Book XI.

32. 1.

a

D

Take in each of the ftraight lines AB, AC, AD any points B, C, D, and join BC, CD, DB. then, becaufe the folid angle at B is contained by the three plane angles CBA, ABD, DBC, any two of a. 20. 11. them are greater than the third; therefore the angles CBA, ABD are greater than the angle DBC. for the fame reafon, the angles BCA, ACD are greater than the angle DCB; and the angles CDA, ADB greater than BDC. wherefore the fix angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles DBC, BCD, CDB. but the three angles DBC, BCD, CDB are equal to two right angles b. therefore the fix 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 right angles, therefore the nine angles of thefe three triangles, viz. the angles CBA, BAC, ACB, ACD, CDA, DAC, ADB, DBA, BAD, are equal to fix right angles. of these the fix angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles. therefore the remaining three angles BAC, DAC, BAD which contain the folid angle at A, are lefs than four right angles.

B

Next, Let the folid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; thefe together are lefs than four right angles.

Let the planes in which the angles are be cut by a plane, and let the common fections of it with thofe

a

planes be BC, CD, DE, EF, FB. and
because the folid angle at B is contained
by three plane angles CBA, ABF, FBC,
of which any two are greater than the
third, the angles CBA, ABF are greater
than the angle FBC. for the fame rea-
fon, the two plane angles at each of the
points C, D, E, F, viz. the angles which
are at the bafes of the triangles having
the common vertex A, are greater than
the third angle at the fame point, which

A

E

D

is one of the angles of the polygon BCDEF. therefore all the angles at the bafes of the triangles are together greater than all the

angles of the polygon. and because all the angles of the triangles Book XI. are together equal to twice as many right angles as there are triangles; that is, as there are fides in the polygon BCDEF; and b. 32. 1. that all the angles of the polygon together with four right angles are likewife equal to twice as many right angles as there are fides

32. I.

in the polygon; therefore all the angles of the triangles are equal c. 1. Cor. to all the angles of the polygon together with four right angles. But all the angles at the bafes of the triangles are greater than all the angles of the polygon, as has been proved, wherefore the remaining angles of the triangles, viz. thofe at the vertex, which contain the folid angle at A, are lefs than four right angles. Therefore every folid angle, &c. Q. E. D.

PRO P. XXII. THEOR.

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

Let ABC, DEF, GHK be three plane angles, whereof every two are greater than the third, and are contained by the equal ftraight 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 ftraight 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 alfo equal', 2. 4. 1. and any two of them greater than the third. but if the angles are

not all equal, let the angle ABC be not lefs than either of the two at E, H; therefore the ftraight line AC is not lefs than either of the

C. 23. 1.

Fock XI. other two DF, GK; and it is plain that AC together with either of the other two must be greater than the third. alfo DF with GK b 4.0r 14.1. are greater than AC. for, at the point B in the ftraight line AB make the angle ABL equal to the angle GHK, and make BL equal to one of the ftraight lines AB, BC, DE, EF, GH, HK, and join AL, LC. then becaufe AB, BL are equal to GH, HK, and the angle ABL to the angle GHK, the bafe AL is equal to the bafe GK. and because the angles at E, H are greater than the angle ABC, of which GHK is equal to ABL, therefore the remaining

d. 24. I.

e. 20. I.

f. 22. I.

See N.

d

angle at E is greater than the angle LBC. and becaufe the two fides LB, BC are equal to the two DE, EF, and that the angle DEF is greater than the angle LBC, the bafe DF is greater than the bafe 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 are greater than AC; much more then are DF and GK greater than AC. Wherefore every two of the ftraight lines AC, DF, GK are greater than the third, and therefore a triangle may be made f the fides of which shall be equal to AC, DF, GK. Q. E. D.

c

PROP. XXIII. PRO B.

To make a folid angle which shall be contained by
Ο
three given plane angles, any two of them being
greater than the third, and all three together lefs 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 lefs than four right angles. It is required, to make a folid angle contained by three plane angles equal to ABC, DEF, GHK, each to each.

209

From the ftraight lines containing the angles, cut off AB, BC, Book XI.
DE, EF, GH, HK all equal to one another; and join AC, DF,
GK. then a triangle may be made of three straight lines equal

a. 22. IA.

A

D

E

៥.
F

K

to AC, DF, GK.

Let this be the triangle LMN b, fo that AC b. 21. 1.

be equal to LM, DF to MN, and GK to LN; and about the tri-
angle LMN defcribe a circle, and find its center X, which will c. 5. 4.
either be within the triangle, or in one of its fides, or without it.

First, Let the center X be within the triangle, and join LX, MX, NX. AB is greater than LX. if not, AB muft either be equal to, or less than LX; firft, 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 bafe AC is, by conftruction, equal to the bafe LM; wherefore the angle ABC is equal to the angle LXM. for the fame reason, the d. 8. 1. angle DEF is equal to the angle MXN, and the angle GHK to the angle NXL. therefore the three an gles ABC, DEF, GHK are equal to the three angles LXM, MXN, NXL. but the three angles LXM, MXN, NXL are equal to four right angles"; therefore alfo the three angles ABC, DEF, GHK are equal to four right angles. but, by the hypothefis, they are lefs than four right angles; which

M

R

L

X

c. 1. Cor. 15.1.

N

is abfurd. therefore AB is not equal to LX. but neither can AB be lefs than LX. for, if poffible, let it be lefs, and upon the straight line LM, on the fide of it on which is the center X, describe the triangle LOM, the fides LO, OM of which are equal to AB, BC; and because the bafe LM is equal to the bafe AC, the angle LOM

d. S. I.

f. 27. r.

L

R

Book XI. is equal to the angle ABC. and AB, that is, LO, by the hypothefis, is lefs than LX; wherefore LO, OM fall within the triangle LXM; for, if they fell upon its fides, or without it, they would be equal to, or greater than LX, XM . therefore the angle LOM, that is, the angle ABC is greater than the angle LXM f. in the fame manner it may be proved, that the angle DEF is greater than the angle MXN, and the angle GHK greater than the angle NXL. therefore the three angles ABC, DEF, GHK are greater than three angles LXM, MXN,NXL; that is, than four right

M

X

N

angles. but the fame angles ABC, DEF, GHK are lefs than four right angles; which is abfurd. therefore AB is not lefs than LX, and it has been proved that it is not equal to LX; wherefore AB is greater than LX.

Next, Let the center X of the circle fall in one of the fides of

the triangle, viz. in MN, and join
XL. in this cafe alfo AB is greater

than LX. if not, AB is either equal

R

L

† 20. 1.

to LX or lefs than it. firft, let it be
equal to LX. therefore AB and BC,
that is, DE and EF, are equal to MX
and XL, that is to MN. but, by the
conftruction, MN is equal to DF;
therefore DE, EF are equal to DF,
which is impoffible +. wherefore AB
is not equal to LX; nor is it lefs;

for then, much more, an abfurdity would follow. therefore AB is greater than LX.

But let the center X of the circle fall without the triangle LMN, and join LX, MX, NX. In this cafe likewife AB is greater than LX. if not, it is either equal to, or lefs than LX. first, let it be equal; it may be proved, in the fame manner, as in the first cafe, that the angle ABC is equal to the angle MXL, and GHK to LXN; therefore the whole angle MXN is equal to the two angles ABC, GHK. but ABC and GHK are together greater than the angle DEF; therefore alfo the angle MXN is greater than DEF.