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PROPOSITION VII.

The traces of the profile angle are perpendicular to the edge of the

dihedral angle.

Before proceding further, let the student refer to Defs. 3, 4, 7, and (Prop. II. Chap. 11.); he will then see that this proposition is only the proper technical form M of the following:

A plane perpendicular to the edge of a dihedral angle cuts the faces in lines which are perpendicular to that edge.

Let MN, PQ be the faces of the dihedral angle MNQP, PN its edge, and RNQS a profile plane whose traces are R' RN, NQ: then RN, NQ will be perpendicular to PN.

For, since PN is perpendicular to the R plane RQ, it is perpendicular to every line in it drawn through N: that is RN, NQ are both perpendicular to PN.

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PROPOSITION VIII.

(1.) The profile traces at all points in the edge of a dihedral angle

are parallel, those in one face to one another, and those in the other

to one another; and (2.) All the profile angles of the same dihedral angle are equal to one another.

(Preceding figure.) (1.) Let RQ, R'Q' be two profile planes to the dihedral angle MNQP; then the trace R'N' is parallel to the trace RN, and the trace N'Q to the trace NQ.

For RN, R'N' being in the same plane, and both perpendicular to PN, they are parallel to one another.

In the same manner N'Q' is parallel to NQ; and that which is proved for any one point N' is true for all other points.

(2.) The profile angle R'N'Q' is equal to the profile angle RNQ.

For, since the two lines R'N', N'Q' which meet in N' are parallel to the two RN, NQ which meet in N, the angle R'N'Q' contained between the first two is equal to the angle RNQ contained between the other two (Prop. vir. Chap. 1.).

Cor. All the profile planes of the same dihedral angle are parallel, that is, the plane R'Q' is parallel to RQ (Prop. vii. Chap. I.),

SCHOLIUM. It might have been inferred from this demonstration that all planes perpendicular to the same straight line are parallel; but this is usually made the subject of a separate proposition, and it is so done in this course.

PROPOSITION IX. (1.) If two dihedral angles be equal, their profile angles are equal ;

and (2.) If the profile angles of two dihedral angles be equal, the dihedral

angles are themselves equal.

(1.) Let PABQ, P'A'B'Q' be two equal dihedral angles, and PQ, P'd the profile planes ; PB, BQ the profile traces of the first dihedral angle, and P'B', B'Q' those of the second: then the angles PBQ, P'B'Q' will be equal.

For let the dihedral angle P'A'B'Q' be transposed to the position such that A'B' shall coincide with AB, the point B' coinciding with B, and the plane A'P' with the plane AP.

Then, since the face A'P' coincides with AP, and the dihedral angles are equal (Hypoth.), the face A'Q' will coincide with the face AQ.

Again, since A'B' coincides with AB, the point B' with the point B, and the face A'P' with the face AP, the perpendiculars B'P', BP in those coincident faces must coincide. In like manner, B'Q' must coincide with BQ; and hence the angles P'B'Q', PBQ will be equal.

(2.) If the profile angles P'B'Q', PBQ be equal, the dihedral angles P'À'B'Q', PABQ will be equal.

For so transpose the dihedral angle P'A'B'Q' that the edge A'B' shall coincide with AB, the point B' with B, and the plane A'B'P' with APB: then if the plane A'Q' do not coincide with AQ, let it take some other position as AS, and B'Qʻ the position BS.

Then, as in the preceding case, B'P' coincides with BP; and since A'B'Q' is a right angle, ABS is a right angle, and BS is in the plane PQ (Prop. III.).

Whence since the dihedral angles PABS, P'A'B'Q' are equal, the profile angles P'B'Q', PBS are equal (preceding case); and P'B'Q' is equal to PBQ by hypothesis. Whence PBS is equal to PBQ, the less to the greater, which is impossible. Whence the dihedral angles P'A'B'Q', PABQ are not unequal ; that is, they are equal.

SCHOLIUM. The properties in this and the preceding proposition are tacitly assumed by Euclid, and by most other writers; but very little reflection is necessary to convince us that such a course is altogether unwarranted under the circumstances of the case. It should be laid down as a general rule in mathematics, that propositions which admit of proof from principles already laid down, ought never to be assumed as independently true. It is tantamount to multiplying our axioms without necessity; and moreover with great risk of involving contradictions amongst them.

PROPOSITION X.

Two dihedral angles have the same ratio to one another that their

profile angles have to one another.

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Let PABM, MBAQ be two dihedral angles, whose profile angles are PBM, MBQ: then

PABM : MABQ :: PBM : MBQ. [In the figure, the dihedral angles are placed contiguously, so as to have one face common to both of them, viz., BM. When this is not so given, we must conceive them to be so placed by transposing one of them nearly as in the preceding proposition.]

For since PB, MB, QB are perpendicular to AB (Def. 4) they are in one plane (Prop. II.). In that plane take any number of angles, PBP, P,BP, etc., each equal to PBM ; and any number QBQı, QiBQ,, etc., each equal to MBQ; and, finally, draw the planes ABP, ABP, etc., and ABQı, ABQ., etc.

Then PBP, is the profile angle of the dihedral angle PBAP, P,BP, of P,BAP, etc.; and as the profile angles were made equal, the dihedral angles are also equal (Prop. ix.). Wherefore, whatever number of profile angles each equal to PBM there be in MBP,,* the same number of equal dihedral angles there will be in MBAP,; that is, equimultiples MBP, and MBAP, have been taken of MBP and MBAP. In like manner, whatever multiple MBQm is of MBQ, the same multiple will MBAQ be of MBAQ.

Again, if the profile angle MBP, be greater than MBQm the dihedral angle MBAP, will be greater than MBAQm; if equal, equal ; and if less, less. Wherefore (Euc. v. Def. 5.)

PBAM : QBAM :: PBM : QBM. SCHOLIUM. It is essential to the acquisition of facility in the application of these principles to give close attention to some special cases that arise.

* This notation is used to signify that there are n dihedral angles with their corresponding profile angles taken, without specifying any particular number. The same with respect to m in the other set of multiples.

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(1.) Let MN, PQ be two planes which cut one another in AB; then four dihedral angles will be formed, viz., NBAP and its opposite, MABQ, and the opposite pair, PABM and QBAN.

Let any profile plane RS be drawn to these two planes, cutting MN in TCS and PQ in VCW.

Then, since the opposite profile angles. VCS, WCT are equal, the corresponding dihedral angles are equal (Prop. ix.), viz., PABN and QBAM. In the same manner the opposite dihedral angles PABM, NBAQ are equal. That is, if two planes cut each other, the opposite dihedral angles are equal.

(2.) If two planes, QA, BP on opposite sides of another plane MN meet in the same line AB and inake the opposite angles (either the dihedral or the profile) equal, those planes are one and the same plane.

(3.) If a plane BỐ meet a plane MN in AB, the dihedral angles PABN, PABM have the same ratio as the profile angles VCS, VČT.

(4.) When the adjacent dihedral angles are equal, each of them is a dihedral right angle, by Def. 4; and in this case each of the adjacent profile angles is a plane right angle. And conversely when the adjacent profile angles are equal, each is a right angle ; and the corresponding dihedral angles are equal, and each a right angle. The one condition leads to the other; and either being given by hypothesis, the other may be inferred in the reasonings towards any specific conclusion.

(5.) The sum of all the dihedral angles that can be formed about a line are together equal to four dihedral right angles.

(6.) It is on account of the proportionality between two dihedral angles and their profile angles that the latter have been used instead of the former in all the reasonings of the ancients, and in those of most of the moderns too. The dihedral angle has, indeed, been scarcely recognised as a subject of geometry at all; and the profile angle has been almost universally substituted for it, under the name of the inclination of one plane to the other. This has been productive of much confusion in the student's mind, and created great difficulty in his appreciating the force of the argument in cases where these angles occur. Even into a simple case of the dihedral right angle, or a plane perpendicular to a plane, it has introduced a degree of confusion that renders the subject difficult to comprehend, and indeed puts the whole inquiry in a shape so vague that very few persons are ever able to acquire a clear and comprehensive view of the subject.

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PROPOSITION XI.

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If from a point without a plane, straight lines be drawn to meet the

plane : then (1.) The perpendicular is the shortest line that can be drawn ; (2.) Those which meet the plane at equal distances from the perpen

dicular are equal ; (3.) That which meets it at the less distance is less than the one that

meets the plane more remotely.

Let A be a given point without the given plane MN; AB perpendicular to MN; AC, AD, lines meeting the plane at equal distances BC, BD from B; and AE meeting MN at a distance BE from B greater than BC: then,

(1.) AB will be less than any one AC of the other lines.

For draw BC:

Then ABC is a right angle; and consequently greater than the angle ACB of the triangle ABC. Wherefore AC opposite the greater angle is greater than AB opposite the less (Euc. 1. 19.). That is, AB is less than AC.

(2.) The lines AC, AD are equal.

For, since AB is perpendicular to the plane MN, the angles ABC, ABD are right angles. Therefore the two sides AB, BC are equal to the two AB, BD, and the included angles are equal; and hence (Euc. 1. 4.) the bases AC, AD are equal.

(3.) Since BE is greater than BC, cut off a part BD equal to BC, and join AD.

Then, since ABD is a right angle, ADE is greater than a right angle (Euc. 1. 17); and hence AED is less than a right angle. Wherefore ADE is greater than AED; and hence AE greater than AD. But AD is equal to AC by the preceding

case; and hence AE is likewise greater than AC, or AC less than AE.

SCHOLIUM. Several corollaries, or rather easy deductions, follow from this simple property. They are put down here as exercises for the student; and they should be established here, as some of them come into use hereafter, and will be quoted by reference to this place.

Cor. 1. Equal lines AC, AD make equal angles with the perpendicular AB; and likewise equal angles with the lines BC, BD drawn to B.

Cor. 2. All equal lines drawn from a point A without the plane MN meet the plane in the circumference of a circle described about B as centre.

Cor. 3. The greater line AE makes a greater angle BAE with the perpendicular AB than the less AC does.

Cor. 4. If lines be drawn from points C, D equidistant from B to the same point A in the perpendicular AB, they will be equal; and they will make equal angles with the perpendicular, and equal angles with the lines drawn to B.

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