Similar Polygons, inscribed in Circles, are to one another as the Squares of the Diameters of the Circles. L ET ABCDE, FGHKL, be Circles, be Diameters of the Circles. I say, as the Square of B M is to the Square of GN, so is the Polygon ABCDE to the Polygon FGHKL. For join BE, AM, GL, FN. Then because the Polygon ABCDE is similar to the Polygon FGHKL, the Angle BAE is equal to the Angle GFL; and BA is to AE, as GF is to FL. Therefore the two Triangles B AE, GFL, have one Angle of the one equal to one Angle of the other, viz. the Angle BAE equal to the Angle GFL, and the Sides about the equal Angles proportional. Wherefore the Triangle ABE is * equiangular to the Triangle FGL;* 6. 6. and +21. 3. 1 31. 3. 4. 6. and so the Angle A E B is equal to the Angle FLG; But the Angle A EB is † equal to the Angle A MB; for they stand on the fame Circumference; and the Angle FLG is t equal to the Angle FNG. Therefore the Angle AM B is equal to the Angle FNG. But the Right Angle BAM is I equal to the Right Angle GFN. Wherefore the other Angle shall be equal to the other Angle. And so the Triangle AMB is equiangular to the Triangle FGN; and consequently * as BM is to GN, so is BA to GF. But the Proportion of the Square -of BM to the Square of GN, is duplicate of the Proportion of BŃ to GN; and the Proportion of the Polygon ABCDE to the Polygon FGHKL, is + duplicate of the Proportion of B A to GF. Wherefore, as the Square of BM is to the Square of GN, so is the Polygon ABCDE to the Polygon F GHKL. Therefore, fimilar Polygons, inscribed in Circles, are to one another as the Squares of the Diameters of the Circles ; which was to be demonstrated, * 20.6. L E M M A. If there be two unequal Mag- D А A greater than its Half; and K F K if from what remains there be again taken a Part H H G ET AB and C be two unequal Magnitudes, be taken a greater Part than half, and from the Part remaining E B is , remaining there be again taken a Part greater than its half, and this be done continually, there will remain a Magnitude at last that shall be less than the Magnitude C. For C being fome Number of Times multiplied, will become greater than the Magnitude AB. Let it be multiplied, and let DE be a Multiple of C greater than AB. Divide DE into Parts DF, FG, GE each equal to C, and take BH a Part greater than half of AB from AB, and again from AH the Part HK greater than half AH, and from AK a Part greater than half AK, and so on, until the Divisions that are in AB are equal in Number to the Divifions in DE. Therefore let the Divisions AK, KH, HB, be equal in Number to the Divisions DF, FG, GE. Then because DE is greater than AB, and the Part EG is taken from EĎ, being less than half thereof, and the Part BH greater than half of AB is taken from it, the Part remaining GD, shall be greater than the Part remaining H.. Again, because GD is greater than HA; and GF being half of GD, is taken from the fame; and HK being greater than half HA, is taken from this likewise, the Part remaining FD, shall be greater than the Part remaining AK; but FD is equal to C. Therefore C is greater than AK; and so the Magnitude AK is lesser than C. Therefore the Magnitude AK being the Part remaining of the Magnitude AB, is less than the lefser propos d Magnitude C; which was to be demonstrated. If the Halves of the Magnitudes fhould have been taken, we demonftrate this after the same Manner. This is the first Proposition of the tenth Book. PROPOSITION II. THEORE M. Circles are to each other as the Squares of their Diameters. ET ABCD, EFGH, be Circles, whose Dia meters are BD, FH. I say, as the Square of BD is to the Square of FH, fo the Circle ABCD to the Circle EFGH, For * 41. 1. For if it be not so, the Sqaure of BD shall be to the Square of FH, as the Circle ABCD is to some Space either less or greater than the Circle EFGH. First let it be to a Space S, less than the Circle EFGH, and let the Square EFGH be described therein. This Square E F GH will be greater than half the Circle EF GH; because if we draw Tangents to the Circle thro' the Points E, F, G, H, the Square EFGH will be half that described about the Circle ; but the Circle is less than the Square described about it. Therefore the Square EFGH is greater than half the Circle EFGH. Let the Circumferences EF, FG, GH, HE, be bisected in the Points K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE. Then each of the Triangles EKF, FLG, GMH, HNE, will be * greater than one half of the Segment of the Circle it stands in. Because if Tangents at the Circle be drawn thro' the Points K, L, M, N, and the Parallelograms that are on the Right Lines EF, FG, GH, HE be compleated, each of the Triangles EKF,FLG, GMH, HNE is half of each of the corresponding Parallelograms; but the Segment is less than the Parallelogram. Wherefore each of the Triangles EKF, FLG, GMH, HNĘ is greater than one half of the Segment of the Circle in which it ftands. Therefore if these Circumferences be again bifected, and Right Lines be drawn joining the Points of Bisection, and you do thus continually, there will at last remain Segments of the Circle, that shall be less than the Excefs, by which the Circle E F G H exceeds the Space S. For it is demonstrated in the foregoing Lemma, that if there be two unequal Magnitudes proposed, and if from the greater a Part greater than half be taken from it, and again from the Part remaining a Part greater than half be taken, and you do this continually; there will at last remain a Magnitude that will be less than the leffer proposed Magnitude. Let the Segments of the Circle EFGH on the Right Lines EK, KF, FL, LG, GM, MH, HN, NE, be those which are less than the Excess, whereby the Circle EFGH exceeds the Space S, and then the remaining Polygon EKFLGMHN shall be greater than the Space S. Alfo describe the Polygon AXBOCPDR in the Circle ABCD, fimilar to the Polygon Polygon EKFLGMHN. Wherefore as the Square of B D is to the Square of FH, so is the Polygon AX BOCPDR to * the Polygon EKFLGMHN. But * 1 of tbiso as the Square of BD is to the Square of FH, so is the Circle ABCD to the Space S. Wherefore as the Circle ABCD is to the Space S, fo is † the Polygon † 11. 5. AXBOCPDR to the Polygon EKFDGMHN. But the Circle ABCD is greater than the Polygon in it. Wherefore the Space S shall be t also greater than the Polygon EKFLGMHN, but it is less ft From that likewise, which is abfurd, which is abfurd. Therefore the Square of Hype BD to the Square of FH, is not as the Circle ABCD to some Space less than the Circle EFGH. After the same manner we likewise demonstrate that the Square of FH to the Square of B D is not as the Circle EFGH, to some Space less than the Circle ABCD. Lastly, I say, the Square of BD to the Square of FH is not as the Circle ABCD, to fome Space greater than the Circle EFGH; for if it be possible, let it be so, and let the Space S be greater than the Circle EFGH; then shall it be (by Inversions) as the Square of FH is to the Square of BD, fo is the Space S to the Circle ABCD. But because S is greater than the Circle EFGH, the Space shall be to the Circle ABCD, as the Circle EFGH is to fome Space less than the Circle ABCD. Therefore, 'as the Square of FH is to the Square of BD, so is the Circle 11. 5. EF GH to some Space less than the Circle ABCD; which has been demonstrated to be impossible. Wherefore the Square of BD to the Square of FH, is not as the Circle ABCD to fome Space greater than the Circle E F G H. But this also has been proved, that the Square of BD to the Square of FH, is not as the Circle ABCD to fome Space less than the Circle EFGH. Wherefore as the Square of BD is to the Square of FH, fo shall the Circle ABCD be to the Circle EFGH. Wherefore Circles are to each other as the Squares of their Diameters; which was to be demonstrated. 7 PRO. |