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Book XI. been made equal to a given solid angle contained by three
plane angles. Which was to be done.
PROP. XXVII. PROB.
To describe from a given straight line a solid parallelepiped similar, and similarly situated to one given.
Let AB be the given straight line, and CD the given solid parallelepiped. It is required from AB, to describe a solid
parallelepiped similar, and similarly situated to CD. a 26. 11. At the point A, of the given straight line AB, make a a solid
angle equal to the solid angle at C, and let BAK, KAH, HAB be the three plane angles which contain it, so that BAK
be equal to the angle ECG, and KAH to GCF, and HAB to b 12. 6. FCE: And as EC to CG, so make 6 BA to AK; and as GC c 22. 5. to CF, so make b KA to AH: wherefore, ex æquali", as EC
to CF, so is BA to AH: Complete the parallelogram BH, and
grams of the solid AL are similar to three of the solid CD; d 24. 11. and the three opposite ones in each solid are equal, and
similar to these, each to each. Also, because the plane angles which contain the solid angles of the figures are equal, each
to each, and situated in the same order, the solid angles are e B. 11. equal<, each to each. Therefore the solid AL is similar f to f 11.def. 11. the solid CD. Wherefore, from a given straight line AB,
a solid parallelepiped AL has been described similar, and similarly situated, to the given one CD. Which was to be done.
PROP. XXVIII. THEOR.
If a solid parallelepiped be cut by a plane passing See N. through the diagonals of two of the opposite planes ; it will be cut into two equal parts.
b 16. 11.
Let AB be a solid parallelepiped, 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, and not in the same plane with it, CD, EF are parallel a : wherefore the diagonals CF, a 9. 11. DE are in the plane in which the parallels are, and are themselves paral
B lels b; and the plane CDEF cuts the solid AB into two equal parts.
F Because the triangle CGF is equal
c 31. 1. to the triangle CBF, and the triangle DAE to DHE; and since the paral
H Н lelogram CA is equal d and similar to
d 24. 11. the opposite one BE; and the parallelogram GE to CH: therefore, the A
E prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal e to the prism contained by the two triangles CBF, e C. 11. 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 solid 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 insisting straight lines of a parallelepiped, men• tioned in the next and some following propositions, are the • sides of the parallelograms betwixt the base and the opposite plane parallel to it.'
PROP. XXIX. THEOR.
Solid parallelepipeds upon the same base, and of the See N. 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 See the
lines AF, AG, LM, LN be terminated in the same straight figures below.
line FN, and CD, CE, BH, BK be terminated in the same straight line DK; the solid AH is equal to the solid 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, passing through the dia
gonals AG, CH of the opposite planes ALGF, CBHD, AH a 28. 11. is cut into two equal parts a by the plane AGHC: There
fore, the solid AH is double
N because the solid AK is cut by the plane LGHB through C
B В the diagonals LG, BH of the opposite planes ALNG, А. L CBKH, the solid AK is double the same prism which is contained betwixt the triangles ALG, CBH; therefore the solid AH is equal to the solid AK.
But, let the parallelograms DM, EN, opposite to the base,
have no common side: Then, because CH, CK, are parallelob 34. 1. grams, CB is equal to each of the opposite sides DH, EK;
wherefore DH is equal to EK: Add, or take away the common
part HE; then DĒ is equal to HK: Wherefore also, the tric 38. 1. angle CDE is equal to the triangle BHK: And the parallelod 36. 1. gram DG is equald to the parallelogram HN: For the same
reason, the triangle AFG is equal to the triangle LMN, and e 24. 11. the parallelogram CF is equal to the parallelogram BM, and
CG to BN; for they are opposite. Therefore, the prism which
is contained by the two triangles AFG, CDE, and the three f c. 11. parallelograms AD, DG, GC is equal to the prism, contain
ed by the two triangles LMN, BHK, and the three parallelograms BM, MK, KL. If therefore the prism LMNBHK be taken from the solid of which the base is the parallelogram Book XI. 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 parallelepiped AH, is equal to the remaining parallelepiped ÅK. Therefore, “ solid parallelepipeds,” &c. Q. E. D.
Solid parallelepipeds upon the same base, and of See 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.
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: Then because the plane LBHM is parallel to the oppo
site plane ACDF, and because 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 because the plane ALNG is that
Book XI. in which are the parallels AL, OPGN, in which also is the
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 parallelepiped : But the solid CM, of which
the base is ACBL, to which FDHM is the opposite paralleloa 29. 11. gram, is equal a to the solid CP, of which the base is the
parallelogram ACBL, to which ORQP is the one opposite : because they are upon the same base, and their insisting straight lines AF, AO, CD, CR; LM, LP, BH, BQ are in the same straight lines FR, MQ: And the solid CP is equal a 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 parallelepipeds,” &c. Q. E. D.
PROP. XXXI. 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.
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,