PROP. I. THEOR.

If from the greater of two unequal magnitudes there be taken away its half, and from the remainder its half: and so on; There will at length remain a magnitude less than the least of the proposed magnitudes.

A

K+

H+

F

Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken away its half, and from the remainder its half, and so on; there shall at length remain a magnitude less than C. For C may be multiplied so as, at length, to become greater than AB. Let DE, therefore, be a multiple of C, which is greater than AB, and let it contain the parts DF, FG, GE, each equal to C. From AB take BH equal to its half, and from the remainder AH, take HK equal to its half, and so on, until there be as many divisions in AB as there are in DE: And let the divisions in AB be AK, KH, HB. And because DE is greater than AB, and EG taken from DE is not greater than its half, but BH taken from AB is equal to its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and GF is not greater than the half of GD, but HK is equal to the half of HA; therefore, the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C. Q. E. D.

PROP. II. THEOR.

BCE

Equilateral polygons, of the same number of sides, inscribed in circles, are similar, and are to one another as the squares of the diameters of the circles.

Let ABCDEF and GHIKLM be two equilateral polygons of the same number of sides inscribed in the circles ABD, and GHK; ABCDEF and GHIKLM are similar, and are to one another as the squares of the diameters of the circles ABD, GHK.

Find N and the centres of the circles; join AN and BN, as also GO and HO, and produce AN and GO till they meet the circumferences in D and K.

Because the straight lines AB, BC, CD, DE, EF, FA, are all equal, the arches AB, BC, CD, DE, EF, FA are also equal (28. 3.). For the same reason, the arches GH, HI, IK, KL, LM; MG are all equal, and they are equal in number to the others; therefore, whatever part the arch AB is of the whole circumference, ABD, the same is the arch GH of the circumference GHK But the angle ANB is the same part of four right angles, that the arch AB is of the circumference ABD (33. 6.); and the angle GOH is the same part of four right angles that the arch GH is of the circumference GHK (33. 6.) therefore the angles ANB, GOH are each of them the same part of four right angles, and therefore they are equal to one another. The isosceles triangles ANB, GOH are therefore equiangular (6. 6.), and the angle ABN equal to the angle GHO; in the same manner, by joining NC, OI, it may be proved that the angles NBC, OHI are

equal to one another, and to the angle ABN. Therefore the whole angle ABC is equal to the whole GHI; and the same may be proved of the angles BCD, HIK, and of the rest. Therefore, the polygons ABCDEF and GHIKLM are equiangular to one another; and since they are equilateral, the sides about the equal angles are proportionals; the polygon ABCD is therefore similar to the polygon GHIKLM (def. 1. 6.). And because similar polygons are as the squares of

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 in H, the angles LHG,KHG are right angles (16. 3.); therefore, there are two angles of the triangle HGK, equal to two angles of the triangle HGL, each to each. But the side GH is common to these triangles; therefore they are equal (26. 1.), and GL is equal to GK. Again, in the triangles KGL, KGN, because GN is equal to GL; and GK common, and also the angle LGK equal to the an

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 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, and 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.

PROP. IV. THEOR

A circle being given, two similar polygons may be found, the one described 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

AF in K, join KG, and draw HL touching the circle in K, and meeting GA and GF produced in H and 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,

because 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 the continual bisection of the arch AB; it is therefore greater than the polygon M. Now, it