ference 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 angle 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 HLand 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 conver sion, 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 poly gon of sixteen, and so on; therefore, the square of AC is greater than any polygon described about the circle by the continual bi section 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. 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 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.

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, therefore 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 polygon 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 not unequal, that is, they are equal to one another.

PROP. V. THEOR.

The area of any circle is equal to the rectangle contained by the semidiameter, 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 side of any equilateral polygon inscribed in the cirele ABC; bisect the circumference AB in &, 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. And as the same is true of all the other triangles having their vertices in D, which make up the inscribed polygon, therefore the whole of the inscribed polygon is less than the rectangle contained by DA, and AK half the perimeter of the polygon. Now, the rectangle DA.AK is less than DA. AH; much more, therefore, is the polygon whose side is AB less than DA. AH; and the rectangle DA.AH is therefore greater than any polygon inscribed in the circle ABC. But the same rectangle DA. AH has been proved to be less than any polygon described about the circle ABC; therefore, the rectangle DA.AH is equal to the circle ABC (2. Cor. 4. 1. Sup.). Now, DA is the semidiameter of the circle ABC, and AH the half of its circumference. Therefore, &c. Q. E. D.

COR. 1. Because DA: AH:: DA: DA. AH (1. 6.), and because by this proposition, DA.AH=the area of the circle, of which DA is the radius: therefore, as the radius of any circle to the semicircumference, or as the diameter to the whole circumference, so is the square of the radius to the area of the circle.

COR. 2. Hence a polygon may be described about a circle, the perimeter of which shall exceed the circumference of the circle by a line that is less than any given line. Let NO be the given line. Take in NO the part NP less than its half, and less also than AD, and let a polygon be described about the circle ABC, so that its excess above ABC may be less than the square of NP (1. Cor. 4. 1. Sup.). Let the side of this polygon be EF. And since, as has been proved, the circle is equal to the rectangle DA. AH, and the polygon to the rectangle DA.AL, the excess of the polygon above the circle is equal to the rectangle DA.HL; therefore the rectangle DA. HL is less than the square of NP; and therefore, since DA is greater than NP, HL is less than NP, and twice HL less than twice NP, wherefore, much more is twice HL less than NO. But HL is the difference between half the perimeter of the polygon whose side is EF, and half the circumference of the circle; therefore, twice HL is the difference between the whole perimeter of the polygon and the whole circumference of the circle (5. 5.). The difference, therefore, between the perimeter of the polygon and the circumference of the circle is less than the given line NO.

COR. 3. Hence also, a polygon may be inscribed in a circle, such that the excess of the circumference above the perimeter of the polygon may be less than any given line. This is proved like the preceding.

PROP. VI. THEOR.

The areas of circles are to one another in the duplicate ratio, or as the squares, of their diameters.

Let ABD and GHL be two circles, of which the diameters are AD and GL; the circle ABD is to the circle GHL as the

square

of

Let ABCDEF and GHKLMN be two equilateral polygons of the same number of sides inscribed in the circles ABD, GHL; and let Q

be such a space that the square of AD is to the square of GL as the circle ABD to the space Q. Because the polygons ABCDEF and GHKLMN are equilateral and of the same number of sides, they are similar (2. 1. Sup.), and their areas are as the squares of the diameters of the circles in which they are inscribed. Therefore AD : GL3:: polygon ABCDEF: polygon GHKLMN; but AD: GL :: circle ABD: Q; and therefore, ABCDEF: GHKLM:: circle ABD:Q. Now, circle ABD>ABCDEF; therefore Q>GHKLMN (14. 5.), that is, Qis greater than any polygon inscribed in the circle GHL.

In the same manner it is demonstrated, that Q is less than any polygon described about the circle GHL; wherefore the space Q is equal to the circle GHL (2. Cor. 4. 1. Sup.). Now, by hypothesis, the circle ABD is to the space Q as the square of AD to the square of GL; therefore the circle ABD is to the circle GHL as the square of AD to the square of GL. Therefore, &c. Q. E. D.

COR. 1. Hence the circumferences of circles are to one another as their diameters.