and because DE, EF are equal to MX, XN, and the bafe DF to Bock XI, the base 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 firft 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, is greater than the angle LXN, the L R X h. 24. 2. therefore alfo DF is greater than AP. Again, becaufe DE, EF are equal to AB, BP, but the bafe DF greater than the base AP, the angle DEF is greater k than the angle ABP. and ABP is equal to k. 25. f. the two angles ABC, CBP, that is to the two angles ABC, GHK; therefore the angle DEF is greater than the two angles Book XI. ABC, GHK; but it is alfo lefs than thefe; which is impoffible. therefore AB is not lefs than LX; and it has been proved that it is not equal to it; therefore AB is greater than LX. 1. 12. 11. R From the point X erect XR at right angles to the plane of the circle LMN. and becaufe 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 join RL, RM, RN. because RX is perpendicular to the plane of the m.3. Def.11. circle LMN, it is perpendicular to n. 47. I. a. 8. 1. m each of the ftraight lines LX, MX, N X 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 cqual n to the fame 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 straight lines BC, DE, EF, GH, HK is equal to AB, and each of the two RM, RN is equal to RZ. wherefore AB, BC, DE, EF, GH, HK are each of them equal to each of the ftraight 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. Book XI. IF PROP. A. THEO R. F each of two folid angles be contained by three See N. plane angles equal to one another, cach 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 FBII, 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 ftraight line KD at right angles to AC, and in the plane CAE the ftraight line KL at right angles to the fame AC. therefore the angle DKL is the inclination of the plane a.6.Def.11. 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 a B K E H as alfo the right angles AKD, BMG, and that the fides AK, PM, adjacent to the equal angles, are equal to one another, therefore KD is equal to MG, and AD to BG. for the fame reafon, in the b. 16. 1. triangles KAL, MBN, KL is equal to MN, and AL to EN. and c in the triangles LAD, NBG, LA, AD are equal to NB, BG, and they contain equal angles; therefore the bafe LD is equal to c. 4. I. the base NG. laftly, in the triangles KLD, MNG, the fides DK, KL are equal to GM, MN, and the bafe 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, Book XI. and the angle GMN is the inclination of the plane FBG to the w plane FBH, which planes have therefore the fame inclination to Def.11. one another. and in the fame 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. QE. D. See N. PROP. B. THEOR. IF 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 HBC. the folid angle at A, is equal to the folid angle at B. B Let the folid angle at A be applied to the fold 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 FBC. and because the inclination of the plane CAE to the . A. 11. plane CAD is equal to the inclination of the plane FBH to the plane FBG, the plane CAE coin a E H D cides with the plane FBH, becaufe the planes CAD, FBG coincide with one another, and becaufe 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 b. 3. Ax. 1. folid angle B, and confequently they are equal to one another. Q. E. D. b PROP. C. THEOR. Book XI. OLID figures contained by the fame number of equal sec N SOL 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 a. 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, first, the plane figure AC being applied to the plane figure KM; the ftraight line AB coin- H 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 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 |