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sections EX, BD are parallela. For the same reason, be- Book XI. cause the two parallel planes

H

a 16. 11. GH, KL are cut hy the plane

C AXFC, the common sections AC, XF are parallel : And because EX is parallel to BD, a side of the triangle ABD;

12. 6. as AE to EB, so is 0 AX to XD. Again, because XF is

EH parallel to AC, a side of the KL triangle ADC; as AX to XD so is CF to FD: And it was

N proved that AX is to XD, as BL

D
AE to EB: Therefore, as M
AE to EB, so is CF to FD. Wherefore, if two straight
lines, &c. Q. E. D.

F
X

c 11. 5.

PROP. XVIII. THEOR.

Ir 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

G

H Н FG in the plane DE at right angles to CE: And because AB is perpendi

K cular to the plane CK, therefore it is also' perpendicular to every straight line in that plane meeting ita : And consequently it

C

B is perpendicular to CE: Wherefore ABF is a right angle; but GFB is likewise a right angle; therefore AB is parallel b to FG. And AB is at right angles to the plane CK; b 28. 1. therefore FG is also at right angles to the same plane. < 3. 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

• 3 Def. 11.

Book XI. their common section, are also at right angles to the other

planed; and any straight line FG in the plane DE, 4 Def. 11.

which is at right angles to CE the conimon section of the planes, has been proved 10 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. XIX. THEOR.

1

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 perpendicuJar 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 By puder because the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD their common section, DE is perpen

E F • 4 Def. 11. dicular to the third planea. In the

same manner, it may be proved, that
DF is perpendicular to the third plane.
Wherefore, from the point D two
straight lines stand at right angles to

D
the third plane, upon the same side ofit,
• 13. 11. which is impossible b: Therefore, from A

the point D there cannot be any straight line at right angles to the third plane, except BD the common section of the planes AB, BC: BD therefore is perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D.

Boox XI.

PROP. XX. THEOR.

Ir 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 he 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 thein 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 a 23. 1. to the angle DAB; and make AE equal to AD, and through E draw BEC cutting AB, AC, in D 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 the angle DAB is equal to the angle EAB: Therefore the base DB is equally to the base BE. And be- B cause BD, DC are greater than CB, and one of them BD « 20. 1. bas been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC. And because DA is equal to AE, and AC common, but the base DC greater than the base EC; therefore the angle DAC is greater d than the angle EAC; and, by the construction, 25. 1. 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.

E C 04.1.

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

Book XI. plane angles BAC, CAD, DAB. These three together are

less than four right angles.

Take in each of the straight lines AB, AC, AD any points B, C, D, and join BC, CD, DB: Then because the solid

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

therefore 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 the

D
three angles DBC, BCD, CDB are
b 32. 1. equal to two right anglesb: There-

fore the six angles CBA, ABD, BCA,
ACD, CDA, ADB are greater than
two right angles: And because the
three angles of each of the triangles

B
ABC, ACD, ADB are equal to two
right angles, therefore the nine angles of these three tri-
angles, viz. the angles CBA, BAC, 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 re-
maining 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 sections of it with those planes beBC,CD,DE,EF, FB: And because the solid angle at B is contained by three plane angles CBA, ABF, FBC, of which any two are greater a than the third, the B angles CBA, ABF, are greater than the angle FBC: for 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 A, are greater than the third angle at the same point, which is one of the angles of the polygon BCDEF : Therefore all the angles at the bases of the triangles are

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together greater than all the angles of the polygon: And Book XI. because all the angles of the triangles are together equal to twice as many right angles as there are triangles b, that is, 32, 1. as there are sides in the polygon BCDEF; and that all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygone; therefore all the angles of the trian- c 1 Cor. gles are equal to all the angles of the polygon together with 32. 1. four right angles. But all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved. Wherefore the remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at A, are less than four right angles. Therefore every solid angle, &c. Q. E. D.

PROP. XXII. THEOR.

Ir every two of three plane angles be greater than See N. the third, and if the straight lines which contain them be all equal; a triangle may be made of the straight lines that join the extremities of those equal straight lines.

Let ABC, DEF, GHK be the three plane angles, whereof every two are greater than the third, and are contained by the equal straight lines, AB, BC, DE, EF, GH, HK; if their extremities be joined by the straight lines AC, DF, GK, a triangle may be made of three straight lines equal to AC, DF, GK; that is, every two of them are together greater than the third.

If the angles at B, E, H, are equal, AC, DF, GK are also equal a, and any two of them greater than the third :* 4. 1 But if the angles are not equal, let the angle ABC be not less than either of the two at E, H; therefore the straight line AC is not less than either of the other two DF, GKb; b 4. or and it is plain that AC, together with either of the other 24. 1. two, must be greater than the third: Also DF with GK are greater than AC: For, at the point B, in the straight

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