EFGH to some space, which must be less (14. 5.) than the circle ABCD, because the space T is greater by hypothesis, than the circle EFGH. Therefore as the square of FH is to the square of BD, so is the circle EFGH to a space less than the 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 is 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 EFGH.* Circles therefore are, &c. Q. E. D. D PROP. III. THEOR. Every pyramid having a triangular base, may be 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 pyra. mid.t 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 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, F, G, H, K, L, and join EH, BG, GH, HK, KL, LH, EK, KF, FG. Because AE is equal to EB, and AH to K HD, HE is parallel (2. 6.) to DB; for the same reason, HK is parallel to AB: therefore HEBK is a parallelogram, and HK equal (34, 1.) to EB: but EB is equal to AE; therefore also AE is equal to HK : and AH is equal to HD; where JEKING fore EA, AH are equal to KH, HD, each to each; and the angle EAH is equal (29. 1.) to ko B F the angle KHD; therefore the base EH is equal – с to the base KD, and the triangle AEH equal (4. 1.) 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 contain equal (10. 11.) 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 * 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. + See Note. EHG equal (4. 1.) and similar to the triangle, KDL; for the same reason the triangle AEG is also equal and similar to the triangle HKL. Therefore the pyramid of which the base is the triangle AEG, and of which the vertex is the point H, is equal (C. 11.) and similar to the pyramid the base of which is the triangle KHL, and vertex the point D: and because, HK is parallel to AB a side of the triangle ADB, the triangle ADB, is equiangular to the triangle HDK, and their sides are proportionals (4. 6.): therefore the triangle ADB is similar to the triangle HDK: and for the same reason, the triangle DBC is similar to the triangle DKL; and the triangle ADC to the tri. angle HDL; and also the triangle ABC to the triangle AEG: but the triangle AEG is similar to the triangle HKL, as before was proved; there. fore the triangle ABC is similar (21. 6.) to the triangle HKL. And the pyramid of which the base is the triangle ABC, and vertex the point D, is therefore similar (B. 11. and 11. def. 11.) to the 1 pyramid of which the base is the triangle HKL, K and vertex the same point D: but the pyramid B F C of which the base is the triangle HKL, and vertex the point D, is similar, 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 similar 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 EBFG is double (41. 1.) of the triangle GFC: but when there are two prisms of the same altitude, of which one has a parallelogram for its base, and the other a triangle that is half of the parallelogram, these prisms are equal (40. 11.) to one another; therefore the prism having the parallelogram EBFG for its base, and the straight line KH 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 same altitude, because they are between the parallel (15. 11.) planes ABC, 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 E, F 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 (C. 11.) to the pyramid the base of which is the triangle AEG, 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 opposite 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 similar 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. PROP. IV. THEOR. If there be two pyramids of the same altitude, upon triangular bases, and each of them be divided into two equal pyramids similar 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.* Let there be two pyramids of the same 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 same 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 (2. 6.) to AB, and the triangle ABC similar to the triangle LXC: for the same reason, the triangle DEF is similar to RVF: and be cause 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 similarly situated rectilineal figures ABC, LXC; and upon EF, FV, in like manner, are described the similar figures DEF, RVF: there. fore, as the triangle ABC is to the triangle LXC, so (22. 6.) is the triangle DEF to the triangle RVF, and, by permutation, as the triangle ABC to the triangle DEF, so is the triangle LXC to the triangle RVF: and because the planes ABC, OMN, as also the planes DEF, STY are parallel (5. 11.), the perpendiculars drawn from the points G, H to the bases ABC, DEF, which, by the hypothesis, are equal to one another, shall be cut each into two equal (17. 11.) parts by the planes OMN, STY, because the straight lines GC, HF are cut into two equal * See Note. parts in the points N, Y by the same planes: therefore the prisms LXCOMN, RVFSTY are of the same altitude; and therefore as the base LXC to the base RVF; that is, as the triangle ABC to the triangle DEF, so (Cor. 32. 11.) is the prism having the triangle LXC for its base, and OMN the triangle opposite to it, to the prism of which the base is the triangle RVF, and the opposite triangle STY: and because the two prisms in the pyramid ABCG are equal to one another, and also the two prisms in the pyramid DEFH equal to one another, as the prism of which the base is the parallelogram KBXL and opposite side MO, to the prism having the triangle LXC for its base, and OMN the triangle opposite to it, so is the prism of which the base (7.5.) is the parallelogram PEVR, and opposite side TS, to the prism of which the base is the triangle RVF, and opposite triangle STY. Therefore componendo, as the prisms KBXLMO, LXCOMN together are unto the prism LXOMN, so are the prisms PEVRTS, RVFSTY, В х E V F to the prism RVFSTY; and permutando, as the prisms KBXLMO, LXCOMN are to the prisms PEVRTS, RVFSTY, so is the prism LXCOMN to the prism RVFSTY: but as the prism LXCOMN to the prism RVFSTY, so is, as has been proved, the base ABC to the base DEF: therefore, as the base ABC to the base DEF, so are the two prisms in the pyramid ABCG to the two prisms in the pyramid DEFH: and likewise if the pyramids now made, for example, the two OMNG, STYH, be divided in the same manner; as the base OMN is to the base STY, so shall the two prisms in the pyramid OMNG be to the two prisms in the pyramid STYH: but the base OMN is to the base STY, as the base ABC to the base DEF; therefore, as the base ABC to the base DEF, so are the two prisms in the pyramid ABCG to the two prisms in the pyramid DEFH; and so are the two prisms in the pyramid OMNG to the two prisms in the pyramid STYH; and so are all four to all four: and the same thing may be shown of the prisms made by dividing the pyramids AKLO and DPRS, and of all made by the same number of divisions. Q. E. D. PROP. V. THEOR. PYRAMIDS of the same altitude, which have triangular bases, are to one another as their bases.* Let the pyramids of which the triangles ABC, DEF are the bases, and of which the vertices are the points G, H, be of the same altitude; as the base ABC, to the base DEF, so is the pyramid ABCG to the pyramid DEFH. For, if it be not so, the base ABC must be to the base DEF, as the pyramid ABCG to a solid either less than the pyramid DEFH, or greater than it.f First, let it be to a solid less than it, viz. to the solid Q: and divide the pyramid DEFH into two equal pyramids, similar to the whole, and into two equal prisms: therefore these two prisms are greater (3. 12.) than the half of the whole pyramid. And again, let the pyramids made by this division be in like manner di. vided, and so on, until the pyramids which remain undivided in the pyramid DEFH be, all of them together, less than the excess of the pyramid DEFH above the solid Q : let these, for example, be the pyramids DPRS, STYH: therefore the prisms, which make the rest of the pyramid DEFH, are greater than the solid Q: divide likewise the pyramid ABCG in the same manner, and into as many parts, as the pyramid DEFH: therefore, as the base ABC to the base DEF, so (4. 12.) are the prisms in the pyramid ABCG to the prisms in the pyramid DEFH: but as the base ABC to the base DEF, so, by hypothesis, is the pyramid ABCG to the solid Q; and therefore, as the pyramid ABCG to the solid Q, so are the prisms in the pyramid ABCG, to the prisms in the pyramid DEFH; but the pyramid ABCG is greater than the prisms contained in it; wherefore (14. 5.) also * See Note. + This may be explained the same way as at the notc † in proposition 2 in the like case. |