allelogram; for a like reason, AE is parallel to BF: hence the angle DAE is equal to the angle CBF, and the planes DAE, CBF, are parallel (Book VI. Prop. XIII.); hence also the parallelogram DAEH is equal to the parallelogram CBFG. In the same way, it might be shown that the opposite parallelograms ABFE, DCGH, are equal and parallel. Cor. 1. Since the parallelopipedon is a solid bounded by six planes, whereof those lying opposite to each other are equal and parallel, it follows that any face and the one opposite to it, may be assumed as the bases of the parallelopipedon. Cor. 2. The diagonals of a parallelopipedon bisect each other. For, suppose two diagonals EC, AG, to be drawn both through opposite vertices: since AE is equal and parallel to CG, the figure AEGC is a parallelogram; hence the diagonals EC, AG will mutually bisect each other. In the same manner, we could show that the diagonal EC and another DF bisect each other; hence the four diagonals will mutually bisect each other, in a point which may be regarded as the centre of the parallelopipedon. Scholium. If three straight lines AB, AE, AD, passing through the same point A, and making given angles with each other, are known, a parallelopipedon may be formed on those nes. For this purpose, a plane must be passed through the extremity of each line, and parallel to the plane of the other two; that is, through the point B a plane parallel to DAE, through D a plane parallel to BAE, and through E a plane parallel to BAD. The mutual intersections of these planes will form the parallelopipedon required. PROPOSITION VII. THEOREM. The two triangular prisms into which a parallelopipedon is divided by a plane passing through its opposite diagonal edges, are equivalent. N* Let the parallelopipedon ABCD-H be divided by the plane BDHF passing through its diagonal edges: then will the triangular prism ABD-H be equivalent to the triangular prism BCD-H. E G a Through the vertices B and F, draw the planes Bade, Fehg, at right angles to the side BF, the former meeting AE, DH, CG, A the three other sides of the parallelopipedon, in the points a, d, c, the latter in e, h, g: the sections Badc, Fehg, will be equal parallelograms. They are equal, because they are formed by planes perpendicular to the same straight line, and consequently parallel (Prop. II.); they are parallelograms, because aB, dc, two opposite sides of the same section, are formed by the meeting of one plane with two parallel planes ABFE, DCGH. B g For a like reason, the figure BaeF is a parallelogram; so also are BFgc, cdhg, adhe, the other lateral faces of the solid Badc-g; hence that solid is a prism (Def. 6.); and that prism is right, because the side BF is perpendicular to its, base. But the right prism Badc-g, is divided by the plane BH into the two right prisms Bad-h, Bcd-h, since they have the same altitude BF, for their bases Bad, Bcd, are halves of the same parallelogram (Prop. V. Cor.). It is now to be proved that the oblique triangular prism ABD-H will be equivalent to the right triangular prism Bad-h; and since those prisms have a common part ABD-h, it will only be necessary to prove that the remaining parts, namely, the solids BaADd, FeEHh, are equivalent. Now, by reason of the parallelograms ABFE, aBFe, the sides AE, ae, being equal to their parallel BF, are equal to each other; and taking away the common part Ae, there remains Aa-Ee. In the same manner we could prove Dd=Hh. Next, to bring about the superposition of the two solids BaADd, FeEHh, let us place the base Feh on its equal Bad: the point e falling on a, and the point h on d, the sides eE, hH, will fall on their equals aA, dD, because they are perpendicular to the same plane Bad. Hence the two solids in question will coincide exactly with each other; hence the oblique prism BAD-H, is equivalent to the right one Bad-h. In the same manner might the oblique prism BCD-H, be proved equivalent to the right prism Bcd-h. But the two right prisms Bad-h, Bcd-h, are equal, since they have the same altitude BF, and since their bases Bad, Bdc, are halves of the same parallelogram (Prop. V. Cor.). Hence the two trian L gular prisms BAD-H, BDC-G, being equivalent to the equal right prisms, are equivalent to each other. Cor. Every triangular prism ABD-HEF is half of the parallelopipedon AG described with the same solid angle A, and the same edges AB, AD, AE. PROPOSITION VIII. THEOREM. If two parallelopipedons have a common base, and their upper bases in the same plane and between the same parallels, they will be equivalent. A B There may be three cases, according as EI is greater, less than, or equal to, EF; but the demonstration is the same for all. In the first place, then we shall show that the triangular prism AEI-MDH, is equal to the triangular prism BFK-LCG. Since AE is parallel to BF, and HE to GF, the angle AEI= BFK, HEI=GFK, and HEA=GFB. Of these six angles the first three form the solid angle E, the last three the solid angle F; therefore, the plane angles being respectively equal, and similarly arranged, the solid angles F and E must be equal (Book VI. Prop. XXI.). Now, if the prism AEI-M is laid on the prism BFK-L, the base AEI being placed on the base BFK will coincide with it because they are equal; and since the solid angle E is equal to the solid angle F, the side EH will fall on its equal FG: and nothing more is required to prove the coincidence of the two prisms throughout their whole extent, for the base AEI and the edge EH determine the prism AEI-M, as the base BFK and the edge FG determine the prism BFK-L (Prop. V.); hence these prisms are equal. But if the prism AEI-M is taken away from the solid AL, there will remain the parallelopipedon BADC-L; and if the prism BFK-L is taken away from the same solid, there will remain the parallelopipedon BADC-G; hence those two parallelopipedons BADC-L, BADC-G, are equivalent. PROPOSITION IX. THEOREM. Two parallelopipedons, having the same base and the same altitude, are equivalent. B Let ABCD be the common base of the two parallelopipedons AG, AL; since they have the same altitude, their upper bases EFGH, IKLM, will be in the same plane. Also the sides EF and AB will be equal and parallel, as well as IK and AB; hence EF is equal and parallel to IK; for a like reason, GF is equal and parallel to LK. Let the sides EF, GH, be produced, and likewise KL, IM, till by their intersections they form the parallelogram NOPQ; this parallelogram will evidently be equal to either of the bases EFGH, IKLM. Now if a third parallelopipedon be conceived, having for its lower base the parallelogram ABCD, and NOPQ for its upper, the third parallelopipedon will be equivalent to the parallelopipedon AG, since with the same lower base, their upper bases lie in the_same_plane and between the same parallels, GQ, FN (Prop. VIII.). For the same reason, this third parallelopipedon will also be equivalent to the parallelopipedon AL; hence the two parallelopipedons AG, AL, which have the same base and the same altitude, are equivalent. N M D Ι P AL PROPOSITION X. THEOREM. Any parallelopipedon may be changed into an equivalent rectangular parallelopipedon having the same altitude and an equivalent base. B Let AG be the par- o allelopipedon proposed. From the points A, B, C, D, draw AI, BK, CL, DM, perpendicular tothe plane of the base; you will thus form the parallelopipedon AL equivalent to AG, and having its lateral faces AK, BL, &c. rectangles. Hence if the base ABCD is a rectangle, AL will be a rectangular parallelopipedon equivalent to AG, and consequently, the parallelopipedon required. But if ABCD is not a rectangle, draw AO and BN perpendicular to CD, and MO OQ and NP perpendicular to the base; you will then have the solid ABNO-IKPQ, which will be a rectangular parallelopipedon : for by construction, the bases ABNO, and IKPQ are rectangles; so also are the lateral faces, the edges AI, OQ, &c. being perpendicular to the plane of the base; hence the solid AP is a rectangular parallelopipedon. But the D two parallelopipedons AP, AL may be conceived as having the same base ABKI and the same altitude AQ: hence the parallelopipedon AG, which was at first changed into an equivalent parallelopipedon AL, is again changed into an equivalent rectangular parallelopipedon AP, having the same altitude AI, and a base ABNO equivalent to the base ABCD. LP B M D A P K PROPOSITION XI. THEOREM. H E Two rectangular parallelopipedons, which have the same base, are to each other as their altitudes. |