EF

a 34. T.

34. I.

Produce AH both ways, and take any number of ftraight Book XI. lines AK, KL each equal to EA, and complete the parallelograms LO, KY, and the folids LP, KR: Then, because the ftraight lines LK, KA, AE are all equal, as alfo KO, AY, are equal a and parallel; the parallelograms LO, KY, AF are equal and fimilar b: For the fame reason, the parallelograms b4.1.8 KX, KB, AG are equal and fimilar; as also the parallelograms LZ, KP, AR: 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 these three are equal and fimilar to them in the feveral folids : and because c 24. 11. the folid angles at A, K, L are contained by three plane angles equal to one another, each to each, the planes in which they are,

have the fame inclination to one another : For the fame reason, d A. 11.

the other planes have the fame inclination: Therefore the folids LP, KR, AV are equal to one another. What multiple foever, e B. 11. therefore, the bafe LF is of the bafe AF, the fame multiple is the folid LV of the folid AV. Take, in like manner, EN the leaft multiple of EH, that is greater than EL,' and complete the parallelogram HS, and the folid HT; and it may be proved, as before, that the folid ET is the fame multiple of the folid ED, that the bafe ES is of the base EC; confequently the bafe HS and the folid HT contain the bafe EC and the folid ED the fame number of times. And if the bafe LF be equal to the base HS, the folid LV is equal to the folid HT; and if LF be not equal to HS, it is greater than HS, but lefs than ES; and the folid LV is greater than the folid HT, but lefs than ET: Therefore the bafes LF, HS contain the bafe EC the fame number of times f, f Def.A. 5. and the folids LV, HT contain the folid ED the fame number of times and the bafe HS and folid HT contain the bafe EC and folid ED equally; therefore the bafe LF and folid LV contain them equally and the bafe LF and folid LV are any equimultiples of the bafe AF and the folid AV; as many times, therefore, as any multiple of the bafe AF contains the bafe EC, fo many times does the fame multiple of the folid AV contain the folid

Bb

BOOK XI. folid ED: Therefore, as the base AF is to the bafe EC, fo is the folid AV to the folid ED. Wherefore, &c. Q. E. D.

g 5. Def. §.

See N.

a II, II.

A

PROP. XXVI. PROB.

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

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 ftraight line AB, a folid angle contained by three plane angles equal to the angles EDC, EDF, FDC.

b

In the ftraight line DF take any point F, from which draw a FG perpendicular to the plane EDC, meeting that plane in G; b 23. 1. join DG, and at the point A in the ftraight line AB, make b 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 c 12. 11. equal to DG, and from the point K erect KH at right angles to the plane BAL; and make KH equal to GF, and join AH: Then the folid angle at A is contained by the three plane angles BAL, BAH, HAL, which are equal, each to each, to the three plane angles EDC, EDF, FDC, which contain the folid angle at D.

Because the angle BAL is equal to EDC, and the angle BAK to EDG, the angle KAL is alfo equal to GDC and

becaufe FG is perpendicular B

to the plane EDC, the angle

d3.Def.11. FGD is a right angle; and the plane FGD is perpendi

e 18. 11. cular to the plane EDC *.

f

K

For the fame reafon, the angle HKA is a right angle, and the plane HAK perpendicular to the plane BAL: and because AK, KH are equal to DG, GF, and they contain right angles; the f 4. 1. bafe AH is equal to FD, and the angle HAK to the angle to GDF: and because the two plane angles BAK, KAH are equal to the two EDG, GDF; and that they are in planes which have the fame inclination to one another; therefore the third Cor. A. angle BAH is equal to the third angle EDF. For the fame reafon, the angle HAL is equal to the angle FDC. Wherefore, at the point A in the ftraight line AB a folid angle has been made,

II.

made, contained by the three plane angles BAL, BAH, HAL, Book XI. which are equal to the three plane angles which contain the folid angle at D, each to each. Which was to be done.

T

given.

PROP. XXVII. PROB.

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

Let AB be the given straight line, and CD the given parallelopiped; it is required from AB to defcribe a parallelopiped, fimilar and fimilarly fituated to CD.

b

At the point A of the given straight line AB, make a a folid a 26. 11. angle contained by the plane angles BAK, KAH, HAB equal to the angles ECG, GCF, FCE, each to each; 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 b KA to AH; wherefore, by equality, as EC to CF, fo is BA to AH: Complete the parallelo- K gram BH, and the folid

I.

b 12.6.

M

D

C 22. S.

AL: and because, as EC

to CG, fo BA to AK,

C

E

the fides about the equal angles ECG, BAK are proportionals; therefore the parallelogram BK is fimilar to EG. For the fame reason, 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, and the d 24. 11. fimilar planes have the fame inclination to one another *, because the folid angles are contained by three plane angles equal to one another, each to each; therefore the folid AL is fimilar f f9.Def.11. to the folid CD. Wherefore, from a given straight line AB, a folid parallelopiped AL has been described fimilar and fimilarly fituated to the given one CD. Which was to be done.

e A. 11.

Book XI.

See N.

a 9. 11.

C 24. II.

IF

PROP. XXVIII. THEOR.

F a parallelopiped be cut by a plane paffing through the diagonals of two of the oppofite planes, it fhall be cut in two equal parts.

Let AB be a parallelopiped, and DE, CF the diagonals which join equal angles of the oppofite parallelograms AH, GB; they are in the fame plane, becaufe CD, EF are parallel, each of them being parallel to AG: the plane CDEF fhall cut the folid AB into two equal parts.

First, Let the ftraight lines AG, DC, EF, HB be at right angles to the planes BG, AH: and because the triangle GCF is

b 34. 1. & equal and fimilar to CBF, and the tri-
1. Def. 6. angle DAE to DHE; as alfo the rect-
angle GD to FH, and GE to CH:
and they have the fame inclination to G
one another, because the solid angles
at G, B are contained by equal plane
angles; and they are fituated in the fame
order; therefore the prifm GCF-ADE
e B. 11. is equal to the prifm BFC-HED,

d A. 11.

A

C

D

B

F

H

E

But, let the ftraight lines AG, DC, EF, HB not be at right angles to the planes BG, AH: At the f 23. 1. point G, in AG, make in the plane GD the angle AGK equal to HBF, and in the plane GE the angle AGL equal to HBC; and make KM, LN each equal to AG, and join KL, AM, AN, MN: and becaufe the angles AGK, AGL are equal to HBF, HBC; and their planes have the fame inclination ", for the folid angles at A, B are contained by equal plane angles; therefore g Cor. A. the third angle KGL is equal to the third angle CBF; and becaufe the angle AGL is equal to

11.

16. I.

HBC.

h

1

And the angle GLF is h 29. 1. equal to AGL, and GFL to k 15. 11. HBC *; the angle GLF is equal to GFL; and therefore the fide GL is equal to GF or CB. For the fame reafon, GK is equal to BF: and the angle KGL is equal to CBF; therefore the triangle KGL is equal and fimilar to m 4. 1. FBC", and AMN to HED; and the parallelograms GM, GN, NK are equal and fimilar to BE, BD, DF: and they are alike fituated,

and

K

H

E

G2

B

C

n

a C. 11.

and have the fame inclination to one another; therefore the prifm Book XI. KGL-MAN is equal to the prism FBC-EHD: But the prifm CGF-DAE is equal to the prifm KGL-MAN, because their c B. 11. parallelograms have the fide AG common, and their fides oppofite to it in the ftraight lines CM, FN: Therefore alfo the prism CGF-DAE is equal to the prism FBC-EHD. Wherefore, &c. Q. E. D.

OLID parallelopipeds upon the fame bafe, and of See N. the fame altitude, the fides of which betwixt the bafe and oppofite plane are terminated in the fame ftraight lines, are equal to one another.

Let the parallelopipeds AH, AK be upon the fame base AB, and of the fame altitude, and let the ftraight lines AF, AG, LM, LN be terminated in the fame ftraight line FN, and CD, CE, BH, BK be terminated in the fame straight line DK: the folid AH is equal to the folid AK.

a

b 29. 1.

Because CH, CK are parallelograms, DC is equal and parallel to HB ; and CE to BK; therefore the angle DCE is equal a 34. 1. to HBK and because the two fides DC, CE are equal to the two HB, BK, and the angle DCE to HBK; therefore the base DE is equal to the bafe HK, and the triangle DCE equal and fimilar to the triangle HBK; and the parallelogram DG to the parallelogram HN d.

C

For

the fame reason, the triangle
AFG is equal and fimilar to
LMN: and the parallelo-
gram CF is equal and fimilar
to BM, and CG to BN,
because they are oppofite:
Therefore the prism AFG- C
DCE is equal to the prifm
LMN-BHK.

f

Take thefe equals from

D

c 4. 1.

d 36. 1.

the whole folid ACBL-FDKN, and the remaining parallelopiped AK is therefore equal to the remaining parallelopiped AH. Wherefore, &c. Q. E. D.

f B. 11

PROP.