« ForrigeFortsett »
OF THE QUADRATURE OF THE CIRCLE.
1. CHORD of an arch of a circle is the straight line joining the ex
II. The perimeter of any figure is the length of the line, or lines, by which it is bounded.
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, that which is contained within the other, if its bounding line or lines be not any where convex toward the base, has the least perimeter.
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 drawntouching 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 there be taken away its
half, and from the remainder its half : and so on; There will at length remain a magnitude less than the least of the proposed magní. tudes,
Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken away its half, and from the remainder its half, and so on ; there
D shall at length remain a magnitude less than C.
A, 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
Kt T let it contain the parts DF, FG, GE, each equal to C. From AB take BH equal to its half, and from the remainder AH, take HK equal to its
H 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. And because DE is greater than AB, and EG taken from DE is not greater than its half, but BH taken from AB is equal to its half ; therefore the remainder GD is
B C E 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 ; therefore, the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK ; that is, AK is less than C. Q. E. D.
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 one another as the squares of the diameters of the circles ABD, GHK.
Find N and O the centres of the circles ; join AN and BN, as also GO and AO, and produce AN and GO till they meet the circumferences in D and K.
Because the straight lines AB, BC, CD, DE, EF, FA, are all equal, the arches AB, BC, CD, DE, EF, FA are also equal (28. 3.). For the same reason, the arches GH, HI, IK, KL, LM, MG are all equal, and they are equal in number to the others; therefore, whatever part the arch AB is of the whole circumference, ABD, the same is the arch GH of the circumference GHK. But the angle ANB is the same part of four right angles, that the arch AB is of the circumfer. ence ABD (33. 6.); and the angle GOH is the same part of four right angles that the arch GH is of the circumference GHK (33. 6.), therefore the angles ANB, GOH are each of them the same part of four right angles, and therefore they are equal to one another. The isosceles triangles ANB, GOH are therefore equiangular (6. 6.), and the angle ABN 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 one another, and to the angle ABN. Therefore the whole angle ABC is equal to the whole GHI ; and the same may be proved of the angles BCD, HIK, and of the rest. Therefore, the polygons. ABCDEF and GHIKLM are equiangular to one another; and since they are equilateral, the sides about the equal angles are proportionals; the polygon ABCD is therefore similar to the polygon GHIKLM (def. 1. 6.). And because similar polygons are as the squares of