Imágenes de páginas
PDF
EPUB

taken from the solid of which the base is the parallelogram AB, and in which FDKN is the one opposite to it; and if from this same solid there be taken the prism AFGCDE, the remaining solid, viz. the parallelopiped AH, is equal to the remaining parallelopiped AK. Therefore, solid parallelopipeds, &c. Q. E. D.

PROP. XXX. THEOR.

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

Produce FD, MH, and NG, KE; and let them meet one another in the points O, P, Q, R; and join AO, LP, BQ, CR: and because the plane LBHM is parallel to the opposite plane ACDF, N K

[blocks in formation]

and that the plane LBHM is that in which are the parallels LB, MHPQ, in which also is the figure BLPQ, and the plane ACDF is that in which are the parallels AC, FDOR, in which also is the figure CAOR; therefore the figures BLPQ, CAOR are in parallel planes; in like manner, because the plane ALNG is parallel to the opposite plane CBKE, and that the plane ALNG is that in which are the parallels AL, OPGN, in which also is the

* See Note.

figure ALPO; and the plane CBKE is that in which are the parallels CB, RQEK, in which also is the figure CBQR; therefore the figures ALPO, CBQR are in parallel planes; and the planes ACBL, ORQP are parallel; therefore the solid CP is a parallelopiped; but the solid CM, of which the base is ACBL, to which FDHM is the opposite parallelogram, is equal (29. 11.) to the solid CP, of which the base is the parallelogram ACBL, to N K

[merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

CR;

which ORQP is the one opposite; because they are upon the same base, and their insisting straight lines AF, AO, CD, LM, LP, BH, BQ are in the same straight lines FR, MQ: and the solid CP is equal (29. 11.) to the solid CN for they are upon the same base ACBL, and their insisting straight lines AO, AG, LP, LN; CR, CE, BQ, BK, are in the same straight lines ON, RK: therefore the solid CM is equal to the solid CN. Wherefore solid parallelopipeds, &c. Q. E. D.

PROP. XXXI. THEOR.

SOLID parallelopipeds which are upon equal bases, and of the same altitude, are equal to one another.*

Let the solid parallelopipeds AE, CF be upon equal bases AB, CD, and be of the same altitude; the solid AE is equal to the solid CF.

First, let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be placed in the same plane,

* See Note.

and so as that the sides CL, LB be in a straight line; therefore the straight line LM which is at right angles to the plane in which the bases are, in the point L, is common (13. 11.) to the 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 (14. 1.). Produce OD, HB, and 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 (7. 5.) the base CD to the same LQ: 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 (25. 11.) the solid 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, so is the solid CF to the solid LR: but as the base AB to the base LQ, so the base CD to the base LQ, as before was proved : therefore as the solid

[ocr errors]

P

F

N

M

[ocr errors]

D

C

L

A S

R

E

B

X

K

H T

AE to the solid LR, so is the solid CF to the solid LR; and therefore the solid AE is equal (9. 5.) to the solid CF.

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 (29. 11.) to the solid SE, of which 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 because the parallelogram AB is equal (35. 1.) to SB, for they are upon the same base LB, and between the same parallels LB, AT;

and that the base SB is equal to the base CD; therefore the base

AB is equal to the

P

F

[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

R

[blocks in formation]

C

[blocks in formation]

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, perpendicular (11. 11.) 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

[blocks in formation]

angles to the same plane, they are parallel (6. 11.) 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 same plane with them, are parallel (15. 11.) 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 at right angles to the bases and the solid EQ is equal (29. or 30. 11.) to the solid

AE; 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.

PROP. XXXII. THEOR.

SOLID parallelopipeds which have the same altitude, 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.

B

D

K

To the straight line FG apply the parallelogram FH equal (Cor. 45. 1.) to AE, so that the angle FGH be equal to the angle 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 the solid AB is equal (31. 11.) to the solid GK, because they are upon equal bases AE, FH, and are of the same altitude: and

P

L

F

because the solid

parallelopiped CK

[blocks in formation]

is cut by the plane DG which is parallel to its opposite planes, the base HF is (25. 11.) to the base FC, as the 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 the base AE is to the base CF; wherefore the prisms,

* See Note.

« AnteriorContinuar »