one BE; and GE to CH: therefore the prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal (XI. C.) to the prism contained by the two triangles CBF, DHE, and the three parallelograms BE, CH, EC; because they are contained by the same number of equal and similar planes, alike situated, and none of their solid angles are contained by more than three plane angles. Therefore, if a parallelepiped, &c. “ N.B. The insisting lines of a parallelepiped, mentioned in some following propositions, are the sides of the parallelograms between the base and the opposite plane." A PROP. XXIX. THEOR. *PARALLELEPIPEDS upon the same base, and of the same altitude, the insisting lines of which are terminated in the same straight lines of the plane opposite to the base, are equal to one another. Let the parallelepipeds AH, AK (2d fig.) be upon the same base AB, and of the same altitude; and let their insisting lines AF, AG, LM, LN, be terminated in the same straight line D H K F N B в A FN, and CD, CE, BH, BK in the same DK; AH is equal to AK. First, let the parallelograms DG, HN, which are opposite to the base AB, have a common side HG. Then because A His cut by the plane AGHC passing through the diagonals AG, CH, of the opposite planes ALGF, CBHD, AH is bisected (XI. 28.) by the plane AGHC. For the same reason, AK is bisected by the plane LGHB through the diagonals LG, BH. Therefore the solids AH, AK are equal, each of them being double of the prism contained between the triangles ALG, CBH. But let the parallelograms DM, EN, opposite to the base, have no common side. Then, because CH, CK are parallelograms, CB is equal (I. 34.) to each of the opposite sides DH, EK; wherefore DH is equal to EK. From DK take separately, EK, DH; then DE is equal to HK: wherefore also (1. 38.) the triangles CDE, BHK are equal : and (I. 36.) the parallelogram DG is equal to HN. For the same reason, the triangle AFG is equal to LMN; and (XI. 24.) the parallelogram CF is equal to BM, and CG to BN; for they are opposite. Therefore (XI. C.) the prism which is contained by the two triangles AFG, CDE, and the three parallelograms AD, DG, GC, is equal to the prism contained by the two triangles LMN, BHK, and the three parallelograms BM, MK, KL. If therefore the prism LMNBHK be taken from the solid of which the base is the parallelogram AB, and in which FDKN is the one opposite to it; and if from the same solid there be taken the prism AFGCDE; the remaining parallelepipeds AH, AK are equal. Therefore parallelepipeds, &c. D E H K F G M B C A L PARALLELEPIPEDS upon the same base and of the same altitude, the insisting lines of which are not terminated in the same straight lines in the plane opposite to the base, are equal to one another. Let the parallelepipeds CM, CN, be upon the same base AB, and of the same altitude, but their insisting lines AF, AG, LM, LN, CD, CE, BH, BK, not terminated in the same straight lines: CM, CN are equal to one another. Produce FD, MH, and NG, KE, and let them meet one another in O, P, Q, R ; and join AO, LP, BQ, CR. Then, because the |