PROPOSITION 14. THEOREM.

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

Let the straight line AB be perpendicular to each of the planes CD and EF: these planes shall be parallel to one another.

For, if not, they will meet one another when produced;

let them meet, then their common section will be a straight line;

let GH be this straight line; in it take any point K, and join AK, BK.

Then, because AB is perpendicular to the plane EF, [Hyp. it is perpendicular to the straight line BK which is in that plane;

[XI. Definition 3.

therefore the angle ABK is a right angle.

K

For the same reason the angle BAK is a right angle. Therefore the two angles ABK, BAK of the triangle ABK are equal to twó right angles;

which is impossible.

[I. 17.

Therefore the planes CD and EF, though produced, do

not meet one another;

that is, they are parallel.

Wherefore, planes &c. Q.E.D.

[XI. Definition 8.

PROPOSITION 15. THEOREM.

If two straight lines which meet one another, be parallel to two other straight lines which meet one another, but are not in the same plane with the first two, the plane passing through these is parallel to the plane passing through the others.

Let AB, BC, two straight lines which meet one another, be parallel to two other straight lines DE, EF, which meet one another, but are not in the same plane with AB, BC: the plane passing through AB, BC, shall be parallel to the plane passing through DE, EF.

From the point B draw BG perpendicular to the plane passing through DE, EF, [XI. 11. and let it meet that plane at G; through G draw GH parallel to ED, and GK parallel to EF. [1.31.

Then, because BG is perpendicular to the plane passing through DE, EF, [Construction.

B

it makes right angles with every straight line meeting it in that plane;

[XI. Definition 3.

but the straight lines GH and GK meet it, and are in that plane ;

therefore each of the angles BGH and BGK is a right angle.

Now because BA is parallel to ED,

and GH is parallel to ED,

[Hypothesis. [Construction.

therefore BA is parallel to GH;

[XI. 9.

therefore the angles ABG and BGH are together equal to two right angles.

[I. 29. And the angle BGH has been shewn to be a right angle; therefore the angle ABG is a right angle.

For the same reason the angle CBG is a right angle.

Then, because the straight line GB stands at right angles to the two straight lines BA, BC, at their point of intersection B,

therefore GB is perpendicular to the plane passing through BA, BC.

[XI. 4. And GB is also perpendicular to the plane passing through DE, EF.

[Construction.

But planes to which the same straight line is perpendicular are parallel to one another;

[XI. 14.

therefore the plane passing through AB, BC is parallel to the plane passing through DE, EF.

Wherefore, if two straight lines &c. Q.E.D.

PROPOSITION 16. THEOREM.

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

Let the parallel planes AB, CD be cut by the plane EFHG, and let their common sections with it be EF, GH: EF shall be parallel to GH.

For if not, EF and GH, being produced, will meet either towards F, H, or towards E, G. Let them be produced and meet towards F, H at the point K.

But they do not meet, since they are parallel by hypothesis. Therefore EF and GH, being produced, do not meet towards F, H.

In the same manner it may be shewn that they do not meet towards E, G.

But straight lines which are in the same plane, and which being produced ever so far both ways do not meet are parallel ;

therefore EF is parallel to GH.

Wherefore, if two parallel planes &c. Q.E.D.

If two straight lines be cut by parallel planes, they shall be cut in the same ratio.

Let the straight lines AB and CD be cut by the parallel planes GH, KL, MN, at the points A, È, B, and C, F, D: AE shall be to EB as CF is to FD.

Join AC, BD, AD; let AD meet the plane KL at the point X; and join EX, XF.

Then, because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are parallel; [XI. 16. and because the two parallel planes GH, KL are cut by the plane AXFC, the common sections AC, XF are parallel. [XI. 16.

And, because EX is parallel to BD, a side of the triangle ABD,

therefore AE is to EB as AX is to XD.

[VI. 2.

Again, because XF is parallel to AC, a side of the triangle

ADC,

therefore AX is to XD as CF is to FD.

[VI. 2.

And it was shewn that AX is to XD as AE is to EB;

therefore AE is to EB as CF is to FD.

[V. 11.

Wherefore, if two straight lines &c. Q.E.D.

If a straight line be at right angles to a plane, every plane which passes through it shall be at right angles to that plane.

Let the straight line AB be at right angles to the plane CK: every plane which passes through AB shall be at right angles to the plane CK.

CK,

[I. 11.

B

K

Then, because AB is at right angles to the plane

[Hypothesis.

therefore it makes right angles with every straight line meeting it in that plane;

but CB meets it, and is in that plane; therefore the angle ABF is a right angle. But the angle GFB is also a right angle ; therefore FG is parallel to AB.

[XI. Definition 3.

[Construction. [I. 28. [Hypothesis.

And AB is at right angles to the plane CK; therefore FG is also at right angles to the same plane. [XI.8.

But one plane is at right angles to another plane, when the straight lines drawn in one of the planes at right angles to their common section, are also at right angles to the other plane; [XI. Definition 4.

and it has been shewn that any straight line FG drawn in the plane DE, at right angles to CE, the common section of the planes, is at right angles to the other plane CK; therefore the plane DE is at right angles to the plane CK. In the same manner it may be shewn that any other plane which passes through AB is at right angles to the plane CK.

Wherefore, if a straight line &c. Q.E.D.

PROPOSITION 19. THEOREM.

If two planes which cut one another be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane.