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Boor XI. 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 so• C. 11. lid LV is equal to the solid NV; and if the base LF be
greater than the base NF, the solid LV is greater than the
O Y F С
S 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, * 5 Def. 5. less : Therefore d 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.
PROP. XXVI. PROB. See N. At a given point in a given straight line, to make
a solid angle equal to a given solid angle contained by three plane angles.
Let AB be a given straight line, A a given point in it, and Da given solid angle contained by the three plane angles EDČ, EDF, FDC: It is required to make at the point A in the straight line AB a solid angle equal to the solid angle D.
In the straight line DF take any point F, from which * 11. 11. drawa FG perpendicular to the plane EDC, meeting that
plane in G; join DG, and at the point A, in the straight 23. 1. line AB, make the angle BAL equal to the angle EDC,
and in the plane BAL make the angle BAK equal to the
angle EDG; then make AK equal to DG, and from the * 19. 11. point K erect KH at right angles to the plane BAL: and
make KH equal to GF, and join AH: Then the solid an- Boor XI.
the equal straight lines AB, DE, and join HB, KB,
F and KL, HL, GC, FC, be joined, since the whole angle BAL is equal to the whole EDC, and the parts of them BAK, EDG are, by the construction, equal; therefore the remaining angle KAL is equal to the remaining angle GDC: And because KA, AL are equal to GD, DC, and contain equal angles, the base KL is equale to the base GC: And KH is equal to GF, so that LK, KH are equal to CG, GF, and they contain right angles; therefore the base HL is equal to the base FC: Again, because HA, AL are equal to FD, DC, and the base HL to the base FC, the angle HAL is equal to the angle FDC: Therefore, because the three plane angles BAL, BAH, HAL, which contain the solid angle at A, are equal to the three plane angles EDC, EDF, FDC, which contain the solid angle at D, each to each, and are situated in the same order, the solid angle at A is equal to the solid angle at D. : B. . Therefore, at a given point in a given straight line, a solid angle has been made equal to a given solid angle contained by three plane angles. Which
was to be done.
See N. To describe from a given straight line a solid pa
rallelopiped similar, and similarly situated, to one
Let AB be the given straight line, and CD the given so-
solid parallelopiped similar and similarly situated to CD.
solid angle equal to the solid angle at C, and, let BAK,
so that BAK be equal to the angle ECG, and KAH to $ 12, 6. GCF, and HAB to FCE: And as EC to CG, so make b
BA to AK; and as GC to CF, so makeb KA to AH; * 22. 5. wherefore, ex æqualic, as EC to CF, so is BA to AH
Complete the parallelogram BH, and the solid AL; And
similar to three of the solid CD; and the three opposite * 24. 11. ones in each solid are equald 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 situ•B. 11. ated in the same order, the solid angles are equale, each to f11 Def. 11. each. Therefore the solid AL is similar to the solid CD.
Wherefore from a given straight line AB a solid parallelo-
PROP XXVIII. THEOR. If a solid parallelopiped be cut by a plane passing See N. through the diagonals of two of the opposite planes; it shall be cut in two equal parts.
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, and not in the same plane with it, CD, FE, are parallel"; where-9.11. fore the diagonals CF, DE, are in
C С the plane in which the parallels are,
B and are themselves parallels b: And
16. 11. the plane CDEF shall cut the solid
F AB into two equal parts.
Because the triangle CGF is equal to the triangle CBF, and the DI
• 34. 1. triangle DAE to DHL; and that the parallelogram CA is equald and
d 24. 11. A
E similar to the opposite one BE; and the parallelogram GE to CH: Therefore the prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equalo to the prison contained. C. 11. 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 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 parallelopiped, mentioned 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 parallelopipeds upon the same base, and of see N. the same altitude, the insisting straight lines of which are terminated in the same straight lines in the planc opposite to the base, are equal to one another.
Book XI. Let the solid parallelopipeds AH, AK, be upon the same
base AB, and of the same altitude, and let their insisting figures be straight lines AF, AG, LM, LN, be terminated in the same low.
straight 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 diagonals AG, CH of the opposite planes * 28. 11. ALGF, CBHD, AH is cut into two equal parts a by the
plane AGHC: Therefore the solid AH' is double of the prism which is contained be
But, let the parallelograms DM, EN, opposite to the
base, have no common side: Then, because CH, CK, are • 34. 1. parallelograms, CB is equalb 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: • 38. 1. Wherefore also the triangle CDE is equal to the triangle d 36. 1. BHK: And the parallelogram DG is equald to the paral
lelogram HN: For the same reason, the triangle AFG is
equal to the triangle LMN, and the parallelogram CF is € 24. 11. equale to the parallelogram BM, and CG to BŇ; for they D
H Е. K D E H K
L are opposite. Therefore the prism which is contained by
the two triangles AFG, CDE, and the three parallelograms fC. 11. AD, DG, GC, is equal to the prism contained by the two
triangles LMN, BHK, and the three parallelograms BM,