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Book XI. with the figure LQ, and the ftraight line CG with MQ, and the point G with the point Q. fince therefore all the planes and fides of the folid figure AG coincide with the planes and fides of the folid figure KQ, AG is equal and fimilar to KQ. and in the fame manner, any ether folid figures whatever contained by the fame number of equal and fimilar planes, alike fituated, and having none of their folid angles contained by more than three plane angles, may be proved to be equal and £milar to one another. Q. E D.

Sce N.

2. 16. II.

C. 4

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PROP. XXIV. THEOR.

Fa folid be contained by fix planes, two and two of which are parallel; the oppofite planes are fimilar and equal parallelograms.

Let the folid CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE. its oppofite planes are fimilar and equal parallelograms.

Because the two parallel planes, BG, CE are cut by the plane AC, their common fections AB, CD are parallel *. again, because the two parallel planes BF, AE are cat by the plane AC, their common fections AD, BC are parallel. and AB is parallel to CD; therefore AC is a parallelogiam. in

like manne it
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be rroved that
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each of the figures CE, TG, Gb, BF,
AE is a parallelogram. join AH, Dr';
and because AB is parallel to DC, and
BH to CF; the two ftraight lines AB,
BH, which meet one another, are pa-
rallel to DC and CF which meet one

another and are not in the fame plane

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b. 19. 1. with the other two; wherefore they contain equal angles; the angle ABH is therefore equal to the angle DCF. and because AB, BH are equal to DC, CF, and the angle ABH equal to the angle DCF, therefore the bafe AH is equal to the bafe DF, and the triangle ABH to the triangle DCF. and the parallelogram BG is double of the triangle ABH, and the parallelogram CE double of the triangle DCF; therefore the parallelogram BG is equal and fimilar to the parallelogram CE. in the fame (manner, it may be proved that the parallelogram AC is equal and fimilar to the pa

d. 34. I.

rallelogram GF, and the parallelogram AE to BF. Therefore if a Book XI. folid, &c. Q. E. D.

IF

PROP. XXV. THEOR.

F a folid parallelepiped be cut by a plane parallel to see N. two of its oppofite planes; it divides the whole into two folids, the base of one of which shall be to the base of the other, as the one folid is to the other.

Let the folid parallelepiped ABCD be cut by the plane EV which is parallel to the oppofite planes AR, HD, and divides the whole into the two folids ABFV, EGCD; as the bafe AEFY of the first is to the bafe EHCF of the other, fo is the folid ABFV to the folid EGCD.

Produce AH both ways, and take any number of ftraight 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 folids LP, KR, HU, MT. then becaufe the ftraight lines LK, KA, AE are all equal, the parallelograms LO, KY, AF are equal . a. 36. 1.

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and likewife the parallelograms KX, KB, AG "; as alfo b the pa- b. 24. II. rallelograms LZ, KP, AR, because they are oppofite planes. for the fame reason, the parallelograms EC, HQ, MS are equal; and the parallelograms HG, HI, IN, as alfo b HD, MU, NT. therefore three plancs of the folid LP, are equal and fimilar to three planes of the folid KR, as alfo to three planes of the folid AV. but the three planes oppofite to thefe three are equal and fimilar to them in the feveral folids, and none of their folid angles are con→ tained by more than three plane angles. therefore the three folds

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LP, KR, AV are equal to one another. for the fame reafon, the c. C. 11. three folids ED, HU, MT are equal to one another. therefore what

Book XI. multiple foever the bafe LF is of the bafe AF, the fame multiple is the folid LV of the folid AV. for the fame reafon, whatever multiple the bafe NF is of the bafe HF, the fame multiple is the folid NV of the folid ED. and if the bafe LF be equal to the bafe NF, the folid LV is equal to the folid NV; and if the bafe LF be greater than the base NF, the folid LV is greater than the folid NV; and if lefs, lefs. fince then there are four magnitudes, viz.

c. C. 11.

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the two bafes AF, FH, and the two folids AV, ED, and of the bafe AF and folid AV, the bafe LF and folid LV are any equimultiples whatever; and of the bafe FH and folid ED, the bafe FN and folid NV are any equimultiples whatever; and it has been proved, that if the bafe LF is greater than the bafe FN, the folid LV is greater than the folid NV; and if equal, equal; and if less, d.5. Def.5. lefs. Therefore as the bafe AF is to the bafe FH, fo is the folid AV to the folid ED. Wherefore if a folid, &c. Q. E. D.

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See N.

a. 11.

A

T a given point in a given straight line, to make a folid angle equal to a given folid angle contained by three plane angles.

Let AB be a given straight line, A a given point in it, and D a given folid 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 folid angle equal to the folid angle D.

In the straight line DF take any point F, from which draw' FG perpendicular to the plane EDC, meeting that plane in G; b. 23. join DG, and at the point A in the ftraight 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

C

to DG, and from the point K erect KH at right angles to the Book XI. plane BAL; and make KH equal to GF, and join AH. then the

folid angle at A which is contained by the three plane angles BAL, c. 12. 11. BAH, HAL is equal to the folid 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 becaufe FG is perpendicular to the plane EDC, it makes right angles with every ftraight line meeting it in that plane. d.3 Def.11. therefore each of the angles FGD, FGE is a right angle. for the fame reason, HKA, HKB are right angles. and because KA, AB are equal to GD, DE, each to each, and contain equal angles, therefore the bafe BK is equal to the bafe EG. and KH is equal e. 4. 1, to GF, and HKB, FGE are right angles, therefore HB is equal to FE. again, because AK, KH are equal to DG, GF, and contain right angles, the bafe AH is equal to the bafe DF; and AB is equal to DE; therefore HA, AB are equal to FD, DE, and the bafe HB is equal to the

bafe FE; therefore the

angle BAH is equal to

the angle EDF. for the fame reafon, the angle HAL is equal to the B angle FDC. because if AL and DC be made equal, and KL, HL,

A

D

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c

GC, FC be joined, fince the whole angle BAL is equal to the whole EDC, and the parts of them BAK, EDG are, by the conftruction, 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 bafe KL is equal to the bafe GC. and KH is equal to GF, fo that LK, KH are equal to CG, GF, and they contain right angles; therefore the bafe HL is equal to the base FC. again, because HA, AL are equal to FD, DC, and the bafe 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 folid angle at A, are equal to the three plane angles EDC, EDF, FDC which contain the folid angle at D, each to each, and are fituated in the fame order; the folid angle at A is equal to the folid angle at D. Therefore at a given g. B. 11.

f

Book XI. point in a given straight line a folid angle has been made equal to a given folid angle contained by three plane angles. Which was to be done.

a. 26. II.

b. 12. 6.

C. 22. S.

.

T

PROP. XXVII. PRO B.

O defcribe from a given ftraight line a folid parallelepiped fimilar, and fimilarly fituated to one

given.

Let AB be the given straight line, and CD the given folid parallelepiped. 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 make 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, to make b KA to AH; wherefore, ex aequali, as EC to CF, fo is BA to AH. complete the parallelogram BH, and the folid AL. and becaufe, as EC to CG, fo

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d. 24. 11.

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GF, and HB to FE. wherefore three parallelograms of the folid AL are fimilar to three of the folid CD; and the three opposite ones in each folid are equal and fimilar to thefe, each to cach. 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 f.11.Def.11, AL is fimilar f to the folid CD. wherefore from a given straight line AB a folid parallelepiped AL has been defcribed fimilar, and Similarly fituated to the given one CD. Which was to be done.

e. B. 11.

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