EF are equal to MX, XN, and the bafe DF to the bafe Book XI. MN, the angle MXN is equal to the angle DEF: And it 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 cafe, the angle ABC is greater than the angle MXL, and the angle GHK greater than the angle LXN. At the point B in the ftraight line CB make the angle CBP equal to the angle GHK, and make BP equal to

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 bafe 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 at the bafe is greater than the angle 8 32. I. ACB at the bafe. For the fame reason, because the angle GHK,

M

R

N

h 24. I.

X

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 MLN
is greater than the angle ACP, the
bafe MN is greater than the base
AP. And MN is equal to 'DF;
therefore alfo DF is greater than
AP. Again, because DE, EF are
equal to AB, BP, but the bafe DF
greater than the base AP, the an-
gle DEF is greater 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 lefs
than thefe; which is impoffible. Therefore AB is not less than

LX;

k 25. I.

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

a 12. II.

R

From the point X erect XR at right angles to the plane of the circle LMN. And because it has been proved in all the cafes, that AB is greater than LX, find a fquare equal to the excefs of the fquare of AB above the fquare of LX, and make RX equal to its fide, and join RL, KM, RN. Because RX is perpendicular to the plane of the circle LMN, it b3. def. 11. is b perpendicular to each of the ftraight lines LX, MX, NX. And because LX is equal to MX, and

C 47. I.

d 8. f.

XR common, and at right angles M

N

X

to each of them, the base RL is e-
qual to the bafe RM. For the fame
reason, RN is equal to each of the
two RL, RM. Therefore the three
ftraight lines RL, RM, RN are all
equal. And because the fquare of
XR is equal to the excefs of the fquare of AB above the fquare
of LX; therefore the fquare of AB is equal to the fquares of
LX, XR. But the fquare of RL is equal to the fame fquares,
because LXR is a right angle. Therefore the fquare of AB
is equal to the fquare of RL, and the ftraight 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. Where-
fore 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 bafe LM to the bafe AC;
the angle LRM is equal to the angle ABC. For the fame
reafon, 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 angles LRM, MRN, NRL, which
are equal to the three given plane angles ABC, DEF, GHK,
each to each. Which was to be done.

PROP.

IF

221

Book XI.

Book

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 fame inclination to one another.

Let there be two folid 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 fame inclination to one another.

In the ftraight line AC CAD from K draw the the to AC, and in plane CAE the ftraight line KL at right angles to the fame AC: Therefore the angle DKL is the inclina

tion of the plane C

CAD to the plane

CAE: In BF take

BM equal to AK, and

take any point K, and in the plane
ftraight line KD at right angles

A

B

K

from the point M draw, in the planes FBG, FBH, the ftraight
lines 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 becaufe in the triangles KAD, MBG, the
angles KAD, MBG are equal, as alfo the right angles AKD,
BMG, and that the fides AK, BM, adjacent to the equal
angles, are equal to one another, therefore KD is equal to b 26. 1.
MG, and AD to BG: For the fame 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 bafe LD is e-
qual to the bafe NG. Laftly, in the triangles KLD, MNG, c4. 1.
the fides DK, KL are equal to GM, MN, and the bafe LD to
the bafe NG; therefore the angle DKL is equal 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-

с

d

tion

1

Book XI. tion of the plane FBG to the plane FBH, which planes have w therefore the fame inclination to one another: And in the a 7. def. 11. same manner it may be demonftrated, that the other planes in which the equal angles are, have the fame inclination to one another. Therefore, if two folid angles, &c. Q. E D.

See N.

IF

PROP. B. THEOR.

F two folid angles be contained, each by three plane angles which are equal to one another, each to each, and alike fituated; thefe folid angles are equal to one another.

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 angle at A is equal to the folid angle at B.

A

B

Let the folid angle at A be applied to the folid angle at B: and, first, the plane angle CAD being applied to the plane angle FBG, fo as the point A may coincide with the point B, and the ftraight line AC with BF; then AD coincides with BG, because the angle CAD is equal to the angle FBG: And becaufe the inclination of the plane CAE to the plane a A. II. CAD is equal a to the inclination of the plane FBH to the plane FBG, the plane CAE coincides with the plane FBH, because the planes CAD, FBG coincide with one another: And because the ftraight 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 folid angle B, and confequently they are e8. A. 6. qual to one another. Q.E. D.

C

E

F

H

D

G

PROR

PROP. C. THEOR.

Book XI,

Book XL

OLID figures contained by the fame number of e. See N. qual and fimilar planes alike fituated, and having none of their folid angles contained by more than three plane angles; are equal and fimilar to one another.

Let AG, KQ be two folid figures contained by the fame number of fimilar and equal planes, alike fituated, viz. let the plane AC be fimilar and equal to the plane KM; the plane AF to KP; BG to LQ, GD to QN; DE to NO; and lastly, FH fimilar and equal to PR: The folid figure AG is equal and fimilar to the folid figure KQ.

Because the folid angle at A is contained by the three plane angles BAD, BAE, EAD, which, by the hypothefs, are equal to the plane angles LKN, LKO, OKN, which contain the folid angle at K, each to each; therefore the folid angle at A is equal to the folid angle at K: In the fame manner, a B. 11. the other folid angles of the figures are equal to one another.

a

If, then, the folid figure AG be applied to the folid figure KQ;

first, the plane fi

gure AC being

figure KM, because they are equal and fimilar: Therefore the traight 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 folid 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 fimilar to one another: Therefore the straight lines AE, EF, FB, coincide with KO, OP, PL; and the points E, F, with the points O, P. In the fame manner, the figure AH coincides with the figure KR, and the ftraight 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 fame manner, that the figure BG coincides with

the