and so as that the sides CL, LB be in a straight line; there- Book XI. fore the straight line LM, which is at right angles to the plane in which the bases are, in the point L, is common a to the a 13. 11. two solids AE, CF; let the other insisting lines of the solids be AG, HK, BE, DF, OP, CN: And first, let the angle ALB be equal to the angle CLD; then, AL, LD are in a straight line. Produce OD, HB, and let them meet in Q, b 14. l. and complete the solid parallelepiped LR, the base of which is the parallelogram LQ, and of which LM is one of its insisting straight lines: Therefore, because the parallelogram AB is equal to CD, as the base AB is to the base LQ, so is c 7. 5. the base CD to the same LQ: And because the solid parallelepiped AR is cut by the plane LMEB, which is parallel to the opposite planes AK, DR; as the base AB is to the base LQ, so is the solid AE to the solid LR: For the same d 25. 11. reason, because the solid parallelepiped CR is cut by the plane LMFD, which is parallel to the opposite planes CP, BR; as the base CD to

the base LQ, so is the solid CF to the solid LR: But as the base AB to

the base LQ, so is

the base CD to the O

D

Q

solid LR: and therefore the solid AE is equal to the solid CF. e 9. 5. But let the solid parallelepipeds SE, CF be upon equal bases SB, CD, and be of the same altitude, and let their insisting straight lines be at right angles to the bases; and place the bases SB, CD in the same plane, so that CL, LB be in a straight line, and let the angles SLB, CLD be unequal, the solid SE is also, in this case, equal to the solid CF : Produce DL, TS until they meet in A, and from B draw BH parallel to DA; and let HB, OD produced meet in Q, and complete the solids AE, LR: Therefore the solid AE, of which the base is the parallelogram LE, and AK the one opposite to it, is equal to the solid SE, of which the base is f 29. 11. LE, and to which SX is opposite, for they are upon the same base LE, and of the same altitude; and their insisting straight lines, viz. LA, LS, BH, BT; MG, MV, EK, EX are in the same straight lines AT, GX: And because the parallelo

Book XI. gram AB is equal to SB, for they are upon the same base LB, and between the same parallels LB, AT: and the base

g 35. 1.

SB is equal to the
base CD; there-

fore, the base AB

is equal to the base

CD, and the angle

ALB is equal to

the angle CLD: 0

was demonstrated; therefore the solid SE is equal to the solid CF.

2do. If the insisting straight lines AG, HK, BE, LM; CN, RS, DF, OP, be not at right angles to the bases AB, CD; in this case likewise the solid AE is equal to the solid CF: From the points G, K, E, M, N, S, F, P, draw the straight lines h 11. 11. GQ, KT, EV, MX; NY, SZ, FI, PU, perpendicular to the plane in which are the bases AB, CD; and let them meet it in the points Q, T, V, X; Y, Z, I, U, and join QT, TV, VX, XQ; YZ, ZI, IU, UY: Then, because GQ, KT are at right

i 6. 11. angles to the same plane, they are parallel to one another: And MG, EK are parallels; therefore the planes MQ, ET, of which one passes through MG, GQ, and the other through EK, ET, which are parallel to MG, GQ, and not in the same k 15. 11. plane with them, are parallel to one another: For the same reason, the planes MV, GT, are parallel to one another: Therefore the solid QE is a parallelepiped: In like manner, it may be proved, that the solid YF is a parallelepiped: But, from what has been demonstrated, the solid EQ is equal to the solid FY, because they are upon equal bases MK, PS, and of the same altitude, and have their insisting straight lines at right

angles to the bases: And the solid EQ is equal' to the solid Book XI. AE; and the solid FY to the solid FC; because they are upon the same bases and of the same altitude: Therefore the 129. or 30. solid AE is equal to the solid CF: Wherefore, "solid parallelepipeds," &c. Q. E. D.

PROP. XXXII. THEOR.

11.

Solid parallelepipeds, which have the same altitude, See N. are to one another as their bases.

Let AB, CD, be solid parallelepipeds of the same altitude: They are to one another as their bases; that is, as the base AE to the base CF, so is the solid AB to the solid CD.

To the straight line FG apply the parallelogram FH, equal a a Cor. 45.1. to AE, so that the angle FGH may be equal to the angle LCG; and complete the solid parallelepiped GK, upon the base FH, one of whose insisting lines is FD, whereby the solids CD, GK

must be of the same altitude: Therefore the solid AB is equal b b 31. 11. to the solid

piped CK is

cut by the plane

DG, which is parallel to its opposite planes, the base HF

C

is to the base FC, as the solid HD to the solid DC: But c 25. 11. the base HF is equal to the base AE, and the solid GK to the solid AB: Therefore, as the base AE to the base CF, so is the solid AB to the solid CD. Wherefore, "solid parallelepipeds," &c. Q. E. D.

COR. From this it is manifest that prisms upon triangular bases of the same altitude are to one another as their bases.

Let the prisms, the bases of which are the triangles AEM, CFG, and NBO, PDQ the triangles opposite to them, have the same altitude; and complete the parallelograms AE, CF, and the solid parallelepipeds AB, CD, in the first of which let MO, and in the other let GQ be one of the insisting lines. And because the solid parallelepipeds AB, CD have the same altitude, they are to one another as the base AE is to the base

P

Book X1. CF; wherefore the prisms, which are their halves, are to one another, as the base AE to the base CF; that is, as the

a 28. 11. triangle AEM to the triangle CFG.

PROP. XXXIII. THEOR.

Similar solid parallelepipeds are one to another in the triplicate ratio of their homologous sides.

Let AB, CD be similar solid parallelepipeds, and the side AE homologous to the side CF: The solid AB has to the solid CD, the triplicate ratio of that which AE has to CF.

Produce AE, GE, HE, and in these produced take EK, equal to CF, EL equal to FN, and EM equal to FR; and complete the parallelogram KL, and the solid KO: Because KE, EL are equal to CF, FN, and the angle KEL equal to the angle CFN, because it is equal to the angle AEG, which is equal to CFN, by reason that the solids AB, CD are similar; therefore the parallelogram KL is similar and equal to the parallelogram CN: For the same reason, the parallelogram MK is simi

lar and equal to
CR, and also OE

to FD. Therefore
three parallelo-
grams of the solid
KO are equal and
similar to three pa-
rallelograms of the
solid CD: And
the three opposite
ones in each solid

b C. 11.

c 1. 6.

B X

D

b

Therefore the solid KO is equal and similar to the solid. CD: Complete the parallelogram GK, and complete the solids EX, LP upon the bases GK, KL, so that EH may be an insisting straight line in each of them, whereby they must be of the same altitude with the solid AB: And because the solids AB, CD are similar, and by permutation, as AE is to CF, so is EG to FN, and so is EH to FR; and FC is equal to EK, and FN to EL, and FR to EM: Therefore, as AE to EK, so is EG to EL, and so is HE to EM: But, as AE to EK, so is the parallelogram AG to the parallelogram

C

GK; and as GE to EL, so is GK to KL, and as HE to Book XI. EM, so is PE to KM: Therefore as the parallelogram AG

to the parallelogram GK, so is GK to KL, and PE to KM: c 1. 6. But as AG to GK, so d is the solid AB to the solid EX; and d 25. 11. as GK to KL, sod is the solid EX to the solid PL; and as PE to KM, sod is the solid PL to the solid KO: And therefore as the solid AB to the solid EX, so is EX to PL, and PL to KO: But if four magnitudes be continual proportionals, the first is said to have to the fourth the triplicate ratio of that which it has to the second: Therefore the solid AB has to the solid KO, the triplicate ratio of that which AB has to EX: But as AB is to EX, so is the parallelogram AG to the parallelogram GK, and the straight line AE to the straight line EK. Wherefore the solid AB has to the solid KO, the triplicate ratio of that which AE has to EK. And the solid KO is equal to the solid CD, and the straight line EK is equal to the straight line CF. Therefore the solid AB has to the solid CD, the triplicate ratio of that which the side AE has to the homologous side CF, &c. Q. E. D.

COR. From this it is manifest, that, if four straight lines be continual proportionals, as the first is to the fourth, so is the solid parallelepiped described from the first to the similar solid similarly described from the second; because the first straight line has to the fourth the triplicate ratio of that which it has to the second.

PROP. D. THEOR.

Solid parallelepipeds contained by parallelograms See N. equiangular to one another, each to each, that is, of which the solid angles are equal, each to each, have to one another the ratio which is the same with the ratio compounded of the ratios of their sides.

Let AB, CD be solid parallelepipeds, of which AB is contained by the parallelograms AE, AF, AG equiangular, each to each, to the parallelograms CH, CK, CL, which contain the solid CD. The ratio which the solid AB has to the solid CD, is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH.