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the points, A, D, C, B, with the points K, N, M, L: and the solid angle at A coincides with * the solid * B. 11. angle at K; wherefore the plane AF coincides with the plane KP, and the figure AF with the figure KP, because they are equal and similar to one another: therefore the straight lines AE, EF, FB, coincide with KO, OP, PL; and the points E, F, with the points O, P. In the same manner, the figure AH coincides with the figure KR, and the straight line DH, with NR, and the point H with the point R. And because the solid angle at B is equal to the solid angle at L, it may be proved, in the same manner, that the figure BG coincides with 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 lid 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.

PROP. XXIV. THEOR.

If a solid be contained by sic planes, two and two of See N.

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 * 16. 11. parallel : again, because the two parallel planes BF, AE are cut by the plane AC, their common sections AD, BC* are parallel: and AB is parallel to CD; * 16. 11. 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 lines AB, BH, which meet one another, are parallel to DC and CF, which meet one another, and are not in the

. 10. 11,

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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. Q. E. D.

PROP. XXV. THEOR.

See N.

If a solid parallelopiped be cut by a plane parallel to two

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

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

E HMN KB, AG: also* the parallelograms LZ, KP, AR, are

Y F equal, because they are opposite planes: for the same reason, the parallelograms EC, HQ, MS, are equal*, and the parallelograms HG, HI, IN: as also* HD, MU, NT: therefore three planes 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*

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to 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 * tó •C. 11. one another: for the 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

* C. 11. equal* to the solid NV; 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, viz., the two bases AF, FH, and the two

B G solids AV, ED; and that of

T the base AF and solid AV,

IELUMN the base LF and solid LV

Y TC Q. S 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; and if less, less; therefore* as the base AF is to the base FH, so is the solid * 5 Def. 5. AV to the solid ED. Wherefore, if a solid, &c. Q. E. D.

PROP. XXVI. PROB.

At a given point in a given straight line, to make a solid See N.

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 D a 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.

In the straight line DF take any point F, from which draw * FG perpendicular to the plane EDC, meeting that plane in G, and join DG: at the point A, in the straight line AB, make * the angle BAL • 23. 1. equal to the angle EDC, and in the plane BĂL make the angle BAK equal to the angle EDG; then make AK equal to DG, and from the point K erect* KH at 12. 11. right angles to the plane BAL, and make KH equal

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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 *3 Def.11. the plane EDC, it 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, DE, each to each, and that

they contain equal angles, therefore the base BK is * 4. 1. equal* to the base EG; and KH is equalt to GF, + Constr.

and HKB, FGE are right angles, therefore HB is * 4. 1.

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. 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 GD, 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, KH are equal to +3 Def, 11. CG, GF, each to each; and they contain rightt

angles; therefore the base HL is
+ 4. 1, equalt to the base FC: again, А

D
because HA, AL are equal to
FD, DC, each to each, and the B

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C base HL to the base FC, the

K angle HAL is equal to the angle

H 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 cach, and are situated in the same order, the solid angle at A is equal * to the solid angle at D. Therefore at a given point in a given straight line a solid angle has been made equal to a given solid angle

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.contained by three plane angles. Which was to be done.

PROP. XXVII. THEOR.

To describe from a given straight line a solid parallelo- See N.

piped similar and similarly situated to one given. Let AB be the given straight line, and CD the given solid parallelopiped. It is required from AB to de scribe a solid parallelopiped similar and similarly situated to CD.

At the point A of the given straight line AB make* * 26. 11. 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 EC to CG, so make * BA to AK; and as GC to CF, so *.12. 6. make * KA to AH; wherefore, ex æquali *, as EC to * 12. 6. CF, so is BA to AH: complete the parallelogram

. 22. 5. 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 parallelogram BK is similar + to EG. †1 Def.6. 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 equal and similar to these, each to H

M each. Also, because the plane angles which contain the solid

K

G! angles of the figures are equal, each to each, and situated in the same order, the solid angles are equal *, each to . B. 11. each. Therefore the solid AL is similar* to the solid < 11 Def. CD. Wherefore from a given straight line AB a solid 11. parallelopiped AL has been described similar and similarly situated to the given one CD. Which was to be done.

PROP. XXVIII. THEOR.
If a solid parallelopiped be cut by a plane passing throu See N.

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

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