their homologous sides (20. 6.), the polygon ABCDEF is to the polygon GHIKLM as the square of AB to the square of GH; but because the triangles ANB, GOH are equiangular, the square of AB is to the square of GH as the square of AN to the square of GO (4. 6.), or as four times the square of AN to four times the square (15. 5.) of GO, that is, as the square of AD to the square of GK (2. Cor. 8. 2.). Therefore also, the polygon ABCDEF is to the polygon GHIKLM as the square of AD to the square of GK ; and they have also been shewn to be similar. Therefore, &c. Q. E. D. Cor. Every equilateral polygon inscribed in a circle is also equiangular : For the isosceles triangles, which have their common vertex in the centre, are all equal and similar ; therefore, the angles at their bases are all equal, and the angles of the polygon are therefore also equal. PROP. III. THEOR. The side of any equilateral polygon inscribed in a circle being given, to find the side of a polygon of the same number of sides described about the circle. Let ABCDEF be an equilateral polygon inscribed in the circle ABD; it is required to find the side of an equilateral polygon of the same number of sides described about the circle. Find G the centre of the circle ; join GA, GB, bisect the arch AB in H ; and through H draw KHL touching the circle in H, and meeting GA and GB produced in K and L ; KL is the side of the polygon required. Produce GF to N, so that GN may be equal to GL ; join KN, and from G draw GM at right angles to KN, join also HG. Because the arch AB is bisected in H, the angle AGH is equal to the angle BGH (27. 3.) ; and because KL touches the circle L in H, the angles LHG, KHG are right angles (16.3.); therefore, there are two angles of the tri- HI, angle HGK, equal to two angles of the triangle HGL, each to each. But the side GH is K G M F the angle LGK equal to the an N gle KGN ; therefore the base KL'is equal to the base KN (4. 1.); But because the triangle KGN is isosceles, the angle GKN is equal to the angle GNK, and the angles GMK, GMN are both right angles by construction ; wherefore, the triangles GMK, GMN have two angles of the one equal to two angles of the other, and they have also the side GM common, therefore they are equal (26. 1.), and the side KM is equal to the side MN, so that KN is bisected in M. But KN is is equal to KL, and therefore their halves KM and KH are also equal. Wherefore, in the triangles GKH, GKM, the two sides GK and KH are equal to the two GK and KM, each to each ; and the angles GKH, GKM, are also equal, therefore GM is equal to GH (4. 1,) ; wherefore, the point M is in the circumference of the circle ; and because KMG is a right angle, KM touches the circle. And in the same manner, by joining the centre and the other angular points of the inscribed polygon, an equilateral polygon may be described about the circle, the sides of which will each be equal to KL, and will be equal in number to the sides of the inscribed polygon. Therefore, KL is the side of an equilateral polygon, described about the circle, of the same number of sides with the inscribed polygon ABCDEF ; which was to be found. Cor. 1. Because GL, GK, GN, and the other straight lines drawn from the centre G to the angular points of the polygon described about the circle ABD are all equal ; if a circle be described from the centre G, with the distance GK, the polygon will be inscribed in that circle ; and therefore, it is similar to the polygon ABCDEF (2. 1.). Cor. 2. It is evident that AB, a side of the inscribed polygon, is to KL, a side of the circumscribed, as the perpendicular from G upon AB, to the perpendicular from G upon KL, that is to the radius of the circle ; therefore also, because magnitudes have the same ratio with their equimultiples (15. 5.), the perimeter of the inscribed polygon is to the perimeter of the circumscribed, as the perpendicular from the centre, on a side of the inscribed polygon, to the radius of the circle. A circle being given, two similar polygons may be found, the one de scribed about the circle, and the other inscribed in it, which shall differ from one another by a space less than any given space. Let ABC be the given circle, and the square of D any given space ; , a polygon may be inscribed in the circle ABC, and a similar polygon described about it, so that the difference between them shall be less than the square of D. In the circle ABC apply the straight line AE equal to D, and let AB be a fourth part of the circumference of the circle. From the circumference AB take away its half, and from the remainder its half, and so on till the circumference AF is found less than the circumference AE (1. 1. Sup.). Find the centre G; draw the diameter AC, as also the straight lines AF and FG ; and having bisected the circumference a AF in K, join KG, and draw HL touching the circle in K, and meeting GA and GF produced in Hand L ; join CF. Because the isosceles triangles HGL and AGF have the common angle AGF, they are equiangular (6. 6.), and the angles GHK, GAF are therefore equal to one another. But the angles GKH, CFA are also equal, for they are right angles ; therefore the triangles HGK, ACF, are likewise equiangular (32. 1.). And because the arch AF was found by taking from the arch AB its half, and from that remainder its half, and so on, AF will be contained a certain number of times, exactly, in the arch AB, and therefore it will also be contained a certain number of times, exactly, in the whole circumference, ABC ; and the straight line AF is therefore the side of an equilateral polygon inscribed in the circle ABC. Wherefore also, HL is the side of an equilateral polygon, of the same number of sides, described about ABC (3. 1. Sup.). Let the polygon described about the circle be called M, and the polygon inscribed be called N ; then, a Decause these polygons are similar (Cor. 3. 1.), they are as the squares of the homologous sides HL and AF (Sup. 3. Cor. 20.6.), that is, because the triangles HLG, AFG are similar, as the square of HG to the square of AG, that is of GK. But the triangles HGK, ACF have been proved to be similar, and therefore the square of AC is to the square of CF as the polygon M to the polygon N ; and, by conversion, the square of AC is to its excess above the square of CF, that is, to the square of AF (47. 1.), as the polygon M to its excess above the polygon N. But the square of AC, that is, the square described about the circle ABC is greater than the equilateral polygon of eight sides described about the circle, because it contains that polygon ; and, for the same reason, the polygon of eight sides is greater than the polygon of sixteen, and so on ; therefore, the square of AC is greater than any polygon described about the circle by continual bisection of the arch AB; it is therefore greater than the polygon M. Now, it has been demonstrated, that the square of AC is to the square of AF as the polygon M to the difference of the polygons ; therefore, since the square of AC is greater than M, the square of AF is greater than the difference of the polygons (14. 5.). The difference of the polygons is therefore less than the square of AF ; but AF is less than D therefore, the difference of the polygons is less than the square of D, that is, than the given space. Therefore, &c R. E. D, Cor. 1. Because the polygons M and N differ from one another more than either of them differs from the circle, the difference between each of them and the circle is less than the given space, viz. the square of D. And therefore, however small any given space may be, a polygon may be inscribed in the circle, and another described about it, each of which shall differ from the circle by a space less than the given space. C Cor. 2. The space B which is greater than any polygon that can be inscribed in the circle A, and less than any polygon that can be described about it, is equal to the circle A. If not, let them be unequal ; and first, let B exceed A by the space C. Then, because the polygons described about the circle A are all greater than B, by hypothesis ; and because B is greater than A by the space C, there, fore no polygon can be described about the circle A, but what must exceed it by a space greater than C, which is absurd. In the same manner, if B be less than A by the space C, it is shewn that no poly. gon can be inscribed in the circle A, but what is less than A by a space greater than C, which is also absurd. Therefore, A and B are hot unequal, that is, they are equal to one another. ܪ PROP. V. THEOR. 1 The area of any circle is equal to the rectangle contained by the semi diameter, and a straight line equal to half the circumference, Let ABC be a circle, of which the centre is D, and the diameter AC ; if in AC produced there be taken AH equal to half the cir. cumference, the area of the circle is equal to the rectangle contained by DA and AH a Let AB be the side of any equilateral polygon inscribed in the circle ABC ; bisect the circumference AB in G, and through G draw EGF touching the circle, and meeting DA produced in E, and DB produced in F ; EF will be the side of an equilateral polygon described about the circle ABC (3. 1. Sup.). In AC produced take AK equal to half the perimeter of the polygon whose side is AB ; and Al equal to half the perimeter of the polygon whose side is EF Then AK will be less, and AL greater than the straight line AH (Ax. 1. Sup.). Now, because in the triangle EDF, DG is drawn perpendicular to the base, the triangle EDF is equal to the rectangle contained by DG and the half of EF (41. 1.); and as the same is true of all the other equal triangles having their vertices in D, which make up the polygon described about the circle ; therefore, the whole polygon is equal to the rectangle contained by DG and AL, half the perimeter of the polygon (1. 2.), or by DA and AL. But AL is greater than AH, therefore the rectangle DA,AL is greater than the rectangle DA.AH ; the rectangle DA AH is therefore less than the rectangle DA.AL, that is, than any polygon described about the circle ABC. Again, the triangle ADB is equal to the rectangle contained by DM the perpendicular, and one half of the base AB, and it is therefore less than the rectangle contained by DG, or DA, and the half of AB. |