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Book XI. given plane is a straight a line passing through A: Let DAE

be their common section: Therefore the straight lines AB, a 3. 11.

AC, DAE are in one plane: And because CA is at right
angles to the given plane, it makes
right angles with every straight B

C
line meeting it in that plane. But
DAE, which is in that plane,
meets CA; therefore CAE is a.
right angle. For the same reason,
BAE is a right angle. Where-
fore, the angle CAE is equal to

D А E
the angle BAE; and they are in
one plane, which is impossible. Also, from a point above a

plane, there can be but one perpendicular to that plane; for, b 6. 11. if there could be two, they would be parallelo to one another,

which is absurd. Therefore, “from the same point,&c. Q. E. D.

PROP. XIV. THEOR.

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, EF; these planes are parallel to one another.

If not, they must meet one another when produced ; let them
meet; their common section must be
a straight line GH, in which take any

G
point K, and join AK, BK: Then,
because AB is perpendicular to the

KI
a 3. def. 11. plane EF, it is perpendicular a to the
straight line BK which is in that C

H Н plane. Therefore ABK is a right

F angle. For the same reason, BAK


is a right angle; wherefore the two
angles ABK, BAK of the triangle
ABK are equal to two right angles,

E
b 17. 1. which is impossible 6 : Therefore the
planes CD, EF, though produced,

D do not meet one another; that is, c 8. def. 11. they are parallel , Therefore,

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

A

Book XI.

PROP. XV. THEOR.

If two straight lines meeting one another, be pa. See N. rallel to two straight lines which also meet one another, but are not in the same plane with the first two; the plane which passes through the first two is parallel to the plane passing through the others.

G

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 a to the plane a 11. 11. which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallelo to ED, and GK pa

b 31. 1. rallel to EF: And because BG is perpendicular to the plane through DE, EF, it must

E make right angles with every

F straight line meeting it in B that plane c. But the straight

K c 3. def. 11. lines GH, GK in that plane meet it. Therefore each of the angles BGH, BGK is a

A

D right angle: And because BA is parallel d to GH (for

H

d 9. 11. each of them is parallel to DE, and they are not both in the same plane with it) the angles GBA, BGH are together equal e to two right angles : e 29. 1. 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 in B; GB is perpendicular f to the f 4. 11. plane through 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

g 14, 11. one another : Therefore, the plane throngh AB, BC is parallel to the plane through DE, EF. Wherefore, if two straight lines," &c. Q. E. D.

Book XI.

PROP. XVI. THEOR.

See N.

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 is parallel to GH.

For, if it is not, EF, GH must 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 that plane; and K

K
is a point in EFK: therefore
K is in the plane AB: For
the same reason, K is also in F

H
the plane CD: Wherefore

B

D the planes AB, CD produced meet one another, but they do not meet, since they are parallel, by the hypothesis : A Therefore the straight 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.

G

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 is CF 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

G

EX, BD are parallela. For the same reason, because the Book XI. two parallel planes GH, KL are cut by the plane AXFC,

a 16. 11.

H Н the common sections AC,

C XF are parallel : And because EX is parallel to BD, a side of the triangle ABD, as AE to EB, so is b AX to

b 2. 6. XD. Again, because XF is

L parallel to AC, a side of the

E

F triangle ADC, as AX to XD, K so is CF to FD: And it was proved, that AX is to XD,

N as AE to EB: Therefore ,

D

cll. 5. as AE to EB, so is CF to MLB FD. Wherefore, if two straight lines,&c. Q. E. D.

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PROP. XVIII. THEOR.

If a straight line be at right angles to a plane, every plane which passes through that line will also be at right angles to the first mentioned plane.

Let the straight line AB be at right angles to the plane CK; every plane which passes through AB will also be at right angles to the plane CK.

Let any plane DE pass through AB, and let CE be the com-
mon section of the planes, DE, CK; take any point F in CE,
from which draw FG in the
plane DE at right angles to

D G A H
CE: And, because AB is
perpendicular to the plane
CK, therefore it is also per-

к pendicular to every straight line in that plane meeting ita : And consequently it is perpendicular to CE: Wherefore ABF is a right angle: С F B E but GFB is likewise a right angle; therefore AB is parallel to FG. And AB is at right b 28. 1. angles to the plane CK: therefore FG is also at right angles to the same plane. But one plane is at right angles to c 8. 11. another plane when the straight lines drawn in one of the planes, at right angles to their common section, are also at

a 3. def.11.

Book XI. right angles to the other plane d; and any straight line FG in

the plane DE, which is at right angles to CE, the common d 4. def. 11. 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 the 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 cutting one another be each of them perpendicular to a third plane ; their common section will also 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 is perpendicular to the third plane.

If it be not, from the point D draw, in the 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 at right angles to CD, the common section of the plane BC with the third plane. And, because the plane AB is perpendicular to B the third plane, and DE is drawn in the plane AB at right angles to AD their com

mon section, DE is perpendicular to the a 4. def. 11. third plane a. In the same manner, it

E F
may be proved, that DF is perpendicular
to the third plane. Wherefore, from the
point D two straight lines stand at right

angles to the third plane, upon the same b 13. 11. side of it, which is impossible 6: There

D
fore, from the point Ď there cannot be
any straight line at right angles to the A

C
third plane except BD the common sec-
tion of the planes AB, BC. BD therefore is perpendicular
to the third plane. Wherefore, “ if two planes," &c. Q. E. D.

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