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BC, DE, EF. But those planes to which the same right line is perpendicular, (by 14. 11.) are parallel : Therefore the planes paffing thro' A B, BC; DE, EP are par. allel.

Therefore if two right lines touching one another be parallel to two right lines touching one another, but not in the same plane; those planes which pass thro' them will be parallel. Which was to be demonstrated.

PRO P. XVI. THEOR. If any plane cuts two parallel planes, their common

seations are also parallel. For let some plane E F G H cut the two parallel planes A B, CD, and let their common sections be E F, Gh: I say E F is parallel to GH.

For if they be not parallel, EF, GH being produced will either meet towards F, H, or towards E, G. Let them first be produced and meet towards F, H, viz, at K: Then because E Fk is in the plane A B, all the points taken in EFK will be in the same plane. But K is one of the points in E FK: Therefore k is in the plane AB: by the same reason K is also in the plane cd: Therefore K

the planes A B, CD being produced will meet. But they are supposed not to meet, because they are par

allel: Therefore the right lines F

H

D

E F, G H being produced will not B

meet towards F, H. In like manner we demonstrate that the right

lines E F, G H being produced, will A

not meet towards E, G. But those E

lines which being both ways pro

duced, do not meet, are [by 35. def. 1.] parallel : Therefore e F, G H are parallel.

Therefore if any plane cuts two parallel planes; their common sections will be parallel. Which was to be de. monstrated.

PROP.

E

PRO P. XVII. THEOR. If two right lines be cut by parallel planes, they will

be cut in the fame ratio. For let two right lines A B, CD, be cut [by the parallel planes GH, KL, MN in the points A, E, B, C, F, D: I fay, as the right line AE is to E B, so is C F to FD.

For join A C, BD, AD, let AD meet the plane k l in the point x, and join ex, XF.

Then because the two parallel planes KL, MN are cut by the plane e BDX; (by 16. 11.) their common sections EX, BD are parallel. By the

K same reason be

M cause the two

A parallel planes GH, KL are

B cut by the plane A XFC, their common secti. ons AC, FX are

H
L

N parallel : But because Ex is drawn parallel to one side BD of the trianglę A B D, [by 2. 6.] as A E is to EB, so will ax be to XD. Again, because x F is drawn parallel to one side AC of the triangle ADC; it will be as Ax to X D, so is cf to Ft. But it has been proved that as a x is to X D, so is A E to EB: Therefore [by 11. 5.] as A e is to E B, so is CF to FD.

Therefore if two right lines be cut by Parallel planes, they will be cut in the same ratio. Which was to be demonstrated.

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F

PROP. XVIII. THEOR. If a right line be at right angles to any plane, all the

planes that pass thro' it will be at right angles to that plane.

For let the right line A B be at right angles to some given plane: I say all planes that pass thro' A B will be at right angles to that given plane,

For

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For let the plane de pass thro’ the right line A B, and let the right line ce be the common fection of the plane DE and the given plane : Take any point F in c E, and

from the same draw Fg in the plane G A H

De at right angles to CE. Then because the right line A B is perpendicular to the given plane, it will also [ by 3. def. 11.] be perpendicular to all right lines lying in the giv

en plane and touching it: Wherefore E

it is also perpendicular to.CE; and F B

so A BF is a right angle. But GFB

is also a right angle: Therefore [by 28. 1.] A B is parallel to FG. But A B is at right angles to the given plane: Wherefore [by 8. 11.] FG also will be at right angles to that fame plane. But [by 4. def. 11.) one plane is at right angles to another plane, when right lines drawn at right angles to their common section in one plane, are at right angles to the other plane: Therefore the right line F G drawn in one plane D E at right angles to the common section ce is proved to be at right angles to the given plane. Consequently the plane D E is at right angles to the given plane. After the same manner we demonstrate that every plane passing thro' A B is perpendicular to the given plane.

If therefore a right line be at right angles to any given plane, all the planes that pass throʻ it will be also at right angles to that given plane. Which was to be demonstrated.

PROP. XIX. THEOR. If two planes cutting one another be at right angles to

fome plane ; their common seation will be at right angles to the fame plane.

For let two plane's A B, BC, cutting one another, be at right angles to some given plane; and let their common fection be BD: I say the right line BD is at right angles to that given plane.

For if it be not, [by 11. 1.] draw the right line D E from the point p in the plane A B at right angles to the right line A D, and in the plane BC draw D F at right an

gles

gles to cd. Then because the plane AB is at right angles to the given plane, and D E is drawn in the plane A B at right angles to their common section A D; DE will be perpendicular to the gi

B ven plane. After the same manner we demonstrate, that D F is also perpendicular to the given plane :

:

F Therefore there are two right lines drawn, on the

A

D same fide, from the same point D at right angles to the given plane, which [by 13.

o 11.). is impossible. Therefore no right line but DB, the common section of the planes A'B, CD, can be drawn from the point D at right angles to the given plane.

Therefore if two planes cutting one another, be at right angles to some plane; their common section will also

be at right angles to that same pláne. Which was to be demonstrated.

PRO P. XX. - THEOR. If a solid angle be contained under three plane angles,

any two of them taken together, are greater than the third.

For let a solid angle at A be contained under the three plane angles B AC, CAD, DAB: I fay any two of these angles BAC, CAD, Dae, taken together, are greater than the third.

For if the angles B'A C, CAD, D A B be equal to one another, it is manifest that any two of them, taken together, are greater than the third. If not, let B ac be the greatest of the three angles; and at the right line A B, and at the point a in it [by 23.1.) make the angle: BA E in the plane passing thro' BAC equal to

D the angle DAB; and (by 3. 1.] make A E equal to AD; also let the right line B E C drawn thro' E cut the right lines A B, A c in the points B, C, and join D B D C.

В

Then

A

Then because DA is equal to A E, and A B is common,, the cwo sides DA, A B arc equal to the two sides A E, A B, and the angle DA B is equal to the angle BAE: Therefore [by 4. 1. ] the base DB is equal to the base B E. And because the two fides DB, DC are greater than BC, but D B has been proved to be equal to BE; the remainder DC. will be greater than the remainder E c. And because DA is equal to A E, and a c is common, and the base Dc is. greater than the base Ec; (by 25. 1.) the angle DAC will be greater than the angle E AC. But the angle DAB has been proved to be equal to the angle B A E; wherefore the angles DAB, DAC, taken together, are greater than the angle B A C. Also after the same manner we demonstrate, if any two other angles be taken, they are both together greater than the third angle remaining.

Therefore if a solid angle be contained under three plane angles, any two of them, taken together, are greater than the third. Which was to be demonstrated.

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PRO P. XXI. THEO R.
Every solid angle is contained under plane angles that

(taken together) are less than four right angles.

Let the solid angle at a be contained under the plane angles B A C, CAD, DAB: I say the angles B AC, CAD, DAB, taken together, are less than four right angles.

For in each of the right lines A B, AC, AD, take any points B, C, D, and join BC, CD, DB. Then because the

solid angle at B is contained under the
three plane angles CBA, ABD, CBD;
any two of them, taken together [by
20. II.) are greater than the third:
Therefore the angles CBA, A B D are
greater than the angle CBD. By the
same reason the angles BCA, A CD,

taken together, are greater than the angle BCD; also the angles CD A, ADB are greater than the angle cDB: Wherefore the fix angles C B A, A B D, BCA, A CD, A DC, A D B are greater than the three angles CBD, BCD, CDB. But [by 32. 1.] the three angles CBD, BCD, CD B are equal to two right angles : Therefore the fix angles CBA, ABD, BCA, ACD, ADC, A D B are greater than two sight angles. But because the three

angles

A

B

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