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equal, and also the straight lines KO, AY, EF, which make equal angles with LK, KA, AE, the parallelograms LO, KY, AF are equal and similar (36. 1. & def. 1. 6.); and likewise the parallelograms KX, KB, AG; as also (2. 3. Sup.) the parallelograms LZ, KP, AR, because they are opposite planes. For the same reason, the parallelograms EC, HQ, MS are equal (36. 1. & def. 1. 6.); and the parallelograms HG, HI, IN, as also (2. 3. Sup.) HD, MU, NT; therefore three planes of the solid LP, are equal and similar to three planes of the solid KR, as also to three planes of the solid AV: but the three planes opposite to these three are equal and similar to them (2. 3. Sap.) in the several solids; therefore the solids LP, KR, AV are contained by equal and similar planes. And because the planes LZ, KP, AR are parallel, and are cut by the plane XV, the inclination of LZ to XP is equal to that of KP to PB; or of AR to BV (15. 2. Sup.): and the same is true of the other contiguous planes, therefore the solids LP, KR, and AV, are equal to one another (1. 3. Sup.). For the same reason, the three solids ED, HU, MT are equal to one another; therefore what multiple soever the base LF is of the base AF, the same multiple is the solid LV of the solid AV; for the same reason, whatever multiple the base NF is of the base HF, the same multiple is the solid NV of the solid ED: And if the base LF be equal to the base NF, the solid LV is equal (1. 3. Sup.) to the solid NV; and if the base LF be greater than the base NF, the solid LV is greater than the solid NV; and if less, less. Since then there are four magnitudes, viz. the two bases AF, FH, and the two solids AV, ED, and of the base AF and solid AV, the base LF and solid LV are any equimultiples whatever; and of the base FH and solid ED, the base FN and solid NV are any equimultiples whatever; and it has been proved, that if the base LF is greater than the base FN, the solid LV is greater than the solid NV; and if equal, equal; and if less, less: Therefore (def. 5. 5.), as the base AF is to the base FH, so is the solid AV to the solid ED. Wherefore, if a solid, &c. Q. E. D.

COR. Because the parallelogram AF is to the parallelogram FH as YF to FC (1. 6.), therefore the solid AV is to the solid ED as YF to FC.

PROP. IV. THEOR.

If a solid parallelopiped be cut by a plane passing through the diagonals of two of the opposite planes, it will be cut into two equal prisms.

Let AB be a solid parallelopiped, and DE, CF the diagonals of the opposite parallelograms AH, GB, viz. those which are drawn betwixt the equal angles in each; and because CD, FE are each of them parallel to GA, though not in the same plane with it, CD, FE are parallel (8. 2. Sup.); wherefore the diagonals CF, DE are in the plane

in which the parallels are, and are themselves parallels (14. 2. Sup.): the plane CDEF cuts the solid AB into two equal parts.

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F

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Because the triangle CGF is equal (34. 1.) to the triangle CBF, and the triangle G DAE to DHE; and since the parallelogram CA is equal (2. 3. Sup.) and similar to the opposite one BE; and the parallelogram GE to CH therefore the planes which contain the prisms CAE, CBE, are equal and similar, each to each; and they are also equally inclined to one another, because the planes AC, EB are parallel, as also AF and BD, and they are cut by the plane CE (15. 2. Sup.). Therefore the prism CAE is equal to the prism CBE (1. 3. Sup.), and the solid AB is cut into two equal prisms by the plane CDEF. Q. E. D.

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N. B. The insisting straight lines of a parallelepiped, mentioned in the following propositions, are the sides of the parallelograms betwixt the base and the plane parallel to it.

PROP. V. THEOR.

Solid parallelepipeds 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 solid parallelepipeds AH, AK be upon the same base AB, and of the same altitude, and let their insisting straight lines AF, AG, LM, LN, be terminated in the same straight line FN, and let the insisting lines CD, CE, BH, BK be terminated in the same straight line DK; the solid AH is equal to the solid AK.

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Because CH, CK are parallelograms, CB is equal (34. 1.) to each of the opposite sides DH, EK; wherefore DH is equal to EK: add, or take away the common part HE; then DE is equal to HK: Wherefore also the triangle CDE is equal (38. 1.) to the triangle BHK and the parallelogram DG is equal (36. 1.) to the parallelogram HN. For the same reason, the triangle AFG is equal to the triangle LMN, and the parallelogram CF is equal (2. 3. Sup.) to the parallelogram BM, and CG to BN; for they are opposite. Therefore the planes which

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contain the prism DAG are similar and equal to those which contain the prism HLN, each to each; and the contiguous planes are- also equally inclined to one another (15. 2. Sup.), because that the parallel planes AD and LH, as also AE and LK, are cut by the same plane DN therefore the prisms DAG, HLN are equal (1. 3. Sup.). If therefore the prism LNH be taken from the solid, of which the base is the parallelogram AB, and FDKN the plane opposite to the base ; and if from this same solid there be taken the prism AGD, the remaining solid, viz. the parallelepiped AH is equal to the remaining parallelepiped AK. Therefore solid parallelepipeds, &c. Q. E. D.

PROP. VI. THEOR.

Solid parallelepipeds 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 parallelepipeds, 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. Because the planes (def. 5. 3. Sup.). LBHM and ACDF are parallel, and because the

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plane LBHM is that in which are the parallels LB, MHPQ (def. 5. 3. Sup., and in which also is the figure BLPQ; and because the plane ACDF is that in which are the parallels AC, FDOR, and in which also is the figure CAOR; therefore the figures BLPQ, CAOR, are in parallel planes. In like manner, because the planes ALNG and CBKE are parallel, and the plane ALNG is that in which are the parallels AL, OPGN, and in which also is the figure ALPO; and the plane CBKE is that in which are the parallels CB, RQEK, and in which also is the

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figure CBQR; therefore the figures ALPO, CBQR are in parallel planes. But the planes ACBL, ORQP are also parallel; therefore the solid CP is a parallelepiped. Now the solid parallelepiped CM is equal (5. 2. Sup.) to the solid parallelepiped CP; because they are upon the same base, and their insisting straight lines AF, AO, CD, CR; LM, LP, BH, BQ are terminated in the same straight lines FR, MQ and the solid CP is equal (5. 2. Sup.) 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 terminated in the same straight lines ON, RK: Therefore the solid CM is equal to the solid CN. Wherefore solid parallelepipeds, &c. Q. E. D.

PROP. VII. THEOR.

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

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

Case 1. Let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be placed in the same plane, 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 (11. 2. Sup.) 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 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 (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

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base LQ, so is (3. 3. Sup.) the solid AE to the solid LR for the same 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 the base CD to the base LQ, as has been proved : therefore as the solid 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 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 plane opposite to it, is equal (5.3. Sup.) to the solid SE, of which the base is LE, and SX the plane 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, MU, 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 because the base SB is equal to the base CD; therefore the base AB is equal to the base CD; but the angle ALB is equal to the angle CLD: therefore, by the 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.

Case 2. If the insisting straight lines AG, HK, BE, LM; CN, RS,

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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. Because solid parallelepipeds on the same base, and of the same altitude, are equal (6. 3. Sup.), if two solid parallelepipeds be constituted on the bases AB and CD of the same altitude with the solids AE and CF, and with their insisting lines perpendicular to their bases, they will be equal to the solids AE and CF; and, by the first case of this proposition, they will be equal to one another; wherefore, the solids AE and CF are also equal. Wherefore, solid parallelepipeds, &c. Q. E. D.

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