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the figure LQ, and the straight line CG with MQ, and the point G with the point Q. Therefore, since all the planes and sides of the solid figure AG, coincide with the planes and sides of the solid figure KQ, AG is equal and similar to KQ. And in the same manner, any other solid figures whatever, contained by the same 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.

PROPOSITION XXIV.

See N.

• 16. 11.

* 16. 11.

B

A

G

Theon.-If a solid be contained by six plancs, tro and Iro

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

Let the solid CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE : its opposite planes shall be similar and equal parallelograms.

Because the two parallel planes BG, CE are cut by the plane AC, their common sections AB, CD are parallel : again, because the two parallel planes BF, AE are cat by the plane AC, their common sections 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 because AB is parallel to DC, and BH to CF; the two straight

F lines AB, BH, which meet one another, are

E parallel to DC and CF, which meet one another, and are not in the same plane with the other two: wherefore they contain * equal angles ; therefore the angle ABH is equal to the angle DCF: and because AB, BH are equal to DC, CF, each to each, and the angle ABH equal to the angle DCF, therefore the base Ah is equal * to the base DF, and the triangle ABH to the triangle DCF: but the parallelogram BG is double * 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 same manner it may be proved, that the parallelogram AC is equal and similar to the parallelogram GF, and the parallelogram AE to BF. Therefore, if a solid, &c.

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Q. E. D.

PROPOSITION XXV.

Theor.-If a solid parallelopiped be cut by a plane parallel See N.

to two of its opposite planes, it divides the whole into two solids, the base of one of which shall be lo the base of the olher, as the one solid is to the other.

• 36. 1.

Let the solid parallelopiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole into the two solids ABFV, EGCD: as the base AEFY of the first is to the base EHCF of the other, so shall the solid ABFV be to the solid EGCD.

Produce Au 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 solids LP, KR, HU, MT. Then, because the straight lines LK, KA, AE, are all equal, the parallelograms LO, KY, AF are * equal; and likewise the parallelograms KX,

B G

R KB, AG: also * the parallelo

T

• 24. 11. grams LZ, KP, AR are equal, be- LI

ELUN cause they are opposite planes: for the same reason, the parallelograms EC, HQ, Ms are equal", and the parallelograms • 36. 1. AG, HI, IN: as also * HD, MU, NT: therefore three planes . 24. 11. of the solid LP are equal and similar to three planes of the solid KR, as also to three planes of the solid AV: but the three planes opposite to these three are equal and similar * to - 24.11. them in the several solids, and none of their solid angles are contained by more than three plane angles; therefore the three solids LP, KR, AV are equal * to one another: for the . c. 11. same reason, the three solids ED, HU, MT are equal to one another: therefore what multiple soever the base LF is of the base AF, the same multiple is the solid LV of the solid AV; and whatever multiple the base NF is of the base HF, the same multiple is the solid NV of the solid ED: and if the base LF be equal to the base NF, the solid LV is equal * to the solid NV; C. 11. and if the base LF be greater than the base NF, the solid LV is greater than the solid NV; and if less, less. Since then there are four magnitudes, the two bases AF, FH, and, the two solids AV, ED; and that of the basc AF and solid,

AV, the base LF and solid LV are any equimultiples whatever; and of the base FH and solid ED, the base FN and solid NV are any equimultiples whatever; and since it has been proved, that if the base LF is greater than the base FN, the

solid LV is greater than the solid NV; and if equal, equal; * 5 Def. 5. and if less, less; therefore * as the base AF is to the base FH,

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

PROPOSITION XXVI.

See N.

PROB.- At a given point in a given straight line, to make a

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

• 12. 11.

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 solid angle equal to the solid angle D. • 11. 11. In the straight line DF take any point F, from which draw

FG perpendicular to the plane EDC, meeting that plane in G, * 23. 1.

and join DG : 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 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: the solid angle at A which is contained by the three plane angles BAL, BAH, HAL, shall be 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 3 Def. 11. makes right angles * with every straight line meeting it in

that plane; 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,

D
DE, each to each, and that they con-

tain equal angles, therefore the base
* 4. 1. BK is equal * to the base EG ; and Ki B

J. E C + Constr. is equal t to GF, and HKB, FGE are K

H

G F 4.1, right angles, therefore HB is equal *

to FE. Again, because AK, KH are equal to DG, GF, each to each, 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, each to each ; and the base HB is equal to the base FE; therefore the angle BAH is equal * to the angle EDF: •8.1. for the same reason, the angle HAL is equal to the angle FDC: because if AL and DC be made 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 go, DC, each to each, and contain equal angles, the base KL is equal * to the base GC; and kh is equal to GF; so that LK, • 4. 1. KH are equal to CG, GF, each to each; and they contain right + + 3 Def. U. angles; therefore the base HL is equal t to the base FC: again, † 4.1. because HA, AL are equal to FD, DC, cach to each, 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 * to the solid angle at D. There- • B. 11. fore, at a given 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.

8. I.

PROPOSITION XXVII.

THEOR.-To describe, from a giren straight line, a solid See N.

parallelopiped similar and similarly siluated to one given.

Let AB be the given straight line, and CD the given solid parallelopiped; it is required from AB, to describe a solid parallelopiped similar and similarly situated to CD.

At the point A of the given straight line AB, make 26. 11. solid angle equal to the solid angle at C, and let BAK, KAH, HAB be

H the three plane angles which con

M tain it, so that BAK be equal to the

K

G angle ECG, and KAH to GCF, and

A B с HAB to FCE: and as EC to CG, so mak * BA to AK; and as GC to CF, so make * KA to AH; * 12. 6. wherefore, ex æquali*, as EC to CF, so is BA to AH: com

12. 6.

• 22. 5.

plete the parallelogram BH, and the solid AL: AL shall be similar and similarly situated to CD.

Because, as EC to GC, so BA to AK, the sides about the

equal angles ECG, BAK, are proportionals; therefore the + 1 Def.6. parallelogram BK is similar † to EG: for the same reason, the

parallelogram KH is similar to GF, and HB to FE; wherefore three parallelograms of the solid AL are similar to three of

the solid CD: and the three opposite ones in each solid are . 24. 11.

equal * and similar 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 same order, the solid • B. 11. angles are equal *, each to each : therefore the solid Al is • 11 Defli. similar * to the solid CD. Wherefore, from a given straight

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

PROPOSITION XXVIII.

See N.

Theor.-If a solid parallelopiped be cut by a plane passing

through the diagonals of two of the opposite planes, it shall be cut into two equal parts.

* 9. 11.

* 16. 11.

F

• 34. I.

Let AB be a solid parallelopiped, 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, FE are * parallel ; wherefore the diagonals CF, DE are in the plane in which the

B parallels are, and are themselves * parallels : and the plane CDEF shall cut the solid AB into two equal parts.

Because the triangle CGF is equal * to the A triangle CBF, and the triangle DAE to DHE; and that the parallelogram CA is equal * and similar to the opposite one BE, and the parallelogram GE to CH; therefore the prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal * 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

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