3. 11.

a

BOOK XI. pass through BA, AC; the common section of this with the given plane is a straight line passing through A: Let DAE be their common section: Therefore the straight lines AB, AC, DAE are in one plane: And because CA is at right angles to the given plane, it shall make right angles with every straight line

C

B

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

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, if there could be 6. 11. two, they would be parallelb 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 shall meet one another when produced: Let them meet; their common section shall be a straight line GH, in which take any point K, and join AK, BK: Then, because AB is perpendicular to the plane

3 Def. 11. EF, it is perpendicular to the
straight line BK which is in that
plane. Therefore ABK is a right
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,
17. 1. which is impossibleb: Therefore

the planes CD, EF, though pro-
duced, do not meet one another;

G

K

H

A

E

D

<8 Def. 11. that is, they are parallel. Therefore planes, &c. Q.E.D.

BOOK XI.

PROP. XV. THEOR.

If two straight lines meeting one another, be pa- See N.
rallel to two 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.

a

From the point B draw BG perpendicular to the plane 11. 11. which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to ED, and GK * 31. 1. parallel to EF: And because BG is perpendicular to the plane through DE, EF, it shall make right angles with

every straight line meeting it
in that plane. But the
straight lines GH, GK in B,
that plane meet it: There-
fore each of the angles BGH,
BGK is a right angle: And

because BA is parallel to A

GH (for each of them is pa

rallel to DE, and they are not

e

both in the same plane with it), the angles GBA, BGH are together equal to two right angles: And BGH is a 29. 1. 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 to the plane 4. 11. 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 parallels to one another: Therefore the plane through AB, BC 14. 11. 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.

K

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 that plane and K is a point in EFK; therefore K is in the plane AB: 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 A lines EF, GH, do not meet when produced on the side of

E

B

D

G

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 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 EX, BD are parallel. For the same reason, be- Book XI.

cause the two parallel planes
GH, KL are cut by the plane
AXFC, the common sections
AC, XF are parallel: 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 to AC, a side of the K
triangle ADC; as AX to XD
so is CF to FD: And it was
proved that AX is to XD, as
AE to EB: Therefore, as M
AE to EB, so is CF to FD.
lines, &c. Q. E. D.

Wherefore, if two straight

PROP. XVIII. THEOR.

Ir 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 a plane CK; every plane which passes through AB shall be at right angles to the plane CK.

D G

A H

K

Let any plane DE pass through AB, and let CE be the common section of the planes DE, 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 ita: And consequently it

B E

is perpendicular to CE: Wherefore ABF is a right angle; but GFB is likewise a right angle; therefore AB is parallel to FG. And AB is at right angles to the plane CK; therefore FG is also at right angles to the same 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

4 Def. 11.

BOOK XI. their common section, are also at right angles to the other planed; and any straight line FG in the plane 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 the planes which pass through AB are at right angles to the plane CK. Therefore, if a straight line, &c. Q. E. D.

a

PROP. XIX. THEOR.

Ir two planes cutting 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 is perpendicular to the third plane.

B

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 the third plane, and DE is drawn in the plane AB at right angles to AD their common section, DE is perpen4 Def. 11. dicular to the third planea. In the same manner, it 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 side of it, 13. 11. which is impossible: Therefore, from A

EF

D

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: BD therefore is perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D.