ELEMENTS OF GEOMETRY. SUPPLEMENT, BOOK I. OF THE QUADRATURE OF THE CIRCLE. DEFINITIONS. I. A CHORD of an arch of a circle is the ftraight line joining Supplement the extremities of the arch; or the ftraight 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 leaft line that can be drawn between two points, is a ftraight 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 leaft perimeter. COR. I. Supplement COR. 1. Hence, the perimeter of any polygon infcribed in a circle is lefs than the circumference of the circle. COR. 2. If from a point two ftraight 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 defcribed about a circle is greater than the cir cumference of the circle. IF F from the greater of two unequal magnitudes there be taken away its half, and from the remainder its half; and fo on: There will at length remain a magnitude less than the least of the propofed magnitudes. D 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 fo on; there fhall at length remain a magnitude less than C. B F +G For C may be multiplied fo 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 divifions in AB as there are in DE: And let the divifions in AB be AK, KH, HB. And becaufe 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 efs than C. Q. E. D. PROP. E & PROP. II. THEOR. QUILATERAL polygons, of the fame number of fides, infcribed in circles, are fimilar, and are to one another as the fquares 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 fquares of the diameters of the circles ABD, GHK. Find N and the centres of the circles; join AN and BN, as also GO and HO, and produce AN and GO till they meet the circumferences in D and K. B Book I. Because the ftraight 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 ABD b; and the angle GOH is the fame 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 part Supplement part of four right angles, and therefore they are equal to one another. The ifofceles triangles ANB, GOH are 6.6. therefore equiangular, and the angle ABN equal to the angle GHO; in the fame 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 reft. 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 d 1. def. 6. ABCD is therefore fimilar to the polygon GHIKLM d. And because fimilar polygons are as the fquares of their homoloe 20. 6. gous fidese, the polygon ABCDEF is to the polygon GHIKLM as the fquare of AB to the fquare of GH; but because the triangles ANB, GOH are equiangular, the fquare of AB is to f 4. 6. the square of GH as the fquare of AN to the square of GOʻ, 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 fquare of AD to the fquare of GK h. There 2. fore alfo, the polygon ABCDEF is to the polygon GHIKLM as the fquare of AD to the fquare of GK; and they have alfo been fhewn to be fimilar. Therefore, &c. Q. E. D. COR. Every equilateral polygon infcribed in a circle is alfo equiangular: For the ifofceles triangles, which have their common vertex in the centre, are all equal and fimilar; therefore, the angles at their bafes are all equal, and the angles of the polygon are therefore alfo equal. PROP. Book I. THE PROP. III. PROB. HE fide of any equilateral polygon infcribed in a circle being given, to find the fide of a polygon of the fame number of fides defcribed about the circle. Let ABCDEF be an equilateral polygon infcribed in the circle ABD; it is required to find the fide of an equilateral polygon of the fame number of fides described about the circle. Find G the centre of the circle; join GA, GB, bifect 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. Produce GF to N, fo that GN may be equal to GL; join KN, and from G draw GM at right angles to K N, join also HG. Because the arch AB is bifected in H, the angle AGH C b 16. 3. is equal to the angle BGH a; and because KL touches the a 27. 3circle in H, the angles LHG, KHG are right angles b; therefore, there are two angles of the triangle HGK, equal to two angles of the triangle HGL, each to each. But the fide GH is common to these triangles; therefore they are equal, and GL is equal to GK. Again, in the triangles KGL, KGN, because GN is equal to GL, and GK common, and alfo the angle LGK K H G c 26. 1. N 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 conftruction; wherefore, the triangles GMK, GMN have two angles of the one equal to two angles of the other, and they have alfo the fide GM common, therefore they are equal c, and c 26. 1. the fide KM is equal to the fide MN, fo that KN is bifected P 2 in |