c. II. 5. Book XII. is to the space S. therefore as the circle ABCD is to the space S, lo is e the polygon AXBOCPDR to the polygon EKFLGMHN. but the circle ABCD is greater than the polygon contained in it; whered. 14. 5. fore the space S is greater than the polygon EKFLGMHN. but it is likewise less, as has been demonstrated; which is impossible. Therefore the square of BD is not to the square of FH, as the circle ABCD is to any space less than the circle EFGH. in the same manner it may be demonstrated that neither is the square of FH to the square of BD, as the circle EFGH is to any space less than the circle ABCD. Nor is the square of BD to the square of FH, as the circle ABCD is to any space greater than the circle EFGH. for, if possible, let it be so to T a space greater than the circle EFGH. therefore, inversely, as the square of FH to the square of BD, fo is the space T to the circle ABCD. but as the space + T is to the circle ABCD, so is the circle EFGH to some space, which must be lefs d than the circle ABCD, because the space T is greater, by hypothesis, than the circle EFGH. therefore as the square of FH is to # For as in the foregoing Note at * manner there can be a fourth proporit was explained how it was possible tional to this other space, named T, there could be a fourth proportional and the circles ABCD, EFGH, and the to the squares of BD, FH and the circle like is to be understood in some of the ABCD, which was named $. fo in like following Propositions, the square of BD, so is the circle EFGH to a space less than the Book XII. circle ABCD, which has been demonstrated to be impossible. therefore the square of BD is not to the square of FH, as the circle ABCD is to any space greater than the circle EFGH. and it has been demonstrated that neither is the square of BD to the square of FH, as the circle ABCD to any space less than the circle EFGH. wherefore as the square of BD to the square of FH, so is the circle ABCD to the circle EFGHỊ. Circles, therefore, are, &c. Q.E.D. E VERY pyramid having a triangular base, may be See N, divided into two equal and similar pyramids having triangular bases, and which are similar to the whole pyramid; and into two equal prisms which together are greater than half of the whole pyramid. Let there be a pyramid of which the base is the triangle ABC and its vertex the point D. the pyramid ABCD may be divided into two equal and similar pyramids having triangular bases, and similar to the DO whole; and into two equal prisms which together are greater than half of the whole pyramid. Divide AB, BC, CA, AD, DB, DC, each into two equal parts in the points E, K. F, G, H, K, L, and join EH, EG, GH, L HK, KL, LH, EK, KF, FG. Because AE is equal to EB, and AH to HD, HE is parallel to DB. for the same reason, HK is E parallel to AB. therefore HEBK is a parallelogram, and HK equal o to EB. but b. 34. I. EB is equal to AE; therefore also AE is B F C equal to HK. and AH is equal to HD; wherefore EA, AH are equal to KH, HD, cach to each; and the angle EAH is equal to the angle KHD; therefore the base EH is c. 29. 1. a. 2. 6. Because as a fourth proportional to the squares of BD, FH and the circle ABCD is possible, and that it can neither be less nor greater than the circle EFGH, it must be equal to it. d. 4. I. 8. 4. 6. Book XII. equal to the bafe KD, and the triangle AEH equal d and similar to the triangle HKD. for the same reason, the triangle AGH is equal and similar to the triangle HLD. and because the two straight lines EH, HG which meet one another are parallel to KD, DL that meet one another, and are not in the same, plane with them, they e. 10. II. contain equal e angles; therefore the angle EHG is equal to the angle KDL. again, because EH, HG are equal to KD, DL, each to each, and the angle EHG equal to the angle KDL; therefore the base EG is equal to the base KL, and the triangle EHG equal d and similar to the triangle KDL. for the same reason, the triangle AEG is also equal and fimilar to the triangle HKL. Therefore the pyramid of which the base is the triangle AEG, and of which f. C. 11. the vertex is the point H, is equal f and D H н K K I G F C the triangle HKL, as before was proved, h. 21. 6. therefore the triangle ABC is similar to the triangle HKL. and the pyramid of which the base is the triangle ABC, and vertex the point i. B. 11. & D, is therefore similar i to the pyramid of which the base is the tri angle HKL, and vertex the same point D. but the pyramid of which the base is the triangle HKL, and vertex the point D, is fimilar, as has been proved, to the pyramid the base of which is the triangle AEG, and vertex the point H. wherefore the pyramid the base of which is the triangle ABC, and vertex the point D, is fimilar to the pyramid of which the base is the triangle AEG and vertex H. therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD. and because BF is equal to FC, the parallelogram 8. 41. 5. EBFG is double k of the triangle GFC. but when there are two prisms of the fame altitude, of which one has a parallelogram for its base, and the other a triangle that is half of the parallelogram, II.Def.11. these prisms are equal to one another; therefore the prism having Book XII. the parallelogram EBFG for its base, and the straight line KH 1. 40. II. opposite to it, is equal to the prism having the triangle GFC for its base, and the triangle HKL opposite to it; for they are of the fame altitude, because they are between the parallel m planes ABC, m. 15. 11. HKL. and it is manifest that each of these prisms is greater than either of the pyramids of which the triangles AEG, HKL are the bases, and the vertices the points H, D; because if EF be joined, the prism having the parallelogram EBFG for its base, and KH the straight line opposite to it, is greater than the pyramid of which the base is the triangle EBF, and vertex the point K; but this pyramid is equal f to the pyramid the base of which is the triangle AEG, f. C. 11. and vertex the point H; because they are contained by equal and similar planes. wherefore the prism having the parallelogram EBFG for its base, and opposite side KH, is greater than the pyramid of which the base is the triangle AEG, and vertex the point H. and the prism of which the base is the parallelogram EBFG, and opposite side KH is equal to the prism having the triangle GFC for its base, and HKL the triangle oppofite to it; and the pyramid of which the base is the triangle AEG, and vertex H, is equal to the pyramid of which the base is the triangle HKL, and vertex D. therefore the two prisms before-mentioned are greater than the two pyramids of which the bases are the triangles AEG, HKL, and vertices the points H, D. Therefore the whole pyramid of which the base is the triangle ABC, and vertex the point D, is divided into two equal pyramids fimilar to one another, and to the whole pyramid; and into two equal prisms; and the two prisms are together greater than half of the whole pyramid. Q. E. D. Book XII. PROP. IV. THEOR. See N. F there be two pyramids of the same altitude, upon triangular bases, and each of them be divided into two equal pyramids fimilar to the whole pyramid, and also into two equal prisms; and if each of these pyramids be divided in the same manner as the first two, and so on. as the base of one of the first two pyramids is to the base of the other, so shall all the prisms in one of them be to all the prisms in the other, that are produced by the same number of divisions. a. 2. 6. Let there be two pyramids of the fame altitude upon the triangular bases ABC, DEF, and having their vertices in the points G, H; and let each of them be divided into two equal pyramids similar to the whole, and into two equal prisms; and let each of the pyramids thus made be conceived to be divided in the like manner, and so on. as the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms in the pyramid DEFH made by the fame number of divisions. Make the same construction as in the foregoing proposition, and because BX is equal to XC, and AL to LC, therefore XL is parallel a to AB, and the triangle ABC similar to the triangle LXC. for the same reason, the triangle DEF is fimilar to RVF. and because BC is double of CX, and EF double of FV, therefore BC is to CX, as EF to FV. and upon BC, CX are described the similar and fimilarly situated rectilineal figures ABC, LXC; and upon EF, FV, in like manner, are described the similar figures DEF, RVF. thereb. 22. 6. fore as the triangle ABC is to the triangle LXC, sob is the triangle DEF to the triangle RVF, and, by permutation, as the triangle RVF. and because the planes ABC, OMN, as also the planes c. 15. 1. DEF, STY are parallel ", the perpendiculars drawn from the points G, H to the bases ABC, DEF, which, by the Hypothesis, are equal d. 17. II. to one another, shall be cut each into two equal parts by the planes OMN, STY, because the straight lines GC, HF are cut into two equal parts in the points N, Y by the fame planes. therefore the prisms LXCOMN, RVFSTY are of the fame altitude; and |