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Book XI. to the base AC, the angle LOM is equal to the angle ABC1:

And AB, that is, LO, by the hypothesis, is less than LX; d 8. 1.

wherefore LO, OM, fall within the triangle LXM; for if they
fell upon its sides, or without it,
they would be equal to, or greater

R
f 21. 1. than LX, XM?: Therefore the
angle LOM, that is, the angle

L
ABC, is greater than the angle
LXMf: In the same manner, it
may be proved, that the angle
DEF is greater than the angle

x
MXN, and the angle GHK great-
er than the angle NXL. There- M

N
fore the three angles ABC, DEF,
GHK, are greater than the 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 less than LX, 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 sides of the triangle, MN, and viz. in

R
join XL: In this case also, AB is
greater than LX. If not, AB is

L
either equal to LX, or less than
it: First, 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 con- M

X

N struction, MN is equal to DF;

therefore DE, EF are equal to DF, + 20. 1. which is impossible t: Wherefore

AB is not equal to LX; nor is it
less; for then, much more, an absurdity would follow : There-
fore AB is greater than LX.

But, let the centre 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 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 are equal to MX, XN, and the base DF to the base Book XI. MN, the angle MXN is equal to the angle DEF: And it has a been proved, that it is greater than DĚF, which is absurd. d 8. 1. Therefore AB is not equal to LX. Nor yet is it less; for then, as has been proved in the first case, the angle ABC is greater than the angle MXL, and the angle GHK greater than the angle LXN. At the point B, in the straight line CB, make the angle CBP equal to the angle GHK, and make BP equal to

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HK, and join CP, AP. And because CB is equal to GH; CB, BP are equal to GH, HK, each to each, and they contain equal angles; wherefore the base CP is equal to the base GK, that is, to LN. And in the isosceles triangles ABC, MXL, because the angle ABC is greater than the angle MXL, therefore the angle MLX at the base is greaters than the angle g 32. 1. ACB at the base. For the same reason, because the angle GHK, or CBP, is greater than the angle LXN, the angle XLN

R is greater than the angle BCP. Therefore the whole angle MLN is greater than the whole angle ACP. And because ML, LN are equal to AC, CP, each to each, but the angle MLN is greater than the angle ACP, the base MN is

M greater b than the base AP.

N

h 24. 1. MN is equal to DF; therefore

X also DF is greater than AP. Again, because DE, EF are equal to AB, BP, but the base DF greater than the base AP, the angle DEF is greater than the angle ABP.

k 25. 1. And ABP is equal to the two angles ABC, CBP, that is, to the two angles ABC, GHK; therefore the angle DEF is

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L

Book XI. greater than the two angles ABC, GHK; but it is also less

than these, which is impossible. Therefore AB is not less than LX; and it has been proved, that it is not equal to it; there

fore AB is greater than LX. a 12. 11. From the point X erect a XR at right angles to the plane

of the circle LMN. And because it has been proved in all
the cases that AB is greater than LX, find a square equal to
the excess of the square of AB
above the square of XL, and make

R
RX equal to its side, and join RL,
RM, RN. Because RX is per-

pendicular to the plane of the circle b 3. def. 11. LMN, it is perpendicular to each

of the straight lines LX, MX,
NX. And because LX is equal
to MX, and XR common, and at
right angles to each of them, the

M

N base RL is equal to the base RM.

X For the same reason, RN is equal to each of the two RL, RM. Therefore the three straight lines RL, RM, RN, are all equal. And because the square of XR is equal to the excess of the square of AB above the square of LX;

therefore the square of AB is equal to the squares of LX, c 47. 1. XR. But the square of RL is equal to the same squares,

because LXR is a right angle. Therefore the square of AB is equal to the square of RL, and the straight line AB to RL. But each of the straight lines BC, DE, EF, GH, HK is equal to AB, and each of the two RM, RN is equal to RL. Wherefore AB, BC, DE, EF, GH, HK, are each of them equal to each of the straight lines RL, RM, RN.

And because RL, RM are equal to AB, BC, and the base d 8. 1. LM to the base AC; the angle LRM is equal d to the angle

ABC. For the same reason, the angle MRN is equal to the angle DEF, and NRL to GHK. Therefore, there is made a solid angle at R, which is contained by three plane angles LRM, MRN, NRL, which are equal to the three given plane angles ABC, DEF, GHK, each to each.

Which was to be done.

Book XI.

PROP. A. THEOR.

If each of two solid angles be contained by three See N. plane angles equal to one another, each to each ; the planes in which the equal angles are, have the same inclination to one another.

Let there be two solid angles at the points A, B; and let the angle at A be contained by the three plane angles CAD, CAE, EAD; and the angle at B, by the three plane angles FBG, FBH, HBG; of which the angle CAD is equal to the angle FBG, and CAE to FBH, and EAD to HBG: The planes in which the equal angles are, have the same inclination to one another.

In the straight line AC take any point K, and in the plane CAD, from K draw the straight line KD at right angles to AC, and in the plane CAE, the straight line KL at right angles to the

A

B same AC. Therefore the angle DKL is the inclination a of the

a 6. def. 11. K L

M

N plane CAD to the plane CAE: In BF,

F take BM equal to AK,

С
D

G and from the point M E

H draw, in the planes FBG, FBH, the straight lines MG, MN at right angles to BF; therefore the angle GMN is the inclination a of the plane FBG to the plane FBH: Join LD, NG; and because in the triangles KAD, MBG, the angles KAD, MBG are equal, as also the right angles AKD, BMG, and the sides AK, BM, adjacent to the equal angles, are equal to one another; therefore KD is equal to MG, and AD to BG: For the same rea

b 26. 1. son, in the triangles KAL, MBN, KL is equal to MN, and AL to BN: And in the triangles LAD, NBG, LA, AD are equal to NB, BG, and they contain equal angles; therefore the base LD is equal o to the base NG. Lastly, in the tri- c 4. 1. angles KLD, MNG, the sides DK, KL, are equal to GM, MN, and the base LD to the base NG; therefore the angle DKL is equal d to the angle GMN: But the angle DKL is d 8. I. the inclination of the plane CAD to the plane CĂE, and the angle GMN is the inclination of the plane FBG to the plane

Book XI. FBH, which planes have therefore the same inclination a to

one another: And in the same manner, it may be demonstraa 7. def. 11. ted, that the other planes in which the equal angles are, have

the same inclination to one another. Therefore, “if two solid angles,&c. Q. E. D.

PROP. B. THEOR.

See N.

If two solid angles de contained, each by three plane angles which are equal to one another, each to each, and alike situated; these solid angles are equal to one another.

Let there be two solid angles at A and B, of which the solid angle at A is contained by the three plane angles CAD, CAE, EAD; and that at B, by the three plane angles FBG, FBH, HBG; of which CAD is equal to FBG; CAE to FBH; and EAD to HBG: The solid angle at A is equal to the solid angle at B.

Let the solid angle at A be applied to the solid angle at B; and, first, the plane angle CAD being applied to the plane angle FBG, so as the point A may coincide with the point B, and the straight line AC with BF; then AD coincides with BG, because the angle CAD

A B.
is equal to the angle FBG:
And because the inclination of

the plane CAE to the plane a A. 11. CAD is equal a to the inclination of the plane FBH to the

H plane FBG, the plane CAE

D coincides with the plane FBH, because the planes CAD, FBG coincide with one another : And because the straight lines AC, Bf coincide, and that the angle CAE is equal to the angle FBH; therefore AE coincides with BH, and AD coincides with BG; wherefore the plane EAD coincides with the plane HBG: Therefore

the solid angle A coincides with the solid angle B, and conb 8. Ax. 1. sequently they are equal to one another, Q. E. D.

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