MN, the angle MXN is equal to the angle DEF. and it has been Book XI. H A к С 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 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 ACB at the 6-31. 1. base. for the same reason, because the R 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 M N angle ACP, the base MN is greater h. 14.1. than the base AP. and MN is equal to X DF; therefore also DF is greater than AP. Again, because DE, EF are equal to AB, BP, but the base DF greater than the base AP, the angle DEF is greater k than the angle ABP. k. sj... 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 inpossible. therefore AB is not less than LX; and it has been pro Book XI. vcd 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 cases, that AB is greater than LX, find a square equal to the excess of the R perpendicular to the plane of the circle m.3. Def.11. LMN, it is perpendicular to each of L N X 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 31. 47.1. squares of LX, XR. but the square of RL is equal n to the same squares, because LXR is a right angle. therefore the square of ABis 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 becanse RL, RM, are equal to AB, BC, 0.8. 1. and the base LM to the base AC; the angle LRM is equal o to the anole 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, cach to each. Which was to be done. PROP Book XI. PRO P. A. THE O R. 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 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, FEET, HBG; of which the angle CAD is equal to the angle FBG, and! CAE to FB4, 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 plane CAE the straight line KL at right B A angies to the same AC. therefore the angle DKL is the inclination N a.5. Duf.ii. M of the plane CAD to с D F the plane CAE. in BF G take BM equal to AK, E H 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 ar gles KAD, MEG are equal, as also the right angles AKD, BMG, and that the files AK, BM, a jacuat to the equal angles, are equal to one another, therefore KD is equais b. 26.1. to MG, and AD to LG. for the fame reason, in the triangles KAL, MBN, KL is equal to MN, and AL to BN. and in the triangles LAD, NBS, LA, AD are equal to NB, BG, and they contain equal angles; therefore the base LD is equal to the bale NG. lahly, c. 4.1. in the triangles KLD, MNG, the files DK, KL are equal to GY, MN, and the base LD to the bale NG; thurifure the argle DKL is equal to the angle GUN. 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 FDH, which places have O 3 therefore Book XI. therefore the same inclination to one another, and in the same man. wner it may be demonstrated, that the other planes in which the equal 3.7.Def.11. angles are, have the same inclination to one another. Therefore if two folid angles, &c. Q. E. D. See N. PROP. B. THEOR. 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 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 solid angle at A, is equal to the folid 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 A B E H a. A. 11. equal a to the inclination of the F G therefore the solid angle A coincides with the solid angle B, and con6. 8. Ax. 1. sequently they are equal b to one another. Q. E. D. С PROP. Book XI. PROP. C. THEOR. SOLID figures contained by the same number of equal Ste N. and similar planes alike fituated, and having none of their folid 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 folid figure AG is equal and similar to the solid fi. gure KO. 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 to the solid angle 4. B. 11, at K. ' in the fame manner, the other solid 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 fi H G R Q gure AC being ap o plied to the plane F P figure KM; the D N M straight line AB COinciding with KL, А B K I the figure AC must coincide with the figure KM, because 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 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 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 |