Sidebilder
PDF
ePub

Then since AB is parallel to MN, it is parallel to PE, the intersection of the plane ABP with MN (Prop. II.); and, similarly, AB is parallel to PF. Whence the two lines PE, PF in the plane ABP are parallel to the same line AB, which is impossible. Whence AB is parallel to the intersection PN of the planes MN, PQ.

M

N

PROPOSITION X. If a plane be parallel to one of two parallel lines, or to one of two

parallel planes, it will be parallel to the other. (1.) Let the plane MN be parallel to AB, one of the two parallel lines AB, CD, it will be parallel to the other.

For since AB, CD are parallel, they are in one plane; and the plane ABCD will either be parallel to MN or intersect it in some line EF.

If the plane MN be parallel to ABCD, it is parallel to every line AB, CD, etc. in ABCD (Prop. viii. Cor. 1.).

Let, on the other hand, the planes ABCD and MN intersect in EF; then, since MN is parallel to AB, the line EF is parallel to AB (Prop. 11.). Also, since CD is parallel to AB (Hypoth.) and AB to EF, the line CD is parallel to a line EF in the plane MN; and hence the line CD and plane MN are parallel (Prop. 1.).

(2.) Let the plane RS be parallel to one of the parallel planes MN, PQ, as MN: it will be parallel to the other, PQ.

In the extreme planes, MN, RS take points K, G, and join KG. This will necessarily cut PQ in some point H.

Through KHG draw any two planes cutting the three planes in KA, HC, GE and KB, HD, GF, respectively.

Then, since MN is parallel to RS, the section AK is parallel to GE (Prop 1); and since MN is parallel to PQ, the section AK is parallel to CH. Whence the lines EG, CH in the plane KE are parallel to AK, and hence they are parallel to one another.

In the same manner it is shown that GF is parallel to KB.

Wherefore the lines AK, KB being parallel to EG, GF, the planes RS, MN drawn through them are parallel (Prop. vii.).

K к

S

PROPOSITION XI. If a line be parallel to a plane, and from points in the line parallel

lines be drawn to meet the plane, these will all be situated in one plane, and equal to one another.

Let AB be parallel to the plane MN, and let parallel lines CD, EF, GH, etc, be drawn from points C, E, G, etc. in AB to meet the

148

GEOMETRY OF PLANES AND SOLIDS.

E

C

B

D

N

B

plane: then CD EF = GH = etc; and all these lines will be situated in one plane.

с (1.) For, since CD, EF are parallel, Ā they are in one plane; and the points C, E being in that plane, the line EC which joins them, and its extension AB are in that plane. In like manner all

H the others, GH, etc. will be in that plane ; viz., the plane which passes through AB and any one of them CD.

(2.) Also since this plane cuts MN in a line DH parallel to AB (Prop. 11.), the figure CDFE is a parallelogram, and hence the opposite sides CD, EF are equal. Similarly all the others, GH, etc. are equal to CD.

PROPOSITION XIE If one of two parallel lines be parallel to a plane, the other is also

parallel to it. Let AB, CD be parallels, and one of A them AB be parallel to the plane PQ; then the other CD will also be parallel to PQ.

For, since AB and CD are parallel, they are in one plane. Let that plane cut PQ in EF, then EF is parallel to AB (Prop. 11.); and EF being drawn in the plane AD of the two parallels, parallel to one of them, it is parallel to the other CD. Whence CD without the plane PQ is parallel to a line EF in the plane, and is therefore parallel to the plane PQ itself (Prop. 111.).

PROPOSITION XIII. If through points in one straight line parallels to another straight line

be drawn, these parallels will all be in the same plane, and that plane will be parallel to the same line that the parallel lines are.

Through points C, D, etc. in the line CD let parallels CE, DF, etc. be drawn to the line AB: then all these parallels will be in one plane, and that plane will be parallel to AB.

For since CE, DF are both parallel to AB, they are parallel to one another (Prop. vi.); and hence they are in the same plane CF; and the points C and D in them are

B also in that plane. But C, D being in the plane CF, the line CD is in that plane; and hence the second, third, etc., parallel to CE are in the plane which contains DC, CE: that is, all the parallels to AB passing through points in CD are in the one plane DE or CF.

Also since AB is parallel to a line CE in the plane CF, it is parallel to the plane itself (Prop. 11.): that is, the plane CF is parallel to the line AB,

B В

PROPOSITION XIV. If a line be parallel to each of two planes, then every plane drawn

through this line to cut them, will cut them in parallel lines. Let the line AB be parallel to

R' each of the planes MN, PQ;

R and any plane RS be drawn through AB to cut them in PR, SN; then PR will be parallel to SN.

For since AB is parallel to both the planes PQ, MN, it is parallel to their common section MQ (Prop. ix.). Whence MQ a line without the plane RS is parallel to the line AB in it; and MQ therefore is parallel to the plane RS (Prop. 111.).

Then, again, since through the line MQ parallel to the plane RS the planes MN, PQ are drawn to cut RS, the sections NS, PR are parallel (Prop. 11.).

M M

N

P

PROPOSITION XV. Parallel lines intercepted between parallel planes are equal. Let AB, CD, EF, etc., be parallel to one another, and limited by the parallel planes MN, PQ; they will be all equal.

For, join AC and BD, CE and DF, etc. Then, since AB, CD are parallel, they are in one plane ; and the intersections AC, BD of this plane with PQ, MN are parallel (Prop. 1.). Wherefore ACDB is a parallelogram, and the opposite sides AB, CD are equal.

In the same way CD and EF are proved equal ; and zo on however many lines there be.

A

PROPOSITION XVI. If two planes which meet one another be parallel to two other planes

which meet one another, the intersection of the first two will be parallel to that of the other two. Let the planes PQ, PM which meet in AP be parallel to NS, NR which meet in NB: then AP will be parallel to NB.

For, produce the plane RN to meet the plane PQ in CQ.

Then since the parallel planes PQ, NS are cut by the plane CR, the sections BN and CQ are parallel (Prop. 1.).

Also, since the parallel planes CR, PM are cut by the plane PQ, the section AP is parallel to the section CQ.

But CQ is also parallel to NB: whence AP is parallel to NB (Prop. vi.).

S

N

M

Cor. If two planes which are parallel meet two others which are parallel, the four lines of intersection will be parallel. Thus, the four planes PQ, PM, CR, DS meet in the four parallel lines CQ, AP, DM, BN.

D

PROPOSITION XVII. If through two parallel lines any planes whatever be drawn to meet a

plane parallel to one of the lines, the sections with that plane will be parallel to each other and to the first two lines.

Let AB, CD be two parallel lines and PQ a plane parallel to one of them ; and through AB, C CD let any planes be drawn to cut PQ in A'B', C'D': these lines will be parallel.

B' For produce the planes AB', CD' to meet in EF. Then since AB, CD are parallel, the line EF is parallel to them (Prop.iv.).

Also since PQ is parallel to one of the parallel lines, AB, it is parallel to the other EF;

and hence the planes through EF (that is, through AB, CD) cut PQ in lines A'B', C'D' which are parallel to EF and hence to AB, CD.

P

E

PROPOSITION XVIII. Through a given point, only one plane can be drawn parallel to two

given lines, except those lines be parallel ; but when they are parallel, innumerable planes can be drawn parallel to both of them.

(1.) Let AB, CD be two lines not parallel (but in the same plane or not) and E be a given point : then one plane can be drawn through E parallel to both the lines AB, CD.

Through E draw a line HF parallel to AB, and a line KG parallel to CD; and through HF and KG, the plane HKFG.

Then since HF in the plane HKFG is parallel to the line AB, the plane itself is parallel to AB (Prop. 111.); and similarly, the plane HKFG is parallel to CD. Whence one plane can be drawn as enunciated.

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

(2.) There can be only one plane drawn through E parallel to both the lines AB, CD.

For if there can be a second plane let it be H'K'F'G'; through AB and E draw a plane to cut it in H'F'; and again through CD and E a plane to cut it in K'G'.

Then H'F' is parallel to AB (Prop. 11.); and likewise (by Constr.) HF is parallel to AB: whence through the same point E two lines HF, H'F' have been drawn parallel to the same line AB; which is impossible.

The same, were it necessary, might be shown by means of KG, K'G'.

One plane only, therefore, can be drawn through E parallel to AB and CD, these lines not being parallel to one another.

(3.) But let AB, CD be parallel, then innumerable planes may be drawn through E parallel to both of them.

For through E draw the line FG parallel F, to AB or CD, and it will be parallel to the other (Prop. vi.); then through FG draw any plane whatever, MN.

Then since the line AB without the plane MN is parallel to FG in the plane, the line AB is parallel to the plane MN (Prop. III.). In the

CD is parallel to the plane MN. The plane MŃ is therefore parallel to both the parallel lines AB, CD; and is any whatever subject only to the condition of passing through FG. The number of such planes, then, is unlimited, as affirmed in the proposition.

A

same way

PROPOSITION XIX.

Through two giren lines, not in the same plane, one pair of parallel planes can be drawn, one plane through one line and the other through the other : and only one such pair of parallels can be drawn.

(1.) Let AB, CD be two lines not parallel, then one pair of planes AGBH, KCLD can be drawn, the first through AB and the second through CD, which shall be parallel to one another.

For take any points E, F in AB and CD; through E draw GH

« ForrigeFortsett »