Abbildungen der Seite
PDF
EPUB

PROPOSITION XVI.

If a line be perpendicular to a plane, every plane drawn through it will be perpendicular to that plane.

Let the line AB be perpendicular to the plane MN: then any plane PQ drawn through AB will be perpendicular to MN.

For let the plane PQ cut MN in BQ, and in MN draw the line BC perpendicular to BQ; and also draw the plane AC through AB, BC.

Then since AB is perpendicular to MN, it is perpendicular to BQ; and BC is perpendicular to BQ by construction. Whence ABC is the profile angle of the dihedral angle PQBN (Def. 4).

M

N

But AB being perpendicular to MN, ABC is a right angle (Prop. II.); and the profile angle of the planes being a right angle, then the dihedral angle is a right angle (Def. 4). That is, the plane PQ is perpendicular to MN.

PROPOSITION XVII.

If two planes be perpendicular to one another, every line drawn in one plane perpendicular to the common section will be perpendicular to the other plane; and every line drawn from a point in one plane perpendicular to the other plane lies wholly in the former, and is perpendicular to the common section.

(1.) Let the plane PQ be perpendicular to the plane MN; and let

[blocks in formation]

the line AB be drawn in one plane FQ perpendicular to MQ the common section of the planes: then AB will be perpendicular to the other plane MN.

For in the plane MN, draw BC perpendicular to the common section MQ. Then since AB, BC are perpendicular to MQ, the plane through them is the profile plane, and ABC the profile angle. But the dihe

dral angle PQMN is a right angle (Hyp.), and hence ABC is a right angle (Prop. x. Scholium.).

But AB is perpendicular to MQ (Hyp.), and hence it is perpendicular to the two lines BC, BM in the plane MN: that is, to the plane MN itself.

(2.) All lines drawn from points, as R, in the plane PQ perpendicular to the plane MN, will be wholly in the plane PQ.

For, if possible, let some one of them RS be without the plane PQ. From R draw RQ perpendicular to the line QM.

Then by the preceding case RQ is perpendicular to MN. But by admission RS is also perpendicular to MN: that is, two lines RQ, RS can be drawn from the same point R perpendicular to the plane MN. This is impossible (Prop. iv.); and hence the perpendicular to MN falls wholly in the plane PQ.

(3.) All perpendiculars from points in PQ to the plane MN are perpendicular to the line MQ.

For if possible let AH be drawn in the plane PQ perpendicular to MN, yet not perpendicular to MQ; and draw AB perpendicular to MQ.

Then reasoning as in the last case, the conclusion follows.

PROPOSITION XVIII.

(1.) If two planes which intersect be perpendicular to a third plane, their intersection will also be perpendicular to this third plane; and (2.) If a plane be perpendicular to the intersection of two other planes, it will be perpendicular to each of those planes.

R

(1.) Let the two planes PQ, RS which intersect in RQ, be both perpendicular to the plane MN; then will RQ be perpendicular to the plane

MN.

For, from any point A in the intersection AQ, draw a line AP in PQ perpendicular to AQ; and similarly a line SB in the plane RS perpendicular to QS.

Then (Prop. XVII.), AP, SB are perpendicular to MN; and hence (Prop. XIV.) they are parallel to one another.

M

A

N

Also, since the planes PQ, RS which intersect in RQ are drawn through the parallel lines AP, BS, the intersection QR is parallel to both of them (Prop. 11. Chap. 1.).

And, finally, since AP is perpendicular to MN, and QR parallel to AP, it follows that QR is also perpendicular to MN.

(2.) Let the plane MN be perpendicular to QR the intersection of the two planes PQ, RS: then it will be perpendicular to the planes PQ, RS themselves.

For, since RQ is perpendicular to MN, the planes PQ, RS passing through it are also perpendicular to MN (Prop. XVI.): that is, MN is perpendicular to PQ and RS.

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. XVII.); 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. 1.).

(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. 11. Chap. 1.); 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.).

PROPOSITION XX.

If two planes intersect, and from any point lines be drawn perpendicular to them, then these lines will contain an angle equal cr supplementary to the profile angle of the planes.

F

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

R

B

A

النار

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.

But

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. these dihedral angles are equal (Hypoth.) and hence the profile angles EGD, DGF are equal. The angles DEG, DFG, also are right angles (Prop. II.); and hence M

N

K

H

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,

[merged small][merged small][merged small][ocr errors][merged small]

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. XVII.). 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 through AB perpendicular to MN.

« ZurückWeiter »