ELEMENTS

OF

GEOMETRY.

SUPPLEMENT.

A

BOOK I.

OF THE QUADRATURE OF THE CIRCLE.

DEFINITIONS.

I.

CHORD of an arch of a circle is

the straight line joining Supplement

the extremities of the arch; or the straight line which

fubtends the arch.

II.

The perimiter 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, 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. I.

Supplement Cor. 1. Hence, the perimeter of any polygon infcribed 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.

F 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 propofed magnitudes.

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 A
half, and so on; there shall at length re-
main a magnitude less than C.

D

BCE

F

G

For C may be multiplied so as, at length, K to become greater than AB. Let DE, therefore, be a multiple of C, which is greater than AB, and let it contain the H 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 half, and fo on, until there be as many divisions in AB as there are in DE: And let the divisions in AB be AK, KH, Hв. 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 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 Dess than C. Q. E. D.

QUILATERAL polygons,

of the fame number of

E fides, infcribed in circles, are fimilar, and are to

one another as the squares of the diameters of the circles.

Let ABCDEF and GHIKLM be two equilateral polygons of the fame number of fides infcribed in the circles ABD, and GHK; ABCDEF and GHIKLM are fimilar, 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 alfo GO and HO, and produce AN and GO till they meet the circumferences in D and K.

Book I.

Because the straight lines AB, BC, CD, DE, EF, FA, are all equal, the arches AB, BC, CD, DE, EF, FA are alfo equal a. For the fame reason, the arches GH, HI, IK, KL, a 28.5. 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 fame is the arch GH of the circumference GHK. But the angle ANB is the fame part of four right angles, that the arch AB is of the circumference ABDb; and the angle GOH is the same part of four b 27.3. right angles that the arch GH is of the circumference GHK b, therefore the angles ANB, GOH are each of them the fame

P

Supplement part of four right angles, and therefore they are equal to one another. The isosceles triangles ANB, GOH are 26.6. therefore equiangulare, and the angle ABN equal to the angle GHO; in the same manner, by joining NC, OE, 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 fame may be proved of the angles BCD, HIK, and of the rest. Therefore, the polygons ABCDEF and GHIKLM are equiangular to one another; and fince they are equilateral, the fides about the equal angles are proportionals; the polygon dr. def. 6. ABCD is therefore fimilar to the polygon GHIKLMd. And because similar polygons are as the squares of their homoloe 20. 6. gous fidese, the polygon ABCDEF is to the polygon GHKLM as the square of AB to the square of GH; but because the triangles ANB, GOH are equiangular, the square of AB is to f 4. 6. the square of GH as the square of AN to the square of GO f, g 15.5. or as four times the square of AN to four times the square g of h 2. cor. 8. GO, that is, as the square of AD to the square of GK h. 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 shewn to be fimilar. Therefore, &c. Q. E. D.

2.

Cor. Every equilateral polygon infcribed in a circle is also equiangular: For the ifofceles triangles, which have their common vertex in the centre, are all equal and fimilar; therefore, the angles at their bases are all equal, and the angles of the polygon are therefore also equal.

PROP.

T

PROP. III. PROB.

HE fide of any equilateral polygon inscribed in a circle being given, to find the fide of a polygon of the fame number of fides described about the circle.

Let ABCDEF be an equilateral polygon inscribed in the circle ABD; it is required to find the fide of an equilateral polygon of the same number of fides described about the circle.

Find G the centre of the circle; join GA, GB, bisect the 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 fide of the polygon required.

Book I.

Produce GF to N, so that GN may be equal to GL; join KN, and from G draw GM at right angles to KN, join also HG. Because the arch AB is bisected in H, the angle AGH is equal to the angle BGHa; and because KL touches the a 27.3.

circle in H, the angles

LHG, KHG are right

angles b; therefore, there

are two angles of the tri

and also the angle LGK

equal to the angle KGN; therefore the base KL is equal to the base KN d. But because the triangle KGN is ifof- d 4.1. celes, the angle GKN is equal to the angle GNK, and the angles GMK, GMN are both right angles by construction; wherefore, the triangles GMK, GMN have two angles of the one equal to two angles of the other, and they have also the fide GM common, therefore they are equal c, and c 26.1. the side. KM is equal to the fide MN, so that KN is bisected