PROPOSITION VI. THEOREM. 532. The plane passed through two diagonally opposite edges of a parallelopiped divides the parallelopiped into two equivalent triangular prisms. H G Let the plane A EGC pass through the opposite edges AE and C G of the parallelopiped A G. We are to prove that the parallelopiped AG is divided into two equivalent triangular prisms, ABC-F, and A DC-H. Let IJKL be a right section of the parallelopiped made by a plane to the edge A E. The intersection IK of this plane with the plane AEG C is the diagonal of the IJK L. ..AIKJ=A I KL. § 133 But prism ABC-F is equivalent to a right prism whose base is IJK and whose altitude is A E, $ 529 (any oblique prism is ≈ to a right prism whose bases are equal to right sections of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism). The prism A D C-H is equivalent to a right prism whose base is IL K, and whose altitude is A E. Now the two right prisms are equal, .. ABC-FA DC-H. $ 529 § 528 Q. E. D. PROPOSITION VII. THEOREM. 533. Two rectangular parallelopipeds having equal bases. are to each other as their altitudes. Let AB and A'B' be the altitudes of the two rectangular parallelopipeds, P, and P', having equal bases. CASE I. When AB and A'B' are commensurable. Find a common measure m, of A B and A' B'. Suppose m to be contained in AB 5 times, and in A' B' 3 times. At the several points of division on AB and A'B' pass planes to these lines. The parallelopiped P will be divided into 5, and P' into 3, parallelopipeds equal, each to each, § 528 (two right prisms having equal bases and altitudes are equal). Let A B be divided into any number of equal parts, and let one of these parts be applied to A' B' as many times as A'B' will contain it. Since A B and A'B' are incommensurable, a certain number of these parts will extend from A' to a point D, leaving a remainder D B' less than one of these parts. Through D pass a plane to A' B', and denote the parallelopiped whose base is the same as that of P', and whose altitude is A' D by. Q. Now, since A B and A' D are commensurable, Q: P = A'DA B. (Case I.) Suppose the number of parts into which AB is divided to be continually increased, the length of each part will become less and less, and the point D will approach nearer and nearer to B'. The limit of Q will be P', and the limit of A' D will be A' B', .. the limit of Q: P will be P' : P, and the limit of A'D: A B will be A'B': A B, Moreover the corresponding values of the two variables Q: P and A'D: A B are always equal, however near these variables approach their limits. ... their limits P': PA' B' : A B. $ 199 Q. E. D. 534. SCHOLIUM. The three edges of a rectangular parallelopiped which meet at a common vertex are its dimensions. Hence two rectangular parallelopipeds which have two dimensions in common are to each other as their third dimensions. 535. Two rectangular parallelopipeds having equal altitudes are to each other as their bases. Let a, b, and c, and a', b', c, be the three dimensions respectively of the two rectangular parallelopipeds Now has the two dimensions b and c in common with P, and the two dimensions a' and c in common with P'. (two rectangular parallelopipeds which have two dimensions in common are to each other as their third dimensions); 536. SCHOLIUM. This proposition may be stated thus: two rectangular parallelopipeds which have one dimension in common are to each other as the products of the other two dimensions. PROPOSITION IX. THEOREM. 537. Any two rectangular parallelopipeds are to each other as the products of their three dimensions. P α 100 αν a Let a, b, c, and a,' b', c', be the three dimensions respectively of the two rectangular parallelopipeds P and P'. Let be a third rectangular parallelopiped whose dimensions are a, b, and c'. (two rectangular parallelopipeds which have two dimensions in common are to each other as their third dimensions); and Q = ахъ a' x b'' $536 (two-rectangular parallelopipeds which have one dimension in common are to each other as the products of their other two dimensions). |