IF PROP. A. THEOR. Book XI. F each of two folid 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 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 FBH, and EAD to HBG. the planes in which the equal angles are, have the fame 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 angles to the fame AC. therefore the angle DKL is the inclination of the plane 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 MG, MN at right an gles to BF; therefore B a.6.Def.11. A as alfo 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 MG, and AD to BG. for the fame reason, in the b. 26. 1. triangles KAL, MBN, KL is equal to MN, and AL to BN. and c in the triangles LAD, NBG, LA, AD are equal to NB, BG, and they contain equal angles; therefore the base LD is equal to c. 4. I. the base NG. lastly, in the triangles KLD, MNG, the fides DK, d 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, e.7.Def.11. e Book XI. and the angle GMN is the inclination of the plane FBG to the plane FBH, which planes have therefore the fame inclination to one another. and in the fame manner it may be demonstrated, that the other planes in which the equal angles are, have the fame inclination to one another. Therefore if two folid angles, &c. Q. E. D. See N. PROP. B. THEOR. F two folid angles be contained, each by three plane angles which are equal to one another, each to each, and alike fituated; these 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 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 3. 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, be 1 a A A E D ៥ cause 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 t. 8. Ax. I. 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 See 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 solid figures contained by the same 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 lastly, FH fimilar and equal to PR. the solid figure AG is equal and fimilar to the folid figure KQ. a 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 angies LKN, LKO, OKN which contain the folid angle at K, each to each; therefore the solid angle at A is equal a 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 solid figure AG be applied to the folid figure KQ, firft, the plane figure AC being applied to the plane figure KM; the straight line AB coin- H 1 lines AD, DC, CB coincide with KN, NM, ML, each with each; Book XI. with the figure LQ, and the straight line CG with MQ, and the point G with the point Q. fince therefore all the planes and fides of the folid figure AG coincide with the planes and fides of the folid figure KQ, AG is equal and fimilar to KQ. and in the same manner, any other folid figures whatever contained by the fame number of equal and fimilar planes, alike fituated, and having none of their folid angles contained by more than three plane angles, may be proved to be equal and fimilar to one another. Q. E. D. See N. IF F a folid be contained by fix planes, two and two of which are parallel; the oppofite planes are fimilar and equal parallelograms. Let the folid CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE. its opposite planes are fimilar and equal parallelograms. Because the two parallel planes, BG, CE are cut by the plane 2. 16. 11. AC, their common sections AB, CD are parallel a. again, because the two parallel planes BF, AE are cut by the plane AC, their common fections AD, BC are parallel a. and AB is parallel to CD; b. 10. 11. with the other two; wherefore they contain equal angles ; the C. 4. I. 4. 34. I. c angle ABH is 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 bafe AH is equal to the bafe DF, and the triangle ABH to the triangle DCF. and the parallelogram BG is doubled of the triangle ABH, and the parallelogram CE double of the triangle DCF; therefore the parallelogram BG is equal and fimilar to the parallelogram CE. in the fame manner, it may be proved that the parallelogram AC is equal and fimilar to the pas rallelogram GF, and the parallelogram AE to BF. Therefore if a Book XI. folid, &c. Q. E. D. [F IF PROP. XXV. THEOR. a folid parallelepiped be cut by a plane parallel to See N. two of its oppofite planes; it divides the whole into two folids, the base of one of which fhall be to the base of the other, as the one folid is to the other. Let the folid parallelepiped ABCD be cut by the plane EV which is parallel to the oppofite planes AR, HD, and divides the whole into the two folids ABFV, EGCD; as the base AEFY of the first is to the bafe EHCF of the other, fo is the folid ABFV to the folid EGCD. Produce AH both ways, and take any number of straight lines HM, MN each equal to EH, and any number AK, KL each equal to EA, and complete the parallelograms LO, KY, HQ, MS, and the folids LP, KR, HU, MT. then because the ftraight lines LK, KA, AE are all equal, the parallelograms LO, KY, AF are equal 2, a. 36. L. b and likewise the parallelograms KX, KB, AG *; as alfo the pa- b. 24. II. rallelograms LZ, KP, AR, because they are oppofite planes. for the fame reason, the parallelograms EC, HQ, MS are equal1; and the parallelograms HG, HI, IN, as alfo b HD, MU, NT. therefore three planes of the folid LP,`are equal and fimilar to three planes of the folid KR, as alfo to three planes of the folid AV. but the three planes opposite to these three are equal and fimilar to them in the feveral folids, and none of their solid angles are contained by more than three plane angles. therefore the three folids c LP, KR, AV are equal to one another. for the fame reason, the c. C. 11. three folids ED, HU, MT are equal to one another. therefore what |