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CHAPTER I.

PARALLELISM OF LINES AND PLANES.

PROPOSITION I. If two parallel planes be cut by a third plane, the sections will be

parallel. Let the parallel planes MN, PQ be cut by the plane AD in the lines AB, CD: these lines, AB, CD, will be parallel.

For since AB, CD are in one plane, if they be not parallel, they will meet in some point, as E.

Then since AB is in the plane MN, the point E is in that plane; and similarly, E being in the line CD, it is also in the plane PQ. The planes MN, PQ therefore meet in E, which is impossible (Def. 1.) since they are by hypothesis parallel to one another.

PROPOSITION II. If a straight line be parallel to a plane, all planes through this line which cut that plane will have their sections with it parallel to the line parallel to it, and to one another.

Let the line AB be parallel to the plane MN, and the planes AD, AD' drawn through AB, cut the plane MN in CD, C'D': then CD, C'D' will be parallel to AB, and to one another.

(1.) Since AB, CD are in one plane, they are either parallel to one another, or they will meet in some point E.

If parallel, the proposition is admitted ; but if not, let AB, CD meet in some point E. Then since E is in the line CD, which is itself in the plane MN, E is in the plane MN; and the line AB meets the plane MN in the point E. But this is impossible, since AB is parallel to MN by the hypothesis. Wherefore since AB, CD are in one plane ABCD and never meet, they are parallel. Similarly, C'D' is parallel to AB.

(2.) The planes AD, AD' intersecting in the line AB, they cannot meet in any point out of that line; and hence the lines CD, C'D' in them, and parallel to AB, can never meet: but they are in one plane MN; and hence they are parallel.

PROPOSITION III. If a line without a plane be parallel to a line in the plane, the first

line will be parallel to the plane.

(Preceding figure.) Let the line AB without the plane MN be parallel to the line CD in it: then AB will be parallel to the plane MN.

For since AB, CD are parallel, they are in one plane, and can never meet ; and since AD and MN meet in the line CD, they cannot meet in any point without CD. Whence AB, lying wholly in the plane ACD, can never meet the plane MN; that is, AB is parallel to MN.

Cor. If through CD one of two parallel lines AB, CD, a plane be drawn; this plane will either wholly include AB, or be parallel to it.

PROPOSITION IV. If through two parallel lines planes be drawn, they will either coalesce,

be parallel, or have their intersection parallel to those two lines.

(1.) The planes may each be drawn through the other line, since parallel lines are in one plane; and hence they may coalesce with the plane which contains the parallels, and therefore with one another.

(2.) They may never meet, since the lines through which they pass never meet; and in this case they would be parallel.

(3.) Let AB, CD be parallel lines, through which are drawn any two planes AF, ED which neither coalesce nor are parallel : they will meet in some straight line EF; and this line is parallel to each of the lines AB, CD.

For since AB is parallel to the line CD lying in the plane ED, it is parallel to the plane ED itself (Prop. 111.); and hence it can never meet the line EF in that plane. But EF is also in the plane AF, since it is the intersection of that plane with the plane ED; and hence AB, EF which lie in the same plane and never meet, are parallel.

In the same way it is proved that EF is parallel to CD.

PROPOSITION V. If two planes intersect, and lines be drawn in those planes parallel to the intersection of the planes, one in each, these lines will be parallel.

Let the two planes AD, DE intersect in CD; and in them respectively let AB, EF be drawn parallel to CD: then AB, EF will be parallel to one another.

(1.) If AB, EF be not in one plane, draw a plane through AB and E to cut the plane CF in EF. Then since through AB and CD the planes AF', CF' are drawn, intersecting in EF', the line EF is parallel to CD (Prop. iv.). But (Hypoth.) EF is parallel to CD; and hence through the point E two lines EF, EF' are drawn parallel to CD, which is impossible.

Wherefore AB, EF are in one plane.

Again, since CD is parallel to the line AB in the plane ABEF, it is parallel to the plane itself (Prop. II.); and since through the line CD which is parallel to the plane ABEF planes AD, DE are drawn to cut it in the lines AB, EF, these lines are themselves parallel to one another (Prop. 11.).

PROPOSITION VI. If each of two lines be parallel to a third line, they will be parallel to

one another.

(Same figure.) Let AB and EF be each parallel to CD; they will be parallel to one another.

For since AB, CD are parallel, they are in one plane ; and similarly EF, CD are in one plane. These planes intersect in CD; and parallel to CD the lines AB, EF are drawn in those planes respectively. Wherefore AB, EF are parallel (Prop. v.).

Cor. 1. If any number of lines be each of them parallel to one line, they will be parallel each to all the others.

Cor. 2. If any number of lines be parallel to one of two parallel lines, and any number parallel to the other, all the lines will be parallel, each to all the others.

For the proof may be extended to a fourth, a fifth, etc., parallel line without limitation as to number. Let the student apply it to an instance or two in detail.

PROPOSITION VII. . If two straight lines which meet be parallel to two others which also

meet, but not in the same plane with them : then (1.) The plane which contains the first two will be parallel to that

which contains the other two; and (2.) The angle contained by the first two will be equal to the angle

contained by the two others.

(1.) Let the lines AB, BC be respectively parallel to DE, EF: then the plane MN through the former pair will be parallel to the plane PQ through the latter pair.

For if not, let the plane MN cut PQ in some line, as GH. Then, since AB is parallel to a line DE in the plane PQ, it is parallel to PQ (Prop. 111.); and since the plane MÑ is drawn through the line AB, which is parallel to PQ to meet PQ in GH, GH is parallel to AB (Prop. 11.); and therefore, again, to DE (Prop. vi.).

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In the same manner BC is parallel to the plane PQ; and the plane MN cuts the plane PQ in a line GH parallel to BC, and therefore to EF. The line GH is therefore parallel to each of two straight lines DE, EF which meet; but this is impossible, and hence the planes MN, PQ cannot meet, that is, they are parallel.

(2.) The angle ABC is equal to the angle DEF. Take AB, BC equal respectively to DE, EF, and join AC, DF, AD, BE, CF.

Then since AB is equal and parallel to DE, the figure ABED is a parallelogram, and the opposite sides AD, BE are equal and parallel (Euc. 33, 1.). Similarly CF is equal and parallel to BE. Whence (Prop. O vi.) AD is equal and parallel to CF; and a therefore, again, AC to DF. The conclusion hence follows from Euc, 8, 1.

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PROPOSITION VIII. Let a plane and a point without it be given : then (1.) Innumerable straight lines can be drawn through the point,

parallel to the plane; . (2.) Only one plane can be drawn through the point parallel to the

plane; (3.) All these lines will be in one plane parallel to the given plane.

(1.) Let A be the given point, and MN the given plane: innumerable lines AC, AD, etc. can be drawn through A parallel to the plane MN.

For take any point B in MN and join AB; draw any lines BE, BF, etc., in the plane MN; draw planes through AB and BE, AB and BF, etc.; and in these planes draw AC parallel to BE, AD to BF, etc.

Then, since AC without the plane MN is parallel to the line BE in that plane, the line AC is parallel to the plane MN. In the same manner AD is parallel to MN; and so on with the innumerable lines which may be drawn in the same manner through A.

(2.) There can be but one plane drawn through A parallel to MN.

For, if possible, let two planes PQ, P'Q' both passing through A be parallel to MN; and let EF be their intersection. Through AB (taken as previously) draw any plane not coinciding with EF, and let it cut MN in BC, PQ in AD, and P'Q' in AD'. .

Then, since the planes PQ, P'Q' intersect in EF, they can have no other point common than in EF, and AD, AD' do not coincide in any other point than A.

Also since the plane ABC meets the parallel planes MN, PQ, the lines BC and AD are parallel (Prop. i.). And since it meets the parallel planes MN, P'Q' the lines BC, AD' are parallel. Whence

VOL. II.

through the point A two lines AD, AD' are drawn parallel to the same line BC in the plane ABC, which is impossible; wherefore only one plane can be drawn through A parallel to 'T MN.

(3.) All lines drawn through A parallel to the plane MN lie in the plane PQ parallel to MN.

For if it be possible, let AD' not in the plane PQ be parallel to MN. Take AB as before, and through D'A, AB, draw the plane D'ABC, meeting MN in BC and PQ in AD.

Then, since AD' is parallel to the plane MN, it is parallel to BC (Prop. II.); and since the plane D'A BC meets the parallel planes PQ, MN in AD, BC, the line AD is parallel to BC (Prop. 1.). Whence in the plane D'ABC there are drawn through the same point A two lines AD, AD' parallel to BC, which is impossible. The line AD' cannot therefore be at the same time parallel to MN and without the plane PQ.

Cor. 1. Any line drawn in one of two parallel planes is parallel to the other plane.

Cor. 2. If a straight line be parallel to a plane, a plane parallel to that plane can always be drawn through the line.

The application of the preceding reasoning to this corollary is left for the student to make.

PROPOSITION IX. (1.) If a straight line be parallel to the intersection of two planes, it

will be parallel to each of those planes ; and (2.) If a straight line be parallel to each of two planes which inter

sect, it will be parallel to their intersection.

(1.) If a straight line AB be parallel to the intersection PN of two planes MN, PQ, it will be parallel to the planes MN, PQ. AS

For since AB is parallel to the line PN in the plane MN, it is parallel to the plane MN itself (Prop. 111.). In the same way it is shown to be parallel to PQ.

(2.) If AB be parallel to each of the planes MN, PQ, it will be parallel to their intersection PN.

Let a plane pass through AB and P, a point in PN, then it may be supposed to coincide with PN or not to coincide.

In the first case, since AB is parallel to the planes MN, PQ, it can never meet either of them; nor, consequently, can it meet PN, their common section. But AB, PN being in one plane, and never meeting, they are parallel.

In the second case, let the plane ABP not coincide with PN, but cut the planes MN, PQ in some other lines PE, PF.

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