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 CD o 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 À, and Da is greater than Do. And, because Co is parallel to ca, Dc: DC :: Da: Do (II. 29.); therefore Dc 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 might be found greater than the circle, such an excess, a rectilineal figure 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. 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 D G 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 (1. 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 A B C D 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 ABCD, 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 A B C D shall be greater than EFGH. com E E Fig. 1. H H Fig. 2. G H' Fig. 3. Join EG, FH: and, as most favourable to the figure EFGH in the with parison ABCD, let one of the angles EFH, EGH, the former for instance, be a right angle; since, FH remaining the 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. NI F G 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 ABC D. 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 F H' 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. of 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 E/F 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 E' F, 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 E H' (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 A D 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. AB cd. Through A draw the diameter A K, and join B K, KC: and upon B c, which is equal to B C, make the triangle B kc equal and similar to the triangle B KC, so that the sides B k, k c may be equal to the sides B K, K C 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 A Bk is not a right angle, and (39.) the triangle BAK is greater than the triangle B Ak. Also, the quadrilateral A D C K, having its three sides A D, D C, C K chords of the semi-circumference upon AK, is greater than any other quadrilateral* Ad ck, 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 de k will be a triangle, not a quadrilateral; but in this case also it is less than AD ̊C K (see note Prop. 40.) 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 A E, CE, and let there be constructed the quadrilateral a bce, such that its sides may be equal to those of A BCE, each to each, and its angles in the circumference of a circle (25. Cor.): and upon the sides a e, 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 a ceg 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 ac eg 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, CDE, EFG, GHA, each to each, in all respects, the whole figure abcdefg will be greater than ABC D E F G, and will have the same perimeter. It appears, therefore, that if a plane figure be not a circle, a greater area than is contained by that figure may be inclosed with the same perimeter. But the area inclosed by a given perimeter cannot exceed a certain limit, which limit, being the greatest possible that can be so inclosed, some figure with the given perimeter must be capable of containing. Therefore the circle only contains the greatest area of all figures having the same perimeter. Cor. In the same manner it may be shown that if a figure is to be inclosed by a given perimeter, of which part is to be a given finite straight line, and if it be not made a circular segment of which the given line is chord, a greater may be inclosed with the same conditions, and therefore that of all figures so inclosed the circular segment is the greatest. PROP. 43. Of all plane figures having the same area, the circle has the least perimeter. Let the circle C have the same area with any other plane figure F: C shall be contained by a less perimeter than F. F Let C' be a second circle, having the same perimeter with F; then by the last proposition, C' has a greater area than F has, that is, than С has. But the areas of circles (33.) are as the squares of their radii; therefore the radius of C' is greater than the radius of C; and the radii of circles (3.3.) are as their circumferences; therefore the circumference of C', or perimeter of F, is greater than the circumference of C. Therefore, &c. sponding to the first condition. But again, the point required must be equidistant from the two given points, that is, it must be in the straight line which bisects the distance of the two given points at right angles; for this, it is easily seen, (I. 6.) is the locus corresponding to this second condition. Therefore, if this straight line be drawn, and intersect the given line, the point of intersection (or any of those points, if there be more than one) will satisfy both conditions, and will be the point required. If there be no point of intersection, the problem is impossible. To take another instance Let it be required "to find a point in a certain plane, which shall be, first, at a given distance from a given point in the plane; and, secondly, at a second given distance from a second given point in the same plane." Here it is evident that the locus corresponding to the first condition is the circumference of a circle described about the given point as a centre with the given distance as radius: and again, that the locus corresponding to the second condition is the circumference of a circle described about the second given point as a centre with the second given distance as radius. Therefore the points which are common to the two circumferences, that is, their points of intersection, if there be any, will either of them be the point required. If the circles do not intersect one another, the problem is impossible. Such is the use of loci in the solution of problems. We have seen also in the above example, that they serve to determine in what cases the solution is possible or impossible. Thus, in the latter example, it will be impossible, if the distances of the point required from the given points differ by more than the mutual distance of those points, or together fall short of that distance: and in the first example it will be impossible, if the given line, being straight, be perpendicular to the line which passes through the two given points, and does not pass through the point which bisects that line; for if it does so pass, the two conditions proposed are identical, and any point in this line will answer them. Every locus is the limit between excess and defect. The points upon one side of it fail by defect, and those upon the other side by excess, of possessing the required property which is possessed |