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ELEMENTS OF GEOMETRY.

SUPPLEMENT.

BOOK I.

OF THE QUADRATURE OF THE CIRCLE.

DEFINITIONS.

I.

A CHORD of an arch of a circle is the straight line joining Suppl. the extremities of the arch, or the straight line which sub

tends the arch.

II.

The perimeter of any figure is the length of the line, or lines, by which it is bounded.

III.

The area of any figure is the space contained within it.

AXIOM.

The least line that can be drawn between two points is a straight line; and if two figures have the same straight line for their base, the figure which is contained within the other, if its bounding line or lines be not any where convex toward the base, has the least perimeter.

Suppl.

COR. 1. Hence the perimeter of any polygon inscribed in a circle is less than the circumference of the circle.

COR. 2. If from a point two straight lines be drawn touch-
ing a circle, these two lines are together greater than the arch
intercepted between them; and hence the perimeter of any
polygon described about a circle is greater than the circum-
ference of the circle.
1

PROP. I. THEOR.

IF from the greater of two unequal magnitudes
its half be taken away, and from the remainder its
half, and so on, there will at length remain a magni-
tude less than the smaller of the proposed magnitudes.

Let AB and C be two unequal magnitudes, of which AB
is the greater.
If from AB its half be

taken away, and from the remainder its half,
and so on, there will at length remain a A
magnitude less than C.

K

D

-F

G

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 H
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 di-
visions in AB as there are in DE; and
let the divisions in AB be AK, KH, HB.
Because DE is greater than AB, and be-
cause EG taken from DE is not greater than its half, but BH
taken from AB is equal to its half, 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, the remainder FD is
greater than the remainder AK. But FD is equal to C,
therefore C is greater than AK, or AK is less than C.
Therefore, if from the greater, &c. Q. E. D.

B C E

4

PROP. II. THEOR.

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 each other as the squares of the diameters of the circles ABD, GHK.

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

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Book I.

Because the straight lines AB, BC, CD, DE, EF, FA are all equal, the arches AB, BC, CD, DE, EF, FA are also equala. For the same reason the arches GH, HI, IK, KL, a 28. 3. LM, MG are all equal. Therefore, whatever part the arch AB is of the whole circumference ABD, the same part is the arch GH of the whole circumference GHK. But the angle ANB is the same part of four right angles that the arch AB is of the circumference ABD, and the angle GOH is the b 27.3. same part of four right angles that the arch GH is of the cir

Suppl.

c 6. 6.

cumference GHK; therefore the angles ANB, GOH are each of them the same part of four right angles, and therefore they are equal to each other. Therefore the isosceles triangles ANB, GOH are equiangulars, and the angle ABN is 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 each other, and to the angle ABN. Therefore the whole angle ABC is equal to the whole GHI. The same may be proved of the angles BCD, HIK, and also of the rest. Therefore the polygons ABCDEF and GHIKLM are equiangular to each other; and since they are equilateral, the sides about the equal angles are proportionals; therefore d Def. 1. 6. the polygon ABCDEF is similar to the polygon GHIKLMd. Again, because the polygons ABCDEF, GHIKLM have been proved to be similar, 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 GOf, or as the square of AD to the square of GK8. 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 shown to be similar. Therefore, equilateral polygons, &c. Q. E. D.

e 20. 6.

f 4. 6. g 15. 5.

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, therefore the angles of the polygon are also equal. /

PROP. III. PROB.

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.

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