B Prop. XXII. THEOR.-Of quadrilaterals and polygons which have their sides equal, each to each, the greatest is that which can be inscribed in a circle. In the rectilineal figures ABCDE, FGHKL, let the sides AB, BC, CD, DE, E A be severally equal to FG, GH, HK, KL, LF, and let AD be inscribed in a circle, but FK not be capable of being inscribed in one : AD is greater than FK. Draw the diameter CM, and join AM, EM; construct the triangle FNL, having the sides FN, NL respectively equal to ÁM, ME, so that (I. 8) the triangles themselves are equal, and join HN. Then, by the preceding proposition, the figures CBĂM, CDEM are respectively greater than HGFN, HKLN; and therefore (I. ax. 4) the polygon ABCDEM is greater than FGHKLN; and, by taking away the equal triangles AME, FNL, there remains the figure ABCDE greater than FGHKL. Cor. From this proposition and the twentieth it follows, that of polygons of the same number of sides, and of equal perimeters, the regular polygon is the greatest; and that of quadrilaterals of equal perimeters, the greatest is the square. M H N E D Prop. XXIII. THEOR.–Of regular polygons which have equal perimeters, that which has the greater number of sides is the greater. Let AB be half the side of the polygon which has the less number of sides, and BC a perpendicular to it, which will evidently pass through the centre of its inscribed or circumscribed circle let C be that centre, and join AC. Then, ACB will be the angle at the centre subtended by the half side AB. Make BCD equal to the angle subtended at the centre of the other polygon by half its side, and from C as centre, with CD as radius, describe an arc cutting AC in E, and CB produced in F. Then, it is plain, that the angle ACB is to four right angles, as AB to the common perimeter; and four right angles are to DCB, as the common perimeter to the half of a side of the other polygon, which, for brevity, call S: then, ex æquo, the angle ACB is to DCB, as AB to S. *But (VI. 33) the angle ACB is to DCB, as the sector ECE to the sector DCF; and consequently (V. 11) the sector ECF is to DCF, as AB to S, and, by division, the sector ECD is to DCF, as AB-S tó s. Now the triangle ACD is greater than the sector CED, and DCB is less than DCF. But (VI. 1) these triangles are as their bases AD, DB ; therefore AD has to DB a greater ratio, than AB-S to S. Hence AB, the H E A GD B sum of the first and second, has to DB, the second, a greater ratio than AB, the sum of the third and fourth, has to s the fourth ;* and therefore (V. 10) S is greater than DB. Let then BG be equal to S, and draw GH parallel to DC, meeting FC produced in H, Then, since the angles GHB, DCB are equal, BH is the perpendicular drawn from the centre of the polygon having the greater number of sides to one of the sides; and since this is greater than BC, the like perpendicular in the other polygon, while the perimeters are equal, it follows that the area of that which has the greater number of sides is greater than that of the other. PROP. XXIV. THEOR. A circle is greater than any regular polygon of the same perimeter. For, if a polygon P similar to the one Q proposed for comparison, be described about the circle C: since the area of the circumscribed polygon is evidently equal to the rectangle under its perimeter and half the radius, and (APP. I. 39) the area of the circle equal to the rectangle under its perimeter and half its radius ; therefore (VI. 1) P is to C, as the perimeter of P to the perimeter of C, or of Q: and (VI. 20) P is to Q in the duplicate ratio of a side of P to a side of Q, or (VI. 20, cor. 4) in the duplicate ratio of the perimeter of P to the perimeter of Q. Hence P is to a mean proportional between P and Q, as the perimeter of P to the perimeter of Q; and therefore (V. 11) P is to C, as P to a mean proportional between P and Q; that is, (V. 9) C is a mean proportional between P and Q; and (V. 14) being less than P, it is greater than Q, the polygon of equal perimeter. Schol. Every thing here proved will evidently hold equally regarding a circle C, as compared with any polygon Q of equal perimeter, which is such that a circle can be inscribed in it; for then a polygon P, similar to Q, can be described about C. Cor. Hence it appears also, that a circle is a mean proportional between any rectilineal figure described about it, and a similar one of equal perimeter with the circle. * This may be proved by dividing the first term by the second, and the third by the fourth, and adding unity to each quotient. See the Supplement to Book V. BOOK III. PROPOSITIONS. A Prop. I. PROB.–Given the base of a triangle, the difference of the sides, and the difference of the angles at the base ; to construct it. Make BC equal to the given base, and CBD equal to half the difference of the angles at the base : from C as centre, at a distance equal to the difference of the sides, describe an arc cutting BD in D : join CD and produce it: make the angle DBA equal to BDA; ÅBC is the required triangle. For (I. 6) AD is equal to AB, and the difference of AC, AD, or of AC, AB, is CD; and (APP. I. 6) since AD is equal to AB, CBD is equal to half the difference of the angles at the base. The triangle ABC, therefore, is the one required, as it has its base equal to the given base, the difference of its sides equal to the given difference, and the difference of the angles at the base equal to the given difference of those angles. Method of Computation. In the triangle BCD, there are given BC, CD, and the angle CBD; whence the angle C can be computed ; and the sum of this and twice CBD is the angle ABC. Then, in the whole triangle ABC, the augles and the side BC are given; whence the other sides may be computed : or, one of them being computed, the other will be found by means of the given difference CD. B Prop. II. PROB.—Given the segments into which the base of a triangle is divided by the line bisecting the vertical angle, and the difference of the sides; to construct the triangle. Construct the triangle CED, having the sides CE, ED equal to the given segments, and CD equal to the given difference of the sides : produce CE, and make EB equal to ED; bisect the angle BED by EA, meeting CD produced in A, and join AB: ABC is the required triangle. For, in the triangles AEB, AED, BE is equal to ED, EA common, and the angle BEA equal to DEA: therefore (I. 4) BA is equal to DĀ, and the angle EAB to EAD. Hence, ABC is the required triangle; for CD, the difference of its sides, B is equal to the given difference, and BE, EC, the segments into A which the base is divided by the line bisecting the vertical angle, are equal to the given segments. Method of Computation. The sides of the triangle CDE are given, and therefore its angles may be computed: one of which, and the supplement of the other are the angles C and B. Then, in the triangle ABC, the angles and BC are given, to compute the remaining sides. OTHERWISE:- Since (VI. 3) CE : EB :: CA: AB, we have, by division, CE – EB : EB :: CA - AB : AB; which, therefore, becomes known, since the first three terms of the analogy are given; and thence AC will be found by adding to As the given difference of the sides. M Р Prop. III. PROB. - Given the base of a triangle, the vertical angle, and the difference of the sides; to construct the triangle. Let MNO be the given vertical angle; produce ON to P, and bisect the angle MÑP by NQ. Then, make BD equal to the difference of the sides, and the angle ADC equal to QNP; from B as centre, with the given base as radius, describe an arc cutting DC in C; and make the angle DCA equal to ADC: ABC is the required triangle. For (const.) the angles ACD, ADC are equal to MNP, and therefore (I. 32, and 13) the angle A is equal to MNO. But (1. 6) AD is equal to AC, and therefore BD is the difference of the sides AB, AC; and the base BC is equal to the given base : wherefore ABC is the triangle required. Method of Computation. In the triangle CBD, the sides BC, BD, and the angle BDC, the supplement of ADC or MNQ are given; whence the other angles can be computed. The rest of the operation will proceed as in the first proposition of this book. D B с PROP. IV. PROB.-Given one of the angles at the base of a triangle, the difference of the sides, and the difference of the segments into which the base is divided by the line bisecting the vertical angle; to construct the triangle. Construct the triangle DBG, having the angle B equal to the given angle, BD equal to the difference of the sides, and BG equal to the difference of the segments; draw DC perpendicular to DG, and meeting BG produced in C; produce BD, and make the angle DCA equal to CDA: ABC is the triangle required. For it has B equal to the given angle, and the difference of its sides BD equal to the given difference: and if AHE be drawn bisecting the angle BAC, it bisects (I. 4) CD, and is perpendicular to it: it is there А D H B E fore parallel to DG, one side of the triangle CDG; and, bisecting CD in H, it also (VI. 2) bisects GC in E. Hence BG, the differ ence of BE, GE, is also the difference of BE, EC, the segments into which the base is divided by the line bisecting the vertical angle. Method of Computation. In the triangle DBG, BD, BG, and the angle B are given ; whence (Trig. 3) we find half the difference of the angles BGD, BDG, which is equal to half the angle C.* Then (by the same proposition) we have in the triangle ABC, tan (C-B) : tan } (C+B) :: -b or BD :c+b; whence the sides c and b become known, and thence BC by the first case. C А Prop. V. PROB. - Given the base of a triangle, the vertical angle, and the sum of the sides; to construct it. Make BD equal to the sum of the sides, and the angle D equal to half the vertical angle; from B as centre, with the base as radius, describe an arc meeting DC in C;t and make the angle DCA equal to D; ABC is the required triangle. For (I. 6) AD is equal to AC, and therefore BA, AC are equal to BD, the given sum. Also (1. 32) the exterior angle BAC is equal to the two D and ACD, or to the double of D), because D and ACD are equal : therefore, since D is half the given vertical angle, BAC is equal to that angle. The triangle ABC, therefore, has its base equal to the given base, its vertical angle equal to the given one, and the sum of its sides equal to the given sum : it is therefore the triangle required. Method of Computation. In the triangle BDC, the angle D, and the sides BC, BD are given: whence the remaining angles can be computed : and then, in the triangle ABC, the angles and the side BC are given, to compute the other sides. B Prop. VI. PROB.—Given the vertical angle of a triangle, and the segments into which the line bisecting it divides the base; to construct it. * For (I. 32) BEA=A+C. and consequently BEA – A=C. But (I. 29) BGD=BEA, and BDG=BAE=qA; and therefore BGD-BDG=C. + Should the circle neither cut nor touch DC, the problem would be impossible with the proposed data. If the circle meet DC in two points, there will be two triangles each of which will answer the conditions of the problem. These triangles, however, will differ only in position, as they will be on equal bases, and will have their remaining sides equal, each to each. This problem might also be solved by describing (III. 33) on the given base BC a segment of a circle containing an angle equal to half the vertical angle; by inscribing a chord BD equal to the sum of the sides ; by joining DC; and then proceeding as before. The construction given above is preferable. |