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PROPOSITION XIX. (1.) If a line and plane be both perpendicular to the same plane,

they will be parallel to one another; and (2.) If a line and plane be parallel, every plane perpendicular to the

line will be also perpendicular to the plane.

(1.) Let the line AB and plane PQ be both perpendicular to the plane MN: then they will be parallel to one another.

For from any point C in the intersection QR of the planes, draw CD in the plane PQ perpendicular to RQ.

Then since the plane PQ is perpencular to MN, the line CD is also perpendicular to MN (Prop. xyii.); and therefore parallel to AB (Prop. xiv.).

Then since AB is parallel to the line CD in the plane PQ, it is parallel to the plane PQ itself (Prop. III. Chap. I.).

(2.) Let AB be parallel to the plane PQ, and perpendicular to the plane MN: then MN will be perpendicular to PQ.

For draw any plane through AB to cut PQ in CD.

Then since AB is parallel to the plane PQ, the line CD is parallel to AB (Prop. II. Chap. I.); and since AB is perpendicular to MN, and CD parallel to AB, CD is also perpendicular to MN (Prop. xiv.).

Wherefore, again, the plane through CD is perpendicular to the plane MN (Prop. xvi.).

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PROPOSITION XX. If two planes inter sect, and from any point lines be drawn perpendicular to them, then these lines will contain an angle equal or supplementary to the profile angle of the planes.

Let the planes PQ, RS intersect in RQ; and from any point A let lines AB, AC be perpendicular to PQ, RS: then the angle BAC will be either equal or supplementary to the profile angle of the planes.

For draw the plane through AB, AC, cutting PQ, RS in BD, CD.

Then since the plane ABDC passes through AB, and AB is perpendicular to PQ, the plane ABDC is perpendicular toPQ(Prop. xvi.). In like manner it is perpendicular to RS ; and hence also to the intersection RQ of PQ, RS (Prop. XVIII.). It is, therefore, the profile plane of the dihedral angle PQRS; and BDC is the profile angle.

Now the four angles of the quadrilateral figure ABDC are together equal to four right angles; and two of them ABD, ACD are right angles (Prop. 11.): whence the other two BDC and BAC are together equal to two right angles ; that is, BAC is the supplement of BDC.

If, however, we produce either of the lines AB or AC to E or F, the adjacent angle CAE or BAF is equal to the profile angle BDC. It is generally understood in this proposition, however, that the angle BAC turned towards the dihedral edge, is that specified.

But when the point A is not between the planes PQ, RS as in the preceding figure; but without the dihedral angle, the profile angle BDC is equal to the angle BAC formed by the perpendiculars.

This will always be decided by joining AD: for when B, C, are on opposite sides of AD, the angle BAC is the supplement of BDC; and when on the same side of AD, BAC will be equal to BDC.

PROPOSITION XXI.

(1.) If a dihedral angle be bisected, any point in the bisecting plane

is equally distant from the faces of the angle; and (2.) All points, each of which is equidistant from the faces of a

dihedral angle are situated in the plane which bisects that angle.

(1.) Let the plane AN bisect the dihedral angle MABP: then any point D in AN will be equidistant from AM and AP.

For draw the perpendiculars DE, DF to AM, AP, and through them the plane DEGF, cutting the faces in EG, GF, the edge in G, and the bisecting plane in DG.

Then, reasoning as in the 'preceding proposition, it is shown that DEGF is the profile plane of the given dihedral angle MABP.

Also, since the plane EGFD is perpendicular to AB, GD is also perpendicular to AB, and EGD, FGD are the profile angles of MABN, NABP. But these dihedral angles are equal (Hypoth.) and hence the profile angles EĞD, DGF are equal. The angles DEG, DFG, also v are right angles (Prop. 11.); and hence M N the triangles are equiangular, and right angled ; and have the common, hypothenuse GD. Wherefore the other sides are equal each to each, namely, those which are opposite to the equal angles; that is, the distance DE is equal to the distance DF.

(2.) Every point equidistant from the faces AM, AP, lies in the bisecting plane AN.

For if it be possible let any point H without the bisecting plane AN be equidistant from the faces AM, AP; and let HE, HK be those distances. Let HE cut AN in D, and draw DF perpendicular to AP, and join HF.

Then by the preceding case ED is equal to DF, and therefore EH (which is equal to ED and DH) is equal to HD and DF together. But HD and DF together are greater than HF; and hence EH is greater than HF.

Again, HKF is a right angle, and hence HFK is less than a right angle; and therefore HF is greater than HK. But HE has been proved to be greater than HF; therefore it is greater than HK. But it has been admitted to be equal to HK, which is contradictory. The point H, then, without the bisecting plane is not equidistant from the faces of the angle; and hence all equidistant points are in the bisecting plane.

CHAPTER III.

OBLIQUE LINE AND PLANE.

PROPOSITION I. Except a line be perpendicular to a plane, only one plane can be drawn

through it perpendicular to the plane.

The two other positions besides perpendicular which a line can have with respect to a plane, are oblique and parallel. If, then, it be possible that through such lines AB, more than one plane perpendicular to the plane MN can be drawn, let there be two, as BAC, BAD,

N . the figures taken in order representing AB oblique to MN, and AB parallel to MN, respectively. And from any point A in AB, draw lines AC, AD perpendicular to CC' and DD' in the planes ACC', ADD'.

Then, because ACC is perpendicular to the plane MN, and AC is drawn in ACC' perpendicular to the common section of the planes CC', it is perpendicular also to the plane MN (Prop. xvi.). In like manner AD is perpendicular to MN. Wherefore from the same point

A, two straight lines have been drawn perpendicular to the plane MN; which (Prop. iv. Chap. 11.) is impossible.

The planes ACC', ADD' cannot therefore be both perpendicular to MN; that is, only one plane ABC can be drawn brough AB perpendicular to MN.

Scholium 1. When AB is perpendicular to MN, it has been already shown (Prop. xvi.) that all planes through it are perpendicular to MŇ.

Scholium 2. The angle ABC which (Def. 6) measures the inclination of the oblique line AB to the plane MŇ, is called the acute profile angle in respect of this line; the perpendicular plane ABC, the profile plane, and BC the profile trace.

the point olique to a plane USITION II.

If a line be oblique to a plane, and lines be drawn in the plane through

the point of intersection making angles with the oblique line :

then (1.) Of all the angles which are so formed, the greatest and least are

the profile angles of the line : - (2.) Only two lines can be drawn to make equal angles with the

oblique line, one on each side of the profile plane: (3.) Of these lines that makes the less angle with the oblique line,

where the corresponding line in the plane makes the less angle with

the acute profile trace, and conversely: (4.) Only one line can be drawn in the plane perpendicular to the

oblique line.

Let the line AB be oblique to the plane MN; and let the profile plane PQ cut MN in GC, and let other lines AD, AE, AF, etc., be drawn in the plane MN: then,

(1.) The acute profile angle BAC is less than another angle BAE, and the obtuse profile angle BAG is greater.

For, about A in the plane MN describe any circle GHC, cutting the several lines drawn from A in that plane; and let BC be drawn in the profile plane ABQ perpendicular to AQ; and join AE, AG.

Then, since BC is perpendicular to ihe plane MN (Prop. 1. Schol. 2), it is less than BE (Prop. xi. Chap. 11.); wherefore the two BA, AC are equal to the two BA, AE each to each ; but the base BC less than the base BE: wherefore (Euc. 1. 25) the angle BAC is less than BAE. That is, BAC is the least angle that can be made under the circumstances.

Again, since CG is the diameter of the circle GHC, it is greater than any other line CE in it (Euc. III. 15); and hence BG is greater than BE (Prop. XI, Chap. II). Wherefore, BA, AG being equal to

BA, AE, but the base BG greater than BE, the angle BAG is greater than BAE (Euc. 1. 25); that is, BAG is the greatest angle that can be made under the circumstances.

(2.) Only two lines AD, AE can be drawn in MN to make equal angles with AB.

For, let the lines AD, AE be drawn to make equal angles with GC, one on each side of it; and join AD, AE, CD, CE.

Then since the angles CAD, CAE at the centre of the circle are equal, the chords CD, CE which subtend them are equal (Euc. III. 29). Wherefore, CD, CE being equal, BD, BE are also equal (Prop. xi. Chap. 11). Whence (Euc. 1. 8) the angle BAD is equal to BAE; and two equal angles can be made.

It remains to show that a third angle cannot be made equal to either of these. For if possible, let BAF be equal to BAD or BAE; and join BF, CF.

Then, since CD, CE are equal, the line CF is greater or less than either of them. Suppose it greater, then BF is greater than BD or BE. Then BA, AF are equal to BA, AD, but BF greater than BD; and hence (Euc. 1. 25) the angle BAF is greater than BAD. Similarly if the line CF be less than CĐ, the angle BAF is less than BAD.

Two such lines, therefore, can be drawn; but not more than two.

(3.) The line BE makes a less angle with BA than BF does, when the angle CAE is less than CAF; and the converse.

These conclusions are virtually contained in the preceding argument; and are left for the student to deduce in form.

(4.) Only one straight line HK can be drawn in the plane MN perpendicular to AB.

For that it may be perpendicular to AB, the angles CAH, CAK must be right angles (Prop. XIII. Chap. 11.); and only one line on each side of GC can be drawn perpendicular to AC from the point A. Whence only one line HK, can be drawn in MN perpendicular to AB.

PROPOSITION III. If a line be oblique to a plane, and about the point in which it meets

the plane a circle be described in that plane; and if, moreover, from any fixed point in that line, lines be drawn to the circum

ference of the circle : then, (1.) The least line is that drawn to the intersection of the acute pro

file trace with the circle; and the greatest, that drawn to the inter

section of the obtuse profile trace to the circle ; (2.) That which meets the circle at a greater circula: distance from

the acute trace is greater than that which meets it at the less ; (3.) Only two such lines can be equal to each other ; (4.) The less line makes a greater angle with the plane than the greater

line does.

Let AB be a line inclined to the plane MN, and GDC any circle described in MN about the centre A; and let the trace of the profile plane PQ cut this circle in GC, the acute profile angle being C and the obtuse one G: then

(1.) The least line that can be drawn from B to the circle is BC and the greatest is BG.

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