Sidebilder
PDF
ePub

The axis of the sphere is the fixed diameter about which the semicircle revolves; and its centre is the same as that of the generating semicircle.

11. The diameter of a sphere is a straight line passing through the centre of the sphere, and terminated at each extremity by the surface.

12. A right cone is a solid described by the revolution of a right-angled triangle about one of the sides containing the right angle, which remains fixed.

The axis of the cone is the fixed line about which the generating triangle revolves; and its base is the circle described by the revolving side containing the right angle.

· 13. A right cylinder is a solid described by the revolution of a rectangle about one of its sides, which remains fixed.

The axis of the cylinder is the fixed line about which the rectangle revolves; and its bases or ends are the circles described by the opposite revolving sides of the rectangle.

14. Similar cones and cylinders are those that have their axes and the diameters of their bases proportional.

It is evident (Def. 12) that the axis of a cone is the straight line joining its vertex and the centre of its base; and (Def. 13) that the axis of a cylinder is the straight line joining the centres of its two ends.

PROPOSITION I. THEOREM.

Any two of the plane angles that form a trihedral angle, are together greater than the third.

Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB. Any two of them are greater than the third.

D

If the angles BAC, CAD, DAB, be all equal, it is evident that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A in the straight line AB, make, in the plane which passes through BA, AC, the angle BAE equal (Pl. Ge. I. 23) to the angle DAB; and make AE equal to AD, and through

B

E

C

E draw BEC cutting AB, AC, in the points B, C, and join DB, DC. And because DA is equal to AE, and AB is common to the two triangles ABD, ABE, and also the angle DAB equal to the angle EAB; therefore the base DB is equal to the base BE. And because BD, DC, are greater (Pl. Ge. I. 20) than CB, and one of them BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC. And because DA is equal to AE, and AC common, but the base DC greater than the base EC; therefore the angle DAC is greater (Pl. Ge. I. 25) than the angle EAC; and, by the construction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC, are together greater than BAE, EAC, that is, than the angle BAC. But BAC is not less than either of the angles DAB, DAC; therefore BAC, with either of them, is greater than the other.

PROPOSITION II. THEOREM.

Every solid angle is contained by plane angles which together are less than four right angles.

First, let the solid angle at A be contained by three plane angles BAC, CAD, DAB. These three together are less than four right angles.

D

Take in each of the straight lines AB, AC, AD, any points B, C, D, and join BC, CD, DB; then, because the solid angle at B is contained by the three plane angles CBA, ABD, DBC, any two of them are greater (II. 1) than the third; therefore the angles CBA, ABD, are greater than the angle DBC. For the B

с

same reason, the angles BCA, ACD, are greater than the angle DCB; and the angles CDA, ADB, greater than BDC; wherefore the six angles CBA, ABD, BCA, ACD, CDA, ADB, are greater than the three angles DBC, BCD, CDB; but the three angles DBC, BCD, CDB, are equal to two right angles (Pl. Ge. I. 32); therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB, are greater than two right angles; and because the three angles of each of the triangles ABC, ACD, ADB, are equal to two right angles, therefore the nine angles of these three triangles,

namely, the angles CBA, BAC, ACB, ACD, CDA, DAC, ADB, DBA, BAD, are equal to six right angles. Of these, the six angles CBA, ACB, ACD, CDA, ADB, DBA, are greater than two right angles; therefore the remaining three angles BAC, DAC, BAD, which contain the solid angle at A, are less than four right angles.

Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; these together are less than four right angles,

D

E

Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB; and because the solid angle at B is contained by three plane angles CBA, ABF, FBC, of which any two are greater (II. 1) than the third, the angles B CBA, ABF, are greater than the angle FBC. For the same reason, the two plane angles at each of the points C, D, E, F, namely, the angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the angles of the polygon BCDEF; therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon; and because all the angles of the triangles are together equal to twice as many right angles as there are triangles (Pl. Ge. I. 32); that is, as there are sides in the polygon BCDEF; and because all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon (Pl. Ge. I. 32, Cor. 1); therefore all the angles of the triangles are equal to all the angles of the polygon, together with four right angles. But all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved. Wherefore, the remaining angles of the triangles, namely, those at the vertex, which contain the solid angle at A, are less than four right angles.

PROPOSITION III. THEOREM.

If two solids be contained by the same number of equal and similar planes, similarly situated, and if the inclination

of any two contiguous planes in the one solid be the same with the inclination of the two equal, and similarly situated planes in the other, the solids themselves are equal and similar.

Let AG and KQ be two solids contained by the same number of equal and similar planes, similarly situated, so that the plane AC is similar and equal to the plane KM, the plane AF to the plane KP, BG to LQ, GD to QN, DE to NO, and FH to PR. Let also the inclination of the plane AF to the plane AC be the same with that of the plane KP to the plane KM, and so of the rest; the solid KQ is equal and similar to the solid AG.

Let the solid KQ be applied to the solid AG, so that the bases KM and AC, which are equal and similar, may coincide, the point N coinciding H with the point D, K with A, L with B, and so on.

And

because the plane KM coincides with the plane AC, and, by hypothesis, the inclination of KR to KM is the same

D

G

R

E

F

[ocr errors]

P

[blocks in formation]

with the inclination of AH to AC, the plane KR will be upon the plane AH, and will coincide with it, because they are similar and equal, and because their equal sides KN and AD coincide. And in the same manner, it is shown that the other planes of the solid KQ coincide with the other planes of the solid AG, each with each; wherefore the solids KQ and AG do wholly coincide, and are equal and similar to one another.

PROPOSITION IV. THEOREM.

If a solid be contained by six planes, two and two 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 are similar and equal parallelograms.

Because the two parallel planes BG, CE, are cut by the plane AC, their common sections AB, CD, are parallel

B

H

(I. 14). 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; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF, AE, is a c parallelogram. Join AH, DF; and because AB is parallel to DC, and BH to

D

E

CF; the two straight lines AB, BH, which meet one another, are parallel to DC and CF, which meet one another; wherefore, though the first two are not in the same plane with the other two, they contain equal angles (I. 9); the angle ABH is therefore equal to the angle DCF. And because AB, BH, are equal to DC, CF, 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. For the same reason, the triangle AGH is equal to the triangle DEF; and 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.

PROPOSITION V. THEOREM.

If a solid parallelopiped be cut by a plane parallel to two of its opposite planes, it will be divided into two solids, which will be to one another as their bases.

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 solids ABFV, EGCD; as the base AEFY to the base EHCF, so is the solid ABFV to the solid EGCD.

Produce AH both ways, and take any

X

B

G

Τ

Z

P

R

D

U T

[blocks in formation]

Y

F

C Q S

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

« ForrigeFortsett »