of any two contiguous planes in the one solid be the same with the inclination of the two equal, and similarly situated planes in the other, the solids themselves are equal and similar. Let AG and KQ be two solids contained by the same number of equal and similar planes, similarly situated, so that the plane AC is similar and equal to the plane KM, the plane AF to the plane KP, BG to LQ, GD to QN, DE to NO, and FH to PR. Let also the inclination of the plane AF to the plane AC be the same with that of the plane KP to the plane KM, and so of the rest; the solid KQ is equal and similar to the solid AG. Let the solid KQ be applied to the solid AG, so that the bases KM and AC, which are equal and similar, may coincide, the point N coinciding with the point D, K with A, L with B, and so on. And because the plane KM coincides with the plane AC, and, DL------by hypothesis, the inclination V of KR to KM is the same A B with the inclination of AH to AC, the plane KR will be upon the plane AH, and will coincide with it, because they are similar and equal, and because their equal sides KN and AD coincide. And in the same manner, it is shown that the other planes of the solid KQ coincide with the other planes of the solid AG, each with each ; wherefore the solids KQ and AG do wholly coincide, and are equal and similar to one another. PROPOSITION IV. THEOREM. 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 CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE; its opposite planes are similar and equal parallelograms. Because the two parallel planes BG, CE, are cut by the plane AC, their common sections AB, CD, are parallel B E (I. 14). Again, because the two parallel planes BF, AE, are cut by the plane AC, their common sections AD, BC. are parallel ; and AB is parallel to CD; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF, AE, is a cl parallelogram. Join AH, DF; and be- V cause AB is parallel to DC, and BH to CF; the two straight lines AB, BH, which meet one another, are parallel to DC and CF, which meet one another ; wherefore, though the first two are not in the same plane with the other two, they contain equal angles (I. 9); the angle ABH is therefore equal to the angle DCF. And because AB, BH, are equal to DC, CF, and the angle ABH equal to the angle DCF; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DCF. For the same reason, the triangle AGH is equal to the triangle DEF; and therefore the parallelogram BG is equal and similar to the parallelogram CE. In the same manner, it may be proved that the parallelogram AC is equal and similar to the parallelogram GF, and the parallelogram AE to BF. PROPOSITION V. THEOREM. If a solid parallelopiped be cut by a plane parallel to two of its opposite planes, it will be divided into two solids, which will be to one another as their bases. Let the solid parallelopiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole into the solids ABFV, EGCD; as the base AEFY to the base x в с EHCF, so is the solid ABFV to the solid EGCD. · Produce AH both MLN ways, and take any v number of straight lines 0 Y F C Q S 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 LË, KR, HU, MT. Then, because By the straight lines LK, KA, AE, are all equal, and also the straight lines KO, AY, EF, which make equal angles with LK, KA, AE, the parallelograms LO, KY, AF, are equal and similar (Pl. Ge. VI. 20); and likewise the parallelograms KX, KB, AG; as also (II. 4) 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 the solid KR, as also to three planes of the solid AV; but the three planes opposite to these three are equal and similar to them in the several solids; therefore the solids LP, KR, AV, are contained by equal and similar planes. And because the planes LZ, KP, AR, are parallel, and are cut by the plane XV, the inclination of LZ to XP is equal to that of KP to PB, or of AR to BV (I. 15); and the same is true of the other contiguous planes ; therefore the solids LP, KR, and AV, are equal to one another (II. 3). For the same reason, the three solids ED, HU, MT, are equal to one another; therefore what multiple soever the base LF is of the base AF, the same multiple is the solid LV of the solid AV; for the same reason, whatever multiple the base NF is of the base HF, the same multiple is the solid NV of the solid ED; and if the base LF be equal to the base NF, the solid LV is equal to the solid NV, and if the base LF be greater than the base NF, the solid LV is greater than the solid NV; and if less, less. Since then there are four magnitudes, namely, the two bases AF, FH, and the two solids AV, ED, and of the base AF and solid AV, the base LF and solid LV are any equimultiples whatever; and of the base FH and solid ED, the base FN and solid NV are any equimultiples whatever ; and it has been proved, that if the base LF is greater than the base FN, the solid LV is greater than the solid NV; and if equal, equal; and if less, less. Therefore (Pl. Ge. V. Def. 10), as the base AF is to the base FH, so is the solid AV to the solid ED. Scholium.- This proposition may be demonstrated by the principle in the twenty-seventh proposition of the additional Fifth Book, thus : Let the parallelopiped AF be cut by a plane GH parallel to either of its sides AE or BF, then ÁG:GB=AK:KB. For, let the bases AK, KB, be commensurable, and hence the sides AH, HB, _ $ I V G V_ W_F are so too. Let AH and HB contain a common measure Z 4 cLQ R . K and 3 times respec X.. tively. Divide AH ! L into 4 and HB into M N H Ö 'B 3 equal parts in L, M, N, O, and P, and through these points let planes LS, MT, &c. pass, parallel to AE, then AG will be divided into 4, and HF into 3, equal parallelopipeds, AS, LT, &c. The figures AS, LT, &c. are parallelopipeds, for their opposite sides are parallel (II. Def. 5), and hence the opposite sides are equal and similar parallelograms (II. 4). Also the parallelograms AQ, LR, are equal, for AL=LM, and AM is parallel to CR. For a similar reason EQ=SR; and also AE=LS. The parallelograms opposite to these are also equal (II. 4); therefore the two parallelopipeds AS, LT, are contained by the same number of equal and similar parallelograms, similarly situated. The parallel planes AE, LS, are cut by the plane EK; therefore the inclination of AE and EQ is equal to that of LS and SR; and the same may be proved of the inclinations of the other sides of AS and LT.“ Hence (II. 3) AS = LT. It may be similarly proved that the parallelopipeds LT, MU, NG, HV, OW, and PF, are equal. Hence AG:GB = 4:3; but AK:KB=4:3; therefore AG:GB= AK:KB. The same proportion is similarly proved, whatever be the number of times that AH:HB contain their common measure, when commensurable ; hence the proportion exists when they are incommensurable (Pl. Ge. Ad. V. 27). . COR. Because the parallelogram AF (former figure) is to the parallelogram FH as YF to FC (Pl. Ge. VI. 1), therefore the solid AV is to the solid ED as YF to FC. PROPOSITION VI. THEOREM. - If a solid parallelopiped be cut by a plane passing through the diagonals of two of the opposite planes, it shall be cut in two equal prisms. Let AB be a solid parallelopiped, and DE, CF, the diagonals of the opposite parallelograms AH, GB, namely, those which are drawn betwixt the equal C. angles in each; and because CD, FE, are each of them parallel to GA, though not GA in the same plane with it, CD, FE, are parallel (I. 8); wherefore the diagonals CF, DE, are in the plane in which the parallels are, and are themselves parallels (I. 14); and the plane CDEF shall cut Ā the solid AB into two equal parts. Because the triangle CGF is equal (Pl. Ge. I. 34) to the triangle CBF, and the triangle DAE to DHE; and that the parallelogram CA is equal (II. 4) and similar to the opposite one BE; and the parallelogram GE to CH ; therefore the planes which contain the prisms CAE, CBE, are equal and similar, each to each; and they are also equally inclined to one another, because the planes AC, EB, are parallel, as also AF and BD, and they are cut by the plane CE; therefore the prism CAE is equal to the prism CBE (II. 3), and the solid AB is cut into two equal prisms by the plane CDEF. Def.-The insisting straight lines of a parallelopiped, mentioned in the following propositions, are the sides of the parallelograms betwixt the base and the plane parallel to it. PROPOSITION VII. THEOREM. Solid parallelopipeds upon the same base, and of the same altitude, the insisting straight lines of which are terminated in the same straight lines in the plane opposite to the base, are equal to one another. Let the solid parallelopipeds AH, AK, be upon the same base AB, and of the same altitude, and let their insisting straight lines AF, AG, LM, LN, be terminated in the same straight line FN; and CD, CE, BH, BK, be terminated in the same straight line DK; the solid AH is equal to the solid AK. Because CH, CK, are parallelograms, CB is equal (Pl. Ge. I. 34) to each of the opposite sides DH, EK ; wherefore DH is equal to EK. Add, or take away the common |