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(1.) The sum of BAD and BAC is greater than BAD; and there. fore greater also than the third angle DAC.

Again, to prove that BAD, DAC are together greater than BAC, make BAE in the plane BAC equal to the angle BAD, and the line AE to AD; draw any line BC through E in the plane ABC, to meet AB, AC in B, C; and join BD, CD.

Then BA, AD are equal to BA, AE and the angle BAD to the angle BAE; hence BD is equal to BE.

But BD and DC are together greater than BC; and hence CD is greater than CE. Whence DA, AC are equal to EA, AC, but the base DC greater than the base EC ; and hence the angle DAC is greater than EAC; and consequently the two BAD and DAC together greater than BAE and EAC together, that is than BAC.

(2.) The difference of BAD, DAC is nothing, they being equal angles; and hence less than the third angle BAC.

Again, it has been shown that EAC, the difference between BAC, and one of the other equal angles BAD, is less than DAC the third angle.

CASE (8.) Let all three angles BAD, DAC, CAB be unequal, and let BAC be the greatest of them.

(1.) Take the most unfavourable case to the truth of the proposition, viz., BAD and DAC compared with BAC. Make the angle BAE in the plane BAC equal to BAD, and 3 AE equal to AD; and complete the construction as in the preceding case.

Then by reasoning exactly similar to that in the preceding case, it is shown that BAD and DAC together are greater than BAE and EAC together; that is than BAC.

If either of the angles BAD, DAC be interchanged with BAC, the sum will be greater than before, and the third angle less; and hence, the sum of those two will be greater than the third angle.

(2.) By similar construction to the preceding, and corresponding reasoning, the difference EAC of the angles BAC and BAD is less than the third angle DAC. And this is equally the case whether BAD be the greater or the less of the two angles BAD, DAC.

PROPOSITION II. If three points be taken in the edges of a trihedral angle equidistant from the vertex, and a plane be described through these points : and if also a perpendicular to this plane be drawn from the summit, it will be the centre of the circle which passes through the three points in the edges.

Let SA, SB, SC be the three edges of a trihedral angle, and these distances be all equal; let a plane MN be drawn through ABC, and a perpendicular SD from S to MN: then D is the centre of the circle through A, B, C, in the plane MN.

For join AD, BD, CD.

Then in the triangles SDA, SDB the two sides SD, SA are equal to the two SD, SB; and the opposite angles to SA, SB right angles; therefore the remaining sides and angles are equal, each to each. That is, DB is equal to DA. In like manner DC is equal to DA; and hence A, B, C are equidistant from D, or D is the centre of the circle ABC.

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Cor. 1. Each of the edges makes equal angles with the plane ABC: for these angles SAD, SBD, SCD are the corresponding angles of equal triangles.

Cor. 2. Each of the edges makes equal angles with the perpendicular SD.

Cor. 3. If AB, BC, CA be joined, and perpendiculars DG, DF, DE be drawn from D to each of them, these perpendiculars bisect those sides.

Cor. 4. If SE, SF, SG be drawn, the angles SED, SFD, SGD will be the profile angles of the dihedral angles formed by the faces of the trihedral angle with the plane MN.

PROPOSITION III. The plane angles of every salient polyhedral angle are together less

than four right angles. (1.) Let the angle be trihedral, and represented by the figures of the preceding proposition. Make the same construction with respect to A, B, C, the plane MN, perpendicular SD, etc.

Then E, F, G will fall between the extremities of the lines AC, CB, BA respectively. Whence (Prop. V., Chap. III.) ADC is greater than ASC, CDB than CSB, and BDA than BSA. Wherefore the three angles at S are together less than the three angles at D together. ,

But the three angles at D are together equal to four right angles ;* and hence the three angles at S are less than four right angles.

(2.) Let the angle at S be tetrahedral, having ASB, BSC, CSD, DSA for its four faces or plane angles : these will together be less than four right angles.

For, produce two of the alternate faces ASB, CSD to meet in SE.

Then (Prop. 1.) the two angles ASE, DSE are together greater than the angle ASD of the trihedral angle S. EDA. Wherefore also, the two angles BSE, CSE E are together greater than BSA, ASD, DSC; and hence again the three angles of the trihedral angle S. BEC at S are greater than the four at S of the tetrahedral angle S. ABCD.

But the angles of S. BEC are less than four right angles by the preceding case; and hence the four angles at S of the tetrahedral angle S. ABCD are together less than four right angles.

GENERALLY. The pentrahedral angle is reduced in the same manner to the tetrahedral, and its plane angles shown to be less; and hence less than those of the tetrahedral angle, and thence again less than four right angles.

PROPOSITION IV. If from any point within a trihedral angle, perpendiculars be drawn

to the faces, these perpendiculars will be the edges of a new trihedral

angle, which has the following relations to the original one :(1.) Its edges will be perpendicular to the faces of the original ; (2.) Its faces will be perpendicular to the edges of the original ; (3.) Its plane angles will be the supplements of the opposite profile

angles of the original; and (4.) Its profile angles will be the supplements of the plane angles of

the original.

(1.) This is but a repetition of the hypothesis, for the purpose of enumerating the connected properties seriatim.

(2.) Let S. ABC be the original trihedral angle ; and from any point s within it let perpendiculars sa, sb, sc to the faces BSC, CSA, ASB be drawn: the planes bsc, csa, asb will be perpendicular to the edges SA, SB, SC, respectively.

For let the plane bsc cut the planes BSA, ASC in c A, 6 A.

Then since sc is perpendicular to the plane BSA, the plane bsc

A . b through it is perpendicular to BSA

(Prop. XVI., Chap. 11.). In like manner the plane bsc through bs is perpendicular to the plane ASC.

Wherefore the plane bsc, being

* In Fig. 2, the angle CDA is equal to two right angles; and in Fig. 3, the angle CDA is the reverse angle.

perpendicular to both the planes BSA, ASC, is perpendicular to SA their common intersection (Prop. XVIII., Chap. 11.).

The two perpendiculars sb, sc determine, therefore, a plane perpendicular to the edge SA ; and in the same manner the other two faces of the trihedral angle s. abc are respectively perpendicular to the other two edges of S. ABC ; viz., bsa to SC and asc to SB.

(3.) Since SA is perpendicular to the plane bsc, the lines Ab, AC are perpendicular to SA; and hence b A c is the profile angle of the dihedral angle BSAC. Also (Prop. xx., Chap. 11.), the angles bAc, bsc are supplementary : that is the plane angle bsc is the supplement of the profile angle b A c of BSA, ASC.

In the same manner it may be proved that csa is the supplement of the profile angle cB a of CSB, BSA; and that asb is the supplement of the profile angle aCb of ASC, CSB.

(4.) Since SA, SB, SC drawn from the point S have been proved to be perpendicular to the faces of the trihedral angle s. abc, it follows from the preceding case, that the plane angles ASB, BSC, CSA are the supplements of the profile angles 6 Ca, a Bc, c A b of the dihedral angles ASC, CSB, of CSB, BSA, and of CSA, ASB; or, merely changing the order, the profile angles of s. abc are the supplements of the opposite plane angles of S. BCA.

SCHOLIUM. The trihedral angles thus related are sometimes called conjugate trihedral angles, from their interchangeability: sometimes supplementary trihedral angles from their properties here proved.

It is easily seen that if s be taken at S instead of within the trihedral angle S. ABC, the new trihedral angle would differ from any other formed according to the enunciation of this proposition, it would be in all respects equal to S. abc. A direct proof might be made, however, of this fact, from the consideration that the three faces of any two trihedral angles thus formed conjugately to S. ABC would be parallel each to each. The details are left for the student to supply.

When S and s are coincident, and a sphere is described about Sto cut the faces of the two trihedral angles, two spherical triangles are formed, the sides and angles of which have an analogous system of relations. They have also another relation (for which see p. 210 of the Spherical Geometry), which causes them to be generally called polar triangles : which relation is of great importance in spherical trigonometry.

PROPOSITION V. The three profile angles of any trihedral angle are together greater than two right angles, but less than six. .

(Figure to preceding Proposition.) Let S. ABC be any trihedral angle, and take a point s within it, and construct the conjugate trihedral angle s. abc as in the preceding proposition. Then c Ab, 6 Ca, a Bc are the profile angles of the trihedral angle S. ABC, as already there proved. These three angles are together greater than two right angles.

For the three pairs of angles, c Ab and csb, 6 C a and bsa, and a Bc and asc, are together equal to six right angles, each pair being equal to two (Prop. xx., Chap. 11.)

But the three asb, bsc, csa are together (Prop. III.) less than four

right angles ; and hence the other three a Bc, c Ab, 6 C a are together greater than two right angles.

Again, the three profile angles are less than six right angles.

For, since the six angles above enumerated are together equal to six right angles, and that the three asb, bsc, csa, are necessarily of some magnitude, the other three together a Bc, c Ab, bCa, are deficient from six right angles, by that magnitude whatever it be.

CHAPTER V. EQUALITY AND SYMMETRY OF TRIHEDRAL ANGLES. In plane triangles an application of one to the other could generally be made, when the conditions of their structure were three parts in one triangle equal to three corresponding parts in the other. This was rendered possible by the fact of both sides of a plane being alike, so that when supraposition became an effective step, the plane of one triangle might be turned over, and its faces thereby inverted.

In the case of trihedral angles this advantage is lost, and it becomes a necessary condition for supraposition that the three parts of the two figures which are to be made coincident shall follow each other in the sime order. Thus, in the two triangles ABC, DEF,

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if the sides AB, AC are equal to the sides ED, DF situated in the same order, and the included angles equal, the inference of equality is made by superposition without reversing the plane of the triangle ABC (Euc. 1. 4.): but if E'D', D'F', were substituted for ED, DF, where the order is inverted, the plane of ABC must be reversed to produce coincidence.

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If, now, there be two trihedral angles V. ABC, and S. DEF; and if the two faces, or plane angles AVB, AVC be equal to DSE, DSF and the included dihedral angles equal, they will be capable of coinci

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