Book XI. 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.

a 12. 11.

11.

R

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 LX, and make RX equal to its side, and join RL, RM, RN. Because RX is perpendicular to the b 3. def plane of the circle LMN, it is b 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 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;

c

M

X

N

therefore the square of AB is equal to the squares of LX, XR, c47. 1. 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 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.

d 8. 1.

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

A

B

tion a of the

a 6.

def. 11.

CAD to the

plane

G

CAE. In BF take

D

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 sides 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 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 c 4. 1. base NG. Lastly, in the triangles 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 to the angle GMN: but the d 8. 1. angle DKL is the inclination of the plane CAD to the plane CAE, and the angle GMN is the inclination of the plane FBG

:

a 7.

ཀ

Book XI. to the plane FBH, which planes have therefore the same incli nation a to one another: and in the same manner it may be demonstrated, that the other planes in which the equal angles are def. 11. have the same inclination to one another. Therefore, if two solid angles, &c. Q. E. D.

See N.

PROP. B. THEOR.

IF two solid angles be 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 is equal to the angle FBG and because the inclination of the plane CAE to the a A. 11. plane CAD is equal to the inclination of the plane FBH to the plane FBG, the plane CAE coincides with the plane FBH, because the planes CAD, FBG co

ΛΑ

G

incide 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. A. 1. sequently they are equal b to one another.

Q. E. D.

Book XI.

PROP. C. THEOR.

SOLID figures contained by the same number of See N. equal and similar planes alike situated, and having none of their solid angles contained by more than three plane angles, are equal and similar to one another.

Let AG, KQ be two solid figures contained by the same number of similar and equal planes, alike situated, viz. let the plane AC be similar 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 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 a a B. 11. to the solid angle at K: in the same manner, the other solid angles of the figures are equal to one another. If, then, the solid figure AG be applied to the solid figure KQ, first, the plane figure AC being ap

plied to the plane H

G

R

figure KM; the

straight line AB co

inciding with KL,

N

D

figure KM, because

A

B

K

P

M

L

they are equal and

:

similar 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 a 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 solid angle at B is equal to the solid angle at L, it may be proved, in the same manner, that the figure BG coincides with the figure

Book XI. LQ, and the straight line CG with MQ, and the point G with the point Q: since, therefore, all the planes and sides of the solid figure AG coincide with the planes and sides of the solid figure KQ, AG is equal and similar to KQ: and, in the same manner, any other solid figures whatever contained by the same number of equal and similar planes, alike situated, and having none of their solid angles contained by more than three plane angles, may be proved to be equal and similar to one another. Q. E. D.

PROP. XXIV. THEOR.

See N.

IF a solid be contained by six planes, two and two of which are parallel; the opposite planes are similar and equal parallelograms.

Let the solid CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE: its opposite planes are similar and equal parallelograms.

B

H

Because the two parallel planes BG, CE are cut by the plane a 16. 11. AC, their common sections AB, CD are parallel 2. Again, because the two parallel planes BF, AE are cut by the plane AC, their common sections AD, BC are parallel: and AB is parallel to CD; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF, AE is a parallelogram: join AH, DF; and because AB is parallel to DC, and BH, to CF; the two straight lines AB, BH which meet one another, are parallel to DC and CF which meet one another, and are not in the same plane with the other two; wherefore they contain b 10. 11. equal angles b; the angle ABH is

c 4. 1.

A

D

C

G

F

E

therefore equal to the angle DCF: and because AB, BH are equal to DC, CF, and the angle ABH equal to the angle DCF; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DCF: and the d 34. 1. parallelogram BG is double d of the triangle ABH, and the parallelogram CE double of the triangle DCF; therefore the parallelogram BG is equal and similar to the parallelogram CE. In the same manner it may be proved, that the parallelogram AC