1. A REGULAR POLYGON is a polygon which is both equilateral and equiangular. PROPOSITION I. THEOREM. Regular polygons of the same number of sides are similar. Let ABCDEF and abcdef be regular polygons of the same number of sides: then will they be similar. For, the corresponding angles in each are equal, because any angle in either polygon is equal to twice as many right angles as the polygon has sides, less four, di F E B C vided by the number of angles (B. I., P. XXVI., C. 4); and further, the corresponding sides are proportional, because all the sides of either polygon are equal (D. 1): hence, the polygons are similar (B. IV., D. 1); which was to be proved. PROPOSITION II. THEOREM. The circumference of a circle may be circumscribed about any regular polygon; a circle may also be inscribed in it. 1o. Let ABCF be a regular polygon: then can the circumference of a circle be circumscribed about it. For, through three consecutive ver tices A, B, C, describe the circumference of a circle (B. III., Problem XIII., S.). Its centre O will lie on PO, drawn perpendicular to BC, at its middle point P; draw OA and OD. Let the quadrilateral OPCD be turned about the line OP, until PC H falls on PB; then, because the angle C is equal to B, the side CD will take the direction BA; and because CD is equal to BA, the vertex D, will fall upon the vertex A; and consequently, the line OD will coincide with OA, and is, therefore, equal to it: hence, the circumference which passes through A, B, and C, will pass through D. In like manner, it may be shown that it will pass through all of the other vertices: hence, it is circumscribed about the polygon; which was to be proved. 2o. A circle may be inscribed in the polygon. For, the sides AB, BC, &c., being equal chords of the circumscribed circle, are equidistant from the centre 0 hence, if a circle be described from O as a centre, with OP as a radius, it will be tangent to all of the sides or the polygon, and consequently, will be inscribed in it; which was to be proved. Scholium. If the circumference of a circle be divided into equal arcs, the chords of these arcs will be sides of a regular inscribed polygon. For, the sides are equal, because they are chords of equal arcs, and the angles are equal, because they are measured by halves of equal arcs. If the vertices A, B, C, &c., of a regular inscribed polygon be joined with the centre O, the triangles thus formed will be equal, because their sides are equal, each to each hence, all of the angles about the point O are equal to each other. E B DEFINITIONS. 1. The CENTRE OF A REGULAR POLYGON, is the common centre of the circumscribed and inscribed circles. 2. The ANGLE AT THE CENTRE, is the angle formed by drawing lines from the centre to the extremities of either side. The angle at the centre is equal to four right angles divided by the number of sides of the polygon. 3. The APOTHEM, is the distance from the centre to either side. The apothem is equal to the radius of the inscribed circle. Scholium. The radius is to the side of the inscribed square as 1 is to √2. If a regular hexagon be inscribed in a circle, any side will be equal to the radius of the circle. Let ABD be a circle, and ABCDEH a regular inscribed hexagon: then will any side, as AB, be equal to the radius of the circle. Draw the radii OA and OB. Then will the angle AOB be equal to one-sixth of four right angles, or to two-thirds of one right angle, because it is an angle at the centre (P. II., D. 2). The sum of the two angles OAB and OBA is, consequently, equal H E to four-thirds of a right angle (B. I., P. XXV., C. 1); but, the angles OAB and OBA are equal, because the opposite sides OB and OA are equal: hence, each is equal to two-thirds of a right angle. The three angles of the triangle AOB are therefore, equal, and consequently, the triangle is equilateral hence, AB is equal to OA; which was to be proved. To inscribe a regular hexagon in a given circle. Let ABE be a circle, and O its centre. Beginning at any point of Cor. 1. If the alternate vertices of the regular hexagon be joined by the lines AC, CE, and EA, the inscribed triangle ACE will be equilateral (P. II., S.). B E Cor. 2. If we draw the radii OA and OC, the figure AOCB will be a rhombus, because its sides are equal: hence (B. IV., P. XIV., C.), we have, AB2 + BC2 + OA2 + 0 02 AC2 + OB2; 1 |