P M. N T E R A B H from any point in the opposite planes EF and MO; the solid AF is to the solid GO in a ratio compounded of the ratios of the base AC to the base GK, and of the perpendicular on AC to the perpendicular on GK. Case 1. When the insisting lines are perpendicular to the bases AC and GK, or when the solids are upright. In GM, one of the insisting lines of the solid GO, take GQ equal to AE, one of the insisting lines of the solid AF, and through Q let a plane pass parallel to the plane GK, meeting the other insisting lines of the solid GO in the points R, S, and T. It is evident that GS is a x solid parallelopiped (II. 4), and that it has the same altitude with AF, namely, GQ or AE. Now, the solid AF is to the solid GO in a ratio compounded of the ratios of the solid AF to the solid GS (Pl. Ge. V. Def. 17), and of the solid GS to the solid GO; but the ratio of the solid AF to the solid GS, is the same with that of the base AC to the base GK (II. 10), because their altitudes AE and GQ are equal; and the ratio of the solid GS to the solid GO, is the same with that of GQ to GM, for they are as their bases GT, GP (II. 5), which are as GQ to GM; therefore, the ratio which is compounded of the ratios of the solid AF to the solid GS, and of the solid GS to the solid GO, is the same with the ratio which is compounded of the ratios of the base AC to the base GK, and of the altitude AE to the altitude GM (Pl. Ge. V. e). But the ratio of the solid AF to the solid GO is that which is compounded of the ratios of AF to GS, and of GS to GO; therefore, the ratio of the solid AF to the solid GO is compounded of the ratios of the base AC to the base GK, and of the altitude AE to the altitude GM.I Case 2. When the insisting lines are not perpendicular to the bases. Let the parallelograms AC and GK be the bases as be fore, and let AE and GM be the altitudes of two parallelo pipeds, Y and Z on these bases. Then, if the upright parallelopipeds AF and GO be constituted on the bases AC and GK, with the altitudes AE and GM, they will be equal to the parallelopipeds Y and Z (II. 9). Now, the solids AF and Go, by the first case, are in the ratio compounded of the ratios of the bases AC and GK, and of the altitudes AE and GM; therefore, also, the solids Y and Z have to one another a ratio that is compounded of these same ratios. Cor. 1.-Hence, two straight lines may be found having the same ratio with the two parallelopipeds AF and GO. TO AB, one of the sides of the parallelogram AC, apply the parallelogram BV equal to GK, having an angle equal to the angle BAD (Pl. Ge. I. 44); and as AE to GM, so let AV be to AX (Pl. Ge. VI. 12), then AD is to AX as the solid AF to the solid GO. For the ratio of AD to AX is compounded of the ratios (Pl. Ge. V. Def. 17) of AD to AV, and of AV to AX; but the ratio of AD to AV is the same with that of the parallelogram AC to the parallelogram BV or GK; and the ratio of AV to AX is the same with that of AE to GM; therefore, the ratio of AD to AX is compounded of the ratios of AC to GK, and of AE to GM (Pl. Ge. V. E). But the ratio of the solid AF to the solid GO compounded of the same ratios ; therefore, as AD to AX, so is the solid AF to the solid GO. CoR. 2.-Hence all prisms are to one another in the ratio compounded of the ratios of their bases, and of their altitudes. For every prism is equal to a parallelopiped of the same altitude with it, and of an equal base (II. 10, Cor. 2). CoR. 3.-The right rectangular parallelopipeds contained by the corresponding lines of three analogies, are pro portional Let A:B=C:D, E:F=G:H, and I:K=L:M, be three analogies, the terms of which are lines, then A.E.I: B.F.K=C.G.L:D.H.M. For let P, Q, R, and S, denote these parallelopipeds ; I, K, L, and M, their respective altitudes; and consequently the rectangles A. E, B F, C.G, and D.H, their bases. Then (PI. Ge. VI. 23, Cor. 1) A É:BF=CG:D.H. But C (II. 11) P:Q=(A.E:B.F, I:K), and R:S=(C•G: D.H, L:M); and the two component ratios of P:Q are equal respectively to those of R:8; therefore (Pl. Ge. V. E.) P:Q=R:8. Cor. 4.-A right rectangular parallelopiped is equal to the cube described on any unit of measure, multiplied by the product of the numbers denoting the number of times that it is contained in any three contiguous edges of the parallelopiped, or its length, breadth, and height. Let A, B, and C, be the three sides; and a, b, and C, the three numbers, whether terminate or interminate, that denote the number of times that a line M, taken as the unit of measure, is contained in these sides respectively; then, considering C and M as the respective altitudes of the parallelopiped Å ·B·C and the cube M3, and A · B and MP as their bases (II. 11), A.B.C:M=(AB:M, C:M). But (Pl. Ge. Qu. 10, and Ad. V. 7) (AB: M, C:M) = (ab:1, 0:1) = abc:1 (Pl. Ge. Ad. V. 26). Therefore A.B.C: M3 = abc : 1, or A •B.C=abc M3. PROPOSITION XII. THEOREM. Solid parallelopipeds which have their bases and altitudes reciprocally proportional, are equal; and parallelopipeds which are equal, have their bases and altitudes reciprocally proportional. Let AG and KQ be two solid parallelopipeds, of which the bases are AC and KM, and the altitudes AE and KO, and let AC be to KM as KO to AE, the solids AG and KQ are equal. As the base AC to the base KM, so let the straight line KO be to the straight line 8. Then, since AC is to KM as KO to S, and also by hypo- 1 thesis, AC to KM as KO to AE, KO has the same ratio to that it has to AE (Pl. Ge. V. 11); D wherefore AE is equal to S. But the solid AG is to the solid KQ in the ratio compounded of the ratios of AE to KO, and of AC to KM (II. 11), that is, in the ratio compounded of the ratios of AE to KO, and of R Q N S KO to S. And the ratio of AE to Sis also compounded of the same ratios (Pl. Ge. V. Def. 17); therefore, the solid AG has to the solid KQ the same ratio that AE has to S. But AE was proved to be equal to S, therefore AG is equal to KQ. Again, if the solids AG and KQ be equal, the base AC is to the base KM as the altitude KO to the altitude AE. Take S, so that AC may be to KM as KO to S, and it will be shown, as was done above, that the solid AG is to the solid KQ as AE to S. Now, the solid AG is, by hypothesis, equal to the solid KQ; therefore, AE is equal to S; but, by construction, AC is to KM as KO is to S; therefore, AC is to KM as KO to AE. CoR.-In the same manner, may it be demonstrated that equal prisms have their bases and altitudes reciprocally proportional, and conversely. PROPOSITION XIII. THEOREM. Similar solid parallelopipeds are to one another in the triplicate ratio of their homologous sides. Let AG, KQ, be two similar parallelopipeds, of which AB and KL are two homologous sides ; the ratio of the solid AG to the solid KQ is triplicate of the ratio of AB to KL. Because the solids are similar, the parallelograms AF, KP, are similar (II. Def. 8), as also the parallelograms AH, KR; therefore, the ratios of AB to KL, of AE to KO, and of AD to KN, are all equal (Pl. Ge. VI. Def.9). But the ratio of the solid D AG to the solid KQ is compounded of the ratios K of AC to KM, and of AE to KO. Now, the ratio of AC to KM, because they are equiangular parallelograms, is compounded (Pl. Ge. VI. 23) of the ratios of AB to KL, and of AD to KN. Wherefore, the ratio of AG to KQ is compounded of the three ratios of AB to KL, AD to KN, and AE to KO; and these three ratios have already been proved to be equal; therefore, the ratio that is compounded G R E 0 N M А B of them, namely, the ratio of the solid AG to the solid KQ, so is m to n, then AB is to n as the solid AG to the the solid AG to the solid KQ. Cor. 2.-As cubes are similar solids, therefore the cube on AB is to the cube on KL in the triplicate ratio of AB to KL, that is, in the same ratio with the solid AG to the solid KQ. Similar solid parallelopipeds are therefore to one another as the cubes on their homolo gous sides. Cor. 3.- In the same manner, it is proved that similar prisms are to one another in the triplicate ratio, or in the ratio of the cubes, of their homologous sides. PROPOSITION XIV. THEOREM. If two triangular pyramids which have equal bases and altitudes be cut by planes that are parallel to the bases, and at equal distances from them, the sections are equal to one another. Let ABCD and EFGH be two pyramids, having equal bases BDC and FGH, and equal altitudes, namely, the perpendiculars AQ and ES drawn from A and E upon the planes BDC and FGH; and let them be cut by planes parallel to BDC and FGH, and at equal altitudes QR and ST above those planes, and let the sections be the triangles KLM, NOP; KLM and NOP Qi are equal to one another. Because the plane ABD B cuts the parallel planes BDC, KLM, the common sections BD and KM are parallel (I. 14). For the same reason, DC and ML are parallel. Since therefore KM and ML are parallel to BD and DC, each to each, though not in the А K к N |