Let the prism AI be applied to the prism ai, so that the equal bases AD and ad may coincide, the point A falling upon a, B upon b, and so on. because the three plane angles which contain the A solid angle B, are equal to the three plane angles

And

which contain the solid angle b, and these planes are similarly situated, the solid angles B and b are equal (Prop. XIX., Sch. B. VII.). Hence the edge BG will coincide with its equal bg: and the point G will coincide with the point g. Now, because the parallelograms AG and ag are equal, the side GF will fall upon its equal gf; and for the same reason, GH wil. fall upon gh. Hence the plane of the base FGHIK will coincide with the plane of the base fghik (Prop. II., B. VII.). But since the upper bases are equal to their corresponding lower bases, they are equal to each other; therefore the base FI will coincide throughout with fi; viz., HI with hi, IK with ik, and KF with kf; hence the prisms coincide throughout, and are equal to each other. Therefore, two prisms, &c.

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 is equal to the rectangle abgf. So, also, the rectangle BGHC is equal to the rectangle bghc; hence the three faces which contain the solid angle B are equal to the three faces which contain the solid angle b consequently, the two prisms are equal.

PROPOSITION IV. THEOREM.

The opposite faces of a parallelopiped are equal and parallel

Let ABGH be a parallelopiped; then will its opposite faces be equal and parallel.

H

G

F

From the definition of a parallelopiped (Def. 8) the bases AC, EG are equal and parallel; and it remains to be proved that the same is true of any two opposite faces, D as AH, BG. Now, because AC is a parallelogram, the side AD is equal and parallel to BC. For the same reason AE is equal and paralle! to BF; hence the angle DAE is equal to the angle CBF

A

B

(Prop. XV., B. VII.), and the plane DAE is parallel to the plane CBF. Therefore also the parallelogram AH is equal to the parallelogram BG. In the same manner, it may be proved that the opposite faces AF and DG are equal and parallel. Therefore, the opposite faces, &c.

Cor. 1. Since a parallelopiped is a solid contained by six faces, of which the opposite ones are equal and parallel, any face may be assumed as the base of a parallelopiped.

Cor. 2. The four diagonals of a parallelopiped bisect each

other.

H

E

G

Draw any two diagonals AG, EC; they will bisect each other. Since AE is equal and parallel to CG, the figure AEGC is a parallelogram; and therefore the diagonals AG, EC bisect each other (Prop. D XXXII., B. I.). In the same manner, it may be proved that the two diagonals BH and DF bisect each other; and hence the four diagonals mutually bisect each other, in a point which may be regarded as the center of the parallelopiped.

A

B

PROPOSITION V. THEOREM.

If a parallelor ped be cut by a plane passing through the diagonals of two opposite faces, it will be divided into two equivalent prisms.

Let AG be a parallelopiped, and AC, EG the diagonals of the opposite parallelograms BD, FH. Now, because AE, CG are each of them parallel to BF, they are parallel to each other; therefore, the diagonals AC, EG are in the same plane with AE, CG; and the plane AEGC divides the solid AG into two equivalent prisms.

F

M

Through the vertices A and E draw the L planes AÏKL, EMNO perpendicular to AE, meeting the other edges of the parallelopiped in the points I, K, L, and in M, N, O. The sections AIKL, EMNO are equal, because they are formed by planes perpendicular to the same straight line, and, consequently, parallel (Prop. II.). They are also parallelograms, because AI, KL, two opposite sides of the same section, are the intersections of two parallel planes ABFE, DCGH, by the same plane.

For the same reason, the figure ALOE is a parallelogram;

But

so, also, are AIME, IKNM, KLON, the
other lateral faces of the solid AIKL-H,
EMNO; hence this solid is a prisın (Def. o
5); and it is a right prism because AE is
perpendicular to the plane of its base.
the right prism AN is divided into two
equal prisms ALK-N, AIK-N; for the D
basis of these prisms are equal, being halves L
cf the same parallelogram AIKL, and they
have the common altitude AE; they are
therefore equal (Prop. III. Cor.).

A

G

Ι

M

Now, because AEHD, AEOL are parallelograms, the sides DH, LO, being equal to AE, are equal to each other. Take away the common part DO, and we have DL equal to HO. For the same reason, CK is equal to GN. Conceive now that ENO, the base of the solid ENGHO, is placed on AKL, the base of the solid AKCDL; then the point O falling on L and N on K, the lines HO, GN will coincide with their equals DL, CK, because they are perpendiculars to the same plane. Hence the two solids coincide throughout, and are equal to each other. To each of these equals, add the solid ADC-N; then will the oblique prism ADC-G be equivalent to the right prism ALK-N.

In the same manner, it may be proved that the oblique prism ABC-G is equivalent to the right prism AIK-N. But the two right prisms have been proved to be equal; hence the two oblique prisms ADC-G, ABC-G are equivalent to each other. Therefore, if a parallelopiped, &c.

Cor. Every triangular prism is half of a parallelopiped having the same solid angle, and the same edges AB, BC, BF. Scholium. The triangular prisms into which the oblique parallelopiped is divided, can not be made to coincide, because the plane angles about the corresponding solid angles are not similarly situated.

PROPOSITION VI. THEOREM.

Parallelopipeds, of the same base and the same altitude, are equivalent.

Case first. When their upper bases are between the same parallel lines.

Let the parallelopipeds AG, AL have the base AC common, and let their opposite bases EG, IL be in the same plane, and between the same parallels EK, HI.; then will the solid AG be equivalent to the sol: AL.

Because AF, AK are parallelograms, EF and IK are each equal to AB, and therefore equal to each other. Hence, if EF and IK be taken away from the same line EK, the remainders EI and D FK will be equal. Therefore the triangle AEI is equal to the triangle BFK. Also, the parallelogram EM is equal to the parallelo gram FL, and AH to BG. Hence the solid angles at E and F are contained by three faces which are equal to each other and similarly situated; therefore the prism AEIM is equal to the prism BFK-L (Prop. III.).

Now, if from the whole solid AL, we take the prism AEI-M, there will remain the parallelopiped AL; and if from the same solid AL, we take the prism BFK-L, there will remain the parallelopiped AG. Hence the parallelopipeds AL, AG are equivalent to one another.

Case second. When their upper bases are not between the same parallel lines.

Let the parallelopipeds AG, AL have the same base AC and the same altitude; then will their opposite bases EG, IL be in the same plane. And, since the sides EF and IK are equal and parallel to AB, they are equal and parallel to each other. For the same reason FG is equal and parallel to KL. Produce the sides EH, FG, as also IK, LM, and let

D

A

P

M

L

G

B

them meet in the points N, O, P, Q; the figure NOPQ is a parallelogram equal to each of the bases EG, IL; and, consequently, equal to ABCD, and parallel to it.

Conceive now a third parallelopiped AP, having AC for its ower base, and NP for its upper base. The solid AP will be equivalent to the solid AG, by the first Case, because they have the same lower base, and their upper bases are in the same plane and between the same parallels, EQ, FP. For the same reason, the solid AP is equivalent to the solid AL: hence the solid AG is equivalen. to he solid AL. There fore, parallelopipeds,. &c

PROPOSITION VII. THEOREM.

Any parallelopiped is equivalent to a right parallelopipea having the same altitude and an equivalent base.

Let AL be any parallelopiped; it is equivalent to a right parallelopiped having the same altitude and an equivalent base.

K

From the points A, B, C, D draw AE, BF, CG, DH, perpendicular to the plane of the lower base, meeting the plane of the upper base in the points E, F, G, H H. Join EF, FG, GH, HE; there will thus be formed the parallelopiped AG, equivalent to AL (Prop. VI.); and its lateral faces AF, BG, CH, DE are rectangles. If the base ABCD is also a rec.angle, AG will be a right parallelopiped, and it is equivalent to the parallel

opiped AL. But if ABCD is not a rectangle, from A and B draw AI, BK perpendicular to CD; and from E and F draw EM, FL perpendicular to GH; and join IM, KL. The solid ABKI-M will be a right parallelopiped. For, by construction, the bases ABKI and EFLM are rectangles; so, also, are the lateral faces, because the edges AE, BF, KL, IM are perpendicular to the plane of D the base. Therefore the solid AL is a right parallelopiped. But the two parallelopipeds AG, AL may be regarded as having the same base AF, and the same altitude AI; they are therefore equivalent. But the parallelopiped AG is equivalent to the first supposed parallelopiped; hence this parallelopiped is equivalent to the righ parallelopiped AL, having the same altitude, and an equiva ent base. Therefore, any parallelopiped, &c.