which satisfies the question; and this polygon will still have the Fig. 179. THEOREM. 309. Among polygons of the same perimeter and the same number of sides the regular polygon is a maximum. Demonstration. According to art. 301, the maximum polygon has all its sides equal; and, according to the preceding theorem, it is such that it may be inscribed in a circle; therefore it is a regular polygon. 310. Two angles at the centre, measured in two different circles, are to each other as the contained arcs divided by their radi; that AB DE Fig. 178 is, the angle C: angle O: the ratio : (fig. 178). AC DO Demonstration. With the radius OF equal to AC, describe the arc FG comprehended between the sides OD, OE, produced; on account of the equal radii AC, OF, LEMMA. AB FG C:0::AB: FG (122), or :: : 311. Of two regular isoperimetrical polygons that is the greater which has the greater number of sides. Demonstration. Let DE (fig. 179), be half of a side of one of these polygons, O its centre, OE a perpendicular let fall from the centre upon one of the sidest; let AB be half of a side of the other polygon, C its centre, CB a perpendicular to the side let fall from the centre. We suppose the centres O and C to be †This perpendicular is called in the original apothême. No English word has been adopted answering to it. situated at any distance OC, and the perpendiculars OE, CB, in the direction OC; thus DOE and ACB will be the semiangles at the centre of the polygons respectively, and as these angles are not equal, the lines CA, OD, being produced, will meet in some point F; from this point let fall upon OC the perpendicular FG; from the points O and C, as centres, describe the arcs GI, GH, terminating in the sides OF, CF. This being done, we have, by the preceding lemma, GI GH 0: C:: : but DE: perimeter of the first polygon :: 0 : four right angles, and AB : perimeter of the second polygon :: C: four right angles; hence, the perimeters of the polygon being equal, DE: AB:: 0 : C DE: AB:: or : Multiplying the antecedents by OG and the consequents by CG,, whence = DEX OG OE × FG ; in like manner OE CB:: GI: GH. or : If therefore it is made evident that the arc GI is greater than the arc GH, it will follow that the perpendicular OE is greater than св. On the other side of CF let there be constructed the figure CK x equal to CG x, so that we may have CK = CG, the angle HCK HCG, and the arc Kxx G; the curve Kx G enclosing the arc KHG will be greater than this arc (283). Hence Gx half of the curve is greater than GH half of the arc; therefore, for a still stronger reason, GI is greater than GH. It follows from this that the perpendicular OE is greater than CB; but the two polygons having the same perimeter are to each other as these perpendiculars (280); therefore the polygon, which has for its half side DE, is greater than that which has for its half side AB. The first has the greater number of sides since its angle at the centre is less; therefore of two regular isoperimetrical polygons, that is the greater which has the greater number of sides. = therefore THEOREM. 312. The circle is greater than any polygon of the same perimeter. Demonstration. It has already been proved that among polygons of the same perimeter and the same number of sides, the regular polygon is the greatest; the inquiry is thus reduced to comparing the circle with regular polygons of the same perimeFig. 180. ter. Let AI (fig. 180) be the half side of any regular polygon, and C its centre. Let there be, in the circle of the same perimeter, the angle DOE ACI, and consequently the arc DE equal to the half side AI; = the polygon P: circle C :: triangle ACI : sector ODE, hence P: C::AI× CI : ¦ DE × OE :: CI: OE. Let there be drawn to the point E the tangent EG meeting OD produced in G; the similar triangles ACI, GOE, give the proportion CI: OE:: AI or DE: GE; P: C:: DE: GE:: DE XOE: GEOE, that is, but the sector is less than the triangle; consequently P is less than C; therefore the circle is greater than any polygon of the same perimeter. PART SECOND. SECTION FIRST. Of Planes and Solid Angles. DEFINITIONS. 313. A STRAIGHT line is perpendicular to a plane, when it is perpendicular to every straight line in the plane which passes through the foot of the perpendicular (326). Reciprocally, the plane, in this case, is perpendicular to the line. The foot of the perpendicular is the point in which the perpendicular meets the plane. 314. A line is parallel to a plane when, each being produced ever so far, they do not meet. Also the plane, in this case, is parallel to the line. * 315. Two planes are parallel when, being produced ever so far, they do not meet. 316. It will be demonstrated art. 324, that the common intersection of two planes, which meet each other, is a straight line. This being premised, the angle or the mutual inclination of two planes is the quantity, whether greater or less, by which they depart from each other; this quantity is measured by the angle contained by two straight lines drawn from the same point perpendicularly to the common intersection, the one being in one of the planes and the other in the other. This angle may be acute, right, or obtuse. 317. If it is right, the two planes are perpendicular to each other. 318. A solid angle is the angular space comprehended between several planes which meet in the same point. Thus the solid angle S (fig. 199) is formed by the meeting of Fig. 199; the planes ASB, BSC, CSD, DSA. It requires at least three planes to form a solid angle. Geom. 14 Fig. 181. THEOREM. 319. One part of a straight line cannot be in a plane and another part without it. Fig. 182. Fig. 183. Demonstration. By the definition of a plane (6) a straight line, which has two points in common with the plane, lies wholly in that plane. 320. Scholium. In order to determine whether a surface is plane, it is necessary to apply a straight line in different directions to this surface and see if it touches the surface in its whole extent. THEOREM. 321. Two straight lines which cut each other are in the same plane, and determine its position. Demonstration. Let AB, AC (fig. 181), be two straight lines which cut each other in A. Conceive a plane to pass through AB, and to be turned about AB, until it passes through the point C; then, two points A and C being in the plane, the whole line AC is in this plane; therefore the position of the plane is determined by the condition of its containing the two lines AB, AC. 322. Corollary 1. A triangle ABC, or three points A, B, C, not in the same straight line determine the position of a plane. 323. Corollary II. Also two parallels AB, CD (fig. 182), determine the position of a plane; for, if the line EF be drawn, the plane of the two straight lines AE, EF, will be that of the parallels AB, CD. THEOREM. 324. If two planes cut each other, their common intersection is a straight line. Demonstration. If among the points common to the two planes there were three not in the same straight line, the two planes in question passing each through these three points would make only one and the same plane, which is contrary to the supposi tion. THEOREM. 325. If a straight line AP (fig. 183) is perpendicular to two others PB, PC, which intersect each other at its foot in the plane MN, it |