In the plane, draw any straight line BC, and from the point,

A, draw AD perpendicular to BC: if then AD be also per- 12.1. pendicular to the plane BH, the thing re

quired is already done: but if it be not,

from the point D, draw, in the plane BH,

the straight line DE at right angles to BC;

and from the point A, draw AF perpendicular to DE: AF shall be perpendicular

11. I.

to the plane BH.

*

31. 1.

4. 11.

8.11.

Through F, draw* GH parallel to BC: and because BC is at right angles to ED and DA, BC is at right angles to the plane passing through ED, DA; and GH is parallel to BC: but, if two straight lines be parallel, one of which is at right angles to a plane, the other is at right angles to the same plane; wherefore GH is at right angles to the plane through ED, DA; and is perpendicular to every straight line meet- 8 Def. 11. ing it in that plane: but AF, which is in the plane through ED, DA, meets it; therefore GH is perpendicular to AF; and consequently AF is perpendicular to GH: and AF is perpendicular to DE; therefore AF is perpendicular to each of the straight lines GH, DE. But if a straight line stand at right angles to each of two straight lines in the point of their intersection, it is also at right angles † to the plane passing through + 4.1. them but the plane passing through ED, GH, is the plane BH; therefore AF is perpendicular to the plane BH: therefore, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane. Which was to be done.

PROPOSITION XII.

PROB. To erect a straight line at right angles to a given plane, from a point given in the plane.

Let A be the point given in the plane; it is required to erect a straight line from the point A, at right angles to the plane.

From any point B above the plane, draw BC perpendicular to it; and from A, draw AD parallel to BC. Because, therefore, AD, CB are two parallel straight lines, and one of them BC is at right angles to the given plane,

D B

11. 11.

31. 1.

the other AD is also at right angles to it: therefore, a *8. 11.

straight line has been erected at right angles to a given plane, from a point given in it. Which was to be done.

PROPOSITION XIII.

THEOR. From the same point in a given plane, there cannot be two straight lines at right angles to the plane, upon the same side of it and there can be but one perpendicular to a plane from a point above the plane.

For, if it be possible, let the two straight lines AB, AC be at right angles to a given plane, from the same point A in the plane, and upon the same side of it. Let a plane 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 † 3 Def. 11. given plane, it makes right angles + with

3. 11.

† 11 Ax.

6. 11.

B

D A

E

every straight 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; wherefore the angle CAE is equal to 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 two, they would be parallel * to one another, which is absurd. Therefore, from the same point, &c. Q. E. D.

PROPOSITION 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 shall be parallel to one another.

If not, they shall meet one another when produced: let them meet; their common section is a straight line GH, in which take any point K, and join AK, BK. Then, because AB is perpendicular to the plane EF, it is perpen

* 3 Def. 11. dicular to the straight line BK which is in

C

G

H

F

E

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; which is impossible: therefore the planes CD, EF, though produced, 17. 1. do not meet one another; that is, they are parallel. There- 8 Def. 11 fore, planes, &c. Q. E. D.

PROPOSITION XV.

THEOR.-If two straight lines meeting one another, be parallel See N.
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.

* 31. 1.

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 parallel to EF. And because BG is perpendicular to the plane through B DE, EF, it makes* right angles with every straight line meeting it in that plane: but the straight lines GH, GK in that plane

A

*

H

F

K

* 8 Def. 11.

meet it; therefore each of the angles BGH, BGK is a right angle: and because BA is parallel * to GH, (for each of them 9. 11. is parallel to DE, and they are not both in the same plane with it,) the angles GBA, BGH are together equal to two right 29. 1. 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 in B, GB is perpendicular to the 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 one another; 14. 11. therefore the plane through AB, BC is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c.

Q. E. D.

*

4. 11.

+ Constr.

See N.

† 1. 11.

PROPOSITION 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 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 hypo

K

H

B

thesis; 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.

16. 11.

PROPOSITION 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 com-
mon sections EX, BD are* parallel: for
the same reason, because the two parallel

G

E

K

F

N

B

M4

planes GH, KL are cut by the plane 4XFC, the common sections AC, XF are parallel: and because EX is parallel to BD,

* 2.6.

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 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 AE to EB, so is CF to 11.5. FD. Wherefore, if two straight lines, &c. Q. E. D.

PROPOSITION 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.

*

† 11. 1.

*

Def. 11.

K

• 28. 1. *8. 11.

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 it, and consequently it 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 † Constr. 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 their common section, are also at right angles* to the other plane; and any straight line FG ⚫ 4 Def. 11. 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 planes which pass through AB, are at right angles to the plane CK. Therefore, if a straight line, &c. Q. E. D.

PROPOSITION 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.