12. Two solids are similar, when they are contained by the same number of similar planes, similarly situated, and having like inclinations to one another.

PROPOSITION I.

THEOREM. Two prisms are equal, when a solid angle in each is contained by three planes, which are equal in both, and similarly situated.

Let the base ABCDE be equal to the base abcde; the parallelogram ABGF equal to the parallelogram abgf, and the parallelogram BCHG equal to the parallelogram bchg: then will the prism ABCI be equal to the prism abci.

For, apply the base ABCDE upon its equal abcde, so that they may 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; that is, ABC=abc, ABG=abg, and GBC=gbc,

and they are also similarly situated; therefore the solid angles B and C are equal, (B. VI, Prop. xxi,) hence BG will coincide with its equal bg. And it is likewise evident, because the parallelograms ABGF and abgƒ are equal, that the side GF will coincide with its equal gf, and in the same manner GH with gh; therefore the upper base FGHIK will coincide with its equal fghik, and the two solids will be identical, since their vertices are the same.

Cor. Two right prisms, which have equal bases and equal altitudes, are equal. For, since the side AB is equal to ab, and the altitude BG to bg, the rectangle ABGF will be equal to abgf, and in the same way, the rectangle BGHC will be equal to bghc; and thus the three planes which form the solid angle B will be equal to the three planes which form the solid angle b: hence the two prisms are equal.

PROPOSITION II.

THEOREM. In every parallelopipedon, the opposite planes are equal and parallel.

H

D

By the definition of this solid, the bases ABCD, EFGH are equal parallelograms, and their sides are parallel: it remains only to show that the same is true of any two opposite lateral faces, such as AEHD, BFGC. Now AD is equal and parallel to BC, because the figure ABCD is a parallelogram ;

A

B

for a like reason, AE is parallel to BF: hence the angle DAE is equal to the angle CBF, and the planes DAE, CBF are parallel; hence also the parallelogram DAEH is equal to the parallelogram CBFG. 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.

PROPOSITION III.

THEOREM. 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.

S

F

H

G

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, &c., of the section NOPQR, are respectively equal to the sides TV, VX, XY, &c., of the section. STVXY; and since the equal sides A are at the same time parallel, it follows that the angles NOP, OPQ, &c.,

E

R

B

T

of the first section are respectively equal to the angles STV, TVX, &c., 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.

PROPOSITION IV.

THEOREM. 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 are equal.

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

Through the vertices B and F, draw the planes Badc, Fehg, at right-angles to the side BF, and meeting AE,

C

DH, CG, the three other sides of the parallelopipedon, in the points a, d, towards one direction, and in e, h, g towards the other: then the sections Badc, 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 op

A

H

posite 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, (Def. 4;) 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