d 4. I. g 4. 6. Buok XII. the triangle AEH equal d and similar to the triangle HKD: w 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 e 10 11. them, they 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 iş also equal and similar 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 and similar to the pyramid the base of which D H н K L E G B F C was proved; therefore the triangle ABC h 21. 6. is similar to the triangle HKL. And the pyramid of which the base is the triangle ABC, and veri B. 11. & tex the point D, is therefore similari to the pyramid of which 11.def. 11. the base is the triangle HKL, and vertex the same point D: But the pyramid 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 simi lar to the whole pyramid ABCD: And because BF is equal k 41. 1. to FC, the parallelogram EBFG is double 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 a 10. 11. that is half the parallelogram, these prisms are equal a to one another, therefore the prism having the parallelogram EBFG Book XII. for its base, and the straight lines 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 b planes ABC, HKL: b 15. 11. 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 vertes the point K; but this pyramid is equal to the pyramid, c c. 11. 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 A EG, 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 the half the whole pyramid. Q. E. D. PROP. IV. THEOR. If there be two pyramids of the same altitude, upon See N. 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 are all the prisms in one of them to all the prisms in the other that are produced by the same number of divisions. Book XII. 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 proposi tion: And because BX is equal to XC, and AL to LC, therea 2. 6. fore XL is parallela to AB, and the triangle ABC similar to the triangle LXC: For the same reason, the triangle DEF is similar to RVF: And because BC is double CX, and EF double 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: Therefore, as the b 22. 6. triangle ABC is to the triangle LXC, sob is the triangle DEF to the triangle RVF, and, alternately, 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, c 15. 11. STY, are parallelo, the perpendiculars drawn from the points G, H to the bases ABC, DEF, which, by the hypothesis, are d 17. 11. equal to one another, are cut each into two equal d 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 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 a Cor. 32. DEF, so a is the prism having the triangle LXC for its base, 11. 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 is the parallelogram PEVR, and opposite side TS, to the prism of which the base is the triangle RVF, and opposite triangle STY. Therefore, by composition, as the prisms KBXLMO, LXCOMN together are to the prism LXCOMN; so are the prisms PEVRTS, RVFSTY to the prism RVFSTY: And alternately, as the pris ms KBXLMO, LXCOMN are to the prisms PEVRTS, b 7. 5. RVESTY; so is the prism LXCOMN to the prism Book XII. RVFSTY: Bat as the prism LXCOMN to the prism RVFSTY, so is, as has been proved, the base ABC to the С B X E V F 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 are the two prisms in the pyramid OMNG 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 trian- Sce N. gular 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. Book XII. 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*. 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 a 3. 12. prisms: Therefore these two prisms are greater a than the half of the whole pyramid. And again, let the pyramids made by this division be in like manner divided, 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 b 4. 12. ABC to the base DEF, so b 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 hiypothesis, 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 c 14. 5. than the prisms contained in it; wherefore c also the solid Q is greater than the prisms in the pyramid DEFH. But it is also less, which is impossible. Therefore the base ABC is not to the base DEF, as the pyramid ABCG to any solid which is This may be explained the same way as in the note at * in Proposition 2, in like case. |