*11 11.

11.

If two straight lines meeting 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 which passes through these is parallel to the plane passing through the others.

Let AB, BC, two straight lines meeting one another, be parallel to DE, EF that meet one another, but are not in the same plane with AB, BC: the planes through AB, BC, and DE, EF shall not meet, though produced.

From the point B draw BG perpendicular to the plane which passes through DE, EF, and let it meet that plane in G; and through G draw 31.1. GH parallel to ED, and GK parallel to EF.

And because BG is perpendicular to the plane *3 Def. through DE, EF, it makes * right angles with every straight line meeting it in that plane: but the straight lines GH, GK in that plane meet it; therefore each of the angles BGH, BGK is a right angle: and

B

9.11. because BA is parallel to GH (for each of them is pa- A rallel to DE, and they are not both in the same plane

E

H

K

with it), the angles GBA, BGH are together *29. 1. equal to two right angles: and BGH is a right angle; therefore also GBA is a right angle, and GB perpendicular to BA. For the same reason, GB is perpendicular to BC. Since therefore the straight line GB stands at right angles to the two straight lines BA, BC that cut one another *4. 11. in B, GB is perpendicular to the plane through +Const, BA, BC: and it is perpendicular to the plane through DE, EF; therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: but planes to which the same straight line

*

*

is perpendicular, are parallel to one another; *14. 1L therefore the plane through AB, BC is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E. D.

PROP. XVI. THEOR.

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

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 it is not, EF, GH shall meet if produced either on the side of FH, or EG. First, let them be produced on the side of FH, and meet in the point K: therefore, since EFK is in the plane AB, every point † in EFK is in +1. 11. that plane and K is a point in EFK; therefore K is in the planeAB: for the same reason, K is also in the plane CD: wherefore the planes AB, CD produced, meet one another: but they do not meet, since they are parallel by the hypothesis; therefore the straight

D

lines EF, GH, do not meet when produced on the side of FH. In the same manner it may be proved, that EF, GH do not meet when produced on the side of EG. But straight lines which are in the same plane, and do not meet, though produced either way, are parallel; therefore EF is parallel to GH. Wherefore, if two parallel planes, &c. Q. E. D.

PROP. XVII. THEOR.

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

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

Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, the common sections * 16. 11. EX, BD are* parallel: for the same reason, be

cause the two parallel planes

GH, KL are cut by the

plane AXFC, the common

:

*2.6.

sections AC, XF are paral-
lel and because EX is
parallel to BD, a side of the
triangle ABD; as AE to
EB, so is* AX to XD:
again, because XF is parallel MA
to AC, a side of the triangle

ADC; as AX to XD, so is CF to FD: and it was proved that AX is to XD, as AE to EB; *11.5. therefore, * as AE to EB, so is CF to FD. Wherefore, if two straight lines, &c. Q. E. D.

PROP. XVIII. THEOR.

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.

Let any plane DE pass through AB, and let
CE be the common section of the planes DE,

to

CK; take any point F in CE, from which draw
FG in the plane DE at
right angles to CE: and
because AB is perpendicular
to the plane CK, therefore*
it is also perpendicular to
every straight line in that
plane meeting it; and con-

DGA H

+11. 1.

*3 Def.

11.

C

F B E

*

sequently it is perpendicular to CE: wherefore ABF is a right angle: but GFB is likewise + a + Const. right angle; therefore AB is parallel to FG: *28/1. and AB is at right angles to the plane CK; therefore FG is also at right angles to the same *8. 11. plane. 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 *4 DeL plane; and any straight line FG in the plane 11. DE, which is at right angles to CE, the common section of the planes, has been proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all planes which pass through AB are at right angles to the plane CK. Therefore, if a straight line, &c. Q. E. D.

PROP. XIX. THEOR.

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.

Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common section of the first two: BD shall be perpendicular to the third plane.

If it be not, from the point D+ draw, in the +11.1. plane AB, the straight line DE at right angles to AD the common section of the plane AB with the third plane; and in the plane BC draw DF

B

at right angles to CD the common section of the plane BC with the third plane. And because the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD, their common *4 Def. section, DE is perpendicular * to the third plane. In the same manner, it may be proved, that DF is perpendicular to the third plane.

11.

*

Wherefore, from the point D two straight lines stand at right angles to the third plane, upon *13. 11. the same side of it, which is impossible: therefore, from the point D there cannot be any straight line at right angles to the third plane, except BD the common section of the planes AB, BC: therefore BD is perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D.

1

PROP. XX. THEOR.

If a solid angle be contained by three plane angles,
any two of them are greater than the third.

Let the solid angle at A be contained by the
three plane angles BAC, CAD, DAB ; any two
of them shall be greater than the third.

If the angles BAC, CAD, DAB be all equal, it is evident, that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A in the straight line AB make, in the plane which passes through BA, 23. 1. AC, the angle BAE equal to the angle DAB; and make AE equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and join DB, DC. And because DA is equal to

*

AE, and AB is common, the two

DA, AB are equal to the two
EA, AB, each to each; and the B

D

E