similar and equiangular (6. 6.); and therefore the angle AEB is equal to the angle FLG: But AEB is equal (21.3.) to AMB, because they stand upon the same circumference; and the angle FLG is, for the same reason, equal to the angle FNG: Therefore also the angle AMB is equal to FNG: And the right angle BAM is equal to the right (31. 3.) angle GFN; wherefore the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one another: Therefore as BM to GN, so (4.6.) is BA to GF; and therefore the duplicate ratio of BM to GN is the same (10. def. 5. & 22. 5.) with the duplicate ratio of BA to GF: But the ratio of the square of BM to the square PROP. II. THEOREM. Circles are to one another as the squares of their diameters. Let ABCD, EFGH be two circles, and BD, FH their diameters: As the square of BD to the square of FH, so is the circle ABCD, to the circle EFGH. For, if it be not so, the square of BD shall be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it. First, let it be to a space S less than the circle EFGH; and in the circle EFGH describe the square EFGH: This square is greater than half of the circle EFGH; because if, through the points E, F, G, H, there be drawn tangents to the circle, the square EFGH is half (41. 1.) of the square described about the circle; and the circle is less than the square described about it; therefore the square EFGH is greater than half of the circle. Divide the circumferences EF, FG, GH, HE, each into two equal parts in the points K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE: Therefore each of the triangles EKF, FLG, GMH, HNE is greater than half of the segment of the circle it stands in; because, if straight lines touching the circle be drawn through the points K, L, M, N, and parallelograms upon the straight lines EF, FG, GH, HE, be completed; each of the triangles EKF, FLG, GMH, HNE shall be the half (41. 1.) of the parallelogram in which it is: But every segment is less than the parallelogram in which it is: Wherefore each of the triangles EKF, FLG, GMH, For there is some square equal to the circle ABCD; let P be the side of it: and to three straight lines BD, FH, and P, there can be a fourth proportional; let this be Q: Therefore the squares of these four straight lines are proportionals; that is, to the squares of BD, FH, and the circle ABCD, it is possible there may be a fourth proportional. Let this be S. And in like manner are to be understood some things in some of the following propositions. this, there will at length remain segments of the circle, which, together, shall be less than the excess of the circle EFGH, above the space S: Because, by the preceding Lemma, if from the greater of two unequal magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than the least of the proposed magnitudes. Let then the segments EK, KF, FL, LG, GM, MH, HN, NE be those that remain and are together less than the excess of the circle EFGH, above S: Therefore the rest of the circle, viz. the polygon EKFLGMHN, is greater than the space S. Describe likewise in the circle ABCD the polygon AXBOCPDR similar to the polygon EKFLGMHN: As therefore, the square of BD is to the square of FH, so (1. 12.) is the polygon AXBOCPDR to the polygon EKFLGMHN; But the square of BD is also to the square of FH, as the circle ABCD is to the space S: Therefore as the circle ABCD is to the space S, so is (11.5.) the polygon AXBOCPDR to the polygon EKFLGMHN: But the circle ABCD is greater than the polygon contained in it: wherefore the space S, is greater (14. 5.) 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 cir cle 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, so is the space T to the circle ABCD. But as the space R. K X E N T is to the circle ABCD, so is the circle 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 • For, as in the foregoing note at, it was explained how it was possible there could be a fourth proportional to the squares of BD, FH, and the circle ABCD, which was named S. So, in like manner, there can be a fourth proportional to this other space, named T, and the circles ABCD, EFGH. And the like is to be understood in some of the following propositions. 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 circ! ABCD to the circle EFGH. Circles therefore are, &c. Q. E. D. PROP. III. THEOREM. 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 pyramid. E D B F C 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 K 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, EG, GH, HK, KL, LH, EK, KF, FG. Because AE is equal to EB, and AH to 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; wherefore EA, AH are equal to KH, HD, each to each; and the angle EAH is equal (29. 1.) to the angle KHD; therefore the base EH is equal 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 D 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 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 K because HK is parallel to AB a side of the triangle ADB, the triangle ADB is equiangu- B lar 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 triangle HDL; and also the triangle ABC to the triangle AEG: But the triangle AEG is similar to the triangle HKL, as before was proved; therefore the triangle ABC is similar (21. 6.) to the 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. triangle HKL. And the pyramid of which the base is the triangle ABC, and vertex the point D, is therefore similar (B. 11. & 11. def. 11.) to the pyramid of which 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 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 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 (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 whols 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. THEOREM. 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 be cause 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 becaase 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: Therefore, 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, 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 parts in the points N, Y by the same planes: Therefore the prisms LXCOMN, RVFSTY are of the same al titude; 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 equa, 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 pristn 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 op posite 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 LXCOMN: so are the prisms PEVRTS, RVFSTY to the prism RVFSTY: And permutando, as the prisms KBXLMÒ, LXCOMN are to the prisms PEVRTS, RVFSTY; so is the prism LXCOMN to the prism RVISTY: 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 prisma in the pyramid DEFH; and so are the two prisms in the pyramid OMNG |