Sidebilder
PDF
ePub

plane, and so as that the sides CL, LB be in a straight line; Book XI. therefore the straight line LM, which is at right angles to the plane in which the bases are, in the point L, is commona to the two solids AE, CF; let the other insisting 13. 11. 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 14. 1. let them meet in Q, and complete the solid parallelopiped 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 the base CD to the same LQ : 7. 5. And because the solid parallelopiped 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 25. 11, AE to the solid LR: For the same reason, because the solid parallelopiped CR is cut by the plane LMFD; which is parallel to the opposite planes CP, BR; as the base CD to the base LQ, p

SO is the solid
CF to the solid

F

R

[blocks in formation]
[blocks in formation]
[ocr errors]

solid AE to the solid LR, so is the solid CF to the solid LR; and therefore the solid AE is equal to the solid CF. 9. 5. But let the solid parallelopipeds 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 29. 11. the base is 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 be

Book XI. cause the parallelogram AB is equals to SB, for they are upon the same base LB, and between the same parallels LB, AT: and that P

€ 35. 1.

the base SB is e

R

qual to the base

[blocks in formation]
[blocks in formation]

first case, the solid AE is equal to the solid CF: But the solid AE is equal to the solid SE, as was demonstrated; therefore the solid SE is equal to the solid CF.

But, 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 GQ, KT, EV, MX; NY, SZ, FI, PU, per11. 11. pendicular 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, be

[blocks in formation]

k

cause GQ, KT, are at right angles to the same plane, they 6. 11. are paralleli to one another: And MG, EK are parallels; therefore the plane MQ, ET, of which one passes through MG, GQ, and the other through EK, KT, which are parallel to MG, GQ, and not in the saine plane with them, 15. 11. are parallel k to one another: For the same reason, the planes MV, GT are parallel to one another: Therefore the solid QE is a parallelopiped: In like manner, it may be proved, that the solid YF is a parallelopiped: 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 129. or 30. at right angles to the bases: And the solid EQ is equal to

11.

the solid AE; and the solid FY to the solid CF; because Book XI. 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.

PROP. XXXII. THEOR.

SOLID parallelopipeds which have the same alti- See N. tude are to one another as their bases.

Let AB, CD be solid parallelopipeds 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 to AE, so that the angle FGH be equal to the angle 'Cor. 45. 1. LCG; and complete the solid parallelopiped GK upon the base FH, one of whose insisting lines is FD, whereby the solids CD, GK must be of the same altitude: Therefore

b

the solid AB is equal to the solid GK, because they are 31. 11.

[blocks in formation]

solid paral

[blocks in formation]
[blocks in formation]

lelopiped CK is cut by the plane DG, which is parallel to its opposite planes, the base HF is to the base FC, as the 25. 11. solid HD to the solid DC: But 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 parallelopipeds, &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 parallelopipeds 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 parallelopipeds AB, CD have the same altitude, they are to one another as

BOOK XI. the base AE is to the base CF: wherefore the prisms, which are their halves, are to one another, as the base AE to the 28. 11. base CF; that is, as the triangle AEM to the triangle CFG.

[ocr errors]

PROP. XXXIII. THEOR.

SIMILAR solid parallelopipeds are one to another in the triplicate ratio of their homologous sides.

Let AB, CD be similar solid parallelopipeds, 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.

X

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 similar and equal to CR, and also OE to FD. Therefore three parallelograms of the solid KO are equal and similar to three parallelograms of the solid

CD: And the

D

H

R

G

M

24. 11. three opposite ones in each solid are equal a and similar to C. 11. these: 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 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: 1. 6. But as AE to EK, so is the parallelogram AG to the pa rallelogram GK; and as GE to EL, so is GK to KL, and

C

1.6. d 25. 11.

as HE to EM, so is PE to KM: therefore as the parallelo- Book XI. gram AG to the parallelogram GK, so is GK to KL, and PE to KM: But as AG to GK, so is the solid AB to the solid EX, and 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 parallelopiped 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 parallelopipeds 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 parallelopipeds, of which AB is contained by the parallelograms AE, AF, AG equiangular, each to each, to the parallelograms CH, CK, CL, which contains 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.

« ForrigeFortsett »