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Book XI. Take in each of the straight lines AB, AC, AD any points

B, C, D, and join BC, CD, DB: Then, because the folid

angle at B is contained by the three plain angles CBA, ABD, * 20. II. DBC, any two of them are greater than the third ; there

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 fix angles CBA, ABD, BCA, ACD,
CDA, ADB are greater than the
three angles DBC, BCD, CDB: But

D
the three angles DBC, BCD, CDB
b 32. I. are equal to two right angleso: There.

fore the fix angles CBA, ABD, BCA, A
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 triangles,
viz. the angles CBA, BAC, ACB, ACD, CDA, DAC, ADB,
DBA, BAD are equal to fix right angles : Of these the fix an-
gles 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 toge-
ther are less 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 those
planes be BC, CD, DE, EF, FB: And A
because the sold angle at B is contain-
ed 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 B

F
the fame 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 C
vertex A, are greater than the third
angle at the same point, which is one

D
of the angles of the polygon BCDEF;
Therefore all the angles at the bases of the triangles are toge-

thes

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E

ther greater than all the angles of the polygon : And because Book XI. all the angles of the triangles are together equal to twice as many right angles as there are triangles; that is, as there are b 32. I. fides 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 fides in the polygon; c I. Cor. therefore all the angles of the triangles are equal to all the an- 34• 1. gles of the polygon together with 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.

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Fevery two of three plain 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 straight lines that join the extremities of those equal straight lines.

Let ABC, DEF, GHK be three plane angles, whereof e. very 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", and any two of them greater than the third : But * 4. to if the angles are not all 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, GKD; and b 4.0r 24550 it is plain that AC, together with either of the other two, must be greater than the third : Also DF with GK are greater than AC: For, at the point B in the straight line AB make the c 33, 1.

angle

O4

Book XI. 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:

E

H
B

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D

F G K
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 LBC, d 24. 1. the base DF is greater than the base LC: And it bas been

proved that GK is equal to AL, therefore DF and GK are c 20. 1. greater than AL and LC: But AL and LC are greater than

AC; much more than are DF and GK greater than AC. Where

fore every two of these straight lines AC, DF, GK are greater f 22. J. than the third ; and, therefore, a triangle may be made, the

fides of which shall be equal to AC, DF, GK. Q. E. D.

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See N.

O make a folid 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 wbich are greater than the third, and all of them together lets 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

From the straight lines containing the angles, cut off AB, Book XI, BC, DE, EF, GH, HK all equal to one another ; and join AC, DF, GK: 'Then a triangle may be made of three straight a 22. II.

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A
С

G

K

F lines equal to AC, DF, GK. Let this be the triangle LMN, . 22. R so that AC be equal to LM, DF to MN, and GK to LN; and about the triangle LMN describe a circle, and find its centre X, which will either be within the triangle, or in one of its fides, or without it.

First, Let the centre X be within the triangle, and join LX, MX, NX: AB is greater than I.X : 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 is equal to the d 8. I. angle MXN, and the angle GHK to the angle NXL: Therefore the

R 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 angles; therefore also

e 2. Cor. the three angles ABC, DEF, GHK

x
are equal to four right angles : But,
by the hypothesis, they are less

N
than four right angles; which is
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 LM is equal to the

base

L

15. I.

M

1

d 8. I.

2

f 21. I.

T

Book XI. base AC, the angle LOM is equal to the angle ABC And
WAB, that is, LO, by the hypothesis, is less than LX; where-

fore 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

R
LX, XMF: Therefore the angle
LOM, that is, the angle ABC, is

L
greater than the angle LXM In
the same manner it may be proved
that the angle DEF is greater than
the angle MXN, and the angle
GHK greater than the angle NXL:

Х
Therefore the three angles ABC,
DEF, GHK are greater than the M

N
three angles LXM, MXF, NXL ;
that is, than four right angles : But
the same angles ABC, DEF, GHK are less than four right
angles; which is absurd : Therefore AB is not less than LS,
and it has been proved that it is not equal to LX ; wherefore
AB is greater than LX.

Next, Let the centre X of the circle fall in one of the fides of the triangle, viz. in MN, and

R
join XL: In this case also AB is
greater than LX. If not, AB is
either equal to LX, or less than it :

L
First, let it be equal to LX: There-
fore AB and BC, that is, DE and
EF, are equal to MX and XL, that
is, to MN: But, by the construction,

N
MN is equal to DF; therefore DE, M

X
EF are equal to DF, which is im
† 20. I.

possible t: Wherefore AB is not e.
qual to LX; nor is it less; for then,
much more, an absurdity would
follow : Therefore AB is greater than LX.

But, let the centre X of the circle fall within the triangle LMN, and join LX, MX, NX. In this case likewise AB is greater than LX: If not, it is either equal to, or less than LX: First, let it be equal ; it may be proved in the same manner, as in the first case, 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 also the angle MXN is greater than DEF. And because DE,

EF

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