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Book XI. base AC, the angle LOM is equal to the angle ABC :: And m AB, that is, LO, by the hypothefis, is less than LX; whered 8. s. fore LO, OM fall within the triangle LXM; for, if they fell

upon its sides, or without it, they
would be equal to, or greater than

R
f 21. 1. LX, XMF: Therefore the angle
LOM, that is, the angle ABC, is

L
greater than the angle LXME: 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:
Therefore the three angles ABC,

M

N DEF, GHK are greater than the three angles LXM, MXF, NXL; that is, than four right angles : But the fame 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 center X of the circle fall in one of the sides 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 lefs 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,

X
EF are equal to LF, which is im-
$ 20. 1. possiblet: 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 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 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,

ET

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
been proved that it is greater than DËF, 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
B

H
P

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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 greater than the angle & 32. Ii ACB at the base. For the same reason, because the angle GHK, or CBP, is greater than the angle

R LXN, the angle XLN is greater than the angle BCP. Therefore the whole angle MLN is greater than the whole angle ACP. And because

IL ML, LN are equal to AC, CP, each to each, but the angle MLN is greater than the angle ACP, the bale MN is greater h than the base M AP. And MN is equal to DF; therefore also DF is greater than AP. Again, because DE, EF are equal to AB, BP, but the bate DF greater than the base AP, the an. gle DEF is greater k than the angle ABP. And ABP is equal to the two angles ABC, CBP, that is, to the two angles ABC, GHK; therefore the angle DEF is greater than the two angles ABC, GHK; but it is also less than these ; which is impossible. Therefore AB is not less than

N h 24. I.

x

k 25.1.

a 12. II.

Book XI. LX; and it has been proved that it is not equal to it; therefore W AB is greater than LX.

From the point X erecta 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

R
the square of LX, and make RX e-
qual to its side, and join RL, RM,
RN. Because RX is perpendicular
to the plane of the circle LMN, it

L b 3. def. 11. is 6 perpendicular to each of the

straight lines LX, MX, NX. And
because LX is equal to MX, and
XR common, and at right angles M

N
to each of them, the base RL is e-
qual to the base RM. For the same

X 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, XR. But the square of RL is equal to the fame 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 LM to the base AC; d 8. I. 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, MRŇ, NRL,'which are equal to the three given plane angles ABC, DEF, GHK, each to each. Which was to be done.

47. I.

PROP.

Book XI.

PRO P. A.

THE O R.

F each of two folid angles be contained by three plane: See N.

angles equal to one another, each to each ; the planes in which the equal angles are, have the same inclination to one another.

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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 draw the straight line KD at right angles to AC, and in the

А

B
planeCAE the straight
line KL at right angles
to the fame AC:
Therefore the angle

K
L

N

M DKL is the inclina

a 6. def, tion of the plane C

IDF CAD to the plane

G CAE: In BF take

E

H BM equal to AK, and from the point M draw, in the planes FBG, FBH, the straight Įines MG, MN at right angles to BF; therefore the angle GMN is the inclination 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, BG, and that the tides AK, BM, adjacent to the equal angles, are equal to one another, therefore KD is equal o to MG, and AD to BG : For the same reason, 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 e qual to the base NG. Lastly, in the triangles KLD, MNG, C 4.5. the fides DK, KL are equal to GM, MN, and the base LD to the bale NG; therefore the angle DKL is equal d to the angle d 8.1. GMN : But the angle DKL is the inclination of the plane CAD to the plane CAE, and the angle GMN is the inclina.

b 26. 1,

Book XI. tion of the plane FBG to the plane FBH, which planes hare m' therefore the same inclination * to one another : And in the * 7. def. 11. same manner it may be demonstrated, that the other planes ir

which the equal angles are, have the same inclination to one another. Therefore, if two solid angle, &c. Q. E. D.

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See N.

F two solid angles be contained, each by three plane

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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 folid angle

at B.

Let the solid angle at A be applied to the folid 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 to the inclina-
tion of the plane FBH to the E

H
plane FBG, the plane CAE C

F
D

G
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 confequently they are e 68. A. 6. qual to one another. Q. E. D.

PROP.

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