PRO P. II. THEOREM S. Similar polygons inscribed in circles, are to each other as the squares of the diameters of those circles. Let ABCDE, FGHKL be two similar polygons, inscribed in the circles ABD, FGK : then will the polygon ABCDE be to the polygon FGHKL as the square of the diameter BM is to the square of the diameter gn. For join the points B, E and A, M, G, L and F, N: Then, because the polygon ABCDE is similar to the polygon FGHKL (by Hyp.), the angle Bae is equal to the angle GFL, and BA is to AE, as GF is to FL (VI. Def. 1.) And, since the angle BAE, of the triangle Abe, is equal to the angle GFL, of the triangle FGL, and the sides about those angles are proportional, the angle AEB will also be equal to the angle FLG (VI. 5.) But the angle AEB is equal to the angle AMB, and the angle Fly to the angle FNG (III. 15.), consequently the angle AMB is also equal to the angle FNG. And fince thefe angles are equal to each other, and the angles BAM, GFN are each of them right angles (III. 16.), the angle MBA will alfo be equal to the angle NGF (1.28. Gar.), and BM will be to GN AS BA is to gr (VI. 5.) But But the polygon ABCDE is to the polygon FGHKL as the fquare of BA is to the square of GF (VI. 17.), therefore the polygon ABCDE is also to the polygon FGHKL as the square of BM is to the square of Gn. Q. E. D. PROP. III. THEOREM. A polygon may be inscribed in a circle that shall differ from it by less than any afligned magnitude whatever. Let ABCD be a circle, and s any given magnitude whatever ; then may a polygon be inscribed in the circle ABCD that shall differ from it by less than the magni tude s. For, let AC, EG be two squares, the one described in the circle ABCD, and the other about it (IV. 6, 7.); and bisect the arcs AB, BC, CD, DA, in the points m, n, r and s (III. 23.); and join Am, MB, Bn, nc, cr, 1D, Ds and sa: Then since the square Ac is half the square EG (I. 32.), and the square EG is greater than the circle ABCD, the square Ac will be greater than half the circle ABCD. In like manner, if tangents be drawn to the circle through the points m, n, r, s, and parallelograms be de fcribed upon the right lines AB, BC, CD, DA, the triangles AMB, BNC, Crd, DSA will each of them be half che parallelogram in which it stands (I. 32.) But every segment is less than the parallelogram which circumscribes it; and therefore each of the triangles AmB, BNC, COD, DSA is greater than half the segment of the circle which contains it. And, if each of the arcs Am, mb, &c. be again divided into two equal parts, and right lines be drawn to the points of bisection, the triangles thus formed, may in like manner, be shewn to be greater than half the fegments which contain them; and so on continually. Since, therefore, the circle ABCD is greater than the space s, and from the former there has been taken more than its half, and from the remainder more than its half, &c. there will at length remain segments which, taken together, shall be less than the excess of the circle ABCD above the space s (VIII. 1.), as was to be thewn. PRO P. IV. THEOREM. A polygon may be circumscribed about a circle that shall differ from it. by less than any assigned magnitude whatever, Let ABCD be the circle, and s any given magnitude whatever ; then may a polygon be circumscribed about the circle ABCD, that shall differ from it by less than the magnitude s. For let the circle ABCD be circumscribed by the square IFGH (IV. 7.), and bifect the arcs AB, BC, CD, DA with the lines oe, of, og and oh ; and to the points of bisection draw the tangents kl, mn, pr, st (III. 1o.) Then since klis a tangent to the circle, and oe is drawn from the centre through the point of contact, the angle Exk is a right angle (III. 12.), and Ek will be greater than kx (I. 17.) or its equal ka. But triangles of the same altitude are to each other as their bases (VI. 1.); whence the base Ek being greater than the base ka, the triangle Exk will also be greater than the triangle kxA. And because the triangle Exk is greater than the triangle kxA, it will also be greater than half the curvelineal space Exa: and the fame may be shewn of any other triangle and the curvelineal space to which it belongs. In like manner, if the arcs AX, XB, &c. be again bifected, and tangents be drawn to the points of bifection, the triangles thus formed will be greater than half the curvelineal spaces to which they belong. Since, therefore, the excess of the square above the circle is greater than the magnitude s, and from the former there has been taken more than its half, and from the remainder more than its half, and so on, there will at length remain spaces, which, taken together, shall be less than the magnitude s (VIII. 1.), as was to be thewn. PRO P. V. THEOR E MS. Circles are to each other as the squares of their diameters. Let ABCD, EFGH be two circles, and BD, fh their diameters: then will the square of BD be to the square of FH as the circle ABCD is to the circle EFGH. For, if they have not this ratio, the square of BD will 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 st less than the circle EFGH; and infcribe the two similar polygons AROPQ, EKLMN so that the circle EFGH may exceed the latter by less than it exceeds the space st (VIII. 3.) Then, fince the circle EFGH exceeds the polygon EKLMN by less than it exceeds the space st, the polygon EKLMN will be greater than the space st. And, because similar polygons, inscribed in circles, are to each other as the squares of their diameters (VIII. 2.), the square of BD will be to the square of Fh as the polygon AROPQ is to the polygon EKLMN. But the square of BD is also to the square of fh as the circle ABCD is to the space st (by Conft.) ; whence the circle ABCD will be to the space ST, as the polygon AROPQ is to the polygon EKLMN. |