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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 subtends 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.

1

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 touching 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 circumference of the circle. /

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 magnitude less than the smaller of the proposed magnitudes.

is the greater.

Let AB and C be two unequal magnitudes, of which AB

If from AB its half be taken away, and from the remainder its half,

D and so on, there will at length remain a A magnitude less than C. For C may be multiplied so as, at length,

K

F to become greater than AB. Let DE, therefore, be a multiple of C, which is greater than ĄB, 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 divisions 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 B C E 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 Ca Therefore, if from the greater, &c. Q. E. D.

PROP. II. THEOR.

Book I.

( 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.

[blocks in formation]

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 AB Db, and the angle GOH is the b 27. 3. same part of four right angles that the arch GH is of the cir

Suppl. cumference GHK; therefore the angles ANB, GOH are

each of them the same part of four right angles, and there

fore they are equal to each other. Therefore the isosceles c 6. 6. 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 e 20. 6.

GHé. 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.

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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