BOOK 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 be contained by the three plane angles, BAC, CAD, DAB. Any two of them are greater than the third.

a

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 ▪ 23. 1. to the angle DAB; and make AE equal to AD, and through E draw BEC 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 the angle DAB is equal to the angle EAB: Therefore the base DB is equal to the base BE. And be- B

D

E C$4.1.

cause BD, DC are greater than CB, and one of them BD < 20. 1. has 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 than the angle EAC; and, by the construction, d 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.

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, a 20. 11. ABD, DBC, any two of them are greater 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 three angles DBC, BCD, CDB are 32. 1. equal to two right anglesb: Therefore 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 ABC, ACD, ADB are equal to two

B

D

right angles, therefore the nine angles of these three triangles, 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 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.

A

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 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 ver

D

E

tex 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

32. 1.

together greater than all the angles of the polygon: And Boox XI.
because all the angles of the triangles are together equal to
twice as many right angles as there are triangles; that is,
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 polygon; therefore all the angles of the trian- 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.

a 4. I

If the angles at B, E, H, are equal, AC, DF, GK are also equal, and any two of them greater than the third 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

1

€ 23. 1.

BOOK XI. line AB, makes the angle ABL equal to the angle GHK, and make BL equal to one of the straight lines AB, BC, DE, EF, GH, HK, and join AL, LC; Then, because AB, BL, are equal to GH, HK, and the angle ABL to the angle GHK, the base AL is equal to the base GK: And because the angles at E, H, are greater than the angle ABC, of which the angle at H is equal to ABL, therefore the remaining angle at E is greater than the angle LBC:

AAA

L.

And because the two sides LB, BC are equal to the two DE, EF, and that the angle DEF is greater than the angle 24. 1. LBC, the base DF is greater than the base LC: And it has been proved that GK is equal to AL; therefore DF and GK are greater than AL and LC: But AL and LC 20. 1. are greater than AC; much more then are DF and GK greater than AC. Wherefore every two of these straight lines AC, DF, GK are greater than the third; and, theref 22.1. fore, a triangle may be made, the sides of which shall be equal to AC, DF, GK. Q. E. D.

PROP. XXIII. PROB.

See N. To make a solid angle which shall be contained by three given plane angles, any two of them being greater than the third, and all three together less than four right angles.

Let the three given plane angles be ABC, DEF, GHK, any two of which are greater than the third, and all of them together less than four right angles. It is required to make a solid angle contained by three plane angles equal to ABC, DEF, GHK, each to each.

From the straight lines containing the angles, cut off Boox XI. AB, BC, DE, EF, GH, HK, all equal to one another;

and join AC, DF, GK: Then a triangle may be made a of ▪ 22. 11.

Ал

D

G

F

H

three straight lines equal to AC, DF, GK. Let this be the triangle LMNb, so that AC be equal to LM, DF to MN, ↳ 22. 1. and GK to LN; and about the triangle LMN describe a .5. 4. circle, and find its centre X, which will either be within the triangle, or in one of its sides, or without it.

R

First, Let the centre X be within the triangle, and join LX, MX, NX: AB is greater than LX: If not, AB must either be equal to, or less than LX: first, let it be equal: Then because AB is equal to LX, and that AB is also equal to BC, and LX to XM, AB and BC are equal to LX and XM, each to each; and the base AC is, by construction, equal to the base LM; wherefore the angle ABC is equal to the angle LXMd. For the same reason, the angle DEF * 8. 1. is equal to the angle MXN, and the angle GHK to the angle NXL: Therefore the three angles ABC, DEF, GHK are equal to the three angles LXM, MXN, NXL: But the three angles LXM, MXN, NXL are equal to four right anglese; therefore also the three angles ABC, DEF, GHK are equal to four right angles: But, M by the hypothesis, they are less than four right angles, which is

N

absurd; therefore AB is not equal to LX. But neither can AB be less than LX: For, if possible, let it be less; and upon the straight line LM, on the side of it on which is the centre X, describe the triangle LOM, the sides LO, OM of which are equal to AB, BC; and because the base P

• 2 Cor. 15. 1.