Book XI. with the figure LQ, and the straight 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 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.

IF

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 3. 16. 11. AC, their common sections AB, CD are parallel. again, becaufe 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

B

H

G

C

D

E

like manner, it may be proved that each of the figures CE, FG, GB, BF, A AE is a parallelogram. join AH, DF ; and becaufe AB is parallel to DC, and BH to CF; the two ftraight lines AB, IH, which meet one another, are parallel to DC and CF which meet one another and are not in the fame plane b. 10. 11. with the other two; wherefore they contain equal angles; the

c

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 c. 4. 1. DCF, therefore the bafe AH is equal to the bafe DF, and the triangle ABH to the triangle DCF. and the parallelogram BG is d. 34. 1. doubled 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 parallelo

gram

gram GF, and the parallelogram AE to BF. Therefore if a fo- Book XI. tid, &c. Q. E. D.

PROP. XXV. THEOR.

Fa 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 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 bafe AEFY of the firft 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 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 folids LP, KR, HU, MT. then because the straight lines LK, KA, AE are all equal, the parallelograms LO, KY, AF are equal. a. 36. 1.

and likewise the parallelograms KX, KB, AG1; as alfo b the pa- b. 24. 17. rallelograms LZ, KP, AR, because they are oppofite planes. for the fame reason, the parallelograms EC, HQ, MS are equal2; and the parallelograms HG, HI, IN, as alfo HD, MU, NT. therefore three planes 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 contained by more than three plane angles. therefore the three folids LP, KR, AV are equal to one another. for the fame reason, the c. C. . three folids ED, HU, MT are equal to one another. therefore what multiple

c

Book XI. multiple foever the base LF is of the base AF, the fame multiple is the folid LV of the folid AV. for the fame reafon, whatever multiple the base NF is of the base HF, the fame multiple is the folid NV of the folid ED. and if the bafe LF be equal to the base NF, c. C. 11. the folid LV is equal to the folid NV; and if the base 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. the two

bafes AF, FH, 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 base FH and folid ED, the base FN and solid 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 fod. s. Def. s. lid NV; and if equal, equal; and if lefs, lefs. Therefore as the bafe AF is to the base FH, fo is the folid AV to the folid ED. Wherefore if a folid, &c. Q. E. D.

See N.

A

PROP, XXVI. PROB.

Ta 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 ftraight 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 ftraight line DF take any point F, from which draw⚫ FG perpendicular to the plane EDC, meeting that plane in G; join DG, b. 23. 1. and at the point A in the straight line AB make the angle BAL e.. qual 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

the point K erect KH at right angles to the plane BAL; and make Book XI. KH equal to GF, and join AH. then the folid angle at A which is contained by the three plane angles BAL, BAH, HAL is equal to c. 12. 11. the folid angle at D contained by the three plane angles EDC, EDF,

FDC.

d

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 with every straight line meeting it in that plane.d.3.De£11, therefore each of the angles FGD, FGE is a right angle. for the same reason, HKA, HKB are right angles. and because KA, AB

and contain right angles,
and AB is equal to DE;
and the bafe HB is equal

are equal to GD, DE, each to each, and contain equal angles, therefore the base BK is equal to the base EG. and KH is equal to GF, e. 4. 1. and HKB, FGE are right angles, therefore HB is equal to FE. again, because AK, KH are equal to DG, GF, the bafe AH is equal to the bafe DF; therefore HA, AB are equal to FD, DE, 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. because if AL and DC be made e

B

A

LE

K

H

G

D

f. 8. 1.

qual, and KL, HL,

e

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 construc tion, 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 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 bafe 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 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 g. B. 12. the folid angle at D. Therefore at a given point in a given straight

Book XI. 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. 5.

To defcribe from a given straight 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 describe a folid parallelepiped fimilar, and fimilarly fituated to CD.

a

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 b BA to AK, and as GC to CF, fo make b KA to AH; wherefore, ex aequali, as EC to CF, so is BA to AH. complete the parallelogram BH, and the solid AL. and becaufe, as EC to

L

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; d. 24. 11. and the three oppofite ones in each folid are equal and fimilar to thefe, each to each. alfo, because the plane angles which contain the folid angles of the figures are equal, each to each, and fituated in e. B. 11. the fame order, the folid angles are equal, each to each. Theref. 11. Def. fore the folid AL is fimilar f to the folid CD. wherefore from a given ftraight line AB a folid parallelepiped AL has been described fimilar, and fimilarly fituated to the given one CD. Which was to be done.

LI.

PROP.