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Find G the centre of the circle; join GA, GB, bisect the Book I. 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, and join HG.

Because the arch AB is bisected in H, the angle AGH is equal to the angle BGHa ; and because KL touches the circle a 27.3. in H, the angles LHG, KHG are right angles,

b 18. 3.

B therefore two angles of

0 the triangle HGK are es

H. qual to two angles of the triangle HGL, each to each; also the side GH is common to both; there- K

D fore the triangles are equalc, and GL is equal

c 36. 1; to GK. Again, in the M triangles KGL, KGN, the side GN is equal to

E GL, and GK common, and the angle LGK equal to the angle KGN; therefore the base KL is equal to the base KNI. Because the triangle KGN is isosceles, the angle d 4. 1. GKN is equal to the angle GNK; the angles GMK, GMN 1 5. 1. are right angles by construction; wherefore the triangles GMK, GMN have two angles of one equal to two angles of the other; and they have also the side GM common; therefore they are equals, and the side KM is equal to the side MN, so that KN is bisected in M. But KN is equal to KL, therefore their halves KM and KH are also equal. 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 GHd; wherefore the point M is in the circumference of the circle ; and because KMG is a right angle, KM touches the circlef. f Cor.16.3. In the same manner, by joining the centre and the other angular points of the inscribed polygon, an equilaterál 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

Suppl. 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 g 2.1. Sup. similar to the polygon ABCDEF8

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 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 circleh.

h 15. 5.

PROP: IV. THEOR.

A CIRCLE being given, two similar polygons may be found, one described about the circle, and the other inscribed in it, which shall differ from each other 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 jnscribed 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 a 1.1, Sup. AF is found less than the circumference AE. Find the

centre G; draw the diameter AC, and the straight lines AF, FG; bisect the arch AF in K, join KG, and. draw HL

touching the circle in K, and meeting GA and GF produced Book I. in H and L; join CF.

Because the isosceles triangles HGL and AGF have the common angle AGF, they are equiangularb, and the angles b 6. 6. GHK, GAF are equal. But the angles GKH, CFA are equal, because they are right angles; therefore the triangles HGK, ACF are equiangulare.

c 32.1. Because the arch AF was found by taking from the arch AB its half, and from the 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; therefore the straight line AF is 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 ABCd. Let

d 3.1. Şup. the polygon. de

B scribed about the circle be called M,

LE and the polygon inscribed be called N; then, because these polygons are similare, they are as H Н the of the

3. 1. Sup. squares homologous sides

f3.Cor. HL and AF, or,

D

20.6. because the triangles HLG, AFG are similar, as the

of HG and squares

E
AG, or GK. But
the triangles HGK, ACF have been proved to be similar;
therefore the

square
of AC is to the

square

of CF as the polygon M to the polygon N; therefore, by conversions, the g D. 5. square of AC is to its excess above the square of CF, that is, to the

square of AFh, as the polygon M to its excess above h 47. 1. 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 sa

KIM

Ce 1. Cor.

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Suppl.

on ; therefore the square of AC is greater than any polygon described about the circle by the continual bisection of the arch AB; therefore it is 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 M and N; therefore the square of AF is greater than the difference of the polygons, that is, the difference of the polygons is 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, or the given space. Therefore, a circle being given, &c, Q. E. D.

i 14. 5.

COR. 1. Because the polygons M and N differ from each other more than either of them differs from the circle, the difference between each of them and the circle is less than the given space, or the square of D. 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 poly,

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gon 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 the polygons described about the circle A are all greater than B by hypothesis, and 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 shown that no polygon can be inscribed Book I.
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 no
unequal, that is, they are equal to each other.

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 AH be taken 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 circle ABC; bisect the arch AB in G, and through G

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draw EGF touching the circle, and meeting DA and DB
produced in E and F, EF will be the side of an equilateral
polygon described about the circle ABCa. In AC produced a 3. 1. Sup.
take AK equal to half the perimeter of the polygon whose
side is AB, and AL equal to half the perimeter of the poly,
gon whose side is EF, Then AK will be less, and AL.

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