(not all situated in the same plane).

PROPOSITION XII. Two straight lines which are each of them parallel to the same straight line, which is not in the same plane with them, are parallel to one another.

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Let AB, CF be each of them || to PQ
(AB, CF, PQ not being all in the same plane).

Then shall AB be || to CF.

In PQ take any point X, and in the plane in which are AB, PQ draw XG 1 to PQ; also in the plane in which are CF, PQ draw XH I to PQ.

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DEFINITION. Planes which never meet, however far they are produced in any direction, are called parallel planes.

PROPOSITION XIII. Planes to which the same straight line is perpendicular are parallel to one another.

Let the straight line AB be I to each of the planes CF, GH.

Then shall CF be || GH.

For if not, if possible, let the planes CF, GH be produced to meet in YX.

In YX take any point Z; join AZ, BZ. Then :: AB is 1 to the plane CF, .. . BAZ is a right 2;

SO . ABZ is a right L.
Hence in a ABZ two of the 2 s are right 2 s;
which is impossible ;

(1. 13) .. planes CF, GH will never meet however far they are produced;

.. CF, GH are parallel.


If two parallel planes are cut by another plane, their common sections with it are parallel.

Let the two parallel planes AB, CF be cut by the plane HK.

Then shall their common sections GH, KL be parallel.

For if not, if possible, let them meet when produced in X.

Then since GH is in plane AB, .. X is in plane AB;

(Prop. 1) so X is in plane CF; .. planes AB, CF will meet if produced; which is impossible, since they are parallel ; .. GH, KL will not meet if produced ;

.. GH, KL are parallel... .

DEDUCTION (A). If two straight lines meeting one another be parallel to two others that meet one another, and are not in the same plane with the first two, the first two and the other two shall contain equal angles, also the plane through the first two shall be parallel to the plane through the other two.

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Let AB, BC two straight lines meeting one another be || to FG, GH, meeting one another, the latter two straight lines not being in the same plane with the former.

Then shall LABC be = 1 FGH,

and the plane ABC be || to plane FGH. Cut off GF-BA and GH =BC; join AC,FH, AF, CH,BG. Then :: GF is rand 1l to AB, .. AF is = and || to BG,

(1. 23) so also is CH = and il to BG; (1. 23) .. AF is = and l to CH; (Prop. 12) .:. CA is = and || to HF.

(1. 23) Then ::: AB, BC, CA are respectively = FG, GH, HF; .. LABC = 1 FGH.

(1. 5) Also plane ABC shall be || to plane FGH; for if not, if possible, let them meet when produced, then their common section will be a straight line.

(Prop. 5) Since AB, BC cannot both be | to the common section one of them will meet it if produced.

Let AB produced meet it in X. Thus X is in the plane containing the parallels AB, FG and also in the plane FGH; and is .. in their common section of which FG is a part. Hence AB, FG meet if produced ; which is impossible.

plane ABC is || to plane FGH. C. G.


DEDUCTION (B). If two straight lines be cut by parallel planes, they shall be cut proportionally.


Let the two straight lines AC, FH be cut by the parallel planes X, Y, Z in A, B, C and F, G, H respectively.

Then AB : BC as FG : GH.

Join AH cutting the plane Y in the point R.

Join AF, CH, BR, RG.

Then ::: the parallel planes Y, Z are cut by the plane ACH, ... their common sections BR, CH are li to one another;

(Prop. 14) .. AB : BC as AR : RH. (VI. I) Similarly it may be shewn that

FG : GH as AR : RH; .. AB : BC as FG : GH. (v. 3)

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