Book XI. the figure LQ, and the ftraight line CG with MQ, and the point G with the point Q: Since, 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 other 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 fimilar to one another. Q. E. D.

See N.

F a folid be contained by fix planes, two and two of

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 a 16. 11. plane AC, their common fections AB, CD, are parallel. Again, because the two parallel planes BF, AE, are cut by the plane AC, their common fections AD, BC, are parallel: And AB is parallel to CD; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF, AE is a parallelogram: Join AH, DF; and becaufe AB is parallel to DC, and BH to CF; the two ftraight lines AB, BH, which meet one another, are parallel to DC and CF which meet one another, and are not in the fame plane with the other two; wherefore they con

C 4. I.

A

D

B

H

G

F

E

b. 10. 11. tain equal angles; the angle ABH is therefore equal to the angle DCF: And becaufe 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 tri34. 1. angle DCF: And the parallelogram BG is doubled of the triangle ABH, and the parallelogram CE double of the triangle DCF; therefore the paralelogram BG is equal and fimilar to the parallelogram CE. In the fame nanner it may be proved, that the parallelogram AC is equal and fi

1

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

~

I'

[F a folid parallelepiped be cut by a plane parallel to two See N. of its oppofite planes; it divides the whole into two folids, the bafe of one of which fhall 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 base 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.

b

YFC Q S

LO, KY, AF are equal: And likewife the parallelograms KX, a 36. 1. KB, AG; as alfo the parallelograms LZ, KP, AR, because b 24. 11, they are oppofite planes: For the fame reafon, the parallelograms EC, HQ, MS are equal; and the parallelograms HG, HI, IN, as alfo HD, MU, NT: Therefore three planes of the folid I.P, are equal and fimilar to three planes of the folid KR, as alfo to three planes of the folid AV: But the three planes opposite to thefe three are equal and fimilar to them in the feveral folids, and none of their folid angles are contained by more than three plane angles: Therefore the three folids LP, KR, AV are equal to one another: For the fame reason, c C. II. the three folids ED, HU, MT are equal to one another: ThereP

fore

Book XI. fore what multiple foever the base LF is of the bafe AF, the fame multiple is the folid LV of the folid AV: For the fame reason, whatever multiple the bafe NF is of the base HF, the fame multiple is the folid NV of the folid ED: And if the base 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 bafe NF, the folid LV is greater than the folid NV; and if lefs, lefs: Since then there are four magnitudes, viz. the two bafes AF, FH,

c C. II.

and the two folids AV, ED, and of the base AF and folid AV, the base LF and folid LV are any equimultiples whatever; and of the bafe FH and folid ED, the base 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 lefs, lefs. d 5. def. 5. Therefore d as the base AF is to the base FH, fo is the solid AV to the folid ED. Wherefore, if a folid, &c. Q. E. D.

Sce N.

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

Let AB be a given ftraight line, A a given point in it, and Da given folid angle contained by the three plane angles EDC, EDF, FDC: It is required to make at the point A in the ftraight line AB a folid angle equal to the folid angle D.

In the ftraight line DF take any point F, from which draw a 11. 11. FG perpendicular to the plane EDC, meeting that plane in G; 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; e 12. 11. then make AK equal to DG, and from the point K erect KH

b. 23. I.

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

с

Take the equal ftraight 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 d3. def. 11. that plane: Therefore each of the angles FGD, FGE is a right angle: For the fame reafon, 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 c 4. 1. bafe EG: And KH is equal to GF, and HKB, FGE, are right angles, therefore HB is equal to FE: Again, becaufe 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 base FE, therefore the angle BAH is equal to the angle EDF: For the fame reason, the angle HAL is equal to the angle FDC. Becaufe if AL and DC be made equal, and KL, HL, GC,

B

A

D

LE

K

G

H

F

FC be joined, fince the whole angle BAL is equal to the whole EDC, and the parts of them BAK, EDG ar, by the construction, equal; therefore the remaining angle KAL is equal to the remaining angle GDC: And becaufe KA, AL are equal to GD, DC, and contain equal angles, the bafe KL is equal to the base GC: and KH is equal to GF, fo that LK, KH are equal to CG, GF, and they contain right angles; therefore the base HL is equal to the bafe FC: Again, becaufe HA, AL are equal to FD, DC, and the bafe HL to the bafe 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 fold 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 point in a given ftraight line, a folid angle has been made equal to a given folid angle contained by three plane angles. Which was to be done.

f 8. I.

g. B II.

Book XI.

a 26. II.

b 12. 6.

€ 22. 5.

T

O defcribe from a given straight line a solid 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.

K

H

L

M

D
F

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, 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 because, as EC to CG, fo BA to AK, the fides about the equal angles ECG, BAK are proportionals; therefore the parallelogram BK is fimilar to EG. For the fame reafon, 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, becaufe 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 pa rallelepiped AL has been defcribed fimilar, and fimilarly fituated to the given one CD. Which was to be done.

鸭典

d 24. XI.

e B. II.

A

B

E

PROF