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lines, and one of them BC is at right angles to the given

plane, the other AD is also at right angles to itt: there- 18. 11. fore a 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

[graphic]

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 sec

* 3. 11.

tion: 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 + 3 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

16 11.

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 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 EF, * 3 Def. 11. it is perpendicular to the straight line BK which is in that plane; therefore ABK is a right angle:

+ 17. 1.

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,

HK G

[graphic]

which is impossiblet: therefore the planes CD, EF, though produced, do not meet one another; that is, they are 18 Def. 11. parallelt. Therefore, planes, &c. Q. E. D.

PROPOSITION XV.

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

*

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 pa-
rallelt to ED, and B
GK parallel to EF.
And because BG
is perpendicular to
the plane through
DE, EF, it shall
make right angles
with
every straight A

line meeting it in

E

[graphic]

F+ 31. 1.

K

that planet. But

the straight lines

GH, GK in that plane meet it; therefore each of the angles
BGH, BGK is a right angle:

3 Def. 11.

And because BA is parallel || to GH, (for each of them is || 9. 11. parallel to DE, and they are not both in the same plane with it,) the angles GBA, BGH are together equal to § 29. 1. 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; GB is perpendicular to the plane through ¶ 4.11. 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 14. 11. another; therefore the plane through AB, BC is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E. D.

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 is 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:

S

* 16. 11.

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 rea-
son, K is also in the

plane CD;

[graphic]
[blocks in formation]

D

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.

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 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 pa-
rallel*:

K
For the same reason, be-
cause the two parallel planes
GH, KL are cut by the plane
AXFC, the common sections
AC, XF are parallel :

C

H

N

M

And because EX is parallel to BD, a side of the triangle ABD; as AE to EB, so ist 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:

† 2. 6.

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 ± 11. 5.

[blocks in formation]

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 a 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

[graphic]

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

H

it*; and consequently it is perpendicular to CE; where- 3 Def. 11.

fore 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‡.

+ 28. 1.

1 8. 11.

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 in the plane || 4 Def. 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.

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