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* 31.1.

* 4. 11.

*

* 8. 11.

E

B

AF perpendicular to DE: AP shall be perpendicular to the plane BH.

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 G# is at right angles to the plane

F H through ED, DA; and is perpen*3 Def. 11. dicular * to every straight line meet

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 † 4. 11.

of their intersection, it is also at right angles t to the plane passing through 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.

PROP. XII. PROB.

To erect a straight line at right angles to a given plane,

from a point given in the plane.

D

* 11. 11. * 31. 1.

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

B A draw * AD parallel to BC. Because, therefore, AD, CB are two parallel straight lines, and one of them BC is at

Ć right angles to the given plane, the other AD is also * at right angles to it: therefore a straight line has been erected at right angles to a given plane, from a point given in it. Which was to be done.

* 8. 11.

PROP. XIII. THEOR.

pass

3. 11.

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 ing 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 makes right angles † with every straight line 73 Def.11. 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 * 6. 11. the same point, &c. Q. E. D.

B

с

A

È † 11 Ax.

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 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 C which takeany point K, and join AK, BK. Then, because AB is perpendicular to the plane EF, it is perpendicular* to the

E

* 3 Def. 11. 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 * 17, 1.

angles, which is* impossible: therefore the planes CD,

EF, though produced, do not meet one another; that + 8 Def. 11. is*, they are parallel. Therefore planes, &c.

Q. E. D.

* 31.1.

PROP. XV. THEOR. See N. If two straight lines meeting one another be parallel to

two other straight lines which mcet 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 pro

duced. * 11. 11. From the point B draw BG perpendicular * to the

plane which passes through DĖ, 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 * 3 Def. 11. perpendicular to the plane 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,

E.
each of the angles BGH, BGK is a
right angle: and because BA is pa-

B

K rallel* to GH (for each of them is parallel to DE, and they are not both A

in the same plane with it), the angles * 29. 1.

GBA, BGH are together 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 in B, * 4. 11.

GB is perpendicular* to the plane through BA, BC: + Constr. and it is perpendiculart 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; therefore the plane through AB, BC is parallel to the plane through DE, EF, "Wherefore, if iwo straight lines, &c. Q. E. D.

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C

* 9. 11.

H

* 14. 11.

PROP. XVI. THEOR.
If two parallel planes be cut by another plane, their com- See N.

mon sections with it are parallels.
Let the parallel planes AB, CD be cut by the plane
EFGH, 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 t + 1.11. in EFK is in that plane: and K is a point in EFK; therefore K is in the

K plane AB : for the same reason K is also in the plane CD: wherefore the

H
planes AB, CD, produced meet one
another: but they do not meet, since
they are parallel by the hypothesis ;
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.

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 EX, BD are * parallel : for the • 16 11. same reason, because the two parallel planes GH, KL, are cut by the

H plane AXFC, the common sections

ПС

G
AC, XF are parallel: and because
EX is parallel to BD, a side of the
triangle ABD; as AE to EB, so is * E

K4
AX to XD: again, because X F is pa-

ZN rallel to AC, a side of the triangle M

B! ADC; as AX to XD, so is CF to FD:

* 2 6.

* 11.5.

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

PROP. XVIII. THEOR.

D

E

* 28. 1. * 8. 11.

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, CK; take any point F in CE, from which draw FG

G A † 11.1.

in the plane DE at rightf angles to
CE: and because AB is perpendicular

K *3 Def. 11. to the plane CK, therefore* it is also

perpendicular to every straight line
in that plane meeting it; and con-

sequently it is perpendicular to CE: wherefore ABF + Constr. is a right angle: but GFB is likewiset 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 their common *4 Def. 11. section, are also at right angles * to the other plane ;

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
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 per-

pendicular 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.

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