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Bouk XI. Take in each of the straight lines AB, AC, AD any points B, C, WD, and join BC, CD, DB. then, because the folid angle at B is

contained by the three plane angles CEA, ABD, DBC, any two of 2. 20. 11. 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 fix angles CBA, ABD,
BCA, ACD, CDA, ADB are greater

D
than the three angles DBC, BCD, CDB.

but the three angles DBC, BCD, CDB b. 32. 1. are equal to two right angles b. thewe

före the fix angles CEA, ABD, BCA,
ACD, CDA, ADB are greater than two
right angles, and because the three angles

B
of each of the triangles 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 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 less
than four right angles.

Next, Let the folid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; these together are lefs rhan 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 be BC, CD, DE, EF, FB. and A
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

1
than the angle FBC. for the same rea-
fon, the two plane angles at each of the
points C, D, E, F, viz. the angles which

E
are at the bases of the triangles liaving
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 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 b; that is, as there are fides in the polygon BCDEF; and b. 32. J. 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 triangles are equal c. to all the angles 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 folid angle, &c. Q. E. D.

I. Cor. 32. I.

PRO P. XXII. THEO R. 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 straight lines that join the extremities of those 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 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', 2. 4. I. and any two of them greater than the third. but if the angles are

A
D

K
I
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

C. 23. 1.

Fock X!. other two DF, GK 5; and it is plain that AC together with either

of the other two must be greater than the third. also DF with GK b 4.0V 24.1. are greater than AC. for, at the point B in the straight line AB

make 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 GHK is equal to ABL, therefore the remaining

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d. 24. I.

A

F G K I angle at E is greaier than the angle LBC. 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, the base DF is greater than the bale 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

greator

than Wherefore every two of the straight lines AC, DF, GK are greater than the third, and therefore a triangle may be made i the sides of which shall be equal to AC, DF, GK. Q. E. D.

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PROP. XXIII. PROB.

Sce N.

O 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, GHIK, any two of which are greater than the third, and all of them together less than four righe angles. It is required, to make a solid angle contained by three plane angles equal to ADC, DEF, GHK, each to each.

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

H

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to AC, DF, GK. Let this be the triangle LMN 5, so that AC 6. 21. 1. be equal to LM, DF to MN, and GK to LN; and about the triangle LMN describe C 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 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 LXM 4. for the fame reason, the d. 8. :, angle DEF is equal to the angle MXN, and the angle GHK to the

R angle NXL. therefore the three an.

L 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 ;

X therefore also the three angles ABC, DEF, GHK are equal to four right

M

N angles. but, by the hypothesis, they are less than four right angles; which is absard. therefore AB is not equal to LX. but neither can AB be less than LX. for, if pollible, let it be less, and upon the straight line LM, on the side of it on which is the center 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 AC, the angle LOM

C. 1. Cor.

is. l,

f. 21.1.

N

Book XI. is equal to the angle ABC d. and AB, that is, LO, by the hypoLove thefis, is less than LX; wherefore LO, OM fall within the triand. $. i.

gle LXM; for, if they fell upon its fides, or without it, they would
be equal to, or greater than LX,
XM1, therefore the angle LOM,

R
that is, the angle ABC is greater

L
than the angle LXMf, in the same
manner it may be proved, that the
angle DEF is greater than the angle
MXN, and the angle GHK greater

X
than the angle NXL. therefore the
three angles ABC, DEF, GHK are

M
greater than three angles LXM,
MXN, 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 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 sides of
the triangle, viz. in MN, and join
XL. in this cafe also AB is greater

R
than LX. if not, AB is either equal
to LX or less than it. first, let it be

L
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

M

N construction, MN is equal to DF;

X
therefore DE, EP are equal to DF,
† 20. 1. which is impossible f. wherefore AB

is not equal to LX; nor is it less;
for then, much more, an absurdity 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 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 fame 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.

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