Book X11. equiangular to one another: therefore as BM to GN, 50 e is more on BA to Gf; and therefore the duplicate ratio of BM to GN, e 4. 6. is the same with the duplicate ratio of BA to GF: but the f 10. def. 5. ratio of the square of BM to the square of GN, is the dupli& 22.5 cate & ratio of that which 'BM has to GN; and the ratio of the g 20. 6. polygon ABCDE to the polygon FGHKL is the duplicate & 6 of that which BA has to GF; therefore, as the square of BM to the square of GN so is the polygon ABCDE, to the polygon FGHKL, Wherefore similar polygons, &c. Q. ED. PROP. II. THEOR. See Note. CERCLES arc 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 *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 he 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. EFGH is half of the square described about the circles and Book XII. the circle is less than the square described about it; therefore the square EFGH is greater than half of the circle. Divide the a 41. 1 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, GMA, 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 halfa of the parallelogram in which it is : but every segment is less than the paralielogram in which it is: wherefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle which contains it : and if these circumferences before named be divided cach into two equal parts, and their extremities be joined by straight lines, by continuing to do this, there will at length rem main segments of the circle, which, together, shall be less than the excess of the circle EFGH above the space S: because, by the preseding 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, sob is the polygon AXBOCPDR to the b 1. 12. polygon EKFLGMHN: but the square of BD is also to the squarc of FH, as the circle ABCD is to the space S: therefore Book XII. as the circle ABCD is to the space S, so is the polygon AXBOCPDR to the polygon EKFLGMHN: but the circle c 11. 5. ABCD is greater than the polygon contained in it :"wherefore d 14. 5. the space S is greater d than the polygon EKFLGMHN : but is likewise less, as has been demonstrated; which is impos- square of FH to the square of BD, as the circle EFGH is A E K N () cle ABCD. But as the spacet T is to the circle ABCD, so is the † For, as in the foregoing note at, it was explained how it was possi, of BD, so is the circle EFGH to a space less than the circle gook XII. 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 EFGH*. Circles Cherefore are, &c. Q. E. D. PROP. III. THEOR, EVERY pyramid having a triangular base, may See Note, 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. 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 pyra. D mids 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, EG, GH, HK, KL, K L. a 2.6. for the same reason HK is parallel to AB: therefore HEBK is a parallelogram, and HK equal-b to EB : but EB is equal to AE; E 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 an-B F с gle EAH is equal c to the angle KHD; c29. cherefore the base EH is equal to the base b 34. * 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, 8 4. 6. Book X11. KD, and the triangle AEH equal d and similar to the triangle moHKD: for the same reason, the triangle AGH is equal and d 4.1. 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 equale anglés; 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 equald 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 1 C, 11. is the point H, is equals and similar to the pyramid the base D HT L was proved ; therefore the triangle ABC is 21. 6. sin ilarh to the triangle HKL. And the py- B i B. 11. & lari to the pyramid of which the base is 11 def. 11. the triangle HKL,and vertex the same point D : but the pyra mid 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 ånd vertex H: therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD: and be. k 41. 1. cause BF is equal to FC, the parallelogram EBFG is double kof the triangle GFC: but when there are two prisms of the same altitude,of which one has a parallelogram forits base and the other a triangle that is half of the parallelogram, these prisms are a 40.11. equal to one another; therefore the prism having the parallelo gram 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 tp it ; for they are of the same alti |