Book XI. with the figure LQ, and the straight line CG with MQ, and

the point G with the point Q. fiace therefore all the planes and sides of the solid figure AG coincide with the planes and sides of the folid figure KQ, AG is equal and similar to KQ. and in the fame manner, any other solid figures whatever contained by the fame number of equal and similar planes, alike situated, and having none of their solid angles contained by more than three plane angles, may be proved to be equal and similar to one another. Q. E D.

PRO P. XXIV. THEO R.

See N.

a solid be contained by fix planes, two and two of

which are parallel ; the opposite planes are similar and equal parallelograms.

G

Iet the folid CDGH be contained by the parallel planes AC, CF; BG, CE; FB, A E. its oppoate planes are fuzilar and equal parallelograms.

Becau'e the two parallel planes, BG, CE are cut by the plane R. 16.11. AC, their common fections AB, CD are parallel *. again, because

the two parallel plores BF, AE are cut by the plane AC, their common fections AD, EC re parallel'. and AB is parallel to CD; tervore AC is a parallelocidin in

H lihe manne, it may be proven that each of the figures CE, TG, Gb, BP, AE is a parallelogram. join AH, DI'; and because AB is parallel to DC, and BH to CF; the two straight lines AB,

E BH, which meet one another, are parallel to DC and CF which meet one

another and are not in the same plane b.19.14. with the other two; wherefore they contain equal angles b; the

angle ABH is therefore equal to the angle DCF. and because AB, EH are equal to DC, CF, and the angle ABH equal to the angle DCF, therefore the base AH is equal to the bafe DF, and the

triangle ABH to the triangle DCF. and the parallelogram BG is d. 34. I.

double d of the triangle ABH, and the parallelogram CE double of the triangle DCF; therefore the parallelogram BG is equal and similar to the parallelogram CE. in the fame manner, it may be proved that the parallelogram AC is equal and similar to the pa

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

PRO P. XXV. THE O R.
F a solid parallelepiped be cut by a plane parallel to See N.

two of its opposite planes ; it divides the whole
into two folids, the base of onc of which shall ta 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 opposite planes AR, HD, and divides the whole into the two folids ABFV, EGCD; as the base AEFY of the first is to the base EHCF of the other, fo 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 LA, 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 are equal a. a. 36. 5.

Y F

ន and likewise the parallelograms KX, KB, AG"; as also b the pa- b. 24. 11. rallelograms LZ, KP, AR, because they are opposite planes. for the same reason, the parallelograms EC, HQ, MS are equal“; and the parallelograms HG, HI, IN, as also b HD, MU, NT. therefore three plancs of the solid LP, are equal and similar to three planes of the folid KR, as also to three planes of the folid AV. but. the three planes opposite to these three are equal and similar to them in the several solids, and none of their folid angles are coiltained by more than three plane angles. therefore the three folds LP, KR, AV are equal to one another. for the same reason, the c. C. 11. three solids ED, HU, MT are equal to one another. thercfore what

Book XI. multiplc soever the base LF is of the base AF, the fame multiple is

the solid LV of the solid AV. for the same reason, 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 bafe 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 solid NV, and if lefs, less. since 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 bife LF and fold LV arc any equimultiples whatever; and of the base FH and folid ED, the base FN and folid NV are any equimultiples whatever ; and it has been proved, that if the base LF is greater than the base FN, the folid

LV is greater than the solid NV; and if equal, equal; and if less, d. s. Def.s. less. 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.

See N.

PRO P. XXVI. PROB.
T a given point in a given straight line, to make a

solid angle equal to a given folid angle contained by three plane angles.

A

Let AB be a given straight line, A a given point in 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. a. 11. 11d. In the straight line DF take any point F, from which draw

FG perpendicular to the plane EDC, meeting that plane in G; 6. 43. So join DG, and at the point A in the straight 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 them
Lolid angle at A which is contained by the three plane angles BAL, C. 12. 11.
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 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 base BK is equal to the base EG. and KH is equal
to GF, and HKB, FGE are right angles, therefore HB is equal o to
FE. again, because AK, KH are equal to DG, GF, and contain
right angles, the base AH is equal to 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 ; therefore the

A angle BAH is equalf to

f. 8. 1. the angle EDF. for the fame reafon, the angle HAL is equal to the B angle FDC. because if AL and DC be made K

H

I 죄 equal, 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 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 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 situated in the same order ; the folid angle at A is equal to the folid angle at D. Therefore at a given z. B. 11.

I E

Book XI. 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.

PROP. XXVII. PROB.

O describe from a given straight line a solid pa

rallelepiped similar, and similarly situated to one given.

To describe

2. 26. II.

b. 12. 6.

C. 22. s.

Let AB be the given straight line, and CD the given solid pa-
railelepiped. It is required froin AB to describe a solid paralel-
epiped fimilar, and similarly situated to CD.

At the point A of the given straight 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, so 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, io 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 be-
cause, as EC to CG, so
BA to AK, the sides

I.
about the equal angles HI

M

D
ECG, BAK are pro-
portionals; therefore
the parallelogram BK
is similar to EG, for the
K

G
fame reason parallelo-
А B

E
gram KH is fimilar to
GF, and HB to FE. wherefore three parallelogramıs of the solid

AL are fimilar to three of the folid CD; and the three opposite d. 24. 11.

ones in each solid are equal d and similar to these, each to cach.
also, because the plane angles which contain the solid angles of

the figures are equal, each to each, and situated in the same order, e. B. 11. the solid angles are equal", each to each. Therefore the solid 1.11. Def.11. AL is funilar f to the folid CD. wherefore from a given straight

line AB a solid parallelepiped AL has been described similar, and
Similarly situated to the given one CD. Which was to be done.