upon its equal, abcde; these two bases will coincide.

But the three plane angles which form the solid angle B are respectively equal to the three plane angles which form the solid angle b, namely, ABC = abc, ABG abg, and GBC = gbc; they are also similarly situated; hence the solid angles B and 6 are equal (Book V., Prop. XXXVIII. ), and therefore the side BG will fall on its equal, bg. It is likewise evident that, by reason of the equal parallelograms, ABGF, abgf, the side GF will fall on its equal, gƒ, and in the same manner, GH on gh; hence, the upper base, FGHIK, will exactly coincide with its equal, fghik, and the two solids will be identical, since they have the same vertices.

PROPOSITION VIII.
THEOREM.

Two right prisms which have equal bases and the same altitude are equal. For, if we make the lower bases coincide, the lateral edges at the coinciding vertices will coincide (Book V., Prop. IV.); and since these are equal to the given altitude, the upper bases of the prisms will also coincide (Book V., Prop. VII.). Therefore, the prisms are equal.

SCHOLIUM. The preceding demonstration applies to the case of two right truncated prisms of the same base, whose corresponding lateral edges are equal. Hence, two right truncated prisms which have equal bases and equal corresponding lateral edges are equal.

PROPOSITION IX.

THEOREM.

Every oblique triangular prism is equivalent to the right triangular prism which has for its base the right section of the oblique prism, and for its altitude the lateral edge of the oblique prism.

right section of the oblique prism, AH, and for its altitude the lateral edge, BG. For since BB' = EE', B'E' is equal to BE. The prism AF = prism A'F', because the part ABCD'E'F' is common to both, and the truncated right prism A'B'C'ABC is equal to the truncated right prism D'E'F'DEF (Prop. VIII., Scholium).

SCHOLIUM. This proposition is equally true of any polygonal oblique prism.

PROPOSITION X.

THEOREM.

The two triangular prisms into which a parallelopipedon is divided by a plane passing through its opposite diagonal edges are equivalent.

First.-If the parallelopipedon is a right one, the proposition is evident (Prop. VIII.), since the two prisms will be right prisms with equal bases and altitudes.

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Secondly.-Let AG be an oblique parallelopipedon. AEGC, containing the opposite edges, AE and GC, divides this parallelopipedon into two triangular prisms, ABCEFG, ACDEGH, which we are to prove equivalent. Draw the right section of the parallelopipedon AG. This section, OKLM, is a parallelogram (Prop. II.,

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Cor. 2), and the two triangles, KLM and KMO, into which it is divided by the diagonal, KM, are respectively the right sections of the prisms ABCEFG, ACDEGH.

The triangular prism ABCEFG is equivalent to the right prism which has KLM for its base and AE for its altitude (Prop. IX.), the triangular prism ACDEGH is equivalent to the right prism which has KMO for its base and AE for its altitude. But these two right prisms are equivalent (Prop. VIII.). Therefore, the two triangular prisms ABCEFG, ACDEGH are equivalent, and each one of them is one half of the parallelopipedon AG.

PROPOSITION XI.

THEOREM.

If two parallelopipedons, AG, AL, have a common base, ABCD, and their upper bases, EFGH, IKLM, in the same plane, and between the same parallels, EK, HL, they will be equivalent.

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There may be three cases, according as EI is greater than, equal to, or less than EF; but the demonstration is the same for all. the first place, then, we shall show that the triangular prism AEIDHM is equal to the triangular prism BFKCGL.

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Since AE is parallel to BF, and HE to GF, the angle AEI = BFK, D HEI= GFK, and HEA = GFB. Of these six angles the first three form the solid angle E, the other three the solid angle F; hence, since the plane angles are respectively equal, and similarly arranged, it follows that the solid angles, E and F, are equal. Now, if the prism AEM is placed on the prism BFL, the base AEI being laid 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 along its equal, FG: nothing more is required to prove that the two prisms will coincide throughout their whole extent (Prop. VII.); for the base AEI and the edge EH determine the prism AEM, as the base BFK and the edge FG determine the prism BFL (Def. 5); hence, the prisms are equal.

But, if the prism AEM is taken away from the solid AL there will remain the parallelopipedon AIL; and if the prism BFL is taken away from the same solid, AL, there will remain the parallelopipedon AEG; hence, the two parallelopipedons, AIL, AEG, are equivalent.

PROPOSITION XII.

THEOREM.

Two parallelopipedons having the same base and the same altitude are equivalent.

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, HG be produced, likewise, LK,

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IM, till, by their intersections they form the parallelogram NOPQ; this parallelogram will evidently q be equal to each of the bases, EFGH, IKLM. Now, if a third N parallelopipedon be conceived, having for its lower base the same, ABCD, and NOPQ for its upper, this third parallelopipedon will be equivalent to the parallelopipedon AG (Prop. XI.), since, with a common lower base, their upper bases lie in the same plane, and between the parallels GQ, FN.

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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 same altitude, are equivalent.

PROPOSITION XIII.

THEOREM.

Any parallelopipedon may be changed into an equivalent rectangular parallelopipedon, having the same altitude and an equivalent base.

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Let AG be the proposed parallelopipedon. From the points A, B, C, D, draw AI, BK, CL, DM, perpendicular to the plane of the base, thus forming the parallelopipedon AL, equivalent to the parallelopipedon AG, and having its lateral faces, AK, BL, etc., rectangles. Hence, if the base, ABCD, is a rectangle, AL will be the rectangular parallelopipedon, equivalent to the proposed parallelopipedon AG.

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But, if ABCD is not a rectan

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gle, draw AO and BN, perpendicular to CD, and OQ and NP, perpendicular to the base; you will then have the solid ABNOIKPQ,

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which will be a rectangular parallelopipedon : for, by construction, the bases ABNO and IKPQ are rectangles; so also are the lateral faces, since the edges, AI, OQ, etc., are perpendicular to the plane of the base; hence, the solid AP is a rectangular parallelopipedon. But the two parallelopipedons, AP, AL, may be considered as having the same base, ABKI, and the same altitude, AO; hence, they are equivalent; hence, the parallelopipedon AG, which was first changed into an equivalent parallelopipedon, AL, is again changed into an equivalent rectangular parallelopipedon, AP, hav- D ing the same altitude, AI, and a base, ABNO, equivalent to the base ABCD.

PROPOSITION XIV.

THEOREM.

Two rectangular parallelopipedons, AG, AL, which have the same base, ABCD, are to each other as their altitudes, AE, AI.

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First.-Suppose that the altitudes, AE, AI, are to each other as two whole numbers, for example, as 15 is to 8. Divide AE into fifteen equal parts, whereof AI will contain eight, and through the points of division, x, y, z, etc., draw planes parallel to the base. These planes will cut the solid AG into fifteen partial parallelopipedons, all equal to each other, because they have equal bases and equal altitudes; equal bases because every section, as MIKL, of a prism, made parallel to its base, ABCD, is equal to this base (Prop. V., Cor.); equal altitudes because these altitudes are the equal divisions, Ax, xy, yz, etc. But, of the fifteen equal parallelopipedons, eight are contained in AL; hence, the solid AG is to the solid AL as 15 is to 8, or, generally, as the altitude AE is to the altitude AI.

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Secondly. If the ratio of AE to AI cannot be expressed in numbers, it can be shown, nevertheless, that

For, if this proportion is not correct, suppose

solid AG solid AL:: AE : AO.