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and because DE, EF are equal to MX, XN, and the base DF to Bock XI. the base MN, the angle MXN is equal d to the angle DEF. and it co has been proved that it is greater than DEF, which is abfurd. d. 8. 1. therefore AB is not equal to LX. nor yet is it lefs ; 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 HK, and join CP, AP.

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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 Ifofceles triangles ABC, MXL, because the angle ABC is greater than the angle MXL, therefore the angle MLX

R at the base is greater & than the angle ACB at the base. for the same reafon, because the angle GHK, or CBP, is greater than the angle LXN, the angle XLN 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

X MLN greater than the angle ACP, the base MN is greater h than the

h. 24. 1. bafe AP. and MN is equal to DF; therefore also DF is greater than AP. Again, because DE, EY are equal to AB, BP, but the bale DF greater than the base AP, the. angle DEF is greater k than the angle ABP. aod ABP is equal to k. 25. î, the two angles ABC, CBP, that is to the two aogles ARC, GHK; therefore the angle DEF is greater than the tio angles

1. 12. 11.

X

Book XI. 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; therefore AB is greater than LX.

From the point X erect I 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

R
LX, and make RX equal to its side,
and join RL, RM, RN. because RX

is perpendicular to the plane of the m.;. Def. 11. circle LMN, it is perpendicular to

cach of the straight lines LX, MX,
NX. and because LX is equal to MX,
and XR conmon, and at right angles M

N
to each of them, the base RL is equal
to the base RM. 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, XR. but the square of RL is cqual n 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 R2, 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; the angle LRM g. 8. . is equal o 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 folid angle at R, which is contained by three plane an-
gles LRM, MRN, NRL, which are equal to the three givea
plane angles ABC, DEF, GHK, each to each. Which was to
be done.

n. 47. I.

!

Book XI.

PROP. A.

THEOR.

IF
F each of two solid angles be contained by three See N.
piane angles equal to one another, cach to each

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

Let there be two olid angles at the points A, B; and let the
angle at A be contained by the three plane angles CAD, CAL,
FAD; 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 FBII, and EAD to HBG, the planes in which the equal
angles are, have the same incl.nation 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 same
AC. therefore the angle DKL is the inclination of the piane a.6.Del.is.
CAD to the plane CAE. in BF take BM equal to AK, and from
the point M draw, in the planes FBG, FBH, the straight lines
MG, MN at right an-

А

B
gles to BF; therefore
the angle GMN is the
inclination of the plane
FBG to the plane FBH. K

M N
join LD, NG; and be-

TE
cause in the triangles C)

D
KAD, MBG, the angles E

H
KAD, MBG are equal,
as also the right angles AKD, BMG, and that the sides AK, PM,
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 b. 26. 1.
triangles KAL, MBN, KL is equal to MN, and AL to DN. and
in the triangles LAD, NBG, LA, AD are equal to NB, BG, and
they contain equal angles; therefore the base D is equal to C. 4. 1.
the base NG. lastly, in the triangles KLD, MNG, the sites DK,
KL are equal to GM, MN, and the base LD to the bale NG;
therefore the angle DKL is equal to the angle GMN. but the d. 8. I:
angle DKL is the inclination of the plane CAD to the plane CAT,

1

Book X1. and the angle GMN is the inclination of the plane FBG to the

plane FBH, which planes have therefore the same inclination to c:7 Def.sl. one another, and in the same manner it may be demonstrated, that

the other planes in which the equal angles are, have the same inclination to one another. Therefore if two folid angles, &c,

9. E. "D.

PROP. B. THEOR.

Sec N.

I F two folid angles be contained, each by three plane

angles which are equal to one another, each to each, and alike situated; these folid angles are equal to one aliocher,

Let there be two folid angles at A and B, of which the folid 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 folid anole at A, is equal to the solid angle at B.

Let the folid angle at A be applied to the fuid argle at B; and first, the plane angi- 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 DF; then AD coincides with BG, because the angle CAD is equal to the angle FBG. and because the in

clination of the plane CAE to the 8. A. 11. plane CAD is equal to the incination of the plane FBH to the

E

IL plane FBG, the plane CAE coin

C D cides 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 folid angle A coincides with the 2. 3. 4x.z. folid angle B, and consequently they are equal o to one another.

Q. E. D.

Book XI.

PROP. C. THE OR.

SOLI
OLID figures contained by the same number of equal gec N.

and similar planes alike situated, and having none of their folid angles contained by more than three plane angles; are equal and similar to one another.

Let AG, KQ be two folid figures contained by the same number of similar and equal planes, alike fituated, viz. let the place AC be similar and equal to the plane KM; the plane AP to KP; BG to LQ; GD to QN; DE to NO; and lastly, FH similar and equal to PR. the solid figure AG is equal and similar to the solid figure KQ.

Because the solid angle at A is contained by the three plane angles BAD, BAE, EAD which, by the hypothesis, are equal to the plane angles LKN, LKO, OKN which contain the solid angle at K, each to each; therefore the solid angle at A is equal to the solid angle a. B. 11, at K. in the fame manner, the other solid angles of the figures are equal to one another. If then the folid figure AG be applied to the solid figure KQ, first, the plane figure AC being applied to the plane figure KM; the straight line AB coin. H

G R ciding with KL, the

E

F figure AC must coincide with the figuren

N

M KM, because they are equal and similar.

А B K L therefore the straight lines AD, DC, CB coincide with KN, NM, ML, each with each; and the points A, D, C, B with the points K, N, M, L. and the solid angle at A coincides with the folid angle at K; wherefore the plane AF coincides with the plane KP, and the figure AF with the figure KP, because they are equal and similar to one another. therefore the straight lines AE, EF, FB coincide with KO, OP, PL; and the points E, F, with the points 0, P. In the same manner, the figure AH coincides with the figure KR, and the straight line DH with NR, and the point H with the point R. and because the folid angle at B is equal to the folid angle at L, it may be proved in the same manner, that the figure BG coincides

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