T

D

M

reason, the parallelogram KH is similar to the parallelogram G F, and HB to FE. Therefore three parallelograms of the solid AL are similar to three of the solid CD: and the three opposite ones in each solid are equal (XI. 24) and similar to these, each to each. Also, because the plane angles which contain the solid angles of the K figures are equal, each to each, and are situated in the same order, the solid angles are equal (XI. B), each to each. Therefore the solid AL is similar (XI. Def. 11) to the solid CD. Wherefore from a given straight line AB a parallelopiped AL has been described similar and similarly situated to the given parallelopiped CD. Q. E. F.

A

PROP. XXVIII. THEOREM.

B

C

If a parallelopiped be cut by a plane passing through the diagonals of two of its opposite planes; it is cut into two equal triangular prisms.

Let A B be a parallelopiped, and DE and CF the diagonals of the opposite parallelograms AH and GB, viz., those which are drawn between the equal angles in each.

C

Because CD and FE are each of them parallel to G A, and not in the same plane with it, CD and FE are (XI. 9) parallel. Therefore the diagonals CF and DE, are in the plane in which the parallels are, and are themselves (XI. 16) parallels. The plane CE cuts the solid AB into two equal triangular prisms.

Because the triangle CGF is equal (I. 34) to the triangle C B F, and the triangle DAE to the triangle DHE; and the parallelogram CA is equal (XI. 24)

B

F

H

E

and similar to the opposite parallelogram BE, and the parallel. ogram GE to the parallelogram CH. The prism contained by the two triangles CGF and DAE, and the three parallelograms CA, GE, and EC is equal (XI. C) to the prism contained by the two triangles CBF and DHE, and the three parallelograms BE, CH and EC; because they are contained by the same number of equal and similar planes, similarly situated, and none of their solid angles are contained by more than three plane angles. Therefore the solid AB is cut into two equal triangular prisms by the plane CE. Q. E. D.

This demonstration is not sufficiently general, as it ought to include oblique as well as right parallelopipeds. The prisms of oblique parallelopipeds, as in the next proposition, are only equal by symmetry. See Legendre's "Geometry."

"N.B.-The insisting straight lines of a parallelopiped, mentioned in the next and some following propositions, are the sides of the parallelograms betwixt the base and the opposite plane parallel to it."

PROP. XXIX. THEOREM.

Parallelopipeds upon the same base, and of the same altitude, the insisting straight lines of which are terminated in the same straight lines in the plane opposite to the base, are equal to one another.

Let the parallelopipeds A H and AK be upon the same base AB,

and

of the same altitude, and let their insisting straight lines A F, AG, LM, and LN be terminated in the same straight line FN; and CD, CE, BH, and BK be terminated in the same straight line DK. The solid AH is equal to the solid A K.

First, let the parallelograms D G and HN, which are opposite to the base A B, have a common side H G.

C

D

F

H

K

N

Because the solid FB is bisected by the plane A H passing through the diagonals AG and CH, of the opposite planes LF and BD (XI. 28), the solid FB is double of the prism ABG. Because the solid AK is bisected by the plane LH passing through the diagonals LG and BH of the opposite planes AN and CK, the solid AK is double of the same prism ABG. Therefore the solid F B is equal (I. Ax. 6) to the solid A K. Next, let the parallelograms DM and EN, which are opposite to the base AB, have no common

side.

Because CH and CK are parallelograms, CB is equal (I. 34) to each of the opposite sides DH and EK. Therefore D H is equal to EK. Add

A

D

H E

K

D

E

H K

or take away the common part HE; and DE is equal (I. Ax. 2 or 3) to HK. Therefore the triangle CDE is equal (I. 38) to the triangle BHK, and the parallelogram DG is equal (1. 36) to the parallelogram HN. For the same reason, the triangle AFG is equal to the triangle LMN. Also the parallelogram CF is equal (X1. 24) to the parallelogram B M, and CG to BN. Therefore the prism which is contained by the two triangles A F G and CD E, and the three parallelograms AD, DG and GC, is equal (XI. C) to the prism contained by the two triangles L M N and B HK, and the three parallelograms B M, M K and K L. If therefore the prism NHL be taken from the whole solid ABND, the remainder is the solid A H; and if from the same solid the prism GDA be taken the remainder is the solid AK. Therefore the parallelopiped AH is equal (I. Ax. 3) to the parallelopiped AK. Therefore parallelopipeds, &c. Q. E.D.

PROP. XXX. THEOREM.

Parallelopipeds upon the the same base, and of the same altitude, the insisting straight lines of which are not terminated in the same straight lines in the plane opposite to the base, are equal to one another.

Let the parallelopipeds CM and CN be upon the same base A B, and of the same altitude, but their insisting straight lines AF, AG, LM, LN, CD, CE, BH, and BK not terminated in the same straight lines. The solids CM and CN are equal to one another.

Produce FD and MH, also NG and KE, and let them meet one another in the points O, P, Q, and R; and ein AO, LP, BQ, and CR.

N

Because the plane LH is parallel to the opposite plane A D, and is that in which are the parallels LB and MQ, and the figure BP; and the plane AD is that in which are the parallels AC and F R, and the figure CO. Therefore the figures BP and CO

are in parallel planes.

Because the plane

A C

N K

GE

R

AN is parallel to the opposite plane CK, forstei and is that in which are the parallels A L, and ON, and the figure AP; and the plane CK is that in which are the paralfels CB and RK, and the figure CQ. Therefore the figures AP and CQ are in parallel planes. But the planes A B and DQ are (Hyp.) parallel. Therefore the solid CP is a parallelopiped. But the solid CM is equal (XI. 29)*o the solid CP, because they are upon the same base AB, and their insisting straight lines AF, A O, CD, and CR; LM, LP, BH, and BQ are terminated in the same straight lines FR and MQ; and the solid CP is equal (XI. 29) to the solid CN, for they are upon the same base AB, and their insisting straight lines AO, AG, LP, and LN; CR, CE, BQ, and BK are terminated in the same straight lines O N and R K. Therefore the solid CM is equal to the solid CN. Wherefore parallelopipeds, &c. Q. E. D.

PROP. XXXI. THEOREM.

Parallelopipeds, which are upon equal bases, and of the same altitude, are equal to one another.

Let the solid parallelopipeds AE and CF be upon equal bases AB and C 1), and of the same altitude. The solid AE is equal to the solid CF.

P

First, let the insisting straight lines be at right angles to the bases A B and C D, and let the bases be placed in the same plane, so that the sides CL and LB may be in a straight line. The straight line LM, which is at right angles to the plane in which the bases are at the point L, is common (XI. 13) to the two solids A E and CF. Let the other insisting lines of the solids be AG, HK, and BE; DF, OP, and CN. Let the angle A L B, in the first case, be equal to the angle CLD. The other sides AL and LD are in a straight line (I. 14). Produce O D and H B; let them meet in Q; and complete the solid parallelopiped LR, of which the base is the parallelogram L Q, and LM is one of its insisting straight lines.

F

R

N

M

ID

A S

HT

Because AB is equal to CD (Hyp.). Therefore A B is to LQ (V. 7), as CD is to LQ. Because the parallelopiped AR is cut by the plane LE, parallel to the opposite planes A K and D) R. Therefore as the base A B is to the base LQ, so is (XI. 25) the slid A E to the solid LR. Because the parallelopiped CR is cut by the plane LF parallel to the opposite planes CP and BR. Therefore as the base CD is to the base LQ, so is

the solid CF to the solid L R. But the base A B is to the base LQ, as the base CD is to the base LQ, as before proved. Therefore, as the solid AE is to the solid LR (V. 11), so is the solid CF to the solid L R. Therefore the solid A E is equal (V. 9) to the solid CF.

In the second case, let the angles SLB and CLD be unequal. The solid SE is equal to the solid C F.

Produce DL and TS until they meet in A, and fromB draw BH parallel to DA; and let H B and OD produced meet in Q, and complete the solids AE and L R.

The solid A E is equal (XI. 29) to the solid S E, because they are upon the same base LE, and of the same altitude, and their insisting straight lines, viz. LA, LS, BH, and BT; MG, MV, EK, and E X, are in the same straight lines A T and G X. Because the parallelogram A B is equal (1.35) to SB, for they are upon the same base L B, and between the same parallels L B and A T; and the base S B is equal to the base C D (Hyp.). Therefore the base A B is equal to the base CD; and the angle ALB is equal to the angle CL D. Wherefore, by the first case, the solid AE is equal to the solid CF; but it was shown that the solid A E is equal to the solid S E. Therefore the solid SE is equal to the solid C F.

Secondly, let the insisting straight lines AG, HK, BE, and LM; CN, RS, DF, and OP be not at right angles to the bases A B and CD; in this case likewise the solid A E is equal to the Solid CF.

From the points G,K, E, and M,N,S, F, and P,

M E

АК

P

F

IN

L

draw the straight lines G Q, KT, E V, and MX; NY, SZ, FI, and P U, perpendicular (XI. 11) to the planes in which are the bases A B and CD; and let them meet these planes in the points Q, T, V, and X ; Y, Z, I, and U. Join QT, T V, VX, and X Q; Ŷ Z, Z I, I Ú, and U Y.

Because G Q and K T are at right angles to the same plane, they are parallel (XI. 6) to one another: and M G and EK are parallels. Therefore the planes MQ and E T, of which one passes through M G and G2, and the other through E K and K T, which are parallel to M G and GQ, and not in the same plane with them, are parallel (XI. 15) to one another. For the same reason, the planes M V and GT are parallel to one another. Therefore the solid Q E is a parallelopiped. In like manner, it may be proved, that the solid Y F is a parallelopiped. But, from what has been demonstrated, the solid E Q is equal to the solid FY, because they are upon equal bases M K and PS, and of the same altitude, and have their insisting straight lines at right angles to the bases. Again, the solid EQ is equal (XI. 29 or 30) to the solid A E, and the solid FY to the solid CF, because they are upon the same bases and of the same altitude. Therefore the solid AE is equal to the solid CF. Wherefore, solid parallelopipeds, &c. Q. E. D.

This proposition is the foundation of the mensuration of solids. As the solid content of a right parallelopiped is found in practice by multiplying the area of its base by its altitude, so the solid content of every oblique parallelopiped is found by multiplying the area of its base by its altitude. And the same rule applies to prisms.

PROP. XXXII. THEOREM.

Parallelopipeds which have the same altitude are to one another as their bases.

Let AB and CD be parallelopipeds of the same altitude. They are to one another as their bases; that is, as the base M N is to the base CF so is the solid A B to the solid CD.

B

To the straight line F G apply the parallelogram F H equal (I. Cor. 45) to MN, so that the angle F GH may be equal to the angle LCG; and upon the base F H complete the parallelopiped GK, one of whose insisting straight lines is FD, so that the solids CD and GK are of the same altitude.

A

D K

0 P

F

M

G H

The solid A B is equal (XI. 31) to the solid GK, because they are upon equal bases MN and FH, and are of the same altitude. Because the parallelopiped CK is cut by the plane DG, which is parallel to its opposite planes, the base H F is (XI. 25) to the base L G, as the solid HD to the solid D C. But the base HF is equal to the base MN, and the solid GK to the solid A B. Therefore as the base M N is to the base CF, so is the solid A B to the solid CD. Wherefore, solid parallelopipeds, &c. Q. E. D.

COROLLARY.-From this it is manifest, that prisms upon triangular bases and of the same altitude, are to one another as their bases. Let the prisms, the bases of which are the triangles AEM and CFG, and NBO and PD Q the triangles opposite to them, have the same altitude: they are to one another as their bases.

Complete the parallelograms AE and CF, and the parallelopipeds A B and CD, in the first of which let M O, and in the other, let GQ be one of the insisting straight lines.

Because the parallelopipeds A B and CD have the same altitude, they are to one another as the base MN is to the base CF. Therefore the prisms, which are their (XI. 28) halves, are to one another, as their bases that is, as the triangle A E M to the triangle CF G.

Corollary-Parallelopipeds which have equal bases are to one another as their

*

altitudes.

PROP. XXXIII.

THEOREM.

Similar parallelopipeds are one to another in the triplicate ratio of their homologous rides or edges.

Let AB and CD be similar parallelopipeds, having the side AE homologous to the side CF. The solid AB has to the solid CD the triplicate ratio of that which A E has to CF.

Produce A E, GE, and HE, and in them take EK equal to CF, EL equal to FN, and EM equal to FR. Complete the parallelogram K L, and the solid K O.

Because the solids A B and CD are similar (Hyp.), the angle AEG is equal to the angle CFN. But the angle A EG is equal to the angle KEL (I. 15). Therefore the angle K E L is equal to the angle CF N. But KE and EL are equal to CF and FN (Const.). Therefore the parallelo