A в с M For, apply the base ABCDE to the equal base LMOPQ; then, the triedral angles B and M, being equal, will coincide, since the plane angles which form these triedral angles are equal each to each, and similarly situated (Prop. XXI. Sch. 2, Bk. VII.); hence the edge BG will fall on its equal M S, and the face BH will coincide with its equal MT, and the face BF with its equal MR. But the upper bases are equal to their corresponding lower bases (Art. 436); therefore the bases FGHIK, RSTVY are equal; hence they coincide with each other. Therefore H I coincides with T V, IK with VY, and KF with YR; and consequently the lateral faces coincide. Hence the two prisms coincide throughout, and are equal. F G K H I L PROPOSITION IV. THEOREM. R S Y 459. Cor. Two right prisms, which have equal bases and equal altitudes, are equal. For, since the side AB is equal to LM, and the altitude BG to MS, the rectangle A B GF is equal to the rectangle L MSR; so, also, the rectangle BG HC is equal to MSTO; and thus the three faces which form the triedral angle B, are equal to the three faces which form the triedral angle M. Hence the two prisms are equal. 460. In every parallelopipedon the opposite faces are equal and parallel. Let ABCD-H be a parallelopipedon; then its opposite faces are equal and parallel. The bases ABCD, EFGH are equal and parallel (Art. 436), and it remains only to be shown that the same is e tre dang tres crrosita lasmi de u BOGRADHE. Now. Hince the lase ABCD is a panliligann dhe she AD 2 spal and pamild a BC a i'm at reason, A E is equal and para..d to BF; bence the ande DAE & etal to the angie CBP Prog. I'L B. VIL), and the planes DAE. CEF are paralel; hence, also, the panelemm BCGFS egal to the parallelogram ADHE. In the same way. it may be shown that the opposite faces ABFE, D C G H are equal and parallel. 1 Let ABCD-H be a parallelopipedon; then its diagonals, as BH, DF, will bisect each other. E H F E 461. Cor. Any two opposite faces of a parallelipipe don may be assumed as its bases, since any fast and the one opposite to it are equal and parallel. 三 PROPOSITION V.-THEOREM. 462. The diagonals of every parallelopipedon bisect each other. I H G B C For, since BF is equal and parallel to DH, the figure BFHD is a parallelogram; hence the diagonals BH, DF bisect each other at the point O (Prop. XXXIV. Bk. I.). In the same manner it may be shown that the two diagonals A G and CE bisect each other at the point 0; hence the several diagonals bisect each other. 463. Scholium. The point at which the diagonals mutually bisect each other may be regarded as the centre of the parallelopipedon. PROPOSITION VI.-THEOREM. 464. Any parallelopipedon may be divided into two equivalent triangular prisms by a plane passing through its opposite diagonal edges. D M Let any parallelopipedon, AB CD-H, be divided into two prisms, ABC-G, ADC-G, by a plane, A CG E, passing H through opposite diagonal edges; then will the two prisms be equivalent. P A Through the vertices A and E, draw the planes A K L M, EN OP, perpendicular to the edge A E, and meeting BF, CG, DH, the three other edges of the parallelopipedon, in the points K, L, M, and in N, O, P. The sections A K L M, ENOP are equal, since they are formed by planes perpendicular to the same straight lines, and hence parallel (Prop. II.). They are parallelograms, since the two opposite sides of the same section, A K, L M, are the intersections of two parallel planes, ABFE, DCGH, by the same plane, AKLM (Prop. XIII. Bk. VII.). For a like reason, the figure AMPE is a parallelogram; so, also, are A K N E, KLON, LMP O, the other lateral faces of the solid AK LM-P; consequently, this solid is a prism (Art. 436); and this prism is right, since the edge AE is perpendicular to the plane of its base. This right prism is divided by the plane ALOE into the two right prisms AK L-0, A ML-0, which, having equal bases, A K L, A M L, and the same altitude, A E, are equal (Prop. III. Cor.). Now, since A EHD, AEPM are parallelograms, the sides DH, MP, being each equal to A E, are equal to each other; and taking away the common part, DP, there remains DM equal to HP. In the same manner it may be shown that CL is equal to GO. E G F N K There may be three cases, according as EI is greater or less than, or equal to, EF; but the demonstration is the same for each. Since AE is parallel to B F, and HE to GF, the plane angle A EI is equal to B F K, HEI to GFK, and HEA to G F B. Of these six plane angles, the three first form the polyedral angle E, the three last the polyedral angle F; consequently, since these plane angles are equal each to each, and similarly situated, the polyedral angles, E, F, must be equal. Now conceive the prism A EI-M to be applied to the prism BFK-L; the base A EI, being placed upon the base BFK, will coincide with it, since they are equal; and, since the polyedral angle E is equal to the polyedral angle F, the side EH will fall upon its equal, F G. But the base A EI and its edge EH determine the prism A EI-M, as the base B F K and its edge F G determine the prism BFK-L (Prop. III.); hence the two prisms coincide throughout, and therefore are equal to each other. Take away, now, from the whole solid A ELC, the prism AEI-M, and there will remain the parallelopipedon AL; and take away from the same solid A L the prism BFK-L, and there will remain the parallelopipedon A G; hence the two parallelopipedons A L, A G are equivalent. PROPOSITION VIII.-THEOREM. 467. Two parallelopipedons having the same base and the same altitude are equivalent. Let the two parallelopipedons AG, AL have the common base ABCD, and the same altitude; then will the two parallelopipedons be equivalent. For, the upper bases EFGH, IKLM being in the same plane, produce the edges EF, HG, L K, IM, till by their intersections they form the parallelogram NOPQ; this parallelogram is equal to either of the bases I L, E G, and |