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every straight line meeting it in that plane.
But DAE, which is in that plane, meets

B

C 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 per

D pendicular to that plane ; for if there

E could be two, they would be parallel (6. 2. Sup.) to one another, which is absurd. Therefore, from the same point, &c. Q. E. D.

PROP. XII, 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 must meet one another when produced, and their common section must be a straight line GH, in

G
which take any point K, and join AK, BK :
Then, because AB is perpendicular to the
plane EF, it is perpendicular (def. 1. 2. Sup.)
to the straight line BK which is in that plane, C
and therefore ABK is a right angle. For the

HI
same reason, BAK is a right angle; where-

E fore the two angles ABK, BAK of the trian

А
gle ABK are equal to two right angles, which
is impossible (17. 1.): Therefore the planes
CD, EF, though produced, do not meet one
another; that is, they are parallel (def. 7. 2.

E
Sup.). Therefore planes, &c. Q. E. D.

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If two straight lines meeting one another, be parallel to two straight

lines which also meet one another, but are not in the same plane with the first two: the plane which passes through the first two is parallel to the plane passing through the others.

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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 (10. 2. Sup.) to the plane which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to ED (31. 1.), and GK parallel to EF :

D

And because BG is perpendicular to the plane through DE, EF, it must make right angles with every straight line meeting it in that plane (1. def. 2. Sup.). But the straight lines В.

I GH, GK in that plane meet it: Therefore each of the

C

K angles BGH, BGK is a right angle : And because BA is parallel (8. 2. Sup.) to GH (for each of them is paral А.

H lel to DE), the angles GBA, BGH are together equal (29. 1.) 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 ; GB is perpendicular (4. 2. Sup.) 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 (12. 2. Sup.) to one another: Therefore the plane through AB, BC, is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E. D.

Cor. It follows from this demonstration, that if a straight line meet two parallel planes, and be perpendicular to one of them, it must be perpendicular to the other also.

PROP. XIV. 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 les their common sections with it be EF, GH; EF is paral

F

H н" lel to GH. For the straight lines EF

B and GH are in the same plane, viz. EFHG, which cuts the planes AB and CD; and they do not meet though produced; for the planes in which they

C are do not meet ; therefore

Α) EF and GH are parallel (def. 30. 1.). Q. E. D.

PROP. XV. THEOR.

If two parallel planes be cut by a third plane, they have the same incli

nation to that plane.

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

Let the straight lines EF and GH be the common section of the plane EH with the two planes AB and CD; and 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 : and through the straight lines KM, KN, let a plane be made to pass, cutting the plane CD in the line LO. And because EF and GH are the common sections of the plane EH with the two parallel planes AB and CD, EF is parallel to GH (14. 2. Sup.). But EF is at right angles to the plane that passes through KN and KM (4. 2. Sup.), because it is at right angles to the lines KM and KN :

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therefore GH is also at right angles to the same plane (7. 2. Sup.), and it is therefore at right angles to the lines LM, LO which it meets in that 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 EH (4. def. 2. Sup.). For the same reason the angle MKN is the inclination of the plane AB to the plane EH. But because KN and LO are parallel, being the common sections of the parallel planes AB and CD with a third plane, the interior angle NKM is equal to the exterior angle OLM (29. 1.); that is, the inclination of the plane AB to the plane EH, is equal to the inclination of the plane CD to the same plane EH. Therefore, &c. Q. E. D.

PROP. XVI. THEOR.

If two straight lines be cut by parallel planes, they must 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 AĒ is to ER, 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

H plane EBDX, the common sections

с EX, BD, are parallel (14. 2. Sup.). For the same reason, because the

G 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

I of the triangle ABD, as AE to EB,

E so is (2. 6.) AX to XD. Again, be K cause XF is parallel AC, a side of the triangle ADC, as AX to XD, so is CF te FD : And it was proved that AX is to XD, as AE to EB : Therefore (11.5.), as AE to EB, so is CF to FD. Wherefore, if two straight lines, &c.

B4 Q. E. D.

M M

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If a straight line be at right angles to a plane, every plane which passes

through that line is at right angles to the first mentioned plané. 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 : And because AB is perpendicular to the plane CK, therefore

A H it is also perpendicular to every straight line meeting it in that plane (1. def. 2. Sup.); and consequently it is perpendicular to CE:

K Wherefore ABF is a right angle ; but GFB is likewise a right angle ; therefore AB is parallel (28. 1.) to FG. And AB is at right angles to the plane CK : therefore FG is

B

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wilso at right angles to the same plane (7. 2. Sup.). 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 (def. 2. 2.); 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,

PROP. XVIII. THEOR.

If two planes cutting one another be each of them perpendicular to a third

plane, their common section is perpendicular to the same plane.

Let the two planes AB, BC be each of them perpendicular to a third plane, and BD be the common section of the first two; 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 AB and ADC; and because the plane AB is at right angles to ADC, DE is at

B right angles to the plane AB (def. 2. 2. Sup.), and therefore also to the straight line BD in that plane (def. 1. 2. Sup.). For the same reason, DF is at right angles to DB. Since BD is therefore at right angles to both the lines DE and DF, it is at right angles to the plane in which DE and DF are, that is, to the plane ADC (4. 2. Sup.). Wherefore, &c. Q. E. D.

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Fwo straight lines not in the same plane being given in position, to draw.

a straight line perpendicular to them both.

Let AB and CD be the given lines, which are not in the same plane ; it is required to draw a straight line which shall be perpendicular both to AB and CD.

In AB take any point E, and through E draw EF parallel to CD, and let EG be drawn perpendicular to the plane which passes through EB, EF (10. 2: Sup.). Through AB and EG let a plane pass, viz. GK,

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