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smaller. The sum of the angles at the centre will accordingly be less than two right angles; thus the extremities of the given sides will not terminate in the extremities of a diameter. The contrary will occur if we take a smaller circle; therefore the polygon under consideration can be inscribed in only one circle. 306. Scholium. We can change at pleasure the order of the sides AB, BC, CD, &c., and the diameter of the circumscribed circle will always be the same, as well as the surface of the polygon; for, whatever be the order of the arcs AB, BC, CD, &c., it is sufficient that their sum makes a semicircumference, and the polygon will always have the same surface, since it will be equal to the semicircle, minus the segments AB, BC, CD, &c., the sum of which is always the same.


307. Of all polygons formed of given sides, the maximum is that which can be inscribed in a circle.

Demonstration. Let ABCDEFG (fig. 177) be an inscribed Fig. 177. polygon, and abcdefg one that does not admit of being inscribed, having its corresponding sides equal to those of the former, namely, ab AB, bc = BC, cd = CD, &c.; the inscribed polygon will be greater than the other.


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Draw the diameter EM, and join AM, MB; upon a b= AB construct the triangle a bm equal to ABM, and join e m.

According to what has just been demonstrated (303), the polygon EFGAM is greater than ef gam, unless this last can also be inscribed in a semicircle having em for its diameter, in which case the two polygons would be equal (305). For the same reason the polygon EDCBM is greater than e dcbm, with the exception of the case in which they are equal. Hence the entire polygon EFGAMBCDE is greater than ef gambcde, unless they should be in all respects equal; but they are not so (161), since one is inscribed in a circle, and the other does not admit of being inscribed; therefore the inscribed polygon is greater than the other. Taking from them respectively the equal triangles ABM, ab m, we have the inscribed polygon ABCDEFG greater than the polygon not inscribed a b c d efg.

308. Scholium. It may be shown, as in art. 305, that there is only one circle, and consequently only one maximum polygon

which satisfies the question; and this polygon will still have the same surface, whatever change be made in the order of the sides.


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 radii; Fig. 178. that is, the angle C: angle O:: the ratio :


AC DO (fig. 179). 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,

C: 0::AB: FG (122), or ::


But, on account of the similar arcs FG, DE,
FG : DE :: FO : DO (288);

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Fig. 179.


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 sides ;* 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.


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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,

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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,

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Multiplying the antecedents by OG, and the consequents by CG, we have DEX OG: AB × CG :: GI: GH.

But the similar triangles ODE, OFG, give

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consequently OE × FG : CB × FG : : GI : GH,

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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 CB.

On the other side of CF let there be constructed the figure CKx 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.

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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 Cits 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 Al;

the polygon P: circle C:: triangle ACI: sector ODE, hence P: CAIX 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


that is,

CI: OE :: AI or DE: GE;

P: C:: DE: GE:: DE XOE: GE × 10E;
P: C: sector DOE triangle GOE;

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.



Of Planes and Solid Angles.


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 be

tween 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.

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