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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 DEF, which is absurd. a 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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L

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 8 32. I. ACB at the base. For the same reason, because the angle GH, or CBP, is greater than the angle LXN, the angle XLN is greater

R 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 greater than the base

h 24. 1. And MN is equal to DF; AP

M

N therefore also DF is greater than

X 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

k 25. 1. 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

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

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

of the circle LMN. And becaule it mbas been proved in all the
cafes, 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
equal to its lide, and join RL, KM,
RN. Because RX is perpendicular

to the plane of the circle LMN, it 63. def. 11. is 6 perpendicular to each of the

L
straight lines LX, MX, NX. And
because LX is equal to MX, and
XR common, and at right angles M
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 C 47. I. LX, XR. But the quare 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 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 base LM to the base AC; d 8. r, the angle LRM is equal 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 cach. Which was to be done.

PROP

Book XI.

PRO P. A.

THEOR.

IF

each of two solid 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.

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

А plane CAE the straight

B
line KL at right angles
to the same AC:
Therefore the angle

İN
K

L
K

M
DKL is the inclina-
tion of the plane C

a 6. def. 11,

DF 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 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 because in the triangles KAD, MBG, the angles KAD, MBG are equal, as also 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. I. 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 equal to the base NG. Lastly, in the triangles KLD, MNG, C4.1. the fides DK, KL are equal to GM, MN, and the base LD to the bafe NG , therefore the angle DKL is equal to the angle d 8. s. GMN: But the angle DKL is the inclination of the plane CAD to the plane CAE, and the angle GMN is the inclina

Book XI. tion of the plane FBG to the plane FBH, which planes have

w therefore the same inclination to one another : And in the a 7. def. II. 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 solid angles, &c. Q. E D.

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IF

two solid angles be contained, each by three plane

angles which are equal to one another, each to each, and alike fituated; these solid angles are equal to one another.

at B.

B

Let there be two folid 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

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
is equal to the angle FBG :
And because the inclination of

the plane CAE to the plane A, 16. CAD is equal a to the inclination of the plane FBH to the

HI
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 folid angle A

coincides with the folid angle B, and consequently they are e8. A. 6, qual to one another. Q. E. D.

E

PRON

Book XI.

PRO P. C.

THEOR.

OLID figures contained by the same number of e. Sce N.

qual and similar planes alike fituated, and having none of their solid angles contained by more than three plane angles ; are equal and fimilar to one another.

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Let AG, KQ be two solid figures contained by the same number of similar 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 similar and equal to PR: The folid 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 hypothers, are equal to the plane angles LKN, LKO, OKN, which contain the folid angle at K, each to each; therefore the solid angle at A is equal a to the solid angle at K: In the same manner,

a B. II. the other folid angles of the figures are equal to one another. If, then, the solid hgure AG be applied to the solid figure KQ; first, the plane fi

.. gure AC being

H G R Q applied to the

0 plane figure KM;

F

P
the straight line
AB coinciding

D
C N

M
with KL, the fi.
gure AC must

А. B K L coincide with the figure KM, because they are equal and similar: Therefore the Atraight 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 solid 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 O, 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 solid angle at L, it may be proped, in the same manner, that the figure BG coincides with

the

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