Supplement in M. But KN is equal to KL, and therefore their halves KM and KH are alfo equal. Wherefore, in the triangles GKH, GKM, the two fides GK and KH are equal to the two GK and KM, each to each; and the angles GKH, GKM are 4.1. alfo equal, therefore GM is equal to GHd; 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 fame manner, by joining the centre and the other angular points of the infcribed polygon, an equilateral polygon may be defcribed about the circle, the fides of which will each be equal to KL, and will be equal in number to the fides of the infcribed polygon. Therefore, KL is the fide of an equilateral polygon, defcribed about the circle, of the fame number of fides with the infcribed polygon ABCDEF; which was to be

found.

COR. 1. Because GL, GK, GN, and the other ftraight lines drawn from the centre G to the angular points of the polygon defcribed about the circle ABD are all equal; if a circle be defcribed from the centre G, with the distance GK, the polygon will be infcribed in that circle; and therefore, it is fimi2. 1. Sup. Iar to the polygon ABCDEF e.

COR. 2. It is evident that AB, a fide of the inscribed polygon, is to KL a fide of the circumfcribed, as the perpendicular from G upon AB, to the perpendicular from G upon KL, that is to the radius of the circle; therefore alfo, because f15. 5 magnitudes have the fame ratio with their equimultiples f, the perimeter of the infcribed polygon is to the perimeter of the circumfcribed, as the perpendicular from the centre, on a fide of the infcribed polygon, to the radius of the circle.

A

CIRCLE being given, two fimilar polygons may be found, the one described about the circle, and the other infcribed in it, which fhall differ from one another by a space lefs than any given space.

Let ABC be the given circle, and the fquare of D any given space; a polygon may be infcribed in the circle ABC, and a fimilar polygon defcribed about it, so that the difference between them fhall be less than the fquare of D.

In

In the circle ABC apply the straight line AE equal to D, Book I. 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 fo on till the circumference AF is found lefs than the circumference AE a. Find the cen- a t. 1. Sup tre G; draw the diameter AC, as also the straight lines AF and FG; and having bifected 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 ifofceles triangles HGL and AGF have the common angie AGF, they are equiangular, and the angles GHK, b 6. 6. 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 likewife equiangular c.

And becaufe the arch AF was found by taking from the arch AB its half, and from that remainder its half, and fo on, AF will be contained a certain number of times, exactly, in the arch AB, and therefore it will alfo be contained a certain number of times, exactly, in the whole circumference ABC; and the ftraight line AF is therefore the fide of an equilateral polygon infcribed in the circle ABC. Wherefore alfo, HL is the fide of an equilateral polygon, of the fame number

fquare of AG, that is of GK. But the triangles HGK, ACF have been proved to be fimilar, and therefore, the square of AC is to the fquare of CF as the polygon M to the polygon N; and, by converfion, the fquare of AC is to its excess above the square

P 3

of

C 32. In

d 3. 1. Sup:

e cor. 3. 1. Sup.

f3. cor. 20. 6.

g 47. I.

Supplement of CF, that is, to the fquare of AF g, as the polygon M to its excefs above the polygon N. But the fquare of AC, that is, the fquare defcribed about the circle ABC is greater than the equilateral polygon of eight fides defcribed about the circle, because it contains that polygon; and, for the fame reason, the polygon of eight fides is greater than the polygon of fixteen, and fo on; therefore, the fquare of AC is greater than any polygon defcribed about the circle by the continual bifection of the arch AB; it is therefore greater than the polygon M. Now, it has been demonftrated, that the fquare of AC is to the fquare of AF as the polygon M to the difference of the polygons; therefore, fince the fquare of AC is greater than M, the fquare of AF is greater than the difference of the h 14.5. polygonsh. The difference of the polygons is therefore less than the fquare of AF; but AF is less than D; therefore, the difference of the polygons is lefs than the fquare of D, that is, than the given fpace. Therefore, &c. Q.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 fquare of D. And therefore, however fmall any given space may be, a polygon may be infcribed in the circle, and another defcribed about it, each of which shall differ from the circle by a space lefs than the given space.

COR. 2. The space B which is greater than any polygon that can be infcribed in the circle A, and lefs than any polygon that

A

B

C

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

are all greater than B, by hypothefis; and because B is great- Book I. er than A by the space C, therefore, no polygon can be described about the circle A, but what muft exceed it by a space greater than C, which is abfurd. In the fame manner, if B be less than A by the space C, it is fhewn that no polygon can be inscribed in the circle A, but what is less than A by a fpace greater than C, which is alfo abfurd. Therefore, A and B are not unequal, that is, they are equal to one another.

TH

HE area of any circle is equal to the rectangle contained by the femidiameter, 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 circumference, the area of the circle is equal to the rectangle contained by DA and AH.

Let AB be the fide of any equilateral polygon inscribed in the circle ABC; bifect the circumference AB in G, and through G drawn EGF touching the circle, and

meeting DA produced in E, and DB produced in F; EF will be the fide of an equilateral polygon described about the circle ABC a. In AC produced take AK equal to a 3. 1, Sup. half the perimeter of the polygon whofe fide is AB; and AL equal to half the perimeter of the polygon whose fide is EF.

P 4

Then

bax. 1.Sup.

c 41. 1.

Supplement Then AK will be lefs, and AL greater than the ftraight line AH. Now, becaufe 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 c; and as the fame is true of all the other equal triangles having their ver tices in D, which make up the polygon defcribed about the circle; therefore, the whole polygon is equal to the rectangle contained by DG and AL, half the perimeter of the polygon, 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 lefs than the rectangle DA.AL, that is, than any polygon described about the circle ABC.

d 1. 2.

Again, the triangle ADB is equal to the rectangle contained by DM the perpendicular, and one half of the bafe AB, and it is therefore lefs than the rectangle contained by DG,

e 2. Cor.

z. Sup.

or DA, and the half of AB. And as the fame is true of all the other triangles having their vertices in D, which make up the infcribed polygon, therefore the whole of the infcribed polygon is lefs than the rectangle contained by DA, and AK half the perimeter of the polygon. Now, the rectangle DA. AK is lefs than DA.AH; much more, therefore, is the polygon whofe fide is AB lefs than DA.AH; and the rectangle DA.AH is therefore greater than any polygon infcribed in the circle ABC. But the fame rectangle DA. AH has been proved to be less than any polygon defcribed about the circle ABC; therefore, the rectangle DA.AH is 4. equal to the circle ABC e. Now, DA is the femidiameter of the