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Book XI.

PROP. XVI. THEOR.

See N.

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; therefore, since EFK is in the plane AB, every point in

K
EFK is in that plane : and K
is a' point in EFK; therefore
K is in the plane AB: for
the same reason K is also in

F

H
the plane CD: wherefore the
planes AB, CD produced

B

D meet one another; but they do not meet since they are

C parallel by the hypothesis :

A therefore the straight lines

E

G 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 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 parallela. For the same reason, because the two Book XI. parallel planes GH, KL are cut by the plane AXFC, the com

C

H a 16. 11. mon sections AC, XF are paral

А lel: and because EX is parallel to BD, a side of the triangle

b 2. 6. ABD, as AE to EB so is b AX to XD. Again, because XF is parallel to AC, a side of the triangle ADC, as AX to XD,

E so is CF to FD: and it was K proved that AX is to XD as AE to EB: therefore c, as AE to EB so is CF to FD.

B

D Wherefore, if two straight lines, M &c. Q. E. D.

XF

N c 1.1. 56

PROP. XVIII. THEOR.

IP 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 planes DE, CK; take any point F in CE, from which draw FG in the plane DE at right angles to D с А н CE: and because AB is perpendicular to the plane CK, therefore it is also perpendi

K cular to every straight line in that plane meeting it a; and

a 3. def,

11. consequently it is perpendicular to CE: wherefore ABF is a right angle; but GFB is

C

F B E likewise a right angle: therefore AB is parallel b to FG. And AB is at right angles to the b 28. 1. plane CK; therefore FG is also at right angles to the same plane c. But one plane is at right angles to another plane when c 8. 11. the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles to the other

Book X1. plane d: and any straight line FG in the plane DE, which is

at right angles to CE the common section of the planes, has d 4. def. been proved to be perpendicular to the other plane CK; there. 11. fore the plane DE is at rig.t 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. XIX. THEOR.

IF two planes cutting one another be each of them perpendicular 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 is perpendicular to the third plane.

If it be not, from the point D draw, in the plane AB, the
straight line DE at right angles to AD the common section of
the plane AB with the third plane; and in the plane BC draw
Df at right angles to CD the common section of the plane BC
with the third plane. And because the
plane AB is perpendicular to the third

B
plane, and DE is drawn in the plane AB
at right angles to AD their common sec-

tion, DE is perpendicular to the third a 4. def. plane a. In the same manner it may be

E | F 11. proved that DF is perpendicular to the

third plane. Wherefore, from the point
D two straight lines stand at right angles

to the third plane, upon the same side of b 13. 11. it, which is impossible b: therefore from the point D there cannot be any straight

D
line at right angles to the third plane,
except BD the common section of the

А
planes AB, BC. BD therefore is perpendicular to the third
plane. Wherefore, if two planes, &c. Q. E. D.

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IF a solid angle be contained by three plane angles, See N. 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, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A, in the straight line AB, make, in the plane which passes through BA, AC, the angle BAE equal a to the angle DAB; and make a 23. 1. AE equal to AD, and through E draw BĒC cutting AB, AC in the points B, C, and join DB, DC. And because DA is equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, and D the angle DAB is equal to the angle EAB: therefore the base DB is equal b

b 4. 1. to the base BE. And because BD, DC are greater c than CB, and one of them

c 20. 1. BD has been proved equal to BE a part

A of CB, therefore the other DC is greater than the remaining part EC. And

B

E C because DA is equal to AE, and AC common, but the base DC greater than the base EC: therefore the angle DAC is greater a than the angle EAC; and, by the d 25. 1. construction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle BAC. But BAC is not less than either of the angles DAB, DAC: therefore BAC, with either of them, is greater than the other. Wherefore, if a solid angle, &c. Q. E. D.

PROP. XXI. THEOR.

EVERY solid angle is contained by plane angles which together are less than four right angles.

First, let the solid angle at A be contained by three plane angles, BAC, CAD, DAB. These three together are less than four right angles.

Book XI. Take in each of the straight lines AB, AC, AD any points B, v C, D, and join BC, CD, DB : then because the solid angle at B

is contained by the three plane angles CBA, ABD, DBC, any a 20. 11. two of them are greater a than the third ; ther::fore the angles

CBA, ABD are greater than the angle DBC: for the same reason, the angles BCA, ACD are greater than the angle DCB; and the angles CDA, ADB greater than BDC: wherefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles DBC, BCD, CDB : but D

the three angles DBC, BCD, CDB are b 32. 1. equal to two right angles b: therefore the six angles CBA, ABD, BCA, ACD,

Α.
CDA, ADB are greater than two right
angles : and because the three angles
of each of the triangles ABC, ACD,

B

с ADB are equal to two right angles, therefore the nine angles of these three triang.es, viz. the angles CBA, BĄC, ACB, ACD, CDA, DAC, ADB, DBA, BAD are equal to six right angles : of these the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles : therefore the remaining three angles BAC, DAC, BAD, which contain the solid angle at A, are less than four right angles.

Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; these together are less than four right angles.

Let the planes in which the angles are be cut by a plane, and
let the common section of it with those
planes be BC, CD, DE, EF, FB : and

A
because the solid angle at B is contain-
ed by three plane angles CBA, ABF,
FBC, of which any two are greater a
than the third, the angles CBA, ABF
are greater than the angle FBC: for

F

B
the same reason, the two plane angles
at each of the points C, D, E, F, viz. the
angles which are at the bases of the
triangles, having the common vertex

E
A, are greater than the third angle at
the same point, which is one of the an-

D gles of the polygon BCDEF: therefore all the angles at the bases of the triangles are together

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