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is equal and similar to the parallelogram GF, and the parallelo-Book XI. gram AE to BF. Therefore, if a solid, &c. Q. E. D.

PROP. XXV. THEOR.

IF a solid parallelepiped be cut by a plane parallel See N. to two of its opposite planes, it divides the whole into two solids, the base of one of which shall be to the base of the other as the one solid is to the other.

Let the solid parallelepiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole into the two solids ABFV, EGCD; as the base AEFY of the first is to the base EHCF of the other, so is the solid ABFV to the solid EGCD.

Produce AH both ways, and take any number of straight lines HM, MN, each equal to EH, and any number AK, KL, each equal to EA, and complete the parallelograms LO, KY, HQ, MS, and the solids LP, KR, HU, MT; then, because the straight lines LK, KA, AE are all equal, the parallelograms LO, KY, AF

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are equal a : and likewise the parallelograms KX, KB, AG a; a 36. 1. as also b the parallelograms LZ, KP, AR, because they are op- b 24. 11. posite planes: for the same reason, the parallelograms EC, HQ, MS are equal a ; and the parallelograms HG, HI, IN, as also b 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 in the several solids, and none of their solid angles are contained by more than three plane angles: therefore the three solids LP, KR, AV are equal < to one c C. 11. another: for the same reason, the three solids ED, HU, MT are equal to one another: therefore, what multiple soever the

Book XI. base LF is of the base AF, the same multiple is the solid LV of be 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 c C. 11. LV is equal c 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

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0 Y F CQ

Q S 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; d 5. def. and if less, less. Therefore d as the base AF is to the base FH, 5. 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 EDC, 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 draw a 11. 11. a FG perpendicular to the plane EDC, meeting that plane in

G; join DG, and at the point A, in the straight line AB, b 23. 1. make b the angle BAL equal to the angle EDC, and in the

plane BAL make the angle BAK equal to the angle EDG; c 12. 11. then make AK equal to DG, and from the point K erect < KH

at right angles to the plane BAL; and make KH equal to Book XI. GF, and join AH: then the solid angle at A, which is contained by the three plane angles BAL, BAH, HAL, is equal to the solid angle at D contained by the three plane angles EDC, EDF, FDC.

Take the equal straight lines AB, DE, and join HB, KB, FE, GE: and because FG is perpendicular to the plane EDC, it makes right angles d with every straight line meeting it in d 3. def, that plane : therefore each of the angles FGD, FGE is a right 11. angle: for the same reason, HKA, HKB are right angles : and because KA, AB are equal to GD, DE, each to each, and contain equal angles, therefore the base BK equal e to the e 4. 1. base EG: and KH is equal to GF, and HKB, FGE are right angles, therefore HB. is equal e to FE: again, because AK, KÉ are equal to DG, GF, and contain right angles, the base AH is equal co the base DF; and AB is equal to DE; therefore HA, AB are equal to FD, DE, and the base HB is equal to the base FE,

А

D therefore the angle BAH is equal f to

f8.1, the angle EDF: for the same reason, the angle HAL is equal B

E

С to the angle FDC. Because if AL and

K к DC be made equal,

H

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 equal e 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 f 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. Therefore, at a given point in 8 B. 11, a given straight linc, a solid angle has been made equal to a given solid angle contained by three plane angles. Which was to be done.

L

Book XI.

PROP. XXVII. PROB.

TO describe from a given straight line a solid pa. rallelepiped similar and similarly situated to one given.

M

Let AB be the given straight line, and CD the given solid pa. rallelepiped; it is required from AB to describe a solid paral

lelepiped similar and similarly situated to CD. a 26. 11. At the point A of the given straight line AB makea 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

FCE: and as EC to CG, so make b BA to AK; and as GC to b 12. 6. CF, so make b KA to AH; wherefore, ex æquali c, as EC to c 22. 5. CF, so is BA to AH: complete the parallelogtam BH, and the solid AL: and

L
because as EC to

H
CG, so BA to AK,

D
the sides about the

F
equal angles ECG,
BAK are propor-
tionals : therefore

K
the parallelogram
BK is similar to

A

B

C E
EG. For the same
reason, the parallelogram KH is similar to GF, and HB to FE.
Wherefore, three parallelograms of the solid AL are similar to

three of the solid CD; and the three opposite ones in each solid d 24. 11. are equal d 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 e B. 11. angles are equal e, each to each. Therefore the solid AL is f 11. def. similar f to the solid CD. Wherefore from a given straight 11. line AB a solid parallelepiped AL has been described, similar

and similarly situated to the given one CD. Which was to be done.

Book XI.

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 shall be cut in two equal parts.

G

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, FE are parallel a; wherefore the diagonals CF, DĘ áre a 9. 11. in the plane in which the parallels are,

C

B and are themselves parallels b: and the

b 16. 11. plane CDEF shall cut the solid AB into two equal parts.

F Because the triangle CGF is equal c

c 34. 1. to the triangle CBF, and the triangle DAE to DHE: and that the parallelogram CA is equal d and similar to D

H d 24. 11. the opposite one BE; and the paral

A lelogram GE to CH: therefore the

E. prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal e e 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 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, 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 parallelepipeds 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 plane opposite to the base, are equal to one another.

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