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Book XI. w
à 26. II.
b 12. 6.
C 22. 5.
d 24. II.
PROP. XXVII. P R O B.
To defcribe from a given straight line a folid parallelepiped fimilar, and fimilarly fituated to one given.
Let AB be the given ftraight line, and CD the given folid pa rallelepiped. It is required from AB to defcribe a folid parallelepiped fimilar, and fimilarly fituated to CD.
At the point A of the given ftraight line AB, make2 a folid angle equal to the folid angle at C, and let BAK, KAH, HAB be the three plane angles which contain it, fo that BAK be equal to the angle ECG, and KAH to GCF, and HAB to FCE: And as EC to CG, fo make BA to AK; and as GC to CF, fo make KA to AH; wherefore, ex aequali, as EC to CF, fo is BA to AH: Complete the parallelogram BH, and
the folid AL: And
CG, fo BA to AK,
is fimilar to EG.
For the fame rea- A fon, the parallelogram KH is fimilar to GF, and HB to FE, Wherefore three parallelograms of the folid AL are fimilar to three of the folid CD; and the three oppofite ones in each folid are equal and fimilar to thefe, each to each. Alfo, because the plane angles which contain the folid angles of the figures are equal, each to each, and fituated in the fame order, the folid angles are equal, each to each. Therefore the folid AL is fimilar f to the f11.def. 11. folid CD. Wherefore from a given ftraight line AB a folid parallelepiped AL has been described fimilar, and fimilarly fituated to the given one CD. Which was to be done.
e B. Ir.
F a folid parallelepiped be cut by a plane paffing thro' See N. the diagonals of two of the oppofite planes; it fhall be cut in two equal parts.
Let AB be a folid parallelepiped, and DE, CF, the diagonals of the oppofite parallelograms AH, GB, viz. thofe which are drawn betwixt the equal angles in each: And becaufe CD, FE are cach of them parallel to GA, and not in the fame plane with it, CD, FF are parallel; wherefore the diagonals CF, a 9, II. DE are in the plane in which the parallels are, and are themselves parallels: And the plane CDEF fhall cut the folid AB into two equal parts.
Because the triangle CGF is equal to the triangle CBF, and the triangle DAE to DHE; and that the parallelogram CA is equal and fimilar to the oppofite one BE; and the parallelogram GE to CH: Therefore the A prifm contained by the two triangles
b 16. 11.
C 34. I.
CGF, DAE, and the three parallelograms CA, GE, EC, is equal to the prifm contained by the two triangles CBF, C. II. DHE, and the three parallelograms BE, CH, EC; becaufe they are contained by the fame number of equal and fimilar planes, alike fituated, and none of their folid angles are contained by more than three plane angles. Therefore the folid AB is cut into two equal parts by the plane CDEF. Q. E. D.
N. B. The infifting ftraight lines of a parallelepiped, mentioned in the next and fome following propofitions, are the fides of the parallelograms betwixt the bafe and the oppofite plane parallel to it.'
SOLID parallelepipeds upon the fame bafe, and of the See N. fame altitude, the infisting straight lines of which are terminated in the fame ftraight lines in the plane oppofite to the bafe, are equal to one another.
Let the folid parallelepipeds AH, AK be upon the fame bafe n AB, and of the fame altitude, and let their infifting ftraight See the fi- lines AF, AG, LM, LN, be terminated in the fame ftraight line gures below. FN, and CD, CE, BH, BK be terminated in the fame straight line DK; the folid AH is equal to the folid AK.
c 38. I.
d 36. I.
C 24. II.
First, Let the parallelograms DG, HN, which are oppofite to the bafe AB, have a common fide HG: Then, because the folid AH is cut by the plane AGHĊ paffing through the diago nals AG, CH of the oppofite planes ALGF, CBHD, AH is cut into two equal parts by the planè AGHC: Therefore the folid AH is double of the prifm which is contained betwixt the triangles ALG, CBH: For the fame reafon, because the folid AK is cut by the plane LGHB through the diagonals LG, BH of the oppofite planes ALNG,
CBKH, the folid AK is double of the fame prifm which is contained betwixt the triangles ALG, CBH. Therefore the folid AH is equal to the folid AK.
But, let the parallelograms DM, EN oppofite to the bafe, have no common fide: Then, because CH, CK are parallelograms, CB is equal to each of the oppofite fides DH, EK; wherefore DH is equal to EK: Add, or take away the common part HE; then DE is equal to HK: Wherefore alfo the triangle CDE is equal to the triangle BHK: And the parallelogram DG is equal to the parallelogram HN: For the fame reafon, the triangle AFG is equal to the triangle LMN, and the parallelogram CF is equal to the parallelogram BM, and
CG to BN; for they are oppofite. Therefore the prifm which is contained by the two triangles AFG, CDE, and the three parallelograms AD, DG, GC is equal f to the prism, contained by the two triangles LMN, BHK, and the three parallelograms BM, MK, KL. If therefore the prifm LMNBHK be
taken from the folid of which the bafe is the parallelogram Book XI. AB, and in which FDKN is the one oppofite to it; and if from this fame folid there be taken the prifm AFGCDE; the remaining folid, viz. the parallelepiped AH, is equal to the remaining parallelepiped AK. Therefore folid parallelepipeds, &c. Q, E. D.
PROP. XXX. THEOR.
SOLID parallelepipeds upon the same base, and of the See Ni
fame altitude, the infifting ftraight lines of which are not terminated in the fame ftraight lines in the plane oppofite to the bafe, are equal to one another.
Let the parallelepipeds CM, CN be upon the fame base AB, and of the fame altitude, but their infifting straight lines AF, AG, LM, LN, CD, CE, BH, BK not terminated in the fame ftraight 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 oppofite
plane ACDF, and that the plane LBHM is that in which are the parallels LB, MHPQ, in which alfo is the figure BLPQ; and the plane ACDF is that in which are the parallels AC, FDOR, in which alfo is the figure CAOR; therefore the figures BLPO, CAOR are in parallel planes: In like manner, because the plane ALNG is parallel to the oppofite plane CBKE, and that the plane ALNG is that in which are the parallels
232 Book XI.
AL, OPGN, in which alfo is the figure ALPO; and the plane CBKE is that in which are the parallels CB, RQEK, in which alfo is the figure CBQR; therefore the figures ALPO, CBQR are in parallel planes: And the planes AČBL, OROP are parallel; therefore the folid CP is a parallelepiped: But the folid CM, of which the bafe is ACBL, to which FDHM is the a 29. 11. oppofite parallelogram, is equal to the folid CP of which, the
bafe is the parallelogram ACBL, to which OROP is the one oppofite; because they are upon the fame bafe, and their infifting ftraight lines AF, AO, CD, CR; LM, LP, BH, BQ are in the fame ftraight lines FR, MQ: And the folid CP is equal to the fold CN; for they are upon the fame bafe ACBL, and their infifting straight lines AO, AG, LP, LN; CR, CE, BO, BK are in the fame ftraight lines ON, RK: Therefore the folia CM is equal to the folid CN. Wherefore folid parallelepipeds, &c. Q. E. D. ·
OLID parallelepipeds which are upon equal bases, and of the fame altitude, are equal to one another.
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 infifting straight lines be at right angles to the bases AB, CD, and let the bafes be placed in the fame plane,