E For, draw BE perpendicular to AB, and draw CF, DK, perpendicular to BE, make FE = FB and KG = KB, join CE and DG. Then the triangles CFE, CFB, have BFFE and FC common, and the angles at F right angles, therefore CE CB. For a similar reason, DGDB. Therefore AC + CE= AC+ CB, and AD + DG = AD+A F D K DB. Now angle ECF = BCF = CBA (I. 29) = CAB (I. 5), therefore ECF + ACFCAB+ ACF = two right angles (I. 29), therefore AC, CE, are in the same straight line. But AD+DG AG, if AG were joined, therefore AGAE, and hence the point G is nearer to the perpendicular AB than E is (1, Cor.), hence GBEB, and therefore KBFB. Therefore the altitude of the triangle ADB being less than that of ACB, ADB ACB (VI. 1, Cor. 2.) PROPOSITION III. THEOREM. E F C Of all isoperimetrical polygons of the same number of sides, the equilateral polygon is a maximum. Let ABCDE be the maximum polygon, then if two of its sides as ED, DC, be not equal, let EF, FC, be drawn equal. Then the triangle EFCEDC (2); and therefore the given polygon is less than ABCFE, which is contrary to hypothesis. Therefore ED DC; and it may be similarly proved that any two adjacent sides are equal. = PROPOSITION IV. THEOREM. B Of all triangles having only two sides given, that is the greatest in which these sides are perpendicular. Let ABC, ABD, be.two triangles having the common base AB and AC: AD, then if AC be perpendicular to AB, ABC > ABD. C D For, draw DE perpendicular to AB, A E then (1) DE <AD, and therefore DE AC; hence the triangle ACD ADB (VI. 1, Cor. 2.) PROPOSITION V. THEOREM. Of all the angles subtended at the centres of different circles by equal chords, that in the least circle is the greatest. Let the same chord AB subtend the angles D, C, at the centres of two circles, of which CB, DB, are the radii, then DC (I. 21.) There is only one way of forming a polygon, all of whose sides are given, except one, and inscribed in a semicircle, of which the unknown side is a diameter. Let the sides of the polygon ABCDEF be all given except AB, and inscribed in a semicircle AEB, of which AB is the diameter. If, now, a greater circle were taken, the angles at the G B is, they would be less than two right angles; and therefore the extremities A, B, of the given sides would not fall at the extremities of a diameter. If a less circle were taken, the sum of the angles would exceed two right angles, and the same consequence would follow. Hence the polygon in question can be inscribed in only one semicircle. Schol. The order of the sides AF, FE, &c. may be altered in any manner, the diameter AB and the area of the polygon remaining the same. For the segments cut off by these side are always the same, and the area of the polygon is equal to that of the semicircle diminished by these segments. PROPOSITION VII. THEOREM. Among all polygons whose sides are all given but one, that is the maximum whose sides can be inscribed in a semicircle, of which the unknown side is the diameter. Let ABCDE be the greatest polygon, which can be E formed with the given sides AE, ED, DC, CB, and the last AB assumed of any magnitude. Join AD, DB; then if the angle ADB were not a right angle, by making it so the triangle ADB would be increased (4), and the parts AED, DCB, remaining the same, the whole A polygon would be increased. But the given polygon, being a maximum, cannot be augmented; therefore the angle ADB must be a right angle. In the same manner, the angles AEB, ACB, must be right angles, and the polygon is inscribed in a semicircle. Of all polygons formed with given sides, that which can be inscribed in a circle is a maximum. B Let ABCDE, abcde, be two polygons, whose corresponding sides are equal, namely, AB=ab, BC=bc, &c. and let the former be inscribed in a circle, but the latter incapable of being so inscribed, the former is the greater. Draw from A the diameter AF; join CF, FD, and make the triangle cfd = CFD, in every respect, so that ef CF, df DF, and join af. = The polygon ABCF→ abcf (7), unless the latter can be inscribed in a semicircle having af for its diameter, in which case the two polygons would be equal. For a similar reason, the polygon AEDF aedf, unless in the case of a similar exception, by which the polygons would be equal. Hence, the whole polygon ABCFDE abcfde, unless the latter can be inscribed in a circle. But it cannot; therefore the former is the greater. Take from both the equal triangles CFD, cfd, and there remains the polygon ABCDE abcde. Schol.-It may be shown, as in the sixth proposition, that there is only one circle in which the polygon can be inscribed, and therefore only one maximum polygon; and this polygon will have the same area in whatever order the sides be arranged. PROPOSITION IX. THEOREM. Of all isoperimetrical polygons, having the same number of sides, the regular polygon is the greatest. For (3) the maximum polygon is equilateral, and (8) it can be inscribed in a circle, it is therefore a regular polygon. Of two isoperimetrical regular polygons, the one having the greater number of sides is the greater. Let DE, AB, be half-sides of these polygons, OE, CB, their apothems, lying in the same straight line, and ACB, DOE, the half-angles at from the centre O describe the arcs GI, MN, with the radius OG and the radius OM = CG; and describe the arc GH from the centre C with the radius CG. The polygons being regular, their perimeters are the same multiples of their half-sides, that four right angles is of their half-angles at their centres, but their perimeters are equal; therefore, by direct equality, angle 0:C=DE:AB. But 0: C=MN:GH (VI. 33), therefore DE: AB = MN : GH; but OG: OM or CGGI: MN (Qu.* VIII. Cor. 1), therefore (VI. 23, Cor. 1) DE OG: AB CG-MN GI:MN GH =GI:GH (VI.1) if MN be the altitudes of these two rectangles. But the triangles OED, OGF, are similar, therefore OE: OGDE:FG (VI. 4), therefore OE FG DE · OG · (VI. 16). Also, from the similar triangles, ABC, FGC, CB: AB CG: FG ; and hence CB FGA B.CG. Therefore OE FG: CB FGGI: GH, or (VI. 1) OE: CB = GI: GH. But (Qu. II. Cor. 2) GK GH, therefore GI is still GH, therefore OECB. But the perimeters of Qu. refers to the preceding book on the Quadrature of the Circle. . the polygons being equal, their areas will be proportional to their apothems; therefore the polygon, of which OE is the apothem, is the greater. But its half-angle O is less than that of the other, or it has a greater number of sides; therefore the polygon having the greater number of sides is the greater. PROPOSITION XI. THEOREM. The circle is greater than any polygon of the same peri meter. G B I D H K E Let AG be a half-side, and C the centre of a regular polygon of the same perimeter as the circle, of which F is the centre, and DE an arc sub- A tending an angle DFE equal to the angle ACB subtended by the side AB; then the triangle ACB and the sector DFE are equal parts of the whole polygon and circle, and AB and DE F are equal parts of their perimeters, and are therefore equal; therefore, if the polygon and circle be respectively denoted by P and C, P:C=triangle ACB: sector DFE AG•GC: DH HF-GC: HF. But if IHK be a side of the corresponding circumscribed polygon, then by the similar triangles ABC, IFK, GC: HF = AB or DE: IK; and therefore P:C=DE: IK. But (Qu. II. Cor. 2) DHIH, and therefore IKDE, hence CP. Now P is a regular polygon, and of all isoperimetrical polygons having the same number of sides, the regular polygon is the greatest; but the circle has been proved to be greater than this polygon; it is therefore greater than any other isoperimetrical polygon. PROPOSITION XII. THEOREM. Of all polygons having the same area and the same number of sides, the regular polygon has its perimeter a mini mum. N', Let A and P be the area and perimeter of a regular polygon B, and N the number of its sides, and A', P', those of an irregular polygon B', if A=A' and N = N', then PP. |