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PROP. XV. THEOR.

Book II.

IF two parallel planes be cut by a third plane, they have the same inclination to that plane.

Let AB, CD be two parallel planes, and EH a third plane cutting them ; the planes AB, CD are equally inclined to EH.

Let the straight lines EF, GH be the common section of the plane EH with the two planes AB, CD; from K, any point in EF, draw in the plane EH the straight line KM at right angles to EF, and let it meet GH in L; draw also KN at right angles to EF in the plane AB; through the straight lines KM, KN let a plane be made to pass, cutting the plane CD in the line LO. Because EF, GH are the common sections of the plane EH with the two parallel planes AB, CD, EF is parallel to GH“. But EF is at right angles to the a 14.2. Sup. plane that passes through KN and KM, therefore

E
G

04. 2. Sup. GH is also at right angles to

с the same plane,

L

K and therefore is

c7. 2. Sup.

M at right angles to the lines LM, B

0 LO, which it

H in that

H Н plane. Therefore, since LM and LO are at right angles to LG, the common section of the two planes CD and EH, the angle OLM is the inclination of the plane CD to the plane EHd. For the same reason the d Def. 4. angle MKN is the inclination of the plane AB to the plane 2. Sup. Eń. But KN and LO, being the common sections of the parallel planes AB and CD. with a third plane, are parallela ; therefore the interior angle NKM is equal to the exterior angle OLM®; that is, the inclination of the plane AB to the e 29. 1.

IN D'

meets

Suppl. plane EH is equal to the inclination of the plane CD to the

same plane EH. Therefore, if two parallel planes, &c. Q. E. D.

PROP. XVI. THEOR.

IF two straight lines be cut by parallel planes they will 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, and C, F, D;
AE : EB:

:: CF:FD.
Join AC, BD, AD; 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

H sections EX, BD are paral

А.

C a 14. 2. Sup. lela; and because the two

parallel planes GH, KL are
cut by the plane AXFC, the
common sections AC, XF

L
are parallel. Because EX
is parallel to BD, a side of

E
К!

X
the triangle ABD, AE : EB
b 2. 6. :: AX : XDb; and because

XF is parallel to AC, a side
of the triangle ADC,
AX : XD :: CF : FD.

IN
Therefore AE : EB :: CF

B

D c 11. 5. : FD. Wherefore, if two

M straight lines, &c. Q. E. D.

PROP. XVII. THEOR.

Book II.

IF a straight line be at right angles to a plane, every plane which passes through the line is also at right angles to that plane.

2. Sup.

Let the straight line AB be at right angles to a plane CK; every plane which passes through AB is 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 in the plane DE at right angles to CE. Because AB is perpendicular to the plane CK, it is also perpendi

D

ΑΗ cular to every straight

a Def. 1. line meeting it in that planea ; consequently it is

K perpendicular to CE; wherefore ABF is a right angle. But GFB is a right angle ; therefore AB is parallelb to FG. But AB is

C

B E b 28. 1. at right angles to the plane

c7.2. Sup. CK; therefore FG is also at right angles to CKC. Now 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 CKd. In like 'manner it d Def. 2.

2. Sup: 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. XVIII. THEOR.

IF two planes which cut each other be perpendicu. lar to a third plane, their common section will be perpendicular to the same plane.

H

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Suppl. Let the two planes AB, BC be perpendicular to a third

plane ADC, and let BD be the common section of AB, BC; BD is perpendicular to the plane ADC.

From D in the plane ADC draw DE perpendicular to AD, and DF to DC. Because DE is perpendicular to AD, the common section of the planes

B
AB and ADC, and because the plane

AB is at right angles to ADC, DE a Def. 2. is at right angles to the plane AB, 2. Sup.

and therefore also to the straight line b Def. 1. BD in that planeb. For the same 2. Sup.

reason DF is at right angles to DB.
! Since BD is at right angles to both

the lines DE and DF, it is at right

angles to the plane in which DE and c 4. 2. Sup. DF are, that is, to the plane ADC®. Wherefore, if two planes, &c. AF

E C Q. E. D.

PROP. XIX. PROB.

TWO straight lines not in the same plane being given in position, to draw a straight line perpendicu. lar to both.

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cular to the plane which passes through EB, EF. Through Book II. AB and EG let a plane GK pass, meeting CD in H; and from H draw HK perpendicular to AB; HK is the line required.

Through H draw HG parallel to AB. Since HK, GE, which are in the same plane, are both at right angles to the straight line AB, they are parallel to each othera. And be- a 28. 1. cause the lines HG, HD are parallel to the lines EB, EF, each to each, the plane GHD is parallel to the plane BEFb; b 9. 2. Sup. therefore EG, which is perpendicular to the plane BEF, is perpendicular also to the plane GHDs. Therefore HK, which c Cor. 13.

2. Sup. is parallel to GE, is also perpendiculard to the plane GHD, and is therefore perpendiculare to HD, which is in that plane. d 7. 2.Sup. But HK is perpendicular to AB; therefore HK is drawn e Def. 1. perpendicular to the two given lines AB, CD. Which was to be done.

2. Sup.

PROP. XX. THEOR.

| IF a solid angle be contained by three plane angles, any two of them are greater than the third.

Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB; any two of them are greater than the third.

If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. But if they are not equal, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB;

D
at the point A, in the straight line
AB, make in the plane which
passes through BA, AC the angle
BAE equals to the angle DAB;
make AE equal to AD, and
through E draw BEC cutting AB, B

E C
AC in the points B, C, and join
DB, DC. Because DA is equal to AE, and AB is common
to the two triangles ABD, ABE, and also the angle DAB

a 23.1.

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