MN, the angle MXN is equal to the angle DEF. and it has been Book XI. proved that it is greater than DEF, which is abfurd. therefore AB

is not equal to LX. nor yet is it lefs; for then, as has been proved d. 8. 1. 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 straight line CB make the angle CBP équal 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 bafe 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 ACB at the g. 31. 1. bafe. for the fame reason, 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 MLN greater than the angle ACP, the base MN is greater than the bafe 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

8

M

k

L

N

h. 14. f.

X

than the base AP, the angle DEF is greater than the angle ABP. k. xj. 1. 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 alfo lefs than thefe; which is impoffible. therefore AB is not lefs than LX; and it has been pros

Book XI. ved that it is not equal to it; therefore AB is greater than LX. From the point X erect! XR at right angles to the plane of the 1. 12. 11. circle LMN. and because it has been proved in all the cafes, that

AB is greater than LX, find a fquare equal to the excess of the fquare of AB above the fquare of LX,

and make RX equal to its fide, and jein RL, RM, RN. becaufe RX is perpendicular to the plane of the circle m.3.Def.11. LMN, it is in perpendicular to each of the ftraight lines LX, MX, NX. and becaufe LX is equal to MX, and XR common, and at right angles to each M of them, the base RL is equal to the bafe RM. for the fame reason, RN is equal to each of the two RL, RM. therefore the three straight lines RL, RM, RN are all equal. and because

R

N

X

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 A.47.1. fquares of LX, XR. but the fquare of RL is equal n to the fame

0.8. I.

fquares, because LXR is a right angle. therefore the fquare of AB is equal to the fquare of RL, and the straight line AB to RL. but each of the ftraight 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 becaufe 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, cach to each. Which was to be done.

PROP.

PRO P. A. THEOR.

Book XI.

IF 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, CAD, EAD; and the angle at B by the three plane angles FBG, FB!!, 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 take any point K, and in the plane CAD from K draw the straight line KD at right angles to AC, and in

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, adjaccat to the equal angles, are equal to one another, therefore KD is equal b. 26. 1. to MG, and AD to EG. for the fame reafon, 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 equal to the bafe NG. lafily, c. 4. 1. in the triangles KLD, MNG, the files DK, KL are equal to GM, MN, and the bafe LD to the bafe NG; therefore the angle DKL is equal to the angle GMN. but the angle DKL is the inclination of d. 8. 1. the plane CAD to the plane CAE, and the angle CMN is the inclination of the plane FBG to the plane FBH, which planes have

03

therefore

Book XI. therefore the fame inclination to one another. and in the fame manner it may be demonftrated, that the other planes in which the equal F. 7.Def.11. angles are, have the fame inclination to one another. Therefore if two folid angles, &c. Q. E. D.

See N.

IF

F two folid angles be contained, each by three planę 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.

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, so as the point A may coincide with the point B, and the straight line

the plane CAE 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 con

b. 8. Ax. 1. fequently they are equal to one another. Q.E. D.

PROP.

PROP. C. THEOR.

Book XI.

SOLID figures contained by the fame number of equal See N. 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 laftly, 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 hypothefis, 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 4. B. 11, at K. in the fame manner, the other folid angles of the figures are equal to one another. If then the folid figure AG be applied to the

folid figure KQ,

2

coincide with the figure KM, because they are equal and fimilar. 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 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 ftraight 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 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 fame manner, that the figure BG coincides with

04