In the same way, it may be shown that the opposite parallelograms ABFE, DCGH are equal and parallel.

Cor. Since the parallelopipedon is a solid bounded by six planes, whereof those lying opposite to each other are equal and parallel, it follows that any face and the one opposite to it may be assumed as the bases of the parallelopipedon.

Scholium. If three straight lines AB, AE, AD, passing through the same point A, and making given angles with each other, are known, a parallelopipedon may be formed on those lines. For this purpose a plane must be extended through the extremity of each line, and parallel to the plane of the other two; that is, through the point B a plane parallel to DAE, through D a plane parallel to BAE, and through E a plane parallel to BAD. The mutual intersections of these planes will form the parallelopipedon required.

THEOREM IV.

In every prism the sections formed by parallel planes are equal polygons.

In the prism ABCI, let the sections NOPQR, STVXY be formed by parallel planes; then will these sections be equal polygons.

A

F

8

N

E

K

I

B

H

For the sides ST, NO are parallel, being the intersections of two parallel planes with a third plane ABGF; moreover the sides ST, NO are included between the parallels NS, OT, which are sides of the prism: hence NO is equal to ST. For like reasons, the sides OP, PQ, QR, etc., of the section NOPQR, are respectively equal to the sides TV, VX, XY, etc., of the section STVXY; and since the equal sides are at the same time parallel, it follows that the angles NOP, OPQ, etc., of the first section are respectively equal to the angles STV, TVX, etc., of the second hence the two sections NOPQR, STVXY are equal polygons.

Cor. Every section in a prism, if made parallel to the base, is also equal to that base.

THEOREM V.

If a plane be made to pass through the diagonal and opposite edges of a parallelopipedon, so as to divide it into two triangular prisms, those prisms will be equal.

Let the parallelopipedon ABCG be divided by the plane BDHF into the two triangular prisms ABDHEF, BCDFGH; then will those prisms be equal.

E

A

a

F

B

H

G

Through the vertices B and F, draw the planes Bade, Fehg at right angles to the side BF, and meeting AE, DH, CG, the three other sides of the parallelopipedon, in the points a, d, e towards one direction, and in e, h, g towards the other: then the sections Bade, Fehg will be equal parallelograms; being equal, because they are formed by planes perpendicular to the same straight line, and consequently parallel; and being parallelograms, because aB, dc, two opposite sides of the same section, are formed by the meeting of one plane with two parallel planes ABFE, DCGH.

For a like reason, the figure BaeF is a parallelogram; so also are BFgc, cdhg, and adhe, the other lateral faces of the solid BadcFehg: hence that solid is a prism (D. IV.), and that prism is a right one, because the side BF is perpendicular to its base.

This being proved, if the right prism Bh be divided by the plane BFHD into two right-triangular prisms, aBdeFh, BdcFhg, it will remain to be shown that the oblique-triangular prism ABDEFH will be equal to the right-triangular prism aBdeFh; and since those two prisms have a part ABDheF in common, it will only be requisite to prove that the remaining parts, namely, the solids BaADd, FeEHh, are equal.

Now, by reason of the parallelograms ABFE, aBFe, the sides AE, ae being equal to their parallel BF, are equal to each other; and taking away the common part Ae, there remains Aa Ee. In the same manner we could prove Dd = Hh.

=

Let us now place the base Feh on its equal Bad; the point e coinciding with a, and the point h with d, the sides eE, H will coincide with their equals aA, dD, because they are perpendicular to the same plane Bad. Hence the two solids in ques

tion will coincide exactly with each other, and the oblique prism BADFEH is therefore equal to the right one Bad Feh.

In the same manner might the oblique prism BDCFHG be proved equal to the right prism BdcFhg. But (T. I.) the two right prisms Bad Feh, BdcFhg are equal, since they have the same altitude BF, and since their bases Bad, Bdc are halves of the same parallelogram. Hence the two triangular prisms BADFEH, BDCFHG, being equal to the equal right prisms, are equal to each other.

Cor. Every triangular prism ABDHEF is half of the parallelopipedon AG described with the same polyedral angle A, and with the same edges AB, AD, AE.

THEOREM VI.

Two parallelopipedons having a common base, and their upper bases in the same plane and between the same parallels, are equivalent to each other.

Let the two parallelopipedons AG, AL have the common base ABCD; and let their upper bases EFGH, IKLM be in the same plane, and between the same parallels EK, HL; then will these parallelopipedons be equivalent.

K

L

M G

H

E

I

D

B

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

Since AE is parallel to BF, and HE to GF, the angle AEI= BFK, HEI=GFK, and HEA = GFB. Of these six angles, the first three form the polyedral angle E, and the last three the polyedral angle F; therefore, the plane angles being respectively equal and similarly arranged, the polyedral angles F and E must be equal. Now if the prism AEM be laid on the prism BFL, the base AEI being placed on the base BFK, will coincide with it, because they are equal; and since the polyedral angle E is equal to the polyedral angle F, the side EH will coincide with its equal FG; and nothing more is required to prove the coinci

dence of the two prisms throughout their whole extent, for (T. II.) the base AEI and the edge EH determine the prism AEM, as the base BFK and the edge FG determine the prism BFL; hence these 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, there will remain the parallelopipedon AEG; hence these two parallelopipedons AIL, AEG are equivalent.

THEOREM VII.

Two parallelopipedons having the same base and same altitude, are equivalent to each other.

N

M

D

I

P

H

B

E

K

Let ABCD be the common Q 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, and likewise LK, IM, till, by their intersections, they form the parallelogram NOPQ: this parallelogram will evidently be equal to either of the bases, EFGH, IKLM. Now, if a third parallelopipedon be conceived, having ABCD for its lower base and NOPQ for its upper, this third parallelopipedon will (T. VI.) be equivalent to the parallelopipedon AG; since, with the same lower base, their upper bases lie in the same plane and between the same parallels GQ, FN. 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 the same altitude, are equivalent.

THEOREM VIII.

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

N

P

H

G

M

L

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

D

A

B

M Q.

LP

K

But if ABCD is not a rectangle, draw AO and BN perpendicular to CD, and OQ and NP perpendicular to the base; then the solid ABNOIKPQ will be a rectangular parallelopipedon; for, by construction, the base ABNO and its opposite IKPQ are rectangles; so also are the lateral faces, the edges AI, OQ, etc., being perpendicular to the plane of the base; hence the solid AP is a rectangular parallelopipedon. But the two parallelopipedons AP, AL may be conceived as having the same base ABKI, and the same altitude AO; hence the parallelopipedon AG, which was at first changed into an equivalent parallelopipedon AL, is again changed into an equivalent rectangular parallelopipedon AP, having the same altitude AI, and a base ABNO equivalent to the base ABCD.

D