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Book XI. PRO P. XXVIII. THEO R. F a solid parallelepiped be cut by a plane pafling Sce N.

thro’ the diagonals of two of the opposite planes; it thall be cut into two equal parts.

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b. 16. 11.

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C. 34. I.

Let AB be a solid parallelepiped, and DE, CF the diagonals of the opputie parallelograms AH, GB, viz. those which are drawn berw xt the equal angles in each. and because CD, FE are each of . them parallel to GA, and not in the same plane with it, CD, EF are paliel"; wherefore the diagonals CF, DE are in the plane in a. 9. 11. winch the parallels are, and are them

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B faires parallels b. and the plane CDEF fall cut the folid AB into two equal paits.

Because the triangle CGF is equal o to the triangle CBF, and the triangle

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H DAE tu DHE ; and that the parallelogram CA is cqual 4 and similar to the

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d. 24. II.

A E pofte one BE ; and the parallelogram GE to CH. therefore the prism cuntained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equalc. C. 11. to the prism contained by the two triangles CBF, DHE, and the three parallelograms BE, CH, EC ; because they are contained by the same number of equal and similar planes, alike situated, and none of their foli: angles are contained by more than three plane angles. Therefore the solid AB is cut into two equal parts by the plane CDEF. Q. E. D.

• N. B. The intifting straight lines of a parallelepiped, men' tioned in the next and some following Propofitions, are the sides ' of the parallelograms betwixt the base and the opposite plane • parallel to it.'

PRO P. XXIX. THEOR.

SOLE
OLID parallelepipeds upon the fame base, and of the Sce N.

fame 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.

Book XI. Let the solid parallelepipeds AH, AK be upon the same bafe MAB, and of the same altitude, aad let their infifting straight lines See the fi- AF, AG, LM, LN; CD, CE, BH, BK be terminated in the same

straight lines FN, DK. the folid AH is equal to the folid AK.

First, Let the parallelograms DG, HN which are opposite to the base AB have a common side HG. then because the solid AH is cut by the plane AGHC paling thro' the diagonals AG, CH of

the opposite planes ALGF, CBHD, AH is cut into two equal 2. 28. 11. parts • by the plane AGHC.

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therefore the folid AH is double

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of the prism which is contained
by the triangles ALG, CBH. for
the same reason, because the fo-

B
lid AK is cut by the plane LGHB
thro' the diagonals LG, BH of A L
the opposite planes ALNG, CBKH, the folid AK is double of
the fame prism which is contained by the triangles ALG, CBH.
Therefore the folid AH is equal to the folid AK.

Put let the parallelograms DM, EN opposite to the base have

no common fide. then because CH, CK are parallelograms, CB is 6. 34. 1. equal 6 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 c. 38. 1. cqual to HK. wherefore allo the triangle CDE is equal to the d. 36. 1. triangle BHK. and the parallelogram DG is equal d' to the paral

lelogram HN. for the fame reason, the triangle AFG is equal to c. 24. 11. the triangle LMN, and the parallelogram CF is equal o to the paralD H E

E H K
M
AG

GC

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F

M

K D

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Α.

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lelogram BM, and CG to BN; for they are opposite. Therefore

the prism which is contained by the two triangles AFG, CDE, and f. C. 11. the three parallelograms AD, DG, GC is equal { to the prism

contained by the two triangles LMN, BHK, and the three parallelograms BM, MK, KL. If therefore the prism LMN, BHK be

taken from the solid of which the base is the parallelogram AB, Book XI. and in which TDKN is the one opposite to it; and if from this fame folid there be taken the prism AFG, CDE; the remaining folid, viz. the parallelepiped AH, is equal to the remaining parallelepiped AK. Therefore solid parallelepipeds, &c. Q. E. D.

PRO P. XXX. THE O R.
ID parallelepipeds upon the same base, and of Soe N.

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.

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Let the parallelepipeds CM, CN be upon the same base AB, and of the same altitude, but their infifting straight lives AF, AG, LM, LN, CD, CE, BH, BK not terminated in the same straight lines. the folids 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,

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and that the plane LBHM is that in which are the parallels LB, MHPQ, in which also is the figure BLPQ; and the plane ACDI 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

Book XI. are the parallels AL, OPGN, in which also is the figure ALPO;

and the plane CBKE is that in which are the parallels CB, ROEK,
in which also is the figure CBOR ; therefore the figures ALPO,
CBOR are in parallel planes. and the planes ACBL, OROP are
parallel; therefore the solid CP is a parallelepiped. but the folid

CM of which the bale is ACBL, to which FDHM is the opposite a. 29. 11. parallelogram, is equal to the solid CP of which the base is the

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parallelogram ACBL, to which ORQP is the one opposite; be-
cause they are upon the saine base, and their insisting straight lines
AF, AO, CD, CR; LM, LP, BH, BQ are in the fame straight
lines FR, MO. and the folid CP is equal to the folid CN, for
they are upon the fame base ACBL, and their infifting straight
linos AO, AG, LP, LN; CR, CE, BQ, BK are in the same
straight lines ON, RK. therefore the folid CM is equal to the
folid CN. Wherefore solid parallelepipeds, &c. Q. E. D.

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PROP. XXXI. THEO R.

See N.

SOLID parallelepipeds which are upon equal bases,

and of the fame altitude, are equal to one another.

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

First, let the insisting straight lines be at right angles to the bases
AB, CD, and let the bases be placed in the fame place, and so as

that the sides CL, LB be in a straight line ; therefore the straight Bock XI: line LM, which is at right angles to the plane in which the bases are, w in the point L, is common ' to the two folids AE, CF ; let the a. 13. 11. 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 b. produce OD, HB, and let them b. 14. 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, fo is © the base c. 7. 5. CD to the fame LQ, and because the folid parallelepiped AR is cut by the plane LMEB which is parallel to the opposite planes AK, DR; as the base AB is to the bale LQ, fo is a the folid d. 25. si: AE to the folid LR. for the same reason, because the folid parallelepiped CR is cut by the plane LMFD which is parallel to the opposite planes CP, BR; as the base CD P

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RO to the base LQ, so

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M is the solid CF to the

X folid LR. but as the

+ base AB to the base LQ, so the base CD to the base LQ ,

I as before was prov

HT ed. therefore as the folid AE to the folid LR, fo is the solid CF to the folid LR; therefore the solid AE is equal to the solid CF.

e. 9. f! But let the folid parallelepipeds SE, CF be upon equal bases SB, CD, and be of the fame altitude, and let their insisting Araight lines be at right angles to the bases; and place the bafes SB, CD in the fame plane, so that CL, LB be in a straight line; and let the angles SLB, CLD be unequal; the folid 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 f to the folid SE, of which the base is LE, 1.2, 17, and to which SX is opposite; for they are upon the same base LE, and of the fame altitude, and their insisting straight lines, viz, LA, LS, BH, BT; MG, MV, EK, EX are in the same straight

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