Book XI. multiple foever the base LF is of the bafe AF, the same multiple is w the folid LV of the solid AV. for the same reason, whatever mul

tiple the base NF is of the base HF, the same multiple is the folid

NV of the folid ED. and if the base LF be equal to the base NF, C. C. ir.

the solid 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.

0 Y F с 9.

S the two bases AF, FH, and the two folids AV, ED, and of the base AF and solid 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 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. 5.Def. 5. less. Therefore d as the base AF is to the base FH, fo is the folid

AV to the solid ED. Wherefore if a solid, &c. Q. E. D.

See N.

AT

T a given point in a given straight line, to make a

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

Let AB be a given straight line, A a given point in it, and Da 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 folid angle equal to the solid 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; D. 23. 1. 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

2. II. II.

C, 12. II.

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 solid angle at A which is contained by the three plane angles BAL, 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.it. therefore each of the angles FGD, FGE is a right angle. for the same 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 e to the base EG. and KH is equal • 4. 8. 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 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

D

f. 8. f. 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 K Қ

H

F equal, and KL, HL, 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 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 e 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 solid angle at A, are equal to the three plane angles EDC, EDF, FDC which contain the solid angle at D, each to each, and are situated in the same order; the solid angle at A is equal 8 to the folid angle at D. Therefore at a given 8. B.11.

LE

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.

o describe from a given straight line a solid pa

given.

Let AB be the given straight line, and CD the given solid parallelepiped. It is required from AB to describe a solid parallel

epiped similar, and similarly situated to CD.
2. 26. II. At the point A of the given straight line AB make a solid

angle equal to the solid 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 b. 12. 6. EC to CG, so make 6 BA to AK, and as GC to CF, so make 6 C. 22. 5. KA to AH; wherefore, ex aequali“, as EC to CF, so 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
about the equal angles

H н
ECG, BAK are pro-

M
portionals; therefore
the parallelogram BK
is similar to EG. for the K
fame reason the parallel-
ogram KH is similar to

B
GF, and HB to FE. wherefore three parallelograms of the solid

AL are similar to three of the folid CD; and the three opposite d. 24. 11. ones in each solid are equal d and fimilar to these, each to each,

also, because the plane angles which contain the solid angles of
the figures are equal, each to each, and situated in the fame order,

the solid angles are equal , each to each. Therefore the solid f.11.Def.11 AL is fimilar f to the folid CD. wherefore from a given straight

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

e. B. II.

Book XI.

PROP. XXVIII.

THE OʻR.

b. 16. II.

с

C. 34. I.

F a folid parallelepiped be cut by a plane passing See N.

thro' the diagonals of two of the opposite planes ; it shall be cut into two equal parts.

Let AB be a solid parallelepiped, and DE, CF the diagonals of the opposite parallelograms AH, GB, viz. those which are drawn betwixt the equal angles in each. and because CD, FE are each of them parallel to GA, and not in the same plane with it, CD, EF are parallel * ; wherefore the diagonals CF, DE are in the plane in 2. g. 11. which the parallels are, and are them

C

B selves parallels b. and the plane CDEF shall cut the solid AB into two equal

G

F parts.

Because the triangle CGF is equal to the triangle CBF, and the triangle

D DAE to DHE; and that the parallelo

H gram CA is equal d and similar to the op

d. 24. 11. posite one BE; and the parallelogram

Α. E GE to CH. therefore the prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal e é. C. ır. to the prism contained by the two triangles CBF, DHE, and the three parallelograms BE, CH, EC; because they are contained by the same number of equal and similar planes, alike situated, and none of their solid angles are contained by more than three plane angles. Therefore the folid AB is cut into two equal parts by the plane CDEF. Q. E. D.

« N. B. The infifting straight lines of a parallelepiped, men« tioned in the next and some following Propositions, are the sides • of the parallelograms betwixt the base and the opposite plane s parallel to it.'

OLID parallelepipeds upon the fame base, and of the See N.

same altitude, the insisting straight lines of which are terminated in the fame straight lines in the plane opposite to the base, are equal to one another.

Book XI. Let the folid parallelepipeds AH, AK be upon the fame base

AB, and of the same altitude, and let their insisting straight lines See the figures be- AF, AG, LM, LN; CD, CE, BH, BK be terminated in the same low. straight lines FN, DK. the folid AH is equal to the folid AK.

First, Let the parallelograms DG, HN which are opposite to the base AB have a common side HG. then because the folid AH is i cut by the plane AGHC passing thro' the diagonals AG, CH of

the opposite planes ALGF, CBHD, AH is cut into two equal a. 28. 11. parts a by the plane AGHC.

D H K therefore the solid AH is double

F of the prism which is contained

G

N
by the triangles ALG, CBH. for
the same reason, because the fo-

B
lid AK is cut by the plane LGHB
thro' the diagonals LG, BH of

А. Li
the opposite planes ALNG, CBKH, the solid AK is double of
the same prism which is contained by the triangles ALG, CBH.
Therefore the solid AH is equal to the solid AK.

But let the parallelograms DM, EN opposite to the base have

no common fide. then because CH, CK are parallelograms, CB is B. 34. 1. equal to each of the opposite sides DH, EK; wherefore DH is

equal to EK. add, or take away the common part HE; then De is C. 38. 1. equal to HK. wherefore also the triangle CDE is equal to the d. 36.1. triangle BHK. and the parallelogram DG is equald to the paral

lelogram HN. for the same reason, the triangle AFG is equal to e. 24. 11. the triangle LMN, and the parallelogram CF is equal o to the paralD Η Ε

Н. К MN

N

G M
F

N

K DE

Α. L

Α.

I lelogram BM, and CG to BN; for they are opposite. Therefore the prism which is contained by the two triangles AFG, CDE, and the three parallelograms AD, DG, GC is equal f to the prism contained by the two triangles LMN, BHK, and the three parallelograms BM, MK, KL. If therefore the prism LMN, BHK be

f. C. II.