because their opposite sides are parallel, each to each (B. VI.,

P. X.); they will also be equal (P. II.): edron Badc-g is a right prism (D. 2, 4), polyedrons Bad-h and Bcd-h.

hence, the poly

as are also the

Place the triangle Feh upon Bad, so that F shall coincide with B, e with a, and h with d; then, because eE, hH, are perpendicular to the plane Feh, and aA, dD, to the plane Bad, the line eE will take the direction aA, and the line h the direction dD. The lines AE and ae are equal, because each is equal to BF (B. I., P. XXVIII.). If we take away from the line aE the part ae, there will remain the part eE; and if from the same line, we take away the part AE, there will remain the part Aa: hence, eE and aA are equal (A. 3); for a like reason hH is equal to dD: hence, the point E will coincide with A, and the point H with D, and consequently, the polyedrons Feh-H and Bad-D will coincide throughout, and are therefore equal.

If from the polyedron Bad-H, we take away the part Bad-D, there will remain the prism BAD-H ; and if from the same polyedron we take away the part Feh-H, there will remain the prism Bad-h: hence, these prisms are equal in volume. In like manner, it may be shown that the prisms BCD-II and Bcd-h are equal in volume.

The prisms Bad-h, and Bcd-h, have equal bases, because these bases are halves of equal parallelograms (B. I., P. XXVIII., C. 1); they have also equal altitudes; they are therefore equal (P. V., C.): hence, the prisms BAD-H and BCD-H are equal (A. 1); which was to be proved.

Cor. Any triangular prism ABD-II, is equal to half of the parallelopipedon AG, which has the same triedral angle A, and the same edges AB, AD, and AE.

PROPOSITION VIII. THEOREM.

If two parallelopipedons have a common lower base, and their upper bases between the same parallels, they are equal in volume.

Let the parallelopipedons AG and AL have the common lower base ABCD, and their upper bases EFGH and IKLM, between the same parallels EK and HL : then will they be equal in volume.

BFK-L, we have the line AE equal and parallel to BF, and EI equal to FK; hence, the face AEI is equal to BFK. In the faces EIMH and FKLG, we have, HE=.GF, EI=FK and HEI=GFK: hence, the two faces are equal (Bk. I. P. xxviii. C. 3): the faces AEHD and BFG C are also equal (P. VI.): hence, the prisms are equal (P. V.)

If from the polyedron ABKE-H, we take away the prism BFK-L, there will remain the parallelopipedon AG; and if from the same polyedron we take away the prism AEI-M, there will remain the parallelopipedon AL: hence, these parallelopipedons are equal in volume (A. 3); which was to be proved.

PROPOSITION IX. THEOREM.

If two parallelopipedons have a common lower base and the same altitude, they will be equal in volume.

Let the parallelopipedons AG and AL have the common lower base ABCD and the same altitude: then will they be equal in volume.

Because they have the same altitude, their upper bases

Now, if a third parallelopipedon be constructed, having for its lower base the parallelogram ABCD, and for its upper base NOPQ, this third parallelopipedon will be equal in volume to the parallelopipedon AG, since they have the same lower base, and their upper bases between the same parallels, QG, NF (P. VIII.). For a like reason, this third parallelopipedon will also be equal in volume to the parallelopipedon AL: hence, the two parallelopipedons AG AL, are equal in volume; which was to be proved.

Cor. Any oblique parallelopipedon may be changed into a right parallelopipedon having the same base and the same altitude; and they will be equal in volume.

PROPOSITION X. PROBLEM.

To construct a rectangular parallelopipedon which shall be equal in volume to a right parallelopipedon whose base is any parallelogram.

Let ABCD-M be a right parallelopipedon, having for its base the parallelogram ABCD.

MQ

Through the edges AI and BK pass the planes AQ and BP, respectively perpendicular to the plane AK, the former meeting the face DL in OQ, and the latter meeting that face produced in NP: then will the polyedron AP be a rectangular parallelopipedon equal to the D given parallelopipedon.

It will be a rect

angular parallelopipedon, because all of its

K

faces are rectangles, and it will be equal to the given parallelopipedon, because the two may be regarded as having the common base AK (P. VI., C. 1), and an equal altitude AO (P. IX.).

Cor. 1. Since any oblique parallelopipedon may be changed into a right parallelopipedon, having the same base and altitude, (P. IX., Cor.); it follows, that any oblique parallelopipedon may be changed into a rectangular parallelopipedon, having an equal base, an equal altitude, and an equal volume.

Cor. 2. An oblique parallelopipedon is equal in volume to a rectangular parallelopipedon, having an equal base and an equal altitude.

Cor. 3. Any two parallelopipedons are equal in volume when they have equal bases and equal altitudes.

PROPOSITION XI. THEOREM.

Troo rectangular parallelopipedons having a common lower base, are to each other as their altitudes.

Let the parallelopipedons AG and AL have the .com mon lower base ABCD: then will they be to each other as their altitudes AE and AI.

1o. Let the altitudes be commensurable, and suppose, for example, that AE is to AI, as 15 is to 8. Conceive AE to be which AI will contain 8; let planes be passed parallel to ABCD.

divided into 15 equal parts, of

through the points of division These planes will

equal altitudes;

divide the parallelopipedon AG into 15 parallelopipedons, which have equal bases (P. II. C.) and hence, they are equal (P. X., Cor. 3).

Now, AG contains 15, and AL 8 of these equal parallelopipedons; hence, AG is to AL, as 15 is to 8, or as AE is to AI. In like manner, it may be shown that AG is to AL, as AE

is to AI, when the altitudes are to each other as any other whole numbers.

able.

2o. Let the altitudes be incommensur

E

m

A

B

Now, if AG is not to AL, as AE is to AI, let us suppose that,

AG AL :: AE AO,

in which AO is greater than AI.

Divide AE into equal parts, such that each shall be less than OI; there will be at least one point of division