133. An Inscribed Polygon is one which has the vertices of all its angles in the circumference of the circle; as the triangle A В С. 134. The circle is then said to be circumscribed about the polygon. 135. A Polygon is circumscribed about a circle when all its sides are tangents to the circumference; as the polygon A B C D E F. 136. The circle is then said to be inscribed in the polygon. THEOREM I. 137. Every diameter divides the circle and its circum ference each into two equal parts. points in the one or the other unequally distant from the center, which is contrary to the definition of the circle. 138. Hence a diameter divides the circle and its circumference into two equal parts. 139. Cor. 1. Conversely, a straight line dividing the circle into two equal parts is a diameter. For, let the line A B divide the circle AEBCF into two equal parts; then, if the center is not in A B, let AC be drawn through it, which is therefore a diameter, and consequently divides the circle into two equal parts; hence the surface AFC is equal to the surface AFCB, a part to the whole, which is impossible. 140. Cor. 2. The arc of a circle, A F E C B whose chord is a diameter, is a semi-circumference, and the included segment is a semicircle.. 45.com Shisei C THEOREM ІІ. 141. In the same circle, e or or in equal circles, equal ares are subtended by equal Hords; and, conversely, equal chords subtend equal arcs. Let ADB and EGF be two equal circles, and let the arc A D be equal to EG; then will the chord AD be equal to the chord E G. semicircle EGF; and the curve line ADB will coincide with the curve line EGF (Theo. I.). But, by hypothesis, the arc A D is equal to the arc EG; hence the point D will fall on G; hence the chord AD is equal to the chord EG (26, Αχ. 10). Conversely, if the chord A Dis equal to the chord EG, the arcs A D, E G will be equal. For, if the radii CD, OG are drawn, the triangles ACD, E OG, having the three sides of the one equal to the three sides of the other, each to each, are themselves equal (Theo. XIII. Bk. I.); therefore the angle A CD is equal to the angle E OG (Theo. XIII. Sch., Bk. I.). If now the semicircle ADB be applied to its equal EGF, with the radius AC on its equal E O, since the angles ACD, E OG are equal, the radius CD will fall on OG, and the point Don G. Therefore the arcs AD and EG coincide with each other; hence they must be equal (26, Ax. 12). THEOREM III. the clay which His 142. In the same circle, or in equal circles, greater arc is subtended by a greater chord; and, conversely, the greater chord subtends the greater arc. In the circle of which Cis the center, let the are A B be greater than the arc AD; then will the chord A B be greater than the chord A D. Draw the radii CA, CD, and CB. The two sides AC, CB in the triangle ACB are equal to the two A C, CD in the triangle ACD, and the angle ACB is greater than the angle A CD; therefore the third side AB is greater than the third side AD (Theo. XI. B D A C Bk. I.); hence the chord which subtends the greater arc is the greater. Conversely, if the chord A B be greater than the chord A D, the arc A B will be greater than the arc A D. For the triangles ACB, ACD have two sides, AC, CB, in the one, equal to two sides, AC, CD, in the other, while the side A B is greater than the side AD; therefore the angle ACB is greater than the angle ACD (Theo. XII. Bk. I.); hence the arc A B is greater than the arc A D. 143. Scholium. The arcs here treated of are each less than the semi-circumference. If they were greater, the contrary would be true; in which case, as the arcs increased, the chords would diminish, and conversely. THEOREM IV. 144. In the same circle, or in equal circles, radii which make equal angles at the center intercept equal arcs on the circumference; and, conversely, if the intercepted arcs are equal, the angles made by the radii are also equal. gles ACB, DCE are equal, the one may be applied to the other; and since their sides, being radii of equal circles, are equal, the point A will coincide with D, and the point B with E. Therefore the arc AB must also coincide with the arc DE, or there would be points in the one or the other unequally distant from the center, which is impossible; hence the arc A B is equal to the arc DE. Second. If the arcs AB and DE are equal, the angles A C B and D CE will be equal. For, if these angles are not equal, let ACB be the greater, and let ACF be taken equal to DCE. From what has been shown, we shall have the arc A F equal to the arc D Е. But, by hypothesis, A B is equal to DE; hence A F must be equal to A B, the part to the whole, which is impossible; hence the angle ACB is equal to the angle DCE. THEOREM V. 145. The radius which is perpendicular to a chord bisects the chord, and also the arc subtended by the chord. Let the radius CE be perpendicular to the chord AB; then will CE bisect the chord at D, and the arc A B at E. Draw the radii CA and CB. Then CA and CB, with respect to the perpendicular CE, are equal oblique lines drawn to the chord AB; therefore their extremities are at equal distances from the perpendicular (Theo. X. Bk. I.); hence AD and DB are equal. angles of as lind fore C D A B Again, since the triangle ACB has the sides A9 and CB equal, it is isosceles; and the line CE bisects the base AB at right angles; therefore CE bisects also the angle Since the angles ACD, DCB are equal, the arcs A E, E B are equal (Theo. IV.); hence the radius CE, which is perpendicular to the chord A B, bisects the arc A B subtended by the chord. ACB (Theo. VI. Book I.).) 146. Cor. 1. Any straight line which joins the center of the circle and the middle of the chord, or the middle of the arc, must be perpendicular to the chord. For, the perpendicular from the center C passes through the middle, D, of the chord, and the middle, E, of the arc subtended by the chord. Now, any two of these three points in the straight line CE, are sufficient to determine its position. 147. Cor. 2. A perpendicular at the middle of a chord passes through the center of the circle, and through the middle of the arc subtended by the chord, bisecting at the center the angle which the arc subtends. THEOREM VI. 148. Through three given points, not in the same straight line, one circumference can be made to pass, and but one. Let A, B, and C be any three points not in the same straight line; one circumference can be made to pass through them, and but one. Join AB and BC; and bisect these straight lines by the |