plying half the sum of its parallel sides by the distance between those sides. 164. The area of any regular polygon may be found by multiplying its perimeter by half the radius of the inscribed circle. Take for example the regular pentagon ABCDE, (fig. 97.) From the centre O of the inscribed circle draw the lines OA, OB, &c., to the vertices of all the angles. As many triangles will thus be formed as the figure has sides. The area of the polygon is equal to the sum of the products of the bases of these triangles, viz., AB, BC, &c., by half their altitudes OF, OG, &c. But these altitudes are radii of the inscribed circle, and therefore are equal; therefore we can multiply the sum of the bases, that is the perimeter of the polygon, by the altitude of one triangle, that is, one half the radius of the inscribed circle. 7. THE CIRCLE. 165. The length of the radius determines the distance of the circumference of a circle from the centre, and consequently determines the size of the circle. Therefore, circles having equal radii or equal diameters are equal. 166. In a circle, or in equal circles, equal angles, the vertices of which are at the centre, will intercept equal arcs upon the circumference, and the chords which subtend those arcs will also be equal; conversely, equal chords subtend equal arcs, and the angles at the centre measured by those arcs are equal. If the arc AB: =arc CD, then the chord AB = CD, (fig. 98.) chord Dem. Draw the radii OA, OB, OC, and OD. Because the angles at the centre are measured by the arcs intercepted by their legs, (59,) equal arcs must be intercepted by equal angles; thus the angle AOB=COD; OA OC, and OB=OD by construction; consequently triangle AOB= COD (126 ;) therefore AB=CD. = = Conversely; AB=CD, OA=OC, OB=OD, consequently triangle COD=AOB, (138 ;) therefore angle AOB COD. Again, the arcs AB and CD belong to the same circle, and therefore have the same curvature; the chord AB = chord CD, by supposition, therefore if the segment AB be laid upon segment CD, so that chord AB shall coincide with chord CD, then the arc AB must coincide with arc CD; therefore they are equal. 167. A straight line drawn from the centre of a circle to the middle of a chord in the same circle is perpendicular to that chord, and bisects the arc subtended by that chord, and the angle at the centre measured by the arc. = = Dem. (Fig. 99.) Draw the radii OA and OB. Now AC CB by supposition, OA=OB by construction, and OCOC; consequently triangle AOC BOC, (138;) therefore the adjacent angles OCA and OCB are equal, and therefore right angles; consequently OC is perpendicular to AB. The angles COA and COB are likewise equal; therefore arc AD= arc DB; consequently the angle O and the arc AB are bisected. 168. A straight line drawn from the centre of a circle perpendicular to a chord of the same circle bisects that chord. Dem. (Fig. 99.) Draw the radii OA and OB. We have now an isosceles triangle of which the centre O is the vertex, and the chord AB is the base. A perpendicular let fall from O upon AB must bisect AB, (133.) 169. Cor. A perpendicular erected upon the middle of a chord passes through the centre of the circle. Therefore we can find the centre of a circle by erecting a perpendicular upon the middle point of each of two chords; the point where these perpendiculars intersect each other will be the centre of the circle. 170. The perpendicular AB erected at the extremity of OA, a radius of a circle, is a tangent to the same circle, (fig. 101.) Dem. In AB take any point C. Draw OC. Angle OAC is a right angle by supposition, therefore angle OCA <1 R. A. (120;) therefore OC>OA (139 ;) But OA is a radius, consequently the point C, and every point in the line AB is more than the length of a radius distant from the centre O, and therefore is without the circle. Thus the line AB touches the circumference only at one point, and therefore is a tangent to the circle. 171. Cor. A radius or a diameter drawn from the point of contact of a tangent to a circle is perpendicular to that tangent; otherwise there might be two perpendiculars to a straight line at one point; that is, two straight lines which shall meet it so as to make the adjacent angles equal; which is impossible. 172. An inscribed angle is one made by 2 chords and having its vertex in the circumference. An angle inscribed in a segment is one made by 2 chords drawn from any point in the arc of the segment to the extremities of its chord or base. An inscribed angle ACB, (fig 102,) has for its measure half of the arc AB intercepted between its sides. Dem. The centre of the circle, O, may be in one of the sides; between the sides; or without the sides. Case 1. (Fig. 102. 1.) Angle AOB=OCB+OBC, (124.) OC=OB, consequently angle OCB=OBC, (129;) therefore angle AOB=2 OCB=2 ACB. Case 2. (Fig. 102. 2.) Draw the diameter CD. Angle AOD=2 ACD; angle BOD=2 BCD; therefore angle AOD+BOD (= AOB) = 2 ACD +2 BCD (=2 ACB.) Case 3. (Fig. 102. 2.) Draw the diameter CD. Angle DOB=2 DCB; angle DOA=2 DCA; therefore angle DOB - DOA (= AOB)=2 DCB-2 DCA=2 DCA +2 ACB-2 DCA (=2 ACB.) In each of these cases the angle AOB is measured by the arc AB, consequently angle ACB = AOB is measured by the half of the arc AB. 173. Cor. 1. All angles inscribed in the same or in equal segments are equal; because each is measured by the half of the same or of equal arcs; for example, angle CED CGD, (fig. 106.) 174. Cor. 2. Every angle, as ACB, ADB, (fig. 103,) inscribed in a semi-circle is a right angle, for it has for its measure the half of a semi-circumference, that is, a quadrant. 175. Cor. 3. Every angle, as ACD, CDB, (fig. 103,) inscribed in a segment less than a semi-circle is 1 an obtuse angle, for it is measured by the half of an arc greater than a semi-circumference. in 176. Cor. 4. Every angle, as ABC, ABD, scribed in a segment greater than a semi-circle is an acute angle, for it is measured by the half of an arc less than a semi-circumference. 177. Cor. 5. In every quadrilateral inscribed in a circle, the sum of the opposite angles = 2 R. A. (Fig. 105,) A+C=2 R. A.; B+D=2 R. A. For the arc DCB angle A is measured by the angle C is mea2 178. The angles made by a tangent and a chord, drawn from the point of contact, are equal to the angles inscribed in the alternate segments of the circle. (fig. 106.) Angle DCB=CED and angle ACD=CHD; that is, each angle made by the tangent and chord is equal to the angle inscribed in the segment on the opposite side of the chord. Angle GCB is a Dem. Draw the diameter CG. right angle (171;) therefore angle GCD+DCB= GCB=1 R. A. Angle GDC is a right angle (172;) therefore angle GCD + DGC-1 R. A. (118;) consequently angles GCD+ DCB=GCD+DGC; therefore angle DCB= DGC. Angle DGC = DEC, therefore DCB DEC. Again, angle CED + CHD = 2 R. A. (177;) angles ACD+DCB=2 R. A., conse = |