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PROPOSITION IV. THEOREM. 529. An oblique prism is equivalent to a right prisi whose bases are equal to right sections of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism.

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B.

Let A D' be an oblique prism, and FI a right section.

Complete the right prism Fl', making its edges equal to those of the oblique prism.

We are to prove oblique prism A D' right prism FI.
In the prismis A I and A'I'
trihedral Z A = trihedral Z A',

§ 492 (two trihedrals are equal when three face É of the one are respectively equal to three face of the other, and are similarly placed). Now face A D= face A' D',

$ 505 (being the two bases of the oblique prism A D'); face A J=face A' J',

Cons. and face A G = face A' .

Cons. .. prism A I = prism A' I',

$ 527 (two truncated prisms are equal when the three faces including a trihedral

of the one are respectively equal to the three faces including a trihedral
of the other, and are similarly placed).
To each of these equal prisms add the prism F D'.
Then oblique prism A D' ^ right prism FI'.

Q. E. D. .

PROPOSITION V. THEOREM. 530. Any two opposite faces of a parallelopiped are equal and parallel.

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Let AG be a parallelopiped.
We are to prove faces A F and D G equal and parallel.
Since A C is a o,

§ 517 A B and D C are equal and il lines. § 125 Also, since A H is a o,

$ 505 A E and D H are equal and || lines. § 125 .. L E AB= _ HDC,

§ 462 (two ts not in the same plane having their sides II and lying in the same

direction are equal).
... face A F=face D G.

§ 140 Moreover, face A F is || to DG

§ 463 (if two & not in the same plane have their sides II and lying in the same

direction their planes are parallel). In like manner we may prove A H and B G equal and parallel.

Q. E. D. 531. SCHOLIUM. Any two opposite faces of a parallelopiped may be taken for bases, since they are equal and parallel parallelograms.

PROPOSITION VI. Theorem. 532. The plane passed through two diagonally opposite edges of a parallelopiped divides the parallelopiped into two equivalent triangular prisms.

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Let the plane A EGC pass through the opposite edges

A E and C G of the parallelopiped A G.

We are to prove that the parallelopiped AG is divided into two equivalent triangular prisms, A B C-F, and A D C-H.

Let IJKL be a right section of the parallelopiped made by a plane I to the edge A E.

The intersection I K of this plane with the plane A EGC is the diagonal of the OIJKL.

AIKJ=A IKL. . § 133 But prism A B C-F is equivalent to a right prism whose base is I J K and whose altitude is A E,

$ 529 (any oblique prism is to a right prison whose. bascs are equal to right sec

tions 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 I L K, and whose altitude is A E.

§ 529 Now the two right prisms are equal, (two right prisms having equal bases and altitudes are equal). .. A B C-F A DC-H.

Q. E. D.

§ 528

PROPOSITION VII. THEOREM. 533. Two rectangular parallelopipeds having equal bases are to each other as their altitudes.

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AB 5

Let A B and A'B' be the altitudes of the two rectangular parallelopipeds, P, and P', having equal bases.

P A B
We are to prove DEA'B'
Case I. When A B and A' B' are commensurable.

Find a common measure m, of A B and A' B'. Suppose m to be contained in A B 5 times, and in A' B' 3 times.

Then we have AB=3

At the several points of division on A B and A' B' pass planes I 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).

P 5
Then

P=3
.P AB
PA'B

Then we have

CASE II. When A B and A' B' are incommensurable.

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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 I to A' B', and denote the parallelopiped whose base is the same as that of P', and whose altitude is À' D by. Q. Now, since A B and A' D are commensurable, Q:P = A' D : A B.

(Case I.) Suppose the number of parts into which A B 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:AB are always equal, however near these variables approach their limits. .. their limits P: P= A' 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.

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