equal to the angle FNG: therefore also the angle AMB is equal to FNG: and the right angle BAM is 31. 3. equal to the right angle GFN; wherefore the remaining angles in the triangles ABM, FGN are equal, * 4. 6. 5. and they are equiangular to one another: therefore as BM to GN, so is BA to GF; and therefore the 10 Def. duplicate ratio of BM to GN, is the same with the 5. & 22. duplicate ratio of BA to GF. But the ratio of the 20. 6. square of BM to the square of GN, is the duplicate* ratio of that which BM has to GN; and the ratio of the polygon ABCDE to the polygon FGHKL is the 20. 6. duplicate* of that which BA has to GF: therefore as the polygon ABCDE to the polygon FGH KL, so is the square of BM to the square of GN. Wherefore similar polygons, &c. Q. E. D. PROP. II. THEOR. 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 shall the circle ABCD be to the circle EFGH. For, if it be not so, the square of BD must 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, if possible, let it be to a space S less than the circle EFGH; and in the circle EFGH inscribe the square EFGH: this square is greater than * 6. 4. 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 of the square de- *4 . 1. scribed 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 in which it stands; 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 is the half** 41. 1. of the parallelogram in which it is: but every segment is less than the parallelogram in which it is: therefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle which contains it. Again, if the circumferences before named be divided each into two equal parts, and their extremities be joined by straight lines by continuing to do this, and there will at length remain segments of the circle which, together, are less than the excess of the circle EFGH above the space S: because, by the preceding lemma, if from the greater of two unequal . 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 * 22. 6. 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. magnitudes there be taken more than its half, and from B R M H о the remainder more than its half, and so on, there at length remains 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; that is, the space S is less than the polygon EKFLGMHN. Describe likewise in the circle ABCD the polygon AXBOCPDR similar to the polygon EKFLGMHN: therefore as the square of BD is to * 1. 12. the square of FH, so 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 11. 5. S, so is the polygon AXBOCPDR to the polygon * Hyp. EKFLGMHN: but the circle ABCD is greater than the polygon contained in it; wherefore the space S is *14. 5. greater than the polygon EKFLGMHN: but this is impossible, because it has been demonstrated to be 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 T is to the circle ABCD, so is the circle EFGH to some space, which must be less than the circle ABCD, because the * 14. 5. 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 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: therefore, as the square of BD to the square of FH, so is the circle ABCD to the circle EFGH.b Therefore, circles are, &c. Q. E. D. For as in the foregoing note, 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. 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. 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 pyramid. D 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 shall be greater than half of the whole pyramid. L 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. cause AE is equal to EB, and AH to HD, HE is * 2.6. parallel to DB. For the same reason, HK is parallel 4. 1. Be F to AB: therefore HEBK is a parallelogram, and HK • 34. 1. equal 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 29. 1. the angle EAH is equal to the angle KHD; therefore the base EH is equal to the base KD, and the triangle AEH equal 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 10. 11. not in the same plane with them, they contain equal 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 |