E วน P B m But, in the second place, if neither of the points a, b lie in a side of the figure, but both of them in the sides produced, take any point m in BC, and join m A. Through B draw Bn parallel to m A, and join mn: the figure nm CDE shall be of equal area with the figure A B C DE, and shall have a less perimeter. The figures are of equal area, because, Bn being parallel to m A, the triangles nAm, BAm are (I. 27.) equal to one another. And because the angle m n A is (I. 8. Cor. 1.) greater than the angle ma A,* much more is it greater than the angle ba A; but the latter angle being, as in the preceding case, equal to a b C', is (I. 8. Cor. 1.) greater than ABC; therefore much more is the angle mn A greater than the angle ABC. And hence it follows, as before, that n A, n m together (35.) are less than BA, Bm together, and that the perimeter of the figure nm CDE is less than that of the figure ABCDE. The two cases in which the sides AE, BC are parallel, are easily demonstrated have not, &c. Cor. 1. Of plane rectilineal figures having the same number of sides, and containing the same area, the regular polygon has the least perimeter. For it is obvious, that a certain area being to be comprehended under a certain number of sides, the perimeter of the containing figure cannot be less than of some certain length depending on the extent of the area; that is, in other words, of figures inclosing the same area, and having the same number of sides, the perimeters cannot, any of them, be less • The line ma is not drawn in the figure. than a certain line, viz. the least possible by which, under the aforesaid condition, the given area can be inclosed. But it is shown in the proposition that, except a figure have all its sides equal and all its angles equal, another may be found inclosing the same area under the same number of sides and with a less perimeter. Therefore, of all the above figures there is one only which is contained by the least possible perimeter, and that one is the regular polygon (def. 11.) Cor. 2. And hence a regular polygon contains a greater area than any other rectilineal figure having the same number of sides and the same perimeter: for a similar polygon which should have the same area with the figure, would have a less perimeter (Cor. 1.), and therefore (30.) a less area than the regular polygon which has the same perimeter. side ab have a greater number of sides than the other. The polygon which has the side ab shall be greater than the other. Let the sides A B, ab, be placed in the same straight line, and so that their middle points may coincide, as at D. Then, because ab is contained a greater number of times than AB in the common perimeter, a b is less than A B, and the point a lies between A and D. From D draw DC at right angles to A B, and therefore (3. Cor. 2. and def. 12.) passing through the centres of both the polygons: and let C be the centre of the polygon which has the side AB, and c the centre of the polygon which has the side ab; it being supposed as yet unknown whether DC or Dc is the greater of the two. Join CA, ca; through C draw Co parallel to ca to meet AD in o; and lastly, with the centre C and radius Co describe the arc m n cutting CA in m, and CD produced in n. Then, because AB is to the common perimeter of the two polygons (II. 17.) as the angle ACB to four right angles, and the cor mon perimeter to ab as four right angles to the angle acb, ex æquali (II. 24), AB is to ab as the angle ACB to the angle acb; and hence, by taking the halves, (II. 17. Cor. 2.) the line AD is to aD as the angle ACD to ac D, or (I. 15.) o CD. Therefore, dividendo (II. 20.) A a is to a D as the angle A Co to o CD, and invertendo (II. 15.) Da is to a A as the angle D Co to o CA. Again, because the sector Cno is greater than the triangle CD o, and the sector Com less than the triangle C o A the sector Cno has, on both accounts, (II. 11.) to the sector Com a greater ratio than the triangle C Do to the triangle Co A: but the angles DC o, o CA have the same ratio as the sectors (13.), and the lines D o, o A the same ratio as the triangles (II. 39): therefore the angle D Co has to the angle o C A a greater ratio than the line Do has to the line o A (II. 12. Cor. 1 and 2.). And it was shown, that Da has to a A the same ratio which the angle D Co has to the angle o CA: therefore Da has to a A a greater ratio than Do to o A, and Da is greater than Do. And, because Co is parallel to ca, Dc: DC :: Da: Do (II. 29.); therefore D c is also greater than D C. But the polygons, being equal, each of them, to half the rectangle under the apothem and perimeter (29.), are to one another as their apothems c D, CD (II. 35.): therefore, the polygon which has the apothem cD and side ab is greater than the other. Therefore, &c. Cor. A regular polygon is greater than any other rectilineal figure having the same perimeter, and the same or a less number of sides (36. Cor, 2.). PROP. 38. A circle is greater than any regular polygon having the same perimeter. For let a similar polygon be circumscribed about the circle, viz. by dividing the circumference (or conceiving it to be divided) into as many equal parts as the polygon is to have sides, and drawing tangents through the points of division. Then, because the area of the polygon is equal (29.) to half the rectangle under its apothem and perimeter, and the area of the circle to half the rectangle under its radius and perimeter (32.); and that the apothem of the circumscribed polygon is equal to the radius of the circle, the circumscribed polygon is to the circle as its perimeter to that of the circle (II. 35.), that is, as its perimeter to the perimeter of the polygon in question, or (II. 17.) as its side to the side of the latter. Again, because the polygons are similar (II. 43.), they are one to another in the duplicate ratio of their sides. Therefore, the circumscribed polygon has to the other the duplicate ratio of that which it has to the circle, and (II. def. 11.) the circle is a mean proportional between the two polygons. But the circumscribed polygon is greater than the circle: therefore, the circle (II. 14.) is greater than the polygon of equal perimeter. Cor. A circle is greater than any plane rectilineal figure of the same perimeter (37. Cor.). It may be inferred also, from the foregoing propositions, that the circle is not less than any curvilineal figure of the same perimeter. For there may be inscribed in the latter a rectilineal figure of less perimeter, yet approaching more nearly to it in area than by any supposed excess of the original figure above the circle, so that, were there such an excess, a rectilineal figure might be found greater than the circle, and yet of less perimeter, which is impossible. This method will not, however, carry us any further. In the following propositions, another view is taken of the subject, and by them it will be made to appear, that the circle is greater than any other figure, curvilineal or otherwise, which has the same perimeter. PROP. 39. equal to the two sides DE, EF of the other, each to each, but the angle ABC a right angle, and the angle DEF greater or less than a right angle; the triangle ABC shall be greater than the triangle DEF. From the point D draw DG perpendicular to E F, or E F produced. Then, because D G is (I. 12. Cor. 3.) less than DE or A B, the rectangle under DG, EF is less than the rectangle under AB, EF, or AB, BC, and therefore (I. 26. Cor.) the triangle DEF is less than the triangle A B C. If two quadrilaterals have three sides of the one equal to three sides of the other, each to each, and the angles of the first lying in a semi-circumference of which the fourth side is diameter, but the angles of the other not so lying, the first quadrilateral shall be greater than the other. For if EFGH (see fig. 1) be that one of the quadrilaterals which has not the angles lying in a semicircumference, of which the fourth side EH is diameter, and if G be an angle which does not so lie; then, joining EG, the angle EGH will not be equal to a right angle, (15 Cor. 3.) and, therefore, if GH' be drawn perpendicular to EG, and equal to GH, and if EH' be joined, the triangle EGH' will be greater than EGH, (39.) and, accordingly, the quadrilateral EFGH', which has its three sides EF, FG, GH' equal to the three EF, FG, GH, each to each, greater than the quadrilateral EFGH. Therefore, if a quadrilateral be inclosed by three given sides and a fourth not given, a greater may be found inclosed by the same three given sides and a fourth not given, except when the angles lie in a semicircumference, of which the fourth side is diameter. But, because the fourth side is (I. 10. Cor. 2.) necessarily less than the sum of the other three, it is evident that there is some certain area, a greater than which cannot be so inclosed, and therefore some quadrilateral which incloses the greatest possible area. Therefore, the quadrilateral ABCD which has its angles lying in a semicircumference, of which the fourth side is diameter, incloses the greatest possible area, and the quadrilateral ABCD is greater than EFGH.* Otherwise. Let A B CD, EFGH be two quadrilaterals which have the sides A B, BC, CD of the one, equal to the sides EF, FG, GH of the other, each to each; and let the angles A, B, C, D of the first lie in the circumference of a circle, of which the side A D is diameter, but the angles E, F, G, H, of the other not lie in the circumference of a circle of which EH is diameter: the quadrilateral ABCD shall be greater than EFGH. com E Fig. 1. G H H Fig. 2. G Join EG, FH: and, as most favourable to the figure EFGH in the parison with ABCD, let one of the angles EFH, EGH, the former for instance, be a right angle; since, FH rethe maining same, the triangle EFH, and therefore the whole figure will be greater (39.) upon this than upon the contrary supposition. It will appear in the demonstration, that it is indifferent whether E G H be supposed less or greater than a right angle: we shall set out with supposing it to be less, and, therefore, (15. Cor. 3.) the point G to be without the semicircle upon E H. E E Fig. 3. F Ꮹ H H Draw G H at right angles to EG, and equal to GH (fig. 1); and join If two of the given sides as FG, GH should be in the same straight line, E F G H would be a triangle, not a quadrilateral: it may be observed, however, that the demonstration is equally applicable to show We that in this case E F G H is less than ABCD. EH'. Then, because EFH is a right angle, EFH is greater than a right angle, and therefore (15. Cor. 3.) the point F falls within the semicircle drawn upon EH', as in fig. 2. Again, draw FE' (fig. 2) at right angles to FH' and equal to E F, and join E'H'. Then, because EG H' is a right angle, E'GH' is less than a right angle, and the point G falls without the semicircle upon E'H' (see fig. 3), as at first. It appears therefore, that if the process be repeated and continued, we shall obtain in this manner a series of figures (fig. 1., fig. 2., fig. 3., &c.), each of which is greater than the preceding (because one of the triangles remaining the same, the other is made to have a right angle), and in which, one of the angles E FH, EGH, being a right angle, the other is greater than a right angle, and less than a right angle alternately. Again, of the bases EH, EH', E' H', &c. each is of a magnitude intermediate between the two preceding. For, because the square of E'H' (fig. 2) is equal (I. 36.) to the squares of EF and FH, and that FH' is greater than FH (fig. 1), (because the two sides FG, G H of the triangle FGH' are equal to the two sides FG,GH of the triangle FGH, and contain a greater angle (I. 11),) the square of E'H' is greater than the squares of EF, FH, or of E F, FH; greater, that is, than (I. 36.) the square of E H (fig. 1), and therefore E'H' (fig. 2 or 3) is greater than EH (fig. 1). But E'H' (fig. 3) is less than EH' (fig. 2), because the two sides E' F, FH' of the triangle E' FH' are equal to the two EF, FH' of the triangle EFH', and contain a less angle, (I. 11.). Therefore, E'H' (fig. 3) is of intermediate magnitude between EH (fig. 1) and EH' (fig. 2). And, in a similar manner, it may be shown that E' H" (fig. 3. or 4) is of intermediate magnitude between EH' (fig. 2) and E'H' (fig. 3); and so on. Now AD is greater than EH, E'H', &c., and less than EH', E' H", &c., because the chords AB, BC, CD, which together subtend the semi-circumference of which AD is diameter, subtend more than a semi-circumference in the circles of which EH, E' H', &c. are diameters, and less than a semi-circumference in those of which EH', E' H", &c. are diameters. Therefore, every successive base EH being alternately greater and less than A D, and lying between the two preceding, approaches more nearly to AD than did the last in the series which was greater or less than AD; that is, AD is the limit to which, in the foregoing process, the bases EH are made to approach. And it has been shown, besides, that the figure EFGH is increased at every step. Therefore, the figure upon the base AD is greater than any of the figures EFGH. Therefore, &c. Cor. Three given finite straight lines with a fourth indefinite, inclose the greatest area, when placed as chords of a semi-circumference, of which the fourth side is diameter. ABcd. Through A draw the diameter AK, and join B K, KC: and upon Bc, which is equal to BC, make the triangle B kc equal and similar to the triangle B KC, so that the sides Bk, kc may be equal to the sides B K, KC respectively; and join A k. 'Then, because the straight lines B c, BC do not coincide, (for if they did, the figures would coincide altogether by I. 7.) the point k does not coincide with the point K: but ABK is a right angle (15. Cor. 1.): therefore ABR is not a right angle, and (39.) the triangle BAK is greater than the triangle B Ak. Also, the quadrilateral ADC K, having its three sides A D, D C, C K chords of the semi-circumference upon AK, is greater than any other quadrilateral* Adck, having three of its sides equal to AD, DC, If dc and c k lie in the same straight line (as is nearly the case in the figure), the figure A dc k will be a triangle, not a quadrilateral; but in this case also it is less than ADC K (see note Prop. 49.) Cor. 1. If a figure A B C D E F is to be inclosed by any number of given sides, and if these sides be not so disposed that the angles may lie in the circumference of a circle, a greater figure may be inclosed by the same sides. For, if the angle E, for instance, do not lie in the circumference which passes through the points A, B, C, join AE, CE, and let there be constructed the quadrilateral a bce, such that its sides may be equal to those of A B C E, each to each, and its angles in the circumference of a circle (25. Cor.): and upon the sides ae, ce, which are equal to AE, CE, respectively, let there be described the figures afe, cde equal to the figures AFE, CDE, respectively. Then, because, by the proposition, the quadrilateral abce is greater than ABCE, the whole figure abcdef is greater than A B C D E F. Cor. 2. And hence, of all figures contained by the same given sides in the same order, that one contains the greatest possible area which has all its angles in the circumference of a circle. For the area inclosed by the given sides 'cannot exceed a certain limit depending upon them, which limit is the greatest possible that can be inclosed by the given sides, and is therefore such as by them can be inclosed. But no figure, so inclosed, contains the greatest possible area, of which the angles do not lie in the circumference of a circle. Therefore, the figure which has its angles in the circumference of a circle contains a greater area than any other figure having the same given sides. Scholium. That a circle may be imagined in which any number of given straight lines shall subtend as chords the whole circumference exactly, is evident from this, that a circle may be imagined in which they shall subtend less than the whole circumference, and a second circle in which they shall subtend more than the whole circumference: for the circle required will be of some magnitude between these two. It may be observed, also, that the order of the sides is indifferent as well to the magnitude of the required circle, as to the magnitude of the figure which is to be inscribed in it; for the same chord will subtend an arc of the same magnitude, at whatever part of the circumference it may be placed (12. Cor. 1); and therefore the arcs subtended by all the chords will be together equal to the whole circumference, whatsoever may be their order. And, because the same chord always cuts off a segment of the same area, the segments cut off by all the chords will amount to the same area, whatsoever may be their order; and therefore the inclosed area, which is the difference between that amount and the area of the circle, will also be the same. From these considerations it appears that Prop. 41. Cor. 2. need not order of the sides. have been qualified by a regard to the as A, C, E, G, which do not lie in the circumference of a circle. Join these points, and let the quadrilateral ac e g be constructed, having its sides equal to those of the quadrilateral ACEG, each to each, and its angles in the circumference of a circle (25. Cor.). Then, because (41.) the quadrilateral a ceg is greater than AC EG, if upon the sides ac, ce, eg, ga, which are equal to A C, CE, EG, GA, respectively, there be constructed the figures abc, cde, efg, gha equal to the figures A B C, C DE, EFG, GHA, each to each, in all respects, the whole figure abcdefg will be greater than ABCDEFG, and will have the same perimeter. It appears, therefore, that if a plane figure be not a circle, a greater area |